Solutions to Introduction to Electric Circuits [9 ed.]


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Table of contents :
Chapter 1
Chapter 2
CH2EX
CH2sec2
CH2sec4
CH2sec5
CH2sec6
ch2sec7
CH2sec8
CH2sec9
CH2sec10
CH2DP
Chapter 3
CH3EX
CH3sec2
CH3sec3
CH3sec4
CH3sec5
CH3sec6
CH3sec7
CH3sec8
CH3DP
Chapter 4
CH4EX
CH4sec2
CH4sec3
CH4sec4
CH4sec5
CH4sec6
CH4sec7
CH4sec8
CH4sec9
CH4sec10
CH4PS
CH4DP
Chapter 5
ch5ex
sec5_2
sec5_3
sec5_4
sec5_5
sec5_6
sec5_8
sec5_9
ch5ps
ch5dp
Chapter 6
ch6ex
sec6_3
sec6_4
sec6_5
sec6_6
sec6_7
sec6_10
ch6ps
ch6dp
Chapter 7
CH7EX
CH7sec2
CH7sec3
CH7sec4
CH7sec5
CH7sec6
CH7sec7
CH7sec8
CH7sec9
CH7sec11
CH7DP
Chapter 8
CH8EX
CH8sec3
CH8sec4
CH8sec5
CH8sec6
CH8sec7
CH8sec10
CH8PS
CH8DP
Chapter 9
CH9EX
CH9sec2
CH9sec3
CH9sec4
CH9sec5
CH9sec6
Ch9sec7
CH9sec8
CH9sec9
Ch9sec10
CH9sec11
CH9PS
CH9DP
Chapter 10
Ch10sec10_2
Ch10sec10_3 updated 10232013
Ch10sec10_4
Ch10sec10_5
Ch10sec10_6
Ch10sec10_7
Ch10sec10_8
Ch10sec10_9
Ch10sec10_10
Ch10sec10_11
Ch10sec10_12
Ch10sec10_14
ch10ps
ch10dp
Chapter 11
CH11EX
CH11sec3
CH11sec4
CH11sec5
CH11sec6
CH11sec7
CH11sec8
CH11sec9
CH11sec10
CH11sec11
CH11PS
CH11DP
Chapter 12
Ch12EX
Ch12sec2
Ch12sec3
Ch12sec4
Ch12sec5
Ch12sec6
Ch12sec7
Ch12sec8
Ch12sec9
Ch12SP
Ch12DP
Chapter 13
Ch13Ex
Ch13_2
Ch13_3
Ch13_4
Ch13_6
Ch13_8
Ch13SP
Ch13DP
Chapter 14
ch14ex
sec14_2
sec14_3
sec14_4
sec14_5
sec14_6
sec14_7
sec14_8
sec14_9
Ch14sec10
sec14_11
sec14_12
ch14sp
ch14dp
Chapter 15
Ch15ex
Ch15sec2
Ch15sec3
Ch15sec5
Ch15sec6
Ch15sec7
Ch15sec9
Ch15sec12
Ch15sec15
Ch15SP
Ch15DP
Chapter 16
Ch16sec3
Ch16sec4
Ch16sec5
Ch16sec7
Ch16SP
Ch16DP
Chapter 17
Ch17Ex
Ch17Sec2
Ch17Sec3
Ch17Sec4
Ch17Sec5
Ch17Sec6
Ch17Sec7
Ch17Sec8
Ch17DP
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Solutions to Introduction to Electric Circuits [9 ed.]

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Chapter 1 – Electric Circuit Variables Exercises Exercise 1.2-1 Find the charge that has entered an element by time t when i = 8t2 – 4t A, t ≥ 0. Assume q(t) = 0 for t < 0. 8 Answer: q(t ) = t 3 − 2t 2 C 3 Solution:

i (t ) = 8 t 2 − 4 t A q(t ) =

t

∫0 i dτ + q(0) =

8 3 8 2 2 t τ − τ τ + = τ − τ = t3 −2 t 2 C (8 4 ) d 0 2 ∫0 0 3 3 t

Exercise 1.2-2 The total charge that has entered a circuit element is q(t) = 4 sin 3t C when t ≥ 0 and q(t) = 0 when t < 0. Determine the current in this circuit element for t > 0. d Answer: i (t ) = 4sin 3t = 12 cos 3t A dt Solution:

i (t ) =

dq d = 4sin 3t = 12 cos 3t dt dt

A

Exercise 1.3-1 Which of the three currents, i1 = 45 μA, i2 = 0.03 mA, and i3 = 25 × 10–4 A, is largest? Answer: i3 is largest. Solution: i1 = 45 μA = 45 × 10-6 A < i2 = 0.03 mA = .03 × 10-3 A = 3 × 10-5 A < i3 = 25 × 10-4 A

1-1

Exercise 1.5-1 Figure E 1.5-1 shows four circuit elements identified by the letters A, B, C, and D. (a) Which of the devices supply 12 W? (b) Which of the devices absorb 12 W? (c) What is the value of the power received by device B? (d) What is the value of the power delivered by device B? (e) What is the value of the power delivered by device D? Answers: (a) B and C, (b) A and D, (c)–12 W, (c) 12 W, (e)–12 W +

4V





2V

+

+

6V





3V

3A

6A

2A

4A

(A)

(B)

(C)

(D)

+

Figure E 1.5-1 Solution: (a) B and C. The element voltage and current do not adhere to the passive convention in B and C so the product of the element voltage and current is the power supplied by these elements.

(b) A and D. The element voltage and current adhere to the passive convention in A and D so the product of the element voltage and current is the power delivered to, or absorbed by these elements. (c) −12 W. The element voltage and current do not adhere to the passive convention in B, so the product of the element voltage and current is the power received by this element: (2 V)(6 A) = −12 W. The power supplied by the element is the negative of the power delivered to the element, 12 W. (d) 12 W (e) –12 W. The element voltage and current adhere to the passive convention in D, so the product of the element voltage and current is the power received by this element: (3 V)(4 A) = 12 W. The power supplied by the element is the negative of the power received to the element, −12 W.

1-2

Problems Section 1-2 Electric Circuits and Current Flow P1.2.1 The total charge that has entered a circuit element is q(t) = 1.25(1–e–5t) when t ≥ 0 and q(t) = 0 when t < 0. Determine the current in this circuit element for t ≥ 0. Answer: i(t) = 6.25e–5t A

i (t ) =

Solution:

d 1.25 (1 − e−5t ) = 6.25 e−5t A dt

P 1.2-2 The current in a circuit element is i(t) = 4(1–e–5t) A when t ≥ 0 and i(t) = 0 when t < 0. Determine the total charge that has entered a circuit element for t ≥ 0. t

0

−∞

−∞

Hint: q (0) = ∫ i (τ ) = ∫ 0dτ = 0 Answer: q(t) = 4t + 0.8e–5t– 0.8 C for t ≥ 0 Solution: t t t t 4 4 q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4 (1 − e−5τ ) dτ + 0 = ∫ 4 dτ − ∫ 4 e−5τ dτ = 4 t + e−5t − C 0 0 0 0 5 5

P 1.2-3 The current in a circuit element is i(t) = 4 sin 3t A when t ≥ 0 and i(t) = 0 when t < 0. Determine the total charge that has entered a circuit element for t ≥ 0. t

0

−∞

−∞

Hint: q (0) = ∫ i (τ )dτ = ∫ 0 dτ = 0 Solution: t

t

0

0

q ( t ) = ∫ i (τ ) dτ + q ( 0 ) = ∫ 4sin 5τ dτ + 0 = −

4 4 4 t cos 3τ 0 = − cos 3 t + C 5 5 5

1-3

⎧0 t < 2 ⎪ ⎪2 2 < t < 4 where the units of current are A P 1.2-4 The current in a circuit element is i (t ) = ⎨ t 1 4 8 − < < ⎪ ⎪⎩0 8 < t and the units of time are s. Determine the total charge that has entered a circuit element for t ≥ 0. Answer:

t 9 V gives i > 3.15 A o s 1 and R = 6Ω, v < 13 V gives i < 3.47 A o s 1 So any 3.15 A < i < 3.47 A keeps 9 V < v < 13 V. s o

P 3.4-6 Figure P 3.4-6 shows a transistor amplifier. The values of R1 and R2 are to be selected. Resistances R1 and R2 are used to bias the transistor, that is, to create useful operating conditions. In this problem, we want to select R1 and R2 so that vb = 5 V. We expect the value of ib to be approximately 10 μA. When i1 ≤ 10ib, it is customary to treat ib as negligible, that is, to assume ib = 0. In that case R1 and R2 comprise a voltage divider.

(a) (b)

Figure P 3.4-6 Select values for R1 and R2 so that vb = 5 V and the total power absorbed by R1 and R2 is no more than 5 mW. An inferior transistor could cause ib to be larger than expected. Using the values of R1 and R2 from part (a), determine the value of vb that would result from ib = 15 μA.

Solution: (a) To insure that ib is negligible we require 15 i1 = ≥ 10 (10 × 10−6 ) = 10−3 R1 + R 2

R1 + R 2 ≤ 150 kΩ

so

To insure that the total power absorbed by R1 and R2 is no more than 5 mW we require 152 R1 + R 2 ≥ 45 kΩ ≤ 5 × 10−3 ⇒ R1 + R 2 Next to cause vb = 5 V we require 5 = vb =

R2 R1 + R 2

15



R1 = 2 R 2

For example, R1 = 40 kΩ, R 2 = 80 kΩ, satisfy all three requirements. (b)

KVL gives KCL gives Therefore Finally

(80 ×10 ) i 3

i1 =

1

+ v b − 15 = 0

vb

+ 15 × 10−6

40 ×10 ⎛ v ⎞ (80 ×103 ) ⎜ 40 ×b103 + 15 ×10−6 ⎟ + v b = 15 ⎝ ⎠ 13.8 3v b + 1.2 = 15 ⇒ vb = = 4.6 V 3 3

P 3.4-7 Determine the value of the current i in the circuit shown in Figure P 3.4-7.

Figure P 3.4-7 Solution: All of the elements of this circuit are connected in parallel. Replace the parallel current sources by a single equivalent 2 – 0.5 + 1.5 = 3 A current source. Replace the parallel 12 Ω and 6 Ω resistors by a 12 × 6 single = 4 Ω resistor. 12 + 6

By current division

12 ⎛ 4 ⎞ i = −⎜ ⎟ 3 = − = −1.714 A 7 ⎝ 3+ 4 ⎠

(checked: LNAP 62/6/07)

P 3.4-8 Determine the value of the voltage v in Figure P 3.4-8.

Figure P 3.4-8 Solution: Each of the resistors is connected between nodes a and b. The resistors are connected in parallel and the circuit can be redrawn like this:

40 & 20 & 40 = 10 Ω

Then

v = 10 ( 0.003) = 0.03 = 30 mV

so

(checked: LNAP 6/21/04)

P 3.4-9 Determine the power supplied by the dependent source in Figure P 3.4-9.

Figure P 3.4-9 Solution: Use current division to get

ia = − so

75 30 × 10−3 ) = −22.5 mA ( 25 + 75

v b = 50 ( −22.5 × 10−3 ) = −1.125 V

The power supplied by the dependent source is given by p = − ( 30 × 10−3 ) ( −1.125 ) = 33.75 mW (checked: LNAP 6/12/04)

. P 3.4-10 Determine the values of the resistances R1 and R2 for the circuit shown in Figure P 3.4-10.

Figure P 3.4-10 Solution: Using voltage division R1 × 24 8= 40 R 2 R1 + R 2 + 40

⇒ Using KVL

R1 ( R 2 + 40 ) 1 = 3 R1 R 2 + 40 ( R1 + R 2 )



R1 R 2 + 40 ( R1 + R 2 ) = 3R1 R 2 + 120 R1 24 = 8 + R 2 (1.6 )

Then R1 =





R1 =

40 R 2 2R 2 + 80

R 2 = 10 Ω

40 (10 ) =4Ω 2 (10 ) + 80

P 3.4-11 Determine the values of the resistances R1 and R2 for the circuit shown in Figure P 3.4-11.

Figure P 3.4-11 Solution: Using KCL

.024 = 0.0192 +

0.384 R2



R2 =

0.384 = 80 Ω 0.0048

Using current division R1 0.384 = × 0.024 R2 R1 + ( R 2 + 80 )



16 =

R1 R 2 R1 + R 2 + 80

=

80 R1 R1 + 160



R1 = 40 Ω

Determine the value of the current measured by the meter in Figure P 3.4-12.

P 3.4-12

Figure P 3.4-12 Solution: Replace the (ideal) ammeter with the equivalent short circuit. Label the current measured by the meter.

Apply KCL at the left node of the VCCS to get 1.2 =

va 10

+ 0.2 v a = 0.3 v a

⇒ va =

1.2 =4V 0.3

Use current division to get im =

30 30 0.2 v a = 0.2 ( 4 ) = 0.6 A 30 + 10 30 + 10 (checked using LNAP 9/11/04)

P3.4-13 Consider the combination of resistors shown in Figure P3.4-13. Let R p denote the equivalent resistance.

(a) Suppose 20 Ω ≤ R ≤ 320 Ω. Determine the corresponding range of values of R p . (b) Suppose instead R = 0 (a short circuit). Determine the value of R p . (c) Suppose instead R = ∞ (an open circuit). Determine the value of R p . (d) Suppose instead the equivalent resistance is R p = 40 Ω. Determine the

Figure P3.4-13

value of R. Solution:

(a) First, when R = 20 Ω then R p = 80 || 20 =

R p = 80 || 320 =

80 ( 20 ) = 16 Ω . Next, when R = 320 Ω then 80 + 20

80 ( 320 ) = 64 Ω . Consequently 80 + 320 16 Ω ≤ R p ≤ 64 Ω.

(b) When R = 0 then R p = 80 || 0 = (c) When R = ∞ then R p =

80 ( 0 ) = 0 Ω. 80 + 0

80 ( ∞ ) 80 = = 80 Ω . 80 + ∞ 80 + 1 ∞

(d) When R p = 40 Ω then 40 = 80 || R =

80 R 80 + R

⇒ 80 + R = 2 R ⇒ R = 80 Ω

P3.4-14 Consider the combination of resistors shown in Figure P3.4-14. Let R p denote the equivalent resistance.

(a) Suppose 40 Ω ≤ R ≤ 400 Ω. Determine the corresponding range of values of R p . (b) Suppose instead R = 0 (a short circuit). Determine the value of R p . (c) Suppose instead R = ∞ (an open circuit). Determine the value of R p . (d) Suppose instead the equivalent resistance is R p = 80 Ω. Determine the value of R. Figure P3.4-14 Solution:

160 ( 80 ) 160 = = 53.33 Ω . Next, when R 160 + 80 3 160 ( 440 ) 16 ( 44 ) = = 117.33 Ω . = 400 Ω then R p = 160 || ( 40 + 400 ) = 160 || 440 = 160 + 440 6 (a) First, when R = 40 Ω then R p = 160 || ( 40 + 40 ) = 160 || 80 =

Consequently

53.33 Ω ≤ R p ≤ 117.33 Ω.

(b) When R = 0 then R p = 160 || ( 0 + 40 ) = 160 || 40 = (c) When R = ∞ then R p = 160 || ( ∞ + 40 ) = 160 || ∞ =

(d) When R p = 80 Ω then 80 = 160 || ( R + 40 ) =

160 ( 40 ) 16 ( 4 ) = = 32 Ω . 160 + 40 2

1 = 160 Ω . 1 1 + 160 ∞

160 ( R + 40 ) 160 ⇒ R + 200 = ( R + 40 ) = 2 R + 80 ⇒ R = 120 Ω . 160 + R + 40 80

P3.4-15 Consider the combination of resistors shown in Figure P3.4-15. Let R p denote the equivalent resistance.

(a) Suppose 50 Ω ≤ R ≤ 800 Ω. Determine the corresponding range of values of R p . (b) Suppose instead R = 0 (a short circuit). Determine the value of R p . (c) Suppose instead R = ∞ (an open circuit). Determine the value of R p . (d) Suppose instead the equivalent resistance is R p = 150 Ω. Determine the value of R.

Figure P3.4-15

Solution:

200 ( 50 ) 200 = 50 + = 90 Ω . Next, when R = 200 + 50 5 200 ( 800 ) 800 800 Ω then R p = 50 + ( 200 || 800 ) = 50 + = 50 + = 210 Ω . 200 + 800 5 (a) First, when R = 50 Ω then R p = 50 + ( 200 || 50 ) = 50 +

Consequently

90 Ω ≤ R p ≤ 210 Ω.

(b) When R = 0 then R p = 50 + ( 200 || 0 ) = 50 +

200 ( 0 ) = 50 Ω . 200 + 0

(c) When R = ∞ then R p = 50 + ( 200 || ∞ ) = 50 +

200 ( ∞ ) 200 = 50 + = 250 Ω . 200 200 + ∞ +1 ∞

(d) When R p = 150 Ω then 150 = 50 + ( 200 || R ) = 50 +

200 R 200 + R

⇒ 100 ( 200 + R ) = 200 R ⇒ R = 200 Ω .

P3.4-16 The input to the circuit shown in Figure P3.4-16 is the source current, i s . The output is the

current measured by the meter, i o . A current divider connects the source to the meter. Given the following observations: A. The input i s = 5 A causes the output to be i o = 2 A .

B. When i s = 2 A the source supplies 48 W. Determine the values of the resistances R1 and R 2 .

Figure P3.4-16 Solution:

⎛ R1 ⎞ R1 2 From current division, i o = ⎜ so i s . When i s = 5 A and i o = 2 A then = ⎟ ⎜ R1 + R 2 ⎟ 5 R1 + R 2 ⎝ ⎠ 2 ( R1 + R 2 ) = 5 R1 or 2 R 2 = 3 R1 . ⎡⎛ R1 R 2 ⎞ ⎤ The power supplied by the source is given by i s ⎢⎜ ⎟ i s ⎥ . When i s = 2 A the source supplies 48 ⎢⎣⎜⎝ R1 + R 2 ⎟⎠ ⎥⎦ ⎡⎛ R1 R 2 ⎞ ⎤ R1 R 2 . W, so 48 = 2 ⎢⎜ 2 ⎥ ⇒ 12 = ⎟ R1 + R 2 ⎢⎣⎜⎝ R1 + R 2 ⎟⎠ ⎥⎦ ⎛3 ⎞ 3 R1 ⎜ R1 ⎟ R1 3 5 2 ⎠ ⎝ 2 Combining these results gives 12 = = = R1 ⇒ R1 = (12 ) = 20 Ω and 5 5 3 ⎛3 ⎞ R1 + ⎜ R1 ⎟ 2 ⎝2 ⎠ 3 R1 = 30 Ω . 2

Figure P3.4-17 P3.4-17. Figure P3.4-17 shows four similar but slightly different circuits. Determine the values of the currents i1 , i 2 , i 3 and i 4 . Solution: Using current division:

⎛ 75 ⎞ ⎛ 45 ⎞ i1 = − ⎜ ⎟ 320 = −240 mA , i 2 = ⎜ ⎟ 300 = 100 mA , ⎝ 75 + 25 ⎠ ⎝ 45 + 90 ⎠ ⎛ 15 ⎞ ⎛ 120 ⎞ i3 = − ⎜ ⎟ 500 = −100 mA and i 4 = ⎜ ⎟ 250 = 200 mA ⎝ 15 + 60 ⎠ ⎝ 120 + 30 ⎠

Figure P3.4-18 P3.4-18. Figure P3.4-18 shows four similar but slightly different circuits. Determine the values of the currents i1 , i 2 , i 3 and i 4 . Solution: Using current division:

⎛ 30 ⎞ ⎛ 60 ⎞ i1 = ⎜ ⎟ 240 = 80 mA , i 2 = − ⎜ ⎟ 240 = −144 mA , ⎝ 30 + 60 ⎠ ⎝ 60 + 40 ⎠ ⎛ 20 ⎞ ⎛ 60 ⎞ i3 = ⎜ ⎟ 240 = 60 mA and i 4 = − ⎜ ⎟ 240 = −192 mA ⎝ 60 + 20 ⎠ ⎝ 60 + 15 ⎠

Figure P3.4-19 P3.4-19. The input to the circuit shown in Figure P3.4-19 is the current source current Is. The output is the current io. The output of this circuit is proportion to the input, that is

io = k I s Determine the value of the constant of proportionality, k. Solution: Replace six resistors at the right of the circuit by an equivalent resistance to get

Using current division: ⎛ 5 ⎞ ⎜ 2R ⎟ 2 i1 = ⎜ Is = Is ⎟ 5 7 ⎜ R+ R⎟ 2 ⎠ ⎝

Return to the original circuit

Using current division: io = Therefore

R 1 1⎛2 ⎞ 1 i1 = i 1 = ⎜ I s ⎟ = I s R+R 2 2⎝7 ⎠ 7 1 k= 7

Figure P3.4-20 P3.4-20. The input to the circuit shown in Figure P3.4-20 is the voltage source voltage Vs. The output is the voltage vo. The output of this circuit is proportion to the input, that is

v o = k Vs Determine the value of the constant of proportionality, k. Solution: Replace six resistors at the right of the circuit by an equivalent resistance to get

Using voltage division: 5 ⎞ ⎛ ⎞ ⎛ 5 ⎜ R || 2 R ⎟ ⎜ 7 R ⎟ 5 v1 = ⎜ Vs ⎜ Vs = Vs ⎟ ⎟ 5 5 19 ⎜ 2 R + R || R ⎟ ⎜ 2 R + R ⎟ 2 ⎠ ⎝ 7 ⎠ ⎝

Return to the original circuit

Using current division: vo =

Therefore

R 2

1 1⎛ 5 ⎞ 1 v1 = v1 = ⎜ I s ⎟ = V s R 5 5 ⎝ 19 ⎠ 19 2R+ 2 1 k= 19

Section 3-5 Series Voltage Sources and Parallel Current Sources P 3.5-1

Determine the power supplied by each source in the circuit shown in Figure P 3.5-1.

. Figure P 3.5-1 Solution: The voltage sources are connected in series and can be replaced by a single equivalent voltage source. Similarly, the parallel current sources can be replaced by an equivalent current source. After doing so, and labeling the resistor currents, we have the circuit shown. Apply KCL at the top node of the current source to get

i1 + 1.75 = i 2 Apply KVL to the outside loop to get 5 + 2 i 2 + 2 i1 = 0 so and

5 + 2 ( i1 + 1.75 ) + 2 i1 = 0 ⇒ i1 = −

8.5 = −2.125 A 4 i 2 = −2.125 + 1.75 = −0.375 A

The power supplied by each sources is: Source 8-V voltage source

Power delivered −8 i1 = 17 W

3-V voltage source

3 i1 = −6.375 W

3-A current source

3 × 2 i 2 = −2.25 W

1.25-A current source

−1.25 × 2 i 2 = 0.9375 W (Checked using LNAP, 9/14/04)

1

P 3.5-2

Determine the power supplied by each source in the circuit shown in Figure P 3.5-2.

Figure P 3.5-2 Solution:

20 × 5 = 4 Ω. 20 + 5 The 7-Ω resistor is connected in parallel with a short circuit, a 0-Ω resistor. The equivalent 0× 7 resistance is = 0 Ω , a short circuit. 0+7

The 20-Ω and 5-Ω resistors are connected in parallel. The equivalent resistance is

The voltage sources are connected in series and can be replaced by a single equivalent voltage source. After doing so, and labeling the resistor currents, we have the circuit shown.

The parallel current sources can be replaced by an equivalent current source. Apply KVL to get −5 + v1 − 4 ( 3.5 ) = 0 ⇒ v1 = 19 V

The power supplied by each sources is: Source 8-V voltage source

Power delivered −2 ( 3.5 ) = −7 W

3-V voltage source

−3 ( 3.5 ) = −10.5 W

3-A current source 0.5-A current source

3 ×19 = 57 W 0.5 ×19 = 9.5 W (Checked using LNAP, 9/15/04)

2

P 3.5-3

Determine the power received by each resistor in the circuit shown in Figure P 3.5-3.

Figure P 3.5-3 Solution: The voltage sources are connected in series and can be replaced by a single equivalent voltage source. Similarly, the parallel current sources can be replaced by an equivalent current source.

After doing so, and labeling the resistor currents, we have the circuit shown. Apply KCL at the top node of the current source to get i1 + 1.75 = i 2

Apply KVL to the outside loop to get 5 + 2 i 2 + i1 = 0 so 5 + 2 ( i1 + 1.75 ) + 2 i1 = 0 ⇒ i1 = −

8.5 = −2.125 A 4

and i 2 = −2.125 + 1.75 = −0.375 A The power supplied by each sources is: Source 8-V voltage source

Power delivered −8 i1 = 17 W

3-V voltage source

3 i1 = −6.375 W

3-A current source

3 × 2 i 2 = −2.25 W

1.25-A current source

−1.25 × 2 i 2 = 0.9375 W (Checked using LNAP, 9/14/04)

3

Section 3-6 Circuit Analysis P 3.6-1 The circuit shown in Figure P 3.6-1a has been divided into two parts. In Figure P 3.61b, the right-hand part has been replaced with an equivalent circuit. The left-hand part of the circuit has not been changed. (a)

Determine the value of the resistance R in Figure P 3.6-1b that makes the circuit in Figure P 3.6-1b equivalent to the circuit in Figure P 3.6-1a.

(b)

Find the current i and the voltage v shown in Figure P 3.6-1b. Because of the equivalence, the current i and the voltage v shown in Figure P 3.6-1a are equal to the current i and the voltage v shown in Figure P 3.6-1b.

(c)

Find the current i2 shown in Figure P 3.61a using current division.

Figure P 3.6-1

Solution:

(a) (b)

(c)

48 ⋅ 24 = 32 Ω 48 + 24 32 ⋅ 32 v = 32 + 32 24 = 16 V ; 32 ⋅ 32 8+ 32 + 32 16 1 i= = A 32 2 48 1 1 i2 = ⋅ = A 48 + 24 2 3 R = 16 +

P 3.6-2 The circuit shown in Figure P 3.6-2a has been divided into three parts. In Figure P 3.6-2b, the rightmost part has been replaced with an equivalent circuit. The rest of the circuit has not been changed. The circuit is simplified further in Figure 3.6-2c. Now the middle and rightmost parts have been replaced by a single equivalent resistance. The leftmost part of the circuit is still unchanged.

Figure P 3.6-2 (a) (b) (c)

(d)

(e)

Determine the value of the resistance R1 in Figure P 3.6-2b that makes the circuit in Figure P 3.62b equivalent to the circuit in Figure P 3.6-2a. Determine the value of the resistance R2 in Figure P 3.6-2c that makes the circuit in Figure P 3.62c equivalent to the circuit in Figure P 3.6-2b. Find the current i1 and the voltage v1 shown in Figure P 3.6-2c. Because of the equivalence, the current i1 and the voltage v1 shown in Figure P 3.6-2b are equal to the current i1 and the voltage v1 shown in Figure P 3.6-2c. Hint: 24 = 6(i1–2) + i1R2 Find the current i2 and the voltage v2 shown in Figure P 3.6-2b. Because of the equivalence, the current i2 and the voltage v2 shown in Figure P 3.6-2a are equal to the current i2 and the voltage v2 shown in Figure P 3.6-2b. Hint: Use current division to calculate i2 from i1. Determine the power absorbed by the 3-Ω resistance shown at the right of Figure P 3.6-2a.

Solution:

3⋅ 6 =6Ω 3+ 6 1 1 1 = + + ⇒ R p = 2.4 Ω then 12 6 6

(a ) R1 = 4 + (b)

1 Rp

R 2 = 8 + R p = 10.4 Ω

(c) KCL: i2 + 2 = i1 and − 24 + 6 i2 + R 2i1 = 0 ⇒ − 24 + 6 (i1 − 2) + 10.4i1 = 0 ⇒

i1 =

36 =2.195 A ⇒ v1 =i1 R 2 =2.2 (10.4)=22.83 V 16.4

1 6 ( d ) i2 = ( 2.195 ) = 0.878 A, 1 1 1 + + 6 6 12 v2 = ( 0.878 ) (6) = 5.3 V (e) i3 =

6 i2 = 0.585 A ⇒ 3+ 6

P = 3 i32 = 1.03 W

P 3.6-3 Find i using appropriate circuit reductions and the current divider principle for the circuit of Figure P 3.6-3.

Figure P 3.6-3 Solution: Reduce the circuit from the right side by repeatedly replacing series 1 Ω resistors in parallel with a 2 Ω resistor by the equivalent 1 Ω resistor

This circuit has become small enough to be easily analyzed. The vertical 1 Ω resistor is equivalent to a 2 Ω resistor connected in parallel with series 1 Ω resistors:

i1 =

1+1 (1.5 ) = 0.75 A 2 + (1 + 1)

P 3.6-4 (a) Determine values of R1 and R2 in Figure P 3.6-4b that make the circuit in Figure P 3.6-4b equivalent to the circuit in Figure P 3.6-4a. (b) Analyze the circuit in Figure P 3.6-4b to determine the values of the currents ia and ib (c) Because the circuits are equivalent, the currents ia and ib shown in Figure P 3.6-4b are equal to the currents ia and ib shown in Figure P 3.6-4a. Use this fact to determine values of the voltage v1 and current i2 shown in Figure P 3.6-4a.

Figure P 3.6-4 Solution:

(a)

1 1 1 1 = + + ⇒ R2 = 4 Ω R2 24 12 8

and

R1 =

(10 + 8) ⋅ 9 = 6Ω 10 + 8 + 9

b

g

(b)

First, apply KVL to the left mesh to get −27 + 6 ia + 3 ia = 0 ⇒ ia = 3 A . Next, apply KVL to the left mesh to get 4 ib − 3 ia = 0 ⇒ ib = 2.25 A . (c)

1 8 i2 = 2.25 = 1125 . A 1 1 1 + + 24 8 12

b gLM b10 +98g + 9 3OP = −10 V Q N

and v1 = − 10

P 3.6-5 The voltmeter in the circuit shown in Figure P 3.6-5 shows that the voltage across the 30-Ω resistor is 6 volts. Determine the value of the resistance R1. Hint: Use the voltage division twice. Answer: R1 = 40 Ω

Figure P 3.6-5 P3.6-5

30 v1 = 6 ⇒ v1 = 8 V 10 + 30

R2 12 = 8 ⇒ R2 = 20 Ω R2 + 10 20 =

b

g

R1 10 + 30 R1 + 10 + 30

b

g

⇒ R1 = 40 Ω

Alternate values that can be used to change the numbers in this problem: meter reading, V 6 4 4 4.8

Right-most resistor, Ω 30 30 20 20

R1, Ω 40 10 15 30

P 3.6-6 Determine the voltages va and vc and the currents ib and id for the circuit shown in Figure P 3.6-6. Answer: va = –2 V, vc = 6 V, ib = –16 mA, and id = 2 mA

Figure P 3.6-6 Solution:

P 3.6-7 Determine the value of the resistance R in Figure P 3.6-7. Answer: R = 28 kΩ

Figure P 3.6-7 Solution: 1× 10−3 =

24 12 ×103 + R p

⇒ R p = 12 × 103 = 12 kΩ

( 21×10 ) R = ( 21×10 ) + R 3

12 × 10 = R p 3

3

⇒ R = 28 kΩ

P 3.6-8 Most of us are familiar with the effects of a mild electric shock. The effects of a severe shock can be devastating and often fatal. Shock results when current is passed through the body. A person can be modeled as a network of resistances. Consider the model circuit shown in Figure P 3.6-8. Determine the voltage developed across the heart and the current Figure P 3.6-8 flowing through the heart of the person when he or she firmly grasps one end of a voltage source whose other end is connected to the floor. The heart is represented by Rh. The floor has resistance to current flow equal to Rf, and the person is standing barefoot on the floor. This type of accident might occur at a swimming pool or boat dock. The upper-body resistance Ru and lower-body resistance RL vary from person to person. Solution:

⎛ ⎞ 130 500 Voltage division ⇒ v = 50 ⎜ 15.963 V ⎜ 130 500 + 200 + 20 ⎟⎟ = ⎝ ⎠ ⎛ 100 ⎞ ⎛ 10 ⎞ ∴v = v ⎜ ⎟ = (15.963) ⎜ ⎟ = 12.279 V h ⎝ 100 + 30 ⎠ ⎝ 13 ⎠ v ∴ i = h = .12279 A h 100

P 3.6-9 Determine the value of the current i in Figure 3.6-9. Answer: i = 0.5 mA

Figure 3.6-9 Solution:

P 3.6-10

Determine the values of ia, ib, and vc in Figure P 3.6-10.

Figure P 3.6-10 Solution:

Req =

15 ( 20 + 10 ) = 10 Ω 15 + ( 20 + 10 )

ia = −

60 ⎛ 30 ⎞ ⎛ 60 ⎞ ⎛ 20 ⎞ = −6 A, ib = ⎜ ⎟⎟ = 4 A, vc = ⎜ ⎟ ⎜⎜ ⎟ ( −60 ) = −40 V Req ⎝ 30 + 15 ⎠ ⎝ Req ⎠ ⎝ 20 + 10 ⎠

P 3.6-11 Find i and Req a–b if vab = 40 V in the circuit of Figure P 3.6-11. Answer: Req a–b = 8 Ω, i = 5/6 A

Figure P 3.6-11 Solution: a)

Req = 24 12 = b)

(24)(12) =8Ω 24 + 12

from voltage division: 100 5 ⎛ 20 ⎞ 100 v = 40 ⎜ = V∴ i = 3 = A ⎟ x x 20 3 3 ⎝ 20 + 4 ⎠ from current division: i = i

5 ⎛ 8 ⎞ A = x ⎜⎝ 8 + 8 ⎟⎠ 6

P 3.6-12 The ohmmeter in Figure P 3.6-12 measures the equivalent resistance, Req, of the resistor circuit. The value of the equivalent resistance, Req, depends on the value of the resistance R.

Figure P 3.6-12

(a) Determine the value of the equivalent resistance, Req, when R = 18 Ω. (b) Determine the value of the resistance R required to cause the equivalent resistance to be Req = 18 Ω. Solution: 9 + 10 + 17 = 36 Ω a.)

b.)

36 R = 12 ⇒ 24 R = (12 )( 36 ) ⇒ R = 18 Ω 36+R

36 ( 9 ) = 7.2 Ω 36+9

P 3.6-13 Find the Req at terminals a–b in Figure P 3.6-13. Also determine i, i1, and i2. Answer: Req = 8 Ω, i = 5 A, i1 = 5/3 A, i2 = 5/2 A

Figure P 3.6-13 Solution:

Req = 2 + 1 + Using current division ⎛ 6 ⎞ i1 = i ⎜ ⎟ = ⎝ 6 + 12 ⎠

( 6 12 ) + ( 2 2 ) = 3 + 4 + 1 = 8 Ω ( 5)

( 13 )

=5

3

A

so

i=

40 40 = =5 A 8 Req

( )

⎛ 2 ⎞ 5 1 and i2 = i ⎜ ⎟ = ( 5) 2 = 2 A ⎝ 2+2⎠

P 3.6-14 All of the resistances in the circuit shown in Figure P 3.6-14 are multiples of R. Determine the value of R.

Figure P 3.6-14 Solution:

4 6 R + R = 2R 5 5 = R + ( 2R & 2R ) = 2R

( R & 4 R ) + ( 2 R & 3R ) =

(

R + 2R & ( R + ( 2R & 2R ))

)

So the circuit is equivalent to

Then

12 = 0.1( R + ( 2 R & 2 R ) ) = 0.1( 2 R )



R = 60 Ω

(checked: ELAB 5/31/04)

P 3.6-15 The circuit shown in Figure P 3.6-15 contains seven resistors, each having resistance R. The input to this circuit is the voltage source voltage, vs. The circuit has two outputs, va and vb. Express each output as a function of the input.

Figure P 3.6-15 Solution: The circuit can be redrawn as

va =

R & ( R + ( R & 2R ))

2R + R & ( R + ( R & 2R ))

vc =

vs =

5 vs 21

R & 2R 2 2 vs = va = vs R + ( R & 2R ) 5 21

vb =

R 1 1 vc = vc = vs R+R 2 21 (Checked using LNAP 5/23/04)

P 3.6-16 The circuit shown in Figure P 3.6-16 contains three 10-Ω, 1/4-W resistors. (Quarter-watt resistors can dissipate 1/4 W safely.) Determine the range of voltage source voltages, vs, such that none of the resistors absorbs more than 1/4 W of power. Figure P 3.6-16 Solution:

vo =

(10 & 10 ) v = 5 v = v s s s 10 + (10 & 10 ) 15 3

vR + vo − vs = 0 iR =

vR 10

⇒ =

vR =

2 vs 3

2 vs 30 2

4 1 ⎛ 2 ⎞ P = ⎜ v s ⎟ (10 ) = v s 2 ≤ 90 4 ⎝ 30 ⎠



vs ≤

90 3 10 = = 2.37 V 16 4 (checked: LNAP 5/31/04)

P 3.6-17 The four resistors shown in Figure P 3.6-17 represent strain gauges. Strain gauges are transducers that measure the strain that results when a resistor is stretched or compressed. Strain gauges are used to measure force, displacement, or pressure. The four strain gauges in Figure P 3.6-17 each have a nominal (unstrained) resistance of 120 Ω and can each absorb 0.2 mW safely. Determine the range of voltage source voltages, vs, such that no strain gauge absorbs more than 0.2 mW of power. Figure P 3.6-17 Solution:

vs

so the current in each strain gauge is

vs

. The power 2 400 v s2 v s ⎛ v s ⎞ v s2 −3 dissipated by each resistor is given by so we require 0.5 × 10 ≤ or = ⎜ ⎟ 800 2 ⎝ 400 ⎠ 800 v s ≤ 0.4 = 0.6325 V . The voltage across each strain gauge is

(checked: LNAP 6/9/04)

P 3.6-18 The circuit shown in Figure P 3.6-18b has been obtained from the circuit shown in Figure P 3.6-18a by replacing series and parallel combinations of resistances by equivalent resistances.

(a)

(b) (c)

Determine the values of the resistances R1, R2, and R3 in Figure P 3.6-18b so that the circuit shown in Figure P 3.6-18b is equivalent to the circuit shown in Figure P 3.6-18a. Determine the values of v1, v2, and i in Figure P 3.618b. Because the circuits are equivalent, the values of v1, v2, and i in Figure P 3.6-18a are equal to the values of v1, v2, and i in Figure P 3.6-18b. Determine the values of v4, i5, i6, and v7 in Figure P 3.6-18a.

Solution: (a) R1 = 10 & ( 30 + 10 ) = 8 Ω ,

R 2 = 4 + (18 & 9 ) = 10 Ω and

R3 = 6 & ( 6 + 6 ) = 4 Ω

(b) i = 1 A , v1 = 8 V and v 2 = 4 V

(c) 10 9 1 8 = −2 V , i 5 = − 1= − A, 10 + 30 9 + 18 3 4 1 ⎛ 1⎞ v 7 = −18 ⎜ − ⎟ = +6 V and i 6 = = A 12 3 ⎝ 3⎠

v4 = −

(checked: LNAP 6/6/04) Figure P 3.6-18

P 3.6-19

Determine the values of v1, v2, i3, v4, v5, and i6 in Figure P 3.6-19.

Figure P 3.6-19 Solution: Replace series and parallel combinations of resistances by equivalent resistances. Then KVL gives ( 20 + 4 + 8 + 16 ) i = 48 ⇒ i = 0.5 A v a = 20 i = 10 V , v b = 16 i = 8 V and v c = 8 i = 4 V

Compare the original circuit to the equivalent circuit to get ⎛ 10 || (10 + 30) ⎞ ⎛ 8 ⎞ v1 = − ⎜ ⎟va = −⎜ ⎟10 = −4 V ⎝ 12 + 8 ⎠ ⎝ 12 + 10 || (10 + 30) ⎠ v 2 = −v c = −4 V

⎛ 20 ⎞ ⎛1⎞ i3 = − ⎜ ⎟ i = − ⎜ ⎟ ( 0.5 ) = −0.1 A ⎝ 20 + 80 ⎠ ⎝5⎠ ⎛ 30 ⎞ ⎛1⎞ v4 = −⎜ ⎟ v1 = − ⎜ ⎟ ( −4 ) = 1 V ⎝ 10 + 30 ⎠ ⎝ 4⎠ 4 ⎛ ⎞ ⎛1⎞ v5 = ⎜ ⎟ v c = ⎜ ⎟ ( 4) = 1 V ⎝ 5+6+6⎠ ⎝ 4⎠ ⎛ ⎞ 16 ⎛1⎞ i 6 = − ⎜⎜ ⎟⎟ i = − ⎜ ⎟ ( 0.5 ) = −0.25 A ⎝ 2⎠ ⎝ 16 + ( 4 + 6 + 6 ) ⎠ (checked: LNAP 6/10/04)

P 3.6-20 Determine the values of i, v, and Req by the circuit model shown in Figure P 3.6-20, given that vab = 18 V.

Figure P 3.6-20

Solution:

Replace parallel resistors by equivalent resistors: 6 || 30 = 5 Ω and 72 || 9 = 8 Ω

A short circuit in parallel with a resistor is equivalent to a short circuit. R eq = 36 || ( 8 + 10 ) = 12 Ω

Using voltage division when vab = 18 V: v=

8 4 v ab = (18 ) = 8 V 8 + 10 9 i=

v =1 A 8 (checked: LNAP 6/21/04)

P 3.6-22 Determine the value of the resistance R in the circuit shown in Figure P 3.6-22, given that Req = 9 Ω. Answer: R = 15 Ω

Figure P 3.6-22 Solution:

Replace parallel resistors by an equivalent resistor: 8 || 24 = 6 Ω A short circuit in parallel with a resistor is equivalent to a short circuit.

Replace series resistors by an equivalent resistor: 4+6 = 10 Ω Now 9 = R eq = 5 + (12 || R ||10 ) so 60 R× 11 ⇒ R = 15 Ω 4= 60 R+ 11 (checked: LNAP 6/21/04)

P 3.6-22 Determine the value of the resistance R in the circuit shown in Figure P 3.6-22, given that Req = 50 Ω.

Figure P 3.6-22

Solution:

R eq = ( R || ( R + R ) || R ) || ( R || ( R + R) || R )

R || ( R + R) || R = 2 R || R eq =

R 2 2 R || R = 5 5 5

R 2 = R 2 5

⇒ R = 5 R eq = 200 Ω (checked: LNAP 6/21/04)

P 3.6-23 Determine the values of r, the gain of the CCVS, and g, the gain of the VCCS, for the circuit shown in Figure P 3.6-23.

Figure P 3.6-23

Solution:

ia =

9.74 = 1.2175 A 8

⎛ 9.74 ⎞ 9.74 − 6.09 = r i a = r ⎜ ⎟ ⎝ 8 ⎠



V ⎛ 9.74-6.09 ⎞ r =⎜ ⎟8 = 3 A ⎝ 9.74 ⎠

v b = 12 − 9.74 = 2.26 V gv b +

6.09 9.74 2.26 + − =0 8 8 8 g=

gv b vb

=



gv b = −1.696 A

−6.696 = −0.75 2.26

(checked: LNAP 6/21/04)

P 3.6-24 The input to the circuit in Figure P 3.6-24 is the voltage of the voltage source, vs. The output is the voltage measured by the meter, vo. Show that the output of this circuit is proportional to the input. Determine the value of the constant of proportionality.

Figure P 3.6-24 Solution: va =

20 & 20 1 vs = vs 20 + ( 20 & 20 ) 3

3 1 ⎛ 12 ⎞ vo = ⎜ ⎟ (10v a ) = × 10 × v s = 2v s 5 3 ⎝ 12 + 8 ⎠ V . So vo is proportional to vs and the constant of proportionality is 2 V

P 3.6-25 The input to the circuit in Figure P 3.6-25 is the voltage of the voltage source, vs. The output is the current measured by the meter, io. Show that the output of this circuit is proportional to the input. Determine the value of the constant of proportionality.

Figure P 3.6-25

Solution: vs ⎛ 40 ⎞ ⎛ 4 ⎞ ⎛ vs ⎞ 4 ia = ⎜ = ⎟ ⎜ ⎟ ⎜ ⎟ = vs ⎝ 40 + 10 ⎠ 2 + ( 40 & 10 ) ⎝ 5 ⎠ ⎝ 10 ⎠ 50

100 ⎛ 4 ⎞ 8 ⎛ 40 ⎞ io = − ⎜ ⎟ ( 50i a ) = − ⎜ ⎟ vs = − vs 3 ⎝ 50 ⎠ 3 ⎝ 20 + 40 ⎠ The output is proportional to the input and the constant of proportionality is −

8 A . 3 V

P 3.6-26

Determine the voltage measured by the voltmeter in the circuit shown in Figure P 3.6-26.

Figure P 3.6-26 Solution:

Replace the voltmeter by the equivalent open circuit and label the voltage measured by the meter as vm. The 10-Ω resistor at the right of the circuit is in series with the open circuit that replaced the voltmeter so it’s current is zero as shown. Ohm’s law indicates that the voltage across that 10-Ω resistor is also zero. Applying KVL to the mesh consisting of the dependent voltage source, 10-Ω resistor and open circuit shows that vm = 8 ia The 10-Ω resistor and 40-Ω resistor are connected in parallel. The parallel combination of these resistors is equivalent to a single resistor with a resistance equal to 40 ×10 =8 Ω 40 + 10

Figure a shows part of the circuit. In Figure b, an equivalent resistor has replaced the parallel resistors. Now the 4-Ω resistor and 8-Ω resistor are connected in series. The series combination of these resistors is equivalent to a single resistor with a resistance equal to 4 + 8 = 12 Ω . In Figure c, an equivalent resistor has replaced the series resistors. Here the same three circuits with the order reversed. The earlier sequence of figures illustrates the process of simplifying the circuit by repeatedly replacing series or parallel resistors by an equivalent resistor. This sequence of figures illustrates an analysis that starts with the simplified circuit and works toward the original circuit.

Consider Figure a. Using Ohm’s law, we see that the current in the 12-Ω resistor is 2 A. The current in the voltage source is also 2 A. Replacing series resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the current in the voltage source must also be 2 A in Figure b. The currents in resistors in Figure b are equal to the current in the voltage source. Next, Ohm’s law is used to calculate the resistor voltages as shown in Figure b. Replacing parallel resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the current in the 4-Ω resistor in Figure c must be equal to the current in the 4Ω resistor in Figure b. Using current division in Figure c are yields ⎛ 40 ⎞ ia = ⎜ ⎟ 2 = 1.6 A ⎝ 40 + 10 ⎠ Finally, v m = 8 i a = 8 ×1.6 = 12.8 V

P 3.6-27

Determine the current measured by the ammeter in the circuit shown in Figure P 3.6-27.

Figure P 3.6-27 Solution:

Replace the ammeter by the equivalent short circuit and label the current measured by the meter as im. The 10-Ω resistor at the right of the circuit is in parallel with the short circuit that replaced the ammeter so it’s voltage is zero as shown. Ohm’s law indicates that the current in that 10-Ω resistor is also zero. Applying KCL at the top node of that 10-Ω resistor shows that i m = 0.8 v a

Figure a shows part of the circuit. The 2-Ω resistor and 4-Ω resistor are connected in series. The series combination of these resistors is equivalent to a single 6-Ω resistor.

In Figure b, an equivalent resistor has replaced the series resistors. Now the 3-Ω resistor and 6-Ω resistor are connected in parallel. The parallel combination of these resistors is equivalent to a single resistor with a resistance equal to 3× 6 =2Ω 3+ 6 In Figure c, an equivalent resistor has replaced the parallel resistors. Here the same three circuits with the order reversed. The earlier sequence of figures illustrates the process of simplifying the circuit by repeatedly replacing series or parallel resistors by an equivalent resistor. This sequence of figures illustrates an analysis that starts with the simplified circuit and works toward the original circuit.

Consider Figure a. Using Ohm’s law, we see that the voltage across the 2-Ω resistor is 6 V. The voltage across the current source is also 6 V. Replacing parallel resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the voltage across the current source must also be 6 V in Figure b. The voltage across each resistor in Figure b is equal to the voltage across the current source. Replacing series resistors by an equivalent resistor does not change the current or voltage of any other element of the circuit, so the voltage across the 3-Ω resistor in Figure c must be equal to the voltage across the 3-Ω resistor in Figure b. Using voltage division in Figure c yields ⎛ 4 ⎞ va = ⎜ ⎟6=4 V ⎝ 2+4⎠ Finally, i m = 0.8 v a = 0.8 × 4 = 3.2 V

P 3.6-28 Determine the value of the resistance R that causes the voltage measured by the voltmeter in the circuit shown in Figure P 3.6-28 to be 6 V.

Figure P 3.6-28

Solution: Use current division in the top part of the circuit to get

⎛ 40 ⎞ ia = ⎜ ⎟ ( −3) = −2.4 A ⎝ 40 + 10 ⎠ Next, denote the voltage measured by the voltmeter as vm and use voltage division in the bottom part of the circuit to get ⎛ R ⎞ ⎛ −5 R ⎞ vm = ⎜ ⎟ ( −5 i a ) = ⎜ ⎟ ia ⎝ 18 + R ⎠ ⎝ 18 + R ⎠ Combining these equations gives: 12 R ⎛ −5 R ⎞ vm = ⎜ ⎟ ( −2.4 ) = 18 + R ⎝ 18 + R ⎠ When vm = 6 V, 6=

12 R 18 + R

⇒ R=

6 × 18 = 18 Ω 12 − 6

P 3.6-29 The input to the circuit shown in Figure P 3.6-29 is the voltage of the voltage source, vs. The output is the current measured by the meter, im.

(a) (b) (c)

Figure P 3.6-29 Suppose vs = 15 V. Determine the value of the resistance R that causes the value of the current measured by the meter to be im = 5 A. Suppose vs = 15 V and R = 24 Ω. Determine the current measured by the ammeter. Suppose R = 24 Ω. Determine the value of the input voltage, vs, that causes the value of the current measured by the meter to be im = 3 A.

Soluton: Use voltage division in the top part of the circuit to get

2 ⎛ 12 ⎞ va = ⎜ ⎟ ( −v s ) = − v s 5 ⎝ 12 + 18 ⎠ Next, use current division in the bottom part of the circuit to get 80 ⎞ ⎛ 16 ⎞ ⎛ im = − ⎜ ⎟ (5 v a ) = ⎜ − ⎟ va ⎝ 16 + R ⎠ ⎝ 16 + R ⎠ Combining these equations gives: 80 ⎞ ⎛ 2 ⎞ ⎛ 32 ⎞ ⎛ im = ⎜ − ⎟ ⎜ − vs ⎟ = ⎜ ⎟ vs ⎝ 16 + R ⎠ ⎝ 5 ⎠ ⎝ 16 + R ⎠ a. When vs = 15 V and im = 12 A 288 ⎛ 32 ⎞ 12 = ⎜ = 24 Ω ⎟ 15 ⇒ 192 + 12 R = 480 ⇒ R = 12 ⎝ 16 + R ⎠ b. When vs = 15 V and R = 80 Ω ⎛ 32 ⎞ im = ⎜ ⎟ 15 = 5 A ⎝ 16 + 80 ⎠ c. When im = 3 A and R = 24 Ω 4 ⎛ 32 ⎞ 3=⎜ ⎟ vs = vs 5 ⎝ 16 + 24 ⎠

⇒ vs =

15 = 3.75 V 4

P 3.6-30 The ohmmeter in Figure P 3.6-30 measures the equivalent resistance of the resistor circuit connected to the meter probes.

Figure P 3.6-31 (a) Determine the value of the resistance R required to cause the equivalent resistance to be Req = 12 Ω. (b) Determine the value of the equivalent resistance when R = 14 Ω. Solution:

R eq = ( ( R + 4 ) || 20 ) + 2 =

(a)

(b)

12 =

( R + 4 ) × 20 + 2 = 20 R + 80 + 2 R + 24 ( R + 4 ) + 20

20 R + 80 20 R + 80 + 2 ⇒ 10 = ⇒ R + 24 = 2 R + 8 ⇒ R = 16 Ω R + 24 R + 24 R eq =

20 (14 ) + 80 + 2 = 11.5 Ω 14 + 24 (Checked: LNAPDC 9/28/04)

P 3.6-31

The voltmeter in Figure P 3.6-31 measures the voltage across the current source.

Figure P 3.6-31

(a) (b)

Determine the value of the voltage measured by the meter. Determine the power supplied by each circuit element.

Solution:

Replace the ideal voltmeter with the equivalent open circuit and label the voltage measured by the meter. Label the element voltages and currents as shown in (b). Using units of V, A, Ω and W:

Using units of V, mA, kΩ and mW:

a.) Determine the value of the voltage measured by the meter.

a.) Determine the value of the voltage measured by the meter.

Kirchhoff’s laws give

Kirchhoff’s laws give

12 + v R = v m and −i R = −i s = 2 mA

12 + v R = v m and −i R = −i s = 2 ×10−3 A

Ohm’s law gives

Ohm’s law gives

(

)

v R = −25 i R

v R = − 25 ×103 i R Then

Then

(

)

(

)(

)

v R = − 25 ×103 i R = − 25 ×103 −2 ×10−3 = 50 V v m = 12 + v R = 12 + 50 = 62 V

v R = −25 i R = −25 ( −2 ) = 50 V v m = 12 + v R = 12 + 50 = 62 V

b.) Determine the power supplied by each element. voltage source

current source resistor

( )

(

12 i s = −12 −2 × 10−3

(

) = 50 ( −2 × 10 )

62 2 ×10−3 = 124 ×10−3 W vR iR

−3

P 3.6-32

0

voltage source

( )

12 i s = −12 ( −2 ) = −24 mW

= −24 ×10−3 W

= −100 × 10 −3 W

total

)

b.) Determine the power supplied by each element.

current source

62 ( 2 ) = 124 mW

resistor

v R i R = 50 ( − 2 ) = −100 mW

total

0

Determine the resistance measured by the ohmmeter in Figure P 3.6-32.

Figure P 3.6-32 Solution:

12 + P 3.6-33

40 × 10 + 4 = 12 Ω 40 + 10

Determine the resistance measured by the ohmmeter in Figure P 3.6-33.

Figure P 3.6-33 Solution:

( 60 + 60 + 60 ) × 60 = 45 Ω ( 60 + 60 + 60 ) + 60

P3.6-34 Consider the circuit shown in Figure P3.6-34. Given the values of the following currents and voltages: i1 = 0.625 A , v 2 = −25 V , i 3 = −1.25 A and v 4 = −18.75 V

Determine the values of R1 , R 2 , R 3 and R 4 .

Figure P3.6-34 Solution:

50 + v 2 − v1 = 0 ⇒ v1 = 50 + ( −25 ) = 25 V

From KVL

R1 =

From Ohm’s law

v1 i1

=

25 = 40 Ω 0.625

From KCL v ⎞ 25 ⎞ ⎛ ⎛ i1 + i 5 + i 2 = 0 ⇒ i 2 = − i1 + i 5 = − ⎜ 0.625 + 1 ⎟ = − ⎜ 0.625 + ⎟ = −1.25 A 40 ⎠ 40 ⎠ ⎝ ⎝ v2 −25 R2 = = = 20 Ω From Ohm’s law i 2 −1.25

(

)

From KCL

v2 v −25 = i 6 + i 3 ⇒ i 6 = −i 3 + 2 = − ( −1.25 ) + = −5 A 4 4 4

From KVL

v 3 + v 4 − 5 i 6 = 0 ⇒ v 3 = −v 4 + 5 i 6 = − ( −18.75 ) + 5 ( −5 ) = −6.25 V

From Ohm’s law

R3 =

v3 i3

=

v 4 −18.75 −6.25 = = 15 Ω = 5 Ω and R 4 = −1.25 −1.25 i3

P3.6-35 Consider the circuits shown in Figure P3.6-35. The equivalent circuit on the right is obtained from the original circuit on the left by replacing series and parallel combinations of resistors by equivalent resistors. The value of the current in the equivalent circuit is is = 0.8 A. Determine the values of R1, R2, R5, v2 and i3.

Figure P3.6-35

Solution: R1 ||18 = 6 ⇒

18 R1 18 + R1

= 6 ⇒ 3 R1 = 18 + R1 ⇒ R1 = 9 Ω

R 2 + 10 = 28 ⇒ R 2 = 18 Ω 40 = ( 6 + 28 + R 5 ) i s



40

= 34 + R 5

⇒ R 5 = 50 − 34 = 16 Ω

v 2 = −10 i s = −10 ( 0.8 ) = −8 V and i 3 = −

32 0.8 is = − = −0.4 A 32 + 32 2

0.8

P3.6-36 Consider the circuit shown in Figure P3.6-36. Given 2 1 3 v 2 = v s , i 3 = i1 and v 4 = v 2 . 3 5 8

Determine the values of R1 , R 2 and R 4 . Figure P3.6-36

2 1 3 Hint: Interpret v 2 = v s , i 3 = i1 and v 4 = v 2 as current and voltage division. 3 5 8 Solution:

From voltage division v 4 = so

R4

50 + R 4

50 + R 4

=

From current division i 3 = so

R4

v2

3 150 ⇒ 8 R 4 = 3 ( 50 + R 4 ) ⇒ R 4 = = 30 Ω . 8 8−3

R2

R 2 + ( 50 + R 4 )

R2 R 2 + 80

=

i1 =

R2 R 2 + 80

i1

1 ⇒ 5 R 2 = R 2 + 80 ⇒ R 2 = 20 Ω . 5

Notice that R 2 || ( 50 + R 4 ) = 20 || ( 50 + 30 ) = 20 || 80 = 16 Ω . From voltage division v1 =

so

R 2 || ( 50 + R 4 )

(

R1 + R 2 || ( 50 + R 4 )

)

vs =

16 vs R1 + 16

16 2 48 − 32 = ⇒ 48 = 2 ( R1 + 16 ) ⇒ R1 = =8 Ω. R1 + 16 3 2

P3.6-37 Consider the circuit shown in Figure P3.6-37. Given

2 2 4 i 2 = i s , v 3 = v1 and i 4 = i 2 . 5 3 5 Determine the values of R1 , R 2 and R 4 . 2 2 4 Hint: Interpret i 2 = i s , v 3 = v1 and i 4 = i 2 as current and voltage division. 5 3 5

Figure P3.6-37 Solution:

From current division i 4 = so

80 4 400 − 320 = ⇒ 400 = 4 ( 80 + R 4 ) ⇒ R 4 = = 20 Ω . 80 + R 4 5 4

From voltage division v 3 =

so

80 i2 80 + R 4

80 || R 4

R 2 + ( 80 || R 4 )

v1 =

80 || 20 16 v1 = v1 R 2 + ( 80 || 20 ) R 2 + 16

16 2 48 − 32 = ⇒ 48 = 2 ( R 2 + 16 ) ⇒ R 2 = =8 Ω. R 2 + 16 3 2

Notice that R 2 + ( 80 || R 4 ) = 8 + ( 80 || 20 ) = 8 + 16 = 24 Ω . From current division i1 =

so

R1 R1 + 24

=

(

R1

R1 + R 2 + ( 80 || R 4 )

)

is =

R1 R1 + 24

is

2 48 ⇒ 5 R1 = 2 ( R1 + 24 ) ⇒ R1 = = 16 Ω 5 3

P3.6-38 Consider the circuit shown in Figure P3.6-38.

Figure P3.6-38 1 (a) Suppose i 3 = i1 . What is the value of the resistance R? 3 (b) Suppose instead v 2 = 4.8 V . What is the value of the equivalent resistance of the parallel resistors? (c) Suppose instead R = 20 Ω. What is the value of the current in the 40 Ω resistor? 1 Hint: Interpret i 3 = i1 as current division. 3

Solution: (a) From current division i 3 =

20 20 1 i1 so = ⇒ 60 = 20 + R ⇒ R = 40 Ω . 20 + R 20 + R 3

(b) From voltage division v 2 =

so

(c)

4.8 =

Rp 40 + R p

Rp 40 + R p

24 ⇒

i1 =

24 4.8 ⎡ 40 + R p ⎤⎦ = R p 24 ⎣

⇒ Rp =

24 24 24 = = = 0.48 A 40 + ( 20 || 20 ) 40 + 10 50

( 0.2 ) 40 = 10 Ω . 1 − 0.2

P3.6-39 Consider the circuit shown in Figure P3.6-39.

Figure P3.6-39 1 (a) Suppose v 3 = v1 . What is the value of the resistance R? 4 (b) Suppose i 2 = 1.2 A . What is the value of the resistance R?

(c) Suppose R = 70 Ω . What is the voltage across the 20 Ω resistor? (d) Suppose R = 30 Ω . What is the value of the current in this 30 Ω resistor? 1 Hint: Interpret v 3 = v1 as voltage division. 4

Solution:

(a) From voltage division v 3 =

(b)

1.2 =

10 10 1 v1 so = ⇒ 40 = 10 + R ⇒ R = 30 Ω . 10 + R 10 + R 4

20 ( 2.4 ) 20 20 2.4 = 2.4 ⇒ R + 30 = = 40 ⇒ R = 10 Ω 20 + ( R + 10 ) R + 30 1.2

(c)

20 || ( 70 + 10 ) =

(d)

i2 =

20 ( 80 ) = 16 Ω so v1 = (16 ) 2.4 = 38.4 V 20 + 80

20 20 20 2.4 = 2.4 = 2.4 = 0.8 A 20 + ( R + 10 ) 20 + ( 30 + 10 ) 60

P3.6-40 Consider the circuit shown in Figure P3.6-40. Given that the voltage of the dependent voltage source is v a = 8 V , determine the values of R1 and v o .

Figure P3.6-40 Solution: First,

vo = −

20 8 = −3.2 V 20 + 30

Next,

⎛ ⎞ ⎜ ⎟ ⎞ 8 40 40 ⎛ 10 40 ⎜ 10 400 400 ⎟= ic = = ib = = ⎜⎜ ⎟⎟ = 40 R1 ⎟ 12 40 + R1 + 40 R1 480 + 52 R1 20 40 + R1 40 + R1 ⎝ 12 + 40 || R1 ⎠ 40 + R1 ⎜ ⎜⎜ 12 + ⎟ 40 + R1 ⎟⎠ ⎝ then 400 ( 20 ) 8 400 1000 − 480 = ⇒ 480 + 52 R1 = = 1000 ⇒ = 10 Ω 20 480 + 52 R1 8 52

(

)

P3.6-41 Consider the circuit shown in Figure P3.6-41. Given that the current of the dependent current source is i a = 2 A , determine the values of R1 and i o .

Figure P3.6-41 Solution: First, io = −

15 2 = −0.5 A 15 + 45

Next,

((

2 25 25 vb = 2 10 || 25 + R1 = vc = 0.2 25 + R1 25 + R1

(

(

then 2 500 = 0.2 35 + R1

)

⎛ 10 25 + R1 ⎜ 1⎜ ⎝ 10 + 25 + R1

) ) = 2550+ R

⇒ 35 + R1 = 50 ⇒ R1 = 15 Ω

)

⎞ 500 ⎟= ⎟ 35 + R1 ⎠

P3.6-42 Determine the values of i a , i b , i 2 , and v1 in the circuit shown in Figure P3.6-42.

Figure P3.6-42 Solution:

Use equivalent resistances to reduce the circuit to

From KCL i b = 4 i a + i a

⇒ i b = −3 i a .

From KVL 12 i a + 10 i b − 6 = 0 ⇒ 12 i a + 10 ( −3 i a ) = 6 So i a = −

1 A , v a = −4 A and i b = 1 A . 3

Returning our attention to the original circuit, notice that i a and i b were not changed when the circuit was reduced. Now v1 = ( 5 || 20 ) i a = ( 4 )( −0.333) = −1.333 V and i 2 =

12 i = 0.333 A . 12 + 24 b

P3.6-43 The Ohmmeter in Figure P3.6-43 measures R eq , the equivalent resistance of the part of the circuit to the left of the terminals. (a) Suppose R eq = 12 Ω. Determine the value of the resistance R. (b) Suppose instead that R = 14 Ω. Determine the value of the equivalent resistance R eq .

Figure P3.6-43

Solution:

R eq = 2 + ( 20 || ( 4 + R ) ) .

(a) When R eq = 12 Ω then 12 = 2 + ( 20 || ( 4 + R ) ) ⇒ 10 =

20 ( 4 + R ) ⇒ 24 + R = 2 ( 4 + R ) ⇒ R = 16 Ω 20 + ( 4 + R )

(b) When R = 14 Ω then

R eq = 2 + ( 20 || ( 4 + 14 ) ) = 2 + ( 20 ||18 ) = 2 +

20 (18 ) = 11.4736 Ω 20 + 18

P3.6-43. Determine the values of the resistance R and current i a in the circuit shown in Figure P3.6-43. Figure P3.6-43 Solution. Replace the series resistors by an equivalent resistor. Then use KVL to write 80 i a + 8 − 24 = 0 ⇒ i a =

24 − 8 = 0.2 A 80

Use KCL to write 24 − 8 8 8 = + ⇒ R 200 80

8 16 8 8 = − = 0.16 ⇒ R = = 50 Ω R 80 200 0.16

Figure P3.6-44 P3.6-44 The input to the circuit shown in Figure P3.6-44 is the voltage of the voltage source, 32 V. The output is the current in the 10 Ω resistor, io. Determine the values of the resistance, R1, and of the gain of the dependent source, G, that cause both the value of voltage across the 12 Ω to be v a = 10.38 V and the value of the output current to be i o = 0.4151 A. Solution: Using voltage division 10.38 = v a =

12 ( 32 ) 12 = 36.9942 ≈ 37 Ω ⇒ R1 = 25 Ω ( 32 ) ⇒ R1 + 12 = R1 + 12 10.38

Using current division 0.4151 = i o =

40 0.4151 A G v a = ( 0.8 ) G (10.38 ) ⇒ G = = 0.05 40 + 10 V ( 0.8)10.38

P3.6-45 The equivalent circuit in Figure 3.6-45 is obtained from the original circuit by replacing series and parallel combinations of resistors by equivalent resistors. The values of the currents in the equivalent circuit are ia = 3.5 A and ib = −1.5 A. Determine the values of the voltages v1 and v2 in the original circuit.

Figure P3.6-45 Solution:

Label the currents in the equivalent circuit that correspond to the give currents in the equivalent circuit:

Use current division:

Using Ohm’s law:

v1 = −35 i a = −35 ( 3.5 ) = −122.5 V and v 2 = −50 ( −0.75 ) = 37.5 V

P3.6-46 Figure 3.6-46 shows three separate, similar circuits. In each a 12 V source is connected to a subcircuit consisting of three resistors. Determine the values of the voltage source currents i1, i2 and i3. Conclude that while the voltage source voltage is 12 V in each circuit, the voltage source current depends on the subcircuit connected to the voltage source.

Figure P3.6-46

Solution: Replace the resistor subcircuit by an equivalent resistor in each circuit:

Using Ohm’s law:

i1 =

12 12 12 = 0.75 mA , i 2 = = 3 mA and i 3 = = 4 mA 16 kΩ 4 kΩ 3 kΩ

Figure P3.6-47 P3.6-47 Determine the values of the voltages, v1 and v2, and of the current, i3, in the circuit shown in Figure P3.6-47. Solution: Replace series and parallel combinations of resistors by equivalent resistors to get

( 4 + ( 80 || 20 ) = 4 + 16 = 20 Ω and 40|| ( 80 + 80 ) = 40 ||160 = 32 Ω. ) Next, apply KVL to the left mesh to get 50 + 30 i a − 20 i a = 0 ⇒ i a =

ib =

Ohm’s law gives

30 i a 32

=

50 = −5 A and 30 i a = −150 V 20 − 10

−150 = −4.6875 A 32

Label ib on the original circuit

Finally and

v1 = ( 80 || 20 ) i a = 16 ( −5 ) = −80 V , v 2 = i3 =

1 ( 30 i a ) = −75 V 2

80 + 80 4 i b = ( −4.6875 ) = −3.75 A 40 + ( 80 + 80 ) 5

Section 3-7 Analyzing Resistive Circuits using MATLAB P3.7-1 Determine the power supplied by each of the sources, independent and dependent, in the circuit shown in Figure P3.7-1. Hint: Use the guidelines given in Section 3.7 to label the circuit diagram. Use MATLAB to solve the equations representing the circuit.

Figure 3.7-1 Solution: We’ll begin by choosing the bottom node to be the reference node. Next we’ll label the other nodes and some element voltages:

Notice that the 8 Ω resistor, the 10 Ω resistor and the two independent current sources are all connected in parallel. Consequently, the element voltages of theses elements can be labeled so that they are equal. Similarly, the 4 Ω resistor and the dependent current source are connected in parallel so their voltages can be labeled so as to be equal. Using Ohm’s Law we see that the current directed downward in the 8 Ω resistor is downward in the 10 Ω resistor is

v1

v1 8

, current directed

, and the current directed from left to right in the 2 Ω resistor is

10 Applying Kirchhoff’s Current Law (KCL) at node a gives 5=

v1 8

+ 2.5 +

v1 10

+

v2 2

⇒ 0.225 v1 + 0.5 v 2 = 2.5

Using Ohm’s Law we see that the current directed downward in the 4 Ω resistor is

v2 2

.

(1) v3 4

. Applying

Kirchhoff’s Current Law (KCL) at node a gives 1

v2

+ 1.5 v1 =

v3

⇒ 1.5 v1 + 0.5 v 2 − 0.25 v 3 = 0 (2) 2 4 Applying Kirchhoff’s Voltage Law (KVL) to the mesh consisting of the 10 Ω resistor, the 2 Ω resistor and the dependent source to get

v 2 + v 3 − v1 = 0

(3)

Equations 1, 2 and 3 comprise a set of three simultaneous equations in the three unknown voltages v1 ,

v 2 and v 3 . We can write these equations in matrix form as 0 ⎤ ⎡ v1 ⎤ ⎡ 2.5⎤ ⎡ 0.225 0.5 ⎢ 1.5 0.5 −0.25⎥ ⎢ v ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −1 1 1 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 0 ⎥⎦ We can solve this matrix equation using MATLAB:

Hence

v1 = −4.1096 V, v 2 = 6.8493 V and v 3 = −10.9589 V

2

The power supplied by the 5 A current source is 5 v1 = 5 ( −4.1096 ) = −20.548 W . The power supplied by the 2.5 A current source is −2.5 v1 = −2.5 ( −4.1096 ) = 10.247 W . The power supplied by the

dependent current source is (1.5 v1 ) v 3 = 1.5 ( −4.1096 )( −10.9589 ) = 67.555 W .

Observation: Changing the order of the 8 Ω resistor, the 10 Ω resistor and the two independent current sources only changes the order of the terms in the KCL equation at node a. We know that addition is commutative, so change the order of the terms will not affect the values of the voltages v1 , v 2 and v 3 .

For example, if the positions of the 2.5 A current source and 8 Ω resistor are swithched:

The KCL equation at node a is 5 = 2.5 +

v1 8

+

v2

+ 2.5 +

v2

+

v1 10

2

⇒ 0.225 v1 + 0.5 v 2 = 2.5

Similarly, when the circuit is drawn as

The KCL equation at node a is 5=

v1 10

+

v1 8

2

⇒ 0.225 v1 + 0.5 v 2 = 2.5

The changes do not affect the values of the voltages v1 , v 2 and v 3 .

3

P3.7-2 Determine the power supplied by each of the sources, independent and dependent, in the circuit shown in Figure P3.7-2. Hint: Use the guidelines given in Section 3.7 to label the circuit diagram. Use MATLAB to solve the equations representing the circuit.

Figure 3.7-2 Solution: We’ll begin by choosing the bottom node to be the reference node. Next we’ll label the other nodes and some element currents:

Notice that two 4 Ω resistors, an 8 Ω resistor and the two independent voltage sources are all connected in series. Consequently, the element currents of theses elements can be labeled so that they are equal. Similarly, a 4 Ω resistor and the dependent voltage source are connected in series so their currents can be labeled so as to be equal. The current in each resistor has been labeled so we can use Ohm’s Law to calculate resistor voltages from the resistor currents and the resistances. Apply Kirchhoff’s Voltage Law (KVL) to the left mesh to get 6 + 8 i1 + 8 i 2 + 4 i1 − 15 + 4 i1 = 0 ⇒ 16 i1 + 8 i 2 = 9

(1)

Apply Kirchhoff’s Voltage Law (KVL) to the right mesh to get 4 i 3 + 5 i1 − 8 i 2 = 0

(2)

Applying Kirchhoff’s Current Law (KCL) at node a to get i 1 = i 2 + i 3 ⇒ − i1 + i 2 + i 3 = 0

(3) 4

Equations 1, 2 and 3 comprise a set of three simultaneous equations in the three unknown voltages v1 , v 2 and v 3 . We can write these equations in matrix form as ⎡16 8 0 ⎤ ⎡ i1 ⎤ ⎡9 ⎤ ⎢ 5 −8 4 ⎥ ⎢i ⎥ = ⎢0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −1 1 1 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣0 ⎥⎦ We can solve this matrix equation using MATLAB:

Hence i1 = 0.4091 A, i 2 = 0.3068 A and i 3 = 0.1023 A The power supplied by the 15 V voltage source is 15 i1 = 15 ( 0.4091) = 6.1365 W . The power supplied by the 6 V voltage source is −6 i1 = −6 ( 0.4091) = −2.4546 W . The power supplied by the dependent

voltage source is − ( 5 i 2 ) i 3 = −5 ( 0.3068 )( 0.1023) = 0.1569 W .

Observation: Changing the order of the two 4 Ω resistors, an 8 Ω resistor and the two independent voltage sources in the left mesh changes the order of the terms in the KVL equation for that mesh. We

5

know that addition is commutative, so change the order of the terms will not affect the values of the currents i1 , i 2 and i 3 . For example, when the circuit is drawn as

The KVL equation for the left mesh is 6 − 15 + 8 i 2 + 4 i1 + 8 i1 + 4 i1 = 0 ⇒ 16 i1 + 8 i 2 = 9 When the circuit is drawn as

The KVL equation for the left mesh is 4 i1 + 4 i1 + 8 i 2 + 6 − 15 + 8 i1 = 0 ⇒ 16 i1 + 8 i 2 = 9 These changes do not affect the values of the currents i1 , i 2 and i 3 .

6

Figure P3.7-3 P3.7-3. Determine the power supplied by each of the independent sources in the circuit shown in Figure P3.7-3. P3.7-3 Label the element currents and voltages as suggested in Table 3.7-1 Guidelines for Labeling Circuit Variables:

Apply KCL at the top left node: Apply KVL to the left mesh: In matrix form:

Solving using MATLAB:

v1 4

+

v1 4

+ 2 + i2 = 0

8 i 2 − 12 + 12 i 2 − v1 = 0 ⎡1 1 ⎤ 1 ⎥ ⎡v1 ⎤ ⎡ −2 ⎤ ⎢4 + 6 = ⎢ ⎥ ⎢⎣ i 2 ⎥⎦ ⎢⎣12 ⎥⎦ ⎣ −1 8 + 12 ⎦ ⎡ v1 ⎤ ⎡ −5.5714 ⎤ ⎢i ⎥ = ⎢ ⎥ ⎣ 2 ⎦ ⎣ 0.3214 ⎦

The current source supplies

−2 v1 = −2 ( −5.5714 ) = 11.1428 W

The voltage source supplies

12 i 2 = 12 ( 0.3214 ) = 3.8568 W

7

Figure P3.7-4 P3.7-4. Determine the power supplied by each of the sources in the circuit shown in Figure P3.7-4. P3.7-4 Label the element currents and voltages as suggested in Table 3.7-1 Guidelines for Labeling Circuit Variables:

Apply KVL to the loop consisting of 30 Ω, 40 Ω and 50 Ω resistors: Apply KVL to the right mesh:

v c = 40 i d + 12 v c

Apply KCL at the top node of the current source:

va 50

Apply KCL at the top node of a 40 Ω resistor: i b =

In matrix form:

Solving using MATLAB:

The current source supplies The VCVS supplies

30 i b + v c − v a = 0

+ i b = 2.4

vc 40

+ id

1 0 ⎤ ⎡v a ⎤ ⎡ 0 ⎤ ⎡ −1 30 ⎢ 0 ⎥ ⎢i ⎥ ⎢ 0 ⎥ 0 11 40 ⎢ ⎥⎢ b ⎥ = ⎢ ⎥ ⎢ 0.02 1 0 0 ⎥ ⎢ v c ⎥ ⎢ 2.4 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ −1 0.025 1 ⎦ ⎢⎣ i d ⎥⎦ ⎣ 0 ⎦ ⎣ 0 ⎡ v a ⎤ ⎡ 41.0526 ⎤ ⎢i ⎥ ⎢ ⎥ ⎢ b ⎥ = ⎢ 1.5789 ⎥ ⎢ v ⎥ ⎢ −6.3158⎥ ⎢ c⎥ ⎢ ⎥ ⎢⎣ i d ⎥⎦ ⎣ 1.7368 ⎦ 2.4 v a = 2.4 ( 41.0526 ) = 98.5262 W

− (12 v c ) i d = −12 ( −6.3158 )(1.7368 ) = 131.6314 W

8

Section 3-8 How Can We Check … P 3.8-1 A computer analysis program, used for the circuit of Figure P 3.8-1, provides the following branch currents and voltages: i1 = –0.833 A, i2 = –0.333 A, i3 = –1.167 A, and v = –2.0 V. Are these answers correct? Hint: Verify that KCL is satisfied at the center node and that KVL is satisfied around the outside loop consisting of the two 6-Ω resistors and the voltage source. Solution:

Figure P 3.8-1

KCL at node a: i = i + i 3 1 2 − 1.167 = −0.833 + ( −0.333) − 1.167 = − 1.166 OK KVL loop consisting of the vertical 6 Ω resistor, the 3 Ω and4Ω resistors, and the voltage source: 6i + 3i + v + 12 = 0 3 2 yields v = −4.0 V not v = −2.0 V

The answers are not correct.

The circuit of Figure P 3.8-2 was assigned as

P 3.8-2

a homework problem. The answer in the back of the textbook says the current, i, is 1.25 A. Verify this answer using current division. Figure P 3.8-2

Solution:

Apply current division to get:

1 ⎛ ⎞ ⎜ ⎟ ⎛1⎞ 20 i=⎜ 5 = ⎜ ⎟ 5 = 1.25 A √ 1 1 1 ⎟ ⎝4⎠ ⎜ ⎟ + + ⎝ 20 20 5 + 5 ⎠

The answer is correct.

P 3.8-3

The circuit of Figure P 3.8-3 was built in the lab and vo was

measured to be 6.25 V. Verify this measurement using the voltage divider principle.

Figure P 3.8-3

Solution: Apply voltage division to get:

320 ⎛ ⎞ ⎛ 320 ⎞ v=⎜ ⎟ 24 = ⎜ ⎟ 24 = 6.4 V ≠ 6.25 V ⎝ 650 + 320 + 230 ⎠ ⎝ 1200 ⎠

The measurement is not correct.

P 3.8-4 The circuit of Figure P 3.8-4 represents an auto’s electrical system. A report states that

iH = 9 A, iB = –9 A, and iA = 19.1 A. Verify that this result is correct. Hint: Verify that KCL is satisfied at each node and that KVL is satisfied around each loop. Figure P 3.8-4

Solution:

KVL bottom loop: KVL right loop: KCL at left node:

− 14 + 0.1i + 1.2i = 0 A H − 12 + 0.05i + 1.2i = 0 B H i +i =i A B H

Solving the three above equations yields: i = 16.8 A i = 10.3 A i = −6.49 A and A H B These are not the given values. Consequently, the report is incorrect.

P 3.8-5 Computer analysis of the circuit in Figure P 3.8-5 shows that ia = –0.5 mA and ib = –2 mA. Was the computer analysis done correctly? Hint: Verify that the KVL equations for all three meshes are satisfied when ia = –0.5 mA and ib = –2 mA.

Figure P 3.8-5

Solution:

1 ⎛ ⎞ Top mesh: 0 = 4 i a + 4 i a + 2 ⎜ i a + − i b ⎟ = 10 ( −0.5 ) + 1 − 2 ( −2 ) 2 ⎝ ⎠ Lower left mesh: vs = 10 + 2 ( i a + 0.5 − i b ) = 10 + 2 ( 2 ) = 14 V

Lower right mesh: vs + 4 i a = 12 ⇒ vs = 12 − 4 (−0.5) = 14 V The KVL equations are satisfied so the analysis is correct.

P 3.8-6 Computer analysis of the circuit in Figure P 3.8-6. shows that ia = 0.5 mA and ib = 4.5 mA. Was the computer analysis done correctly? Hint: First, verify that the KCL equations for all five nodes are satisfied when ia = 0.5 mA and ib = 4.5 mA. Next, verify that the KVL equation for the lower left mesh (a-e-d-a) is satisfied. (The KVL equations for the other meshes aren’t useful because each involves an unknown voltage.)

Solution: Apply KCL at nodes b and c to label the circuit as Apply KCL to get

i c + 4 = 4.5 ⇒ i c = 0.5 A i a + i c = 1 ⇒ 0.5 + 0.5 = 1 √ 6 = 1 + i a + i b = 1 + 0.5 + 4.5 √

(node d) (node a) (node e)

Apply KVL to get to mesh (a-e-d-a)

−3 ( 0.5 ) + 2 ( 4.5 ) − 5 ( 0.5 ) = −1.5 + 9 − 2.5 ≠ 0 The given currents do not satisfy these five Kirchhoff’s laws equations and therefore are not correct.

P 3.8-7 Verify that the element currents and voltages shown in Figure P 3.8-7 satisfy Kirchhoff’s laws: (a)

Verify that the given currents satisfy the KCL equations corresponding to nodes a, b, and c.

(b)

Verify that the given voltages satisfy the KVL equations corresponding to loops a-b-d-c-a and a-b-c-d-a.

Figure P 3.8-7 Solution: (a)

7 + ( −3) = 4

4 + ( −2 ) = 2 − 5 = − 2 + ( −3 )

(b)

−1 − ( −6 ) + ( −8 ) + 3 = 0 −1 − 2 − ( −8 ) − 5 = 0

(node a ) (node b) (node c) (loop a - b - d - c - a ) (loop a - b - c - d - a )

The given currents and voltages satisfy these five Kirchhoff’s laws equations

P 3.8-8 Figure P 3.8-8 shows a circuit and some corresponding data. The tabulated data provides values of the current, i, and voltage, v, corresponding to several values of the resistance R2. (a) Use the data in rows 1 and 2 of the table to find the values of vs and R1. (b) Use the results of part (a) to verify that the tabulated data are consistent. (c) Fill in the missing entries in the table.

Figure P 3.8-8 Solution: i=

(a)

R1 + R 2

2.4 =

from row 1

1.2 =

from row 2 so

vs vs R1 vs R1 + 10

2.4 R1 = v s = 1.2 ( R1 + 10 )

then

(b)



R1 = 10 Ω

v s = 2.4 (10 ) = 24 V i=

24 R 2 24 and v = 10 + R 2 10 + R 2

24 480 = 0.8 A and v = = 16 V . 30 30 720 When R2 = 30 Ω then v = = 18 V . 40 24 When R2 = 40 Ω the i = = 0.48 A . 50

When R2 = 20 Ω then i =

24 = 0.6 A . 40 960 When R2 = 40 W then v = = 19.2 V . 50

(c) When R2 = 30 Ω then i =

(checked: LNAP 6/21/04)

P 3.8-9 Figure P 3.8-9 shows a circuit and some corresponding data. The tabulated data provide values of the current, i, and voltage, v, corresponding to several values of the resistance R2. (a) Use the data in rows 1 and 2 of the table to find the values of is and R1. (b) Use the results of part (a) to verify that the tabulated data are consistent. (c) Fill in the missing entries in the table.

Figure P 3.8-9 Solution: i=

(a)

R1 R1 + R 2

is

From row 1

R1 4 = is 3 R1 + 10



4R1 + 40 = 3R1i s

From row 2

R1 6 = is 7 R1 + 20



6R1 + 120 = 7 R1i s



28R1 + 280 = 18 R1 + 360



is = 3 A

So

4 R1 + 40 3R1

= is =

When R2 = 40 Ω then i =

7 R1

4 8 = is 3 8 + 10

Then

(b)

6 R1 + 120

i=



R1 = 8 Ω

24 R 2 8 24 and v = R 2 i = ( 3) = 8 + R2 8 + R2 8 + R2

24 960 = 0.5 A and v = = 20 V . These are the values in the table so tabulated 48 48

data is consistent. (c) When R2 = 80 Ω then i =

24 ( 80 ) 240 24 3 = A and v = = V. 88 11 88 11 (checked: LNAP 6/21/04)

Design Problems DP 3-1 The circuit shown in Figure DP 3.1 uses a potentiometer to produce a variable voltage. The voltage vm varies as a knob connected to the wiper of the potentiometer is turned. Specify the resistances R1 and R2 so that the following three requirements are satisfied: 1. The voltage vm varies from 8 V to 12 V as the wiper moves from one end of the potentiometer to the other end of the potentiometer. 2. The voltage source supplies less than 0.5 W of power. 3. Each of R1, R2, and RP dissipates less than 0.25 W.

Figure DP 3.1

Solution: Using voltage division:

vm =

R 2 + aR p

R1 + (1 − a ) R p + R 2 + aR p vm = 8 V when a = 0 ⇒

vm = 12 V when a = 1 ⇒

24 =

R 2 + aR p R1 + R 2 + R p

R2 R1 + R 2 + R p R2 + R p R1 + R 2 + R p

=

1 3

=

1 2

24

The specification on the power of the voltage source indicates 242 1 ≤ ⇒ R1 + R 2 + R p ≥ 1152 Ω R1 + R 2 + R p 2 Try Rp = 2000 Ω. Substituting into the equations obtained above using voltage division gives 3R 2 = R1 + R 2 + 2000 and 2 ( R 2 + 2000 ) = R1 + R 2 + 2000 . Solving these equations gives R1 = 6000 Ω

and R 2 = 4000 Ω . With these resistance values, the voltage source supplies 48 mW while R1, R2 and Rp dissipate 24 mW, 16 mW and 8 mW respectively. Therefore the design is complete.

DP 3-2 The resistance RL in Figure DP 3.2 is the equivalent resistance of a pressure transducer. This resistance is specified to be 200 Ω ± 5 percent. That is, 190 Ω ≤ RL ≤ 210 Ω. The voltage source is a 12 V ± 1 percent source capable of supplying 5 W. Design this circuit, using 5 percent, 1/8-watt resistors for R1 and R2, so that the voltage across RL is

vo = 4V ± 10% (A 5 percent, 1/8-watt 100-Ω resistor has a resistance between 95 and 105 Ω and can safely dissipate 1/8-W continuously.)

Figure DP 3.2 Solution: Try R1 = ∞. That is, R1 is an open circuit. From KVL, 8 V will appear across R2. Using voltage 200 division, 12 = 4 ⇒ R 2 = 400 Ω . The power required to be dissipated by R2 R 2 + 200

82 1 = 0.16 W < W . To reduce the voltage across any one resistor, let’s implement R2 as the series is 400 8 combination of two 200 Ω resistors. The power required to be dissipated by each of these resistors is 42 1 = 0.08 W < W . 200 8 Now let’s check the voltage: 190 210 11.88 < v < 12.12 0 190 + 420 210 + 380 3.700 < v0 < 4.314 4 − 7.5% < v0 < 4 + 7.85% Hence, vo = 4 V ± 8% and the design is complete.

DP 3-3 A phonograph pickup, stereo amplifier, and speaker are shown in Figure DP 3.3a and redrawn as a circuit model as shown in Figure DP 3.3b. Determine the resistance R so that the voltage v across the speaker is 16 V. Determine the power delivered to the speaker.

Figure DP 3.3 Solution: Vab ≅ 200 mV 10 10 120 Vab = (120) (0.2) 10 + R 10 + R 240 ⇒ R=5Ω let v = 16 = 10 + R 162 = 25.6W ∴P= 10 v=

DP 3-4 A Christmas tree light set is required that will operate from a 6-V battery on a tree in a city park. The heavy-duty battery can provide 9A for the four-hour period of operation each night. Design a parallel set of lights (select the maximum number of lights) when the resistance of each bulb is 12 Ω. Solution:

N 1 N ⎛1⎞ i = G v = v where G = ∑ = N⎜ ⎟ T T R ⎝ R⎠ n = 1 Rn ∴N=

iR ( 9 )(12 ) = = 18 bulbs 6 v

DP 3-5 The input to the circuit shown in Figure DP 3.5 is the voltage source voltage, vs. The output is the voltage vo. The output is related to the input by

vo =

R2 vs = gvs R1 + R2

The output of the voltage divider is proportional to the input. The constant of proportionality, g, is called the gain of the voltage divider and is given by

R2 R1 + R2 The power supplied by the voltage source is ⎛ vs ⎞ vs2 vs2 p = vsis = vs ⎜ = ⎟= ⎝ R1 + R2 ⎠ R1 + R2 Rin where Rin = R1 + R2 is called the input resistance of the voltage divider. g=

(a) (b)

Figure DP 3.5

Design a voltage divider to have a gain, g = 0.65. Design a voltage divider to have a gain, g = 0.65, and an input resistance, Rin = 2500 Ω.

Solution:

R2

⇒ g R1 = (1 − g ) R 2 R1 + R 2 Thus either resistance can be determined from the other resistance and the gain of the voltage divider. Also R2 R2 g= = ⇒ R 2 = g R in R1 + R 2 R in Notice that

Consequently

g=

g R1 = (1 − g ) R 2 = (1 − g ) g R in

⇒ R1 = (1 − g ) R in

(a) The solution of this problem is not unique. Given any value of R1, we can determine a value of R2 that will cause g = 0.65. Let’s pick a convenient value for R1, say

R1 = 100 Ω Then (b)

and

g R1 = (1 − g ) R 2

⇒ R2 =

g R1 1− g

=

0.65 ×100 = 186 Ω 1 − 0.65

R 2 = g R in = 0.65 × 2500 = 1625 Ω R1 = (1 − g ) R in = (1 − 0.65 ) 2500 = 875 Ω

DP 3-6 The input to the circuit shown in Figure DP 3.6 is the current source current, is. The output is the current io. The output is related to the input by R1 io = is = gis R1 + R2 The output of the current divider is proportional to the input. The constant of proportionality, g, is called the gain of the current divider and is given by R1 g= R1 + R2 The power supplied by the current source is ⎡ ⎛ R R ⎞⎤ RR p = vsis = ⎢is ⎜ 1 2 ⎟ ⎥ is = 1 2 is2 = Rin is2 R1 + R2 ⎣ ⎝ R1 + R2 ⎠ ⎦ RR where Rin = 1 2 R1 + R2 is called the input resistance of the current divider.

(a) (b)

Figure DP 3.6

Design a current divider to have a gain, g = 0.65. Design a current divider to have a gain, g = 0.65, and an input resistance, Rin = 10000 Ω.

Solution:

Notice that

g=

R1

⇒ g R 2 = (1 − g ) R1

R1 + R 2

Thus either resistance can be determined from the other resistance and the gain of the current divider. Also R1 R in R in = ⇒ R2 = g= R1 + R 2 R 2 g Consequently

(1 − g ) R1 = g R 2 = R in

⇒ R1 =

R in

(1 − g )

Thus specified values of g and Rin uniquely determine the required values of R1 and R2. (a) The solution of this problem is not unique. Given any value of R1, we can determine a value of R2 that will cause g = 0.65. Let’s pick a convenient value for R1, say

R1 = 100 Ω Then

g R 2 = (1 − g ) R1 ⇒ R 2 =

R2 =

(b)

and

R1 =

R in

R in g

(1 − g )

=

=

(1 − g ) R1 (1 − 0.65 ) ×100 = = 54 Ω g

10000 = 15385 Ω 0.65

10000 = 28571 Ω (1 − 0.65 )

0.65

DP 3-7 Design the circuit shown in Figure DP 3-7 to have an output vo = 8.5 V when the input is vs = 12 V. The circuit should require no more than 1 mW from the voltage source.

Figure DP 3.7 Solution: vo

g=

The required gain is

8.5 = 0.7083 12

=

vs

0.7083 R1 = (1 − 0.7083) R 2

Consequently,

⇒ R 2 = 2.428 R1

It is also required that 122 0.001 ≥ R1 + R 2

⇒ R1 + R 2 ≥ 144000 ⇒ 3.428 R1 ≥ 144000 ⇒ R1 ≥ 42007 Ω

R1 = 45 kΩ and R 2 = 109.26 kΩ

For example,

DP 3-8 Design the circuit shown in Figure DP 3.8 to have an output io = 1.8 mA when the input is is = 5 mA. The circuit should require no more than 1 mW from the current source.

Figure DP 3.8 Solution: g=

The required gain is

io is

=

1.8 = 0.36 5

0.36 R 2 = (1 − 0.36 ) R1 ⇒ R 2 = 1.778 R1

Consequently, It is also required that

0.001 ≥ 0.005 2

R1 R 2 R1 + R 2



R1 + R 2 R1 R 2

⇒ R1 ≤

For example,

=

2.778 R1 1.778 R12



0.005 2 = 0.025 0.001

2.778 = 62.5 Ω 1.778 ( 0.025 )

R1 = 60 Ω and R 2 = 106.7 Ω

DP 3.9 A thermistor is a temperature dependent resistor. The thermistor resistance, RT, is related to the temperature by the equation β 1 T −1 T o ) RT = RT e (

where T has units of °K and R is in Ohms. R0 is resistance at temperature T0 and the parameter β is in °K. For example, suppose that a particular thermistor has a resistance R0 = 620 Ω at the temperature T0 = 20 °C = 293 °K and β = 3330 °K. At T = 70 °C = 343 °K the resistance of this thermistor will be 3330 ( 1 342 −1 293 )

R T = 620 e

= 121.68 Ω

In Figure DP 3-9 this particular thermistor in used in a voltage divider circuit. Specify the value of the resistor R that will cause the voltage vT across the thermistor to be 4 V when the temperature is 100 °C.

Solution

At T = 373°K

R T = 620 e

3330 ( 1 372 −1 293 )

= 54.17 Ω

Using voltage division 4 = vT =

54.17 54.17 ( 40 ) ⇒ R + 54.17 = ( 40 ) 4 R + 54.17 R = 541.7 − 54.17 = 487.54 Ω

Figure DP 3-9

DP3-10 The circuit shown in Figure DP 3-10 contains a thermistor that has a resistance R0 = 620 Ω at the temperature T0 = 20 °C = 293 °K and β = 3330 °K. (See problem DP 3-9.) Design this circuit (that is, specify the values of R and Vs) so that the thermistor voltage is vT = 4 V when T = 100 °C and vT = 20 V when T = 0 °C.

Figure DP 3-10 Solution

3330 ( 1 273 −1 293 )

= 1425.6 Ω

3330 ( 1 373 −1 293 )

= 54.17 Ω

At T = 0°C = 273°K

R T = 620 e

At T = 100°C = 373°K

R T = 620 e

Using voltage division 54.17 54.17 V s ) ⇒ R + 54.17 = ( (Vs ) 4 R + 54.17 1425.6 1425.6 20 = v T = V s ) ⇒ R + 1425.6 = ( (Vs ) 20 R + 1425.6 4 = vT =

In matrix form:

Solving gives

⎡ 54.17 ⎢ 4 ⎢ ⎢1425.6 ⎢⎣ 20

⎤ −1⎥ ⎡V s ⎤ ⎡ 54.17 ⎤ ⎥⎢ ⎥ = ⎢ R ⎦ ⎣1425.6 ⎥⎦ ⎣ ⎥ −1 ⎥⎦

V s = 23.7528 V and R = 267.5029 Ω

DP3-11 The circuit shown in Figure DP 3-11 is designed help orange grower protect their crops against frost by sounding an alarm when the temperature falls below freezing. It contains a thermistor that has a resistance R0 = 620 Ω at the temperature T0 = 20 °C = 293 °K and β = 3330 °K. (See problem DP 3-9.) The alarm will sound when the voltage at the − input of the comparator is less than the voltage at the + input. Using voltage division twice, we see that the alarm sounds whenever

R2 RT + R2


0.25 1305 + 475.2 so the alarm is off.

Chapter 4 Exercises

Exercise 4.2-1 Determine the node voltages, va and vb, for the circuit of Figure E 4.2-1. Answer: va = 3 V and vb = 11 V

Figure E 4.2-1 Solution: KCL at a:

v −v a a b + + 3 = 0 ⇒ 5 v − 3 v = −18 a b 3 2

KCL at b:

v −v b a − 3 −1 = 0 ⇒ v − v = 8 b a 2

v

Solving these equations gives:

va = 3 V and vb = 11 V

Exercise 4.2-2 Determine the node voltages, va and vb, for the circuit of Figure E 4.2-2. Answer: va = –4/3 V and vb = 4 V

Figure E 4.2-2 Solution:

v −v a a b + + 3 = 0 ⇒ 3 v − 2 v = −12 a b 4 2 v v −v b a b − −4= 0 ⇒ − 3 v + 5 v = 24 a b 3 2 v

KCL at a: KCL at b: Solving:

va = −4/3 V and vb = 4 V

Exercise 4.3-1 Find the node voltages for the circuit of Figure E 4.3-1. Hint: Write a KCL equation for the supernode corresponding to the 10-V voltage source. Answer:

2+

vb + 10 vb + = 5 ⇒ vb = 30 V and va = 40 V 20 30 Figure E 4.3-1

Solution: Apply KCL to the supernode to get v + 10 v 2+ b + b =5 20 30

Solving:

v = 30 V and v = v + 10 = 40 V b a b

Exercise 4.3-2 Find the voltages va and vb for the circuit of Figure E 4.3-2. Answer:

(vb + 8) − (−12) vb + = 3 ⇒ vb = 8 V and va = 16 V 10 40

Figure E 4.3-2 Solution:

( vb + 8) − ( −12) + vb = 3 10

40

⇒ v = 8 V and v = 16 V b a

Exercise 4.4-1 Find the node voltage vb for the circuit shown in Figure E 4.4-2. Hint: Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into vb = 4ia and solve for vb. 6 v v Answer: − + b − b = 0 ⇒ vb = 4.5 V 8 4 12 Figure E 4.4-2 Solution: Apply KCL at node a to express ia as a function of the node voltages. Substitute the result into vb = 4 ia and solve for vb .

6 vb + =i 8 12 a

⎛ 9 + vb ⎞ ⇒ v = 4i = 4 ⎜ ⎟ ⇒ v = 4.5 V b a b ⎜ 12 ⎟ ⎝ ⎠

Exercise 4.4-2 Find the node voltages for the circuit shown in Figure E 4.4-2. Hint: The controlling voltage of the dependent source is a node voltage, so it is already expressed as a function of the node voltages. Apply KCL at node a. Answer:

va − 6 va − 4va + = 0 ⇒ va = −2 V 20 15 Figure E 4.4-2

Solution: The controlling voltage of the dependent source is a node voltage so it is already expressed as a function of the node voltages. Apply KCL at node a. v −6 v −4v a a = 0 ⇒ v = −2 V + a a 20 15

Exercise 4.5-1 Determine the value of the voltage measured by the voltmeter in Figure E 4.5-1.

Answer: –1 V

Figure E 4.5-1 Solution:

Mesh equations: −12 + 6 i + 3 ⎛⎜ i − i ⎞⎟ − 8 = 0 ⇒ 9 i − 3 i = 20 1 1 2 ⎝ 1 2⎠

8 − 3 ⎛⎜ i − i ⎞⎟ + 6 i = 0 ⇒ − 3 i + 9 i = −8 2 1 2 ⎝ 1 2⎠

Solving these equations gives: 13 1 i = A and i = − A 1 6 2 6 The voltage measured by the meter is 6 i2 = −1 V.

Exercise 4.6-1 Determine the value of the voltage measured by the voltmeter in Figure E 4.6-1. Hint:

Write and solve a single mesh equation to determine the current in the 3-Ω resistor.

Answer: –4 V

Figure E 4.6-1 Solution:

−12 ⎛ 3⎞ 9 + 3 i + 2 i + 4 ⎜ i + ⎟ = 0 ⇒ ( 3 + 2 + 4 ) i = −9 − 3 ⇒ i = A 9 ⎝ 4⎠ The voltmeter measures 3 i = −4 V

Mesh equation:

Exercise 4.6-2 Determine the value of the current measured by the ammeter in Figure E 4.6-2. Hint:

Write and solve a single mesh equation.

Answer: –3.67 A

Figure E 4.6-2 Solution:

Mesh equation:

15 + 3 i + 6 ( i + 3) = 0 ⇒

( 3 + 6 ) i = −15 − 6 ( 3)

⇒ i=

−33 2 = −3 A 9 3

Section 4-2 Node Voltage Analysis of Circuits with Current Sources

P 4.2-1

The node voltages in the circuit of Figure P 4.2-1 are v1 = –4 V and v2 = 2 V.

Determine i, the current of the current source. Answer: i = 1.5 A

Figure P 4.2-1 Solution:

v −v −4 − 4 − 2 1 1 2 0= + +i = + + i = −1.5 + i ⇒ i = 1.5 A 8 6 8 6 (checked using LNAP 8/13/02) v

KCL at node 1:

P 4.2-2 Determine the node voltages for the circuit of Figure P 4.2-2. Answer: v1 = 2 V, v2 = 30 V, and v3 = 24 V

Figure P 4.2-2 Solution: KCL at node 1: KCL at node 2: KCL at node 3:

v −v v 1 2 1 + + 1 = 0 ⇒ 5 v − v = −20 1 2 20 5 v −v v −v 1 2 2 3 +2= ⇒ − v + 3 v − 2 v = 40 1 2 3 20 10 v −v v 2 3 3 +1 = ⇒ − 3 v + 5 v = 30 2 3 10 15

Solving gives v1 = 2 V, v2 = 30 V and v3 = 24 V.

(checked using LNAP 8/13/02)

Figure P4.2-3 P4.2-3 The encircled numbers in the circuit shown Figure P4.2-3 are node numbers. Determine the values of the corresponding node voltages, v1 and v2. Solution: First, express the resistor currents in terms of the node voltages:

Apply KCL at node 1 to get

v1 15

+ 0.025 +

v1 − v 2 40

=0

Multiply both sides by 40 and simplify to get

Apply KCL at node 2 to get

11 v 1 − v 2 = −1 3 v1 − v 2 v 2 0.025 + = 40 25

Multiply both sides by 40 and simplify to get −v1 + Solving, we get

13 v2 = 1 5

v1 = −0.1875 V and v 2 = 0.3125 V

P 4.2-4 Consider the circuit shown in Figure P 4.2-4. Find values of the resistances R1 and R2 that cause the voltages v1 and v2 to be v1 = 1 V and v2 = 2 V.

Figure P 4.2-4 Solution:

−.003 + Write the node equations:

− When v1 = 1 V, v2 = 2 V

v1 v1 − v2 + =0 R1 500

v1 − v2 v2 + − .005 = 0 500 R2

1 1 −1 + = 0 ⇒ R1 = = 200 Ω 1 R1 500 .003 + 500 2 −1 2 − + − .005 = 0 ⇒ R2 = = 667 Ω 1 500 R2 .005 − 500

−.003 +

(checked using LNAP 8/13/02)

P 4.2-5 Find the voltage v for the circuit shown in Figure P 4.2-5. Answer: v = 21.7 mV

Figure P 4.2-5 Solution:

Write node equations:

Solving gives: Finally:

v1 v − v 2 v1 − v3 + 1 + =0 500 125 250 v − v3 v − v2 − 1 − .001 + 2 =0 125 250 v − v3 v1 − v3 v3 − 2 − + =0 250 250 500 v1 = 0.261 V, v2 = 0.337 V, v3 = 0.239 V v = v1 − v3 = 0.022 V (checked using LNAP 8/13/02)

P 4.2-6 Simplify the circuit shown in Figure P 4.2-6 by replacing series and parallel resistors with equivalent resistors; then analyze the simplified circuit by writing and solving node equations. (a) Determine the power supplied by each current source. (b) Determine the power received by the 12-Ω resistor.

Figure P 4.2-6

Solution: Replacing series and parallel resistors with equivalent resistors we get 12 Ω + ( 40 Ω & 10 Ω ) = 20 Ω

60 Ω & 120 Ω = 40 Ω

The node equations are

3 × 10−3 = 2 × 10−3 + v2 − v3

v1 − v 2

20 v1 − v 2

+

= 20 v1 − v 3

v1 − v 3

20 v 2 − v3 10 v3



0.06 = 2v1 − ( v 2 − v 3 )



0.04 = −v1 + 3v 2 − 2v 3



0 = − ( 2v1 + 4v 2 ) + 7v 3

⎡ 2 −1 −1⎤ ⎡ v1 ⎤ ⎡.06 ⎤ ⎢ −1 3 −2 ⎥ ⎢v ⎥ = ⎢.04 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −2 −4 +7 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 0 ⎥⎦

⎡ v1 ⎤ ⎡0.244 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢v 2 ⎥ = ⎢ 0.228⎥ ⎢ v 3 ⎥ ⎢⎣ 0.200 ⎥⎦ ⎣ ⎦

10

+

20

=

40

Solving, e.g. using MATLAB, gives ⇒

(a) The power supplied by the 3 mA current source is ( 3 ×10−3 ) ( 0.244 ) = 0.732 mW. The power

supplied by the 2 mA source is ( 2 ×10−3 ) ( 0.228 ) = 0.456 mW.

(b) The current in the 12 Ω resistor is equal to the current i =

power received by the 12 Ω resistor is ( 0.8 ×10

v1 − v 2

20

) (12 ) = 7.68 ×10

−3 2

−b

=

0.244 − 0.228 = 0.8 mA so the 20

= 7.68 μ W.

(checked: LNAP and MATLAB 5/31/04)

P 4.2-7 The node voltages in the circuit shown in Figure P 4.2-7 are va = 7 V and vb = 10 V. Determine values of the current source current, is, and the resistance, R.

Figure P 4.2-7 Solution Apply KCL at node a to get

2=

va R

+

va

4

+

va − vb

2

=

7 7 7 − 10 7 1 + + = + ⇒ R=4Ω R 4 R 4 2

Apply KCL at node b to get is +

va − vb

2

=

vb

8

+

vb

8

= is +

7 − 10 10 10 = + ⇒ is = 4 A 2 8 8 (checked: LNAP 6/21/04)

Figure P4.2-8 P4.2-8 The encircled numbers in the circuit shown Figure P4.2-8 are node numbers. The corresponding node voltages are v1 and v2.The node equation representing this circuit is ⎡ 0.225 −0.125⎤ ⎡ v1 ⎤ ⎡ −3⎤ ⎢ −0.125 0.125 ⎥ ⎢v ⎥ = ⎢ 2 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦

a) Determine the values of R and Is in Figure P4.2-8 b) Determine the value of the power supplied by the 3 A current source.

Solution: a) The node equations representing the circuit are

v1 R

+

v1 − v 2

8

+ 3 = 0 and

v1 − v 2

8

+ Is = 0

In matrix form, we have 1 ⎡ ⎢ 0.125 + R ⎢ ⎣ 0.125

⎤ −0.125⎥ ⎡ v1 ⎤ ⎡ −3 ⎤ = ⎥ ⎢⎣v 2 ⎥⎦ ⎢⎣ − I s ⎥⎦ −0.125⎦

After multiplying the 2nd row by -1, we have 1 ⎡ ⎢ 0.125 + R ⎢ ⎣ −0.125

⎤ −0.125⎥ ⎡ v1 ⎤ ⎡ −3⎤ = ⎥ ⎢⎣v 2 ⎥⎦ ⎢⎣ I s ⎥⎦ 0.125 ⎦

Comparing to the given node equation, we see that 1 = 0.1 ⇒ R = 10 Ω and I s = 2 A R b) Solving the given node equations, we get v1 = −10 V and v 2 = 6 V

The power supplied by the 3 A current source is given by −v1 ( 3) = − ( −10 )( 3) = 30 W

Section 4-3 Node Voltage Analysis of Circuits with Current and Voltage Sources P 4.3-1 The voltmeter in Figure P 4.3-1 measures vc, the node voltage at node c. Determine the value of vc.

Figure P 4.3-1 Solution:

Express the voltage of the voltage source in terms of its node voltages: 0 − va = 6 ⇒ va = −6 V KCL at node b:

va − vb v −v +2= b c 6 10 KCL at node c:

Finally:



−6 − vb v −v +2= b c 6 10

vb − vc vc = 10 8

⇒ −1−

⇒ 4 vb − 4 vc = 5 vc

⎛9 ⎞ 30 = 8 ⎜ vc ⎟ − 3 vc ⎝4 ⎠

vb v −v +2= b c 6 10

⇒ vb =

⇒ 30 = 8 vb − 3 vc

9 vc 4

⇒ vc = 2 V (checked using LNAP 8/13/02)

P 4.3-2 The voltages va, vb, vc, and vd in Figure P 4.3-2 are the node voltages corresponding to nodes a, b, c, and d. The current i is the current in a short circuit connected between nodes b and c. Determine the values of va, vb, vc, and vd and of i. Answer: va = –12 V, vb = vc = 4 V, vd = –4 V, i = 2 mA

Figure P 4.3-2 Solution:

Express the branch voltage of each voltage source in terms of its node voltages to get: va = −12 V, vb = vc = vd + 8 KCL at node b: vb − va = 0.002 + i ⇒ 4000

vb − ( −12 ) = 0.002 + i ⇒ vb + 12 = 8 + 4000 i 4000

KCL at the supernode corresponding to the 8 V source: v 0.001 = d + i ⇒ 4 = vd + 4000 i 4000 so

vb + 4 = 4 − vd



( vd + 8 ) + 4 = 4 − vd

Consequently vb = vc = vd + 8 = 4 V and i =

⇒ vd = −4 V

4 − vd = 2 mA 4000 (checked using LNAP 8/13/02)

Figure P4.3-3 P4.3-3. Determine the values of the power supplied by each of the sources in the circuit shown in Figure P4.3-3. Solution: First, label the node voltages. Next, express the resistor currents in terms of the node voltages.

Identify the supernode corresponding to the 24 V source

Apply KCL to the supernode to get 12 − ( v a − 24 ) 10

+ 0.6 =

v a − 24 40

+

va 40

⇒ 196 = 6 v a

⇒ v a = 32 V

The 12 V source supplies

⎛ 12 − ( v a − 24 ) ⎞ ⎛ 12 − ( 32 − 24 ) ⎞ ⎟ = 12 ⎜ 12 ⎜ ⎟ = 4.8 W ⎜ ⎟ 10 10 ⎝ ⎠ ⎝ ⎠

The 24 V source supplies

va ⎞ ⎛ 32 ⎞ ⎛ 24 ⎜ −0.6 + ⎟ = 24 ⎜ −0.6 + ⎟ = 4.8 W 40 ⎠ 40 ⎠ ⎝ ⎝

The current source supplies

0.6 v a = 0.6 ( 32 ) = 19.2 W

Figure P4.3-4 P4.3-4. Determine the values of the node voltages, v1, v2 and v3 in the circuit shown in Figure P4.3-13. Solution:

First, express the resistor currents in terms of the node voltages:

Apply KCL to the supernode to get −15 − v1 20

+

v 2 − ( v1 + 15 )

Apply KCL at node 2 to get In matrix form: Solving using MATLAB:

50 v1 − v 2 25

=

=

v2 40

v1 − v 2 25 +

+

v1 + 5 10

v 2 − ( v1 + 15 ) 50

⇒ 0.21 v1 − 0.06 v 2 = −1.55 ⇒ − 0.06 v1 + 0.085 v 2 = 0.3

⎡ 0.21 −0.06 ⎤ ⎡ v1 ⎤ ⎡ −1.55⎤ ⎢ −0.06 0.085 ⎥ ⎢v ⎥ = ⎢ 0.3 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦

v1 = −7.9825 V and v 2 = −2.1053 V

P 4.3-5 The voltages va, vb, and vc in Figure P 4.3-5 are the node voltages corresponding to nodes a, b, and c. The values of these voltages are:

va = 12 V, vb = 9.882 V, and vc = 5.294 V Determine the power supplied by the voltage source.

Figure P 4.3-5 Solution:

The power supplied by the voltage source is ⎛v −v v −v ⎞ ⎛ 12 − 9.882 12 − 5.294 ⎞ va ( i1 + i 2 ) = va ⎜ a b + a c ⎟ = 12 ⎜ + ⎟ 6 ⎠ 4 6 ⎝ ⎠ ⎝ 4 = 12(0.5295 + 1.118) = 12(1.648) = 19.76 W (checked using LNAP 8/13/02)

P 4.3-6 The voltmeter in the circuit of Figure P 4.3-6 measures a node voltage. The value of that node voltage depends on the value of the resistance R. (a)

Determine the value of the resistance R that will cause the voltage measured by the voltmeter to be 4 V.

(b)

Determine the voltage measured by the voltmeter when R = 1.2 kΩ = 1200 Ω.

Answer: (a) 6 kΩ (b) 2 V

Figure P 4.3-6 Solution: Label the voltage measured by the meter. Notice that this is a node voltage.

Write a node equation at the node at which the node voltage is measured.

⎛ 12 − v m ⎞ v m v −8 −⎜ + 0.002 + m =0 ⎟+ 3000 ⎝ 6000 ⎠ R That is 6000 ⎞ 6000 ⎛ ⎜3 + ⎟ v m = 16 ⇒ R = 16 R ⎠ ⎝ −3 vm

(a) The voltage measured by the meter will be 4 volts when R = 6 kΩ. (b) The voltage measured by the meter will be 2 volts when R = 1.2 kΩ.

P 4.3-7 Determine the values of the node voltages, v1 and v2, in Figure P 4.3-7. Determine the values of the currents ia and ib.

Figure P 4.3-7 Solution:

Apply KCL at nodes 1 and 2 to get 10 − v1 1000 10 − v 2

v1

=

+

v1 − v 2

3000 5000 v1 − v 3 v3 + = 4000 5000 2000



23v1 − 3v 2 = 150



-4v1 + 19v 3 = 50

Solving, e.g. using MATLAB, gives ⎡ 23 −3⎤ ⎡ v1 ⎤ ⎡150 ⎤ ⎢ −4 19 ⎥ ⎢ v ⎥ = ⎢ 50 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦

ib =

Then

v1 − v 2 5000

=



v1 = 7.06 V and v1 = 4.12 V

7.06 − 4.12 = 0.588 mA 5000

Apply KCL at the top node to get ia =

v 1 − 10 1000

+

v 2 − 10 4000

=

7.06 − 10 4.12 − 10 + = −4.41 mA 1000 4000 (checked: LNAP 5/31/04)

P 4.3-8 The circuit shown in Figure P 4.3-8 has two inputs, v1 and v2, and one output, vo. The output is related to the input by the equation

vo = av1 + bv2 where a and b are constants that depend on R1, R2 and R3. (a) (b)

Determine the values of the coefficients a and b when R1 = 10 Ω, R2 = 40 Ω and R3 = 8 Ω. Determine the values of the coefficients a and b when R1 = R2 and R3 = R1 || R2.

Figure P 4.3-8 Solution:

vo R3

+

v o − v1 R1

+

vo − v2 R2

=0



vo =

v1 v2 + R R R R 1+ 1 + 1 1 + 2 + 2 R 2 R3 R1 R 3

(a) When R 1 = 10 Ω, R 2 = 40 Ω and R 3 = 8 Ω

vo =

v1 v2 + = 0.4v1 + 0.1v 2 1 5 1+ 4 + 5 1+ + 4 4

So a = 0.4 and b = 0.1. (b) When R 1 = R 2 and R 3 = R1 & R 2 = R1 / 2

vo =

v1

+

v2

1+1+ 2 1+1+ 2

= 0.25v1 + 0.25v 2

So a = 0.25 and b = 0.25. (checked: LNAP 5/31/04)

P 4.3-9

Determine the values of the node voltages of the circuit shown in Figure P 4.3-9.

Figure P 4.3-9

Solution: Express the voltage source voltages as functions of the node voltages to get

v 2 − v1 = 5 and v 4 = 15 Apply KCL to the supernode corresponding to the 5 V source to get 1.25 =

v1 − v 3 8

+

v 2 − 15 20

=0



80 = 5v1 + 2v 2 − 5v 3

Apply KCL at node 3 to get v1 − v 3 8

=

v3 40

+

v 3 − 15 12



− 15v1 + 28v 3 = 150

Solving, e.g. using MATLAB, gives ⎡ −1 1 0 ⎤ ⎡ v1 ⎤ ⎡ 5 ⎤ ⎢ 5 2 −5⎥ ⎢v ⎥ = ⎢ 80 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −15 0 28 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣150 ⎥⎦



⎡ v1 ⎤ ⎡ 22.4 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢v 2 ⎥ = ⎢ 27.4 ⎥ ⎢ v 3 ⎥ ⎢⎣17.4 ⎥⎦ ⎣ ⎦

So the node voltages are: v1 = 22.4 V, v 2 = 27.4 V, v 3 = 17.4 V, and v 4 = 15 (checked: LNAP 6/9/04)

P 4.3-10 Figure P 4.3-10 shows a measurement made in the laboratory. Your lab partner forgot to record the values of R1, R2, and R3. He thinks that the two resistors were 10-kΩ resistors and the other was a 5-kΩ resistor. Is this possible? Which resistor is the 5-kΩ resistor?

Figure P 4.3-10 Solution: Write a node equation to get

⎛ 12 − 4.5 ⎞ 4.5 4.5 − 6 7.5 4.5 1.5 −⎜ + + =0 ⇒ − + − =0 ⎜ R1 ⎟⎟ R 3 R R R R 2 1 3 2 ⎝ ⎠ Notice that

7.5 4.5 is either 0.75 mA or 1.5 mA depending on whether R1 is 10 kΩ or 5 kΩ. Similarly, R1 R3

is either 0.45 mA or 0.9 mA and

1.5 is either 0.15 mA or 0.3 mA. Suppose R1 and R2 are 10 kΩ R2

resistors and R3 is a 5 kΩ resistor. Then −

7.5 4.5 1.5 + − = −0.75 + 0.9 − 0.15 = 0 R1 R 3 R 2

It is possible that two of the resistors are 10 kΩ and the third is 5 kΩ. R3 is the 5 kΩ resistor. (checked: LNAP 6/9/04)

Figure P4.3-11 P4.3-11. Determine the values of the power supplied by each of the sources in the circuit shown in Figure P4.3-11. Solution: First, label the node voltages. Next, express the resistor currents in terms of the node voltages.

Identify the supernode corresponding to the 10 V source

Apply KCL to the supernode to get 15 − ( v a + 10 ) 15 − v a v a + 10 v a + = + 4 6 8 3

⇒ 60 = 21 v a

⇒ v a = 2.857 V

The 15 V source supplies ⎛ 15 − ( v a + 10 ) 15 − v a + 15 ⎜ ⎜ 4 6 ⎝

⎞ 15 − 12.857 15 − 2.857 ⎞ ⎟ = 15 ⎛⎜ + ⎟ = 15 ( 2.56 ) = 38.4 W ⎟ 4 6 ⎝ ⎠ ⎠

⎛ 15 − v a v a ⎞ ⎛ 15 − 2.857 2.857 ⎞ The 10 V source supplies 10 ⎜ + ⎟ = 10 ⎜ + ⎟ = 10 (1.071) = 10.71 W 6 3 6 3 ⎝ ⎠ ⎝ ⎠

P 4.3-12

Determine the values of the node voltages of the circuit shown in Figure P 4.3-12.

Figure P 4.3-12 Solution: Express the voltage source voltages in terms of the node voltages:

v 2 − v1 = 8 and v 3 − v1 = 12 Apply KVL to the supernode to get v2 10 so

+

v1 4

+

v3 5

=0



2v 2 + 5v1 + 4v 3 = 0

2 ( 8 + v1 ) + 5v1 + 4 (12 + v1 ) = 0



v1 = −

64 V 11

The node voltages are

v1 = −5.818 V v 2 = 2.182 V v 3 = 6.182 V (checked: LNAP 6/21/04)

Figure P4.3-13 P4.3-13. Determine the values node voltages, v1 and v2, in the circuit shown in Figure P4.3-13. Solution: Apply KCL at node 1 to get v1 50

+

v1 − v 2 65

v1 − 60

+

80

1 1 ⎞ 60 ⎛ 1 ⎛ 1 ⎞ = 0 ⇒ ⎜ + + ⎟ v1 − ⎜ ⎟ v 2 = 80 ⎝ 50 65 80 ⎠ ⎝ 65 ⎠

Apply KCL at node 2 to get 0.1 =

v 2 − v1 65

+

v 2 − 60 75

1 ⎞ ⎛ 1 ⎞ ⎛ 1 = ⇒ − ⎜ ⎟ v1 + ⎜ + ⎟ v 2 = 0.1 ⎝ 65 ⎠ ⎝ 65 75 ⎠

In matrix form

1 1 ⎡1 ⎢ 50 + 65 + 80 ⎢ 1 ⎢ − ⎢⎣ 65

1 ⎤ ⎡ 60 ⎤ 65 ⎥ ⎡ v1 ⎤ ⎢ ⎥ ⎥ ⎢ ⎥ = 80 1 1 ⎥ ⎣v 2 ⎦ ⎢ ⎥ + ⎣0.1⎦ 65 75 ⎥⎦

Solving, we get

v1 = 13.2356 V and v2 = 22.3456 V



Figure P4.3-14 P4.3-14. The voltage source in the circuit shown in Figure P4.3-14 supplies 83.802 W. The current source supplies 17.572 W. Determine the values of the node voltages v1 and v2. Solution: From the power supplied by the current source we calculate

17.572 = v 2 ( 0.25 ) ⇒ v 2 = i3 =

Using Ohm’s law

17.572 = 70.288 V 0.25

70.288 − 80 = −0.4856 A 20

From the power supplied by the voltage source we calculate 83.802 = 80 i 6

⇒ i6 =

83.802 = 1.0475 V 80

i1 = − ( i 3 + i 6 ) = − ( −0.4856 + 1.0475 ) = −0.5619 A Using Ohm’s law In summary

−0.5619 =

v1 − 80 50

⇒ v1 = 80 + 50 ( −0.5619 ) = 51.905 V

v1 = 51.905 V and v2 = 70.288 V

Section 4-4 Node Voltage Analysis with Dependent Sources

P 4.4-1 The voltages va, vb, and vc in Figure P 4.4-1 are the node voltages corresponding to nodes a, b, and c. The values of these voltages are: va = 8.667 V, vb = 2 V,

and vc = 10 V

Determine the value of A, the gain of the dependent source.

Figure P 4.4-1 Solution: Express the resistor currents in terms of the node voltages:

va − vc = 8.667 − 10 = −1.333 A and 1 v −v 2 − 10 = −4 A i 2= b c = 2 2 i 1=

Apply KCL at node c: i1 + i 2 = A i1 ⇒ − 1.333 + ( −4 ) = A (−1.333) ⇒

A=

−5.333 =4 −1.333

(checked using LNAP 8/13/02)

P 4.4-2

Find ib for the circuit shown in Figure P 4.4-2.

Answer: ib = –12 mA

Figure P 4.4-2 Solution:

Write and solve a node equation:

va − 6 v v − 4va + a + a = 0 ⇒ va = 12 V 1000 2000 3000

ib =

va − 4va = −12 mA 3000 (checked using LNAP 8/13/02)

P 4.4-3 Determine the node voltage vb for the circuit of Figure P 4.4-3. Answer: vb = 1.5 V

Figure P 4.4-3 Solution:

First express the controlling current in terms of the node voltages: 2 − vb i = a 4000 Write and solve a node equation: −

2 − vb v ⎛ 2 − vb ⎞ + b − 5⎜ ⎟ = 0 ⇒ vb = 1.5 V 4000 2000 ⎝ 4000 ⎠ (checked using LNAP 8/14/02)

P 4.4-4 The circled numbers in Figure P 4.4-4 are node numbers. The node voltages of this circuit are v1 = 10 V, v2 = 14 V, and v3 = 12 V.

(a)

Determine the value of the current ib.

(b)

Determine the value of r, the gain of the CCVS.

Answers: (a) –2 A (b) 4 V/A

Figure P 4.4-4 Solution:

Apply KCL to the supernode of the CCVS to get 12 − 10 14 − 10 1 + − + i b = 0 ⇒ i b = −2 A 4 2 2 Next 10 − 12 1⎫ =− ⎪ V −2 =4 4 2⎬ ⇒ r = 1 A − r i a = 12 − 14 ⎪⎭ 2

ia =

(checked using LNAP 8/14/02)

P 4.4-5 Determine the value of the current ix in the circuit of Figure P 4.4-5. Answer: ix = 2.4 A

Figure P 4.4-5 Solution:

First, express the controlling current of the CCVS in v2 terms of the node voltages: i x = 2 Next, express the controlled voltage in terms of the node voltages: v2 24 12 − v 2 = 3 i x = 3 V ⇒ v2 = 2 5 so ix = 12/5 A = 2.4 A. (checked using ELab 9/5/02)

Figure P4.4-6 P4.4-6 The encircled numbers in the circuit shown Figure P4.4-6 are node numbers. Determine the value of the power supplied by the CCVS . P4.4-6

First, express the contolling current of the CCVS in terms of the node voltages: ia =

v2 20

⎛ v2 ⎞ v1 = 12 V and v 3 = 40 i a = 40 ⎜ ⎟ = 2 v 2 ⎝ 20 ⎠ Next, express the resistor currents in terms of the node voltages:

Notice that

Apply KCL at node 2 to get

Then The CCVS supplies

12 − v 2

+

v2

+

v2 − 2v2

= 0 ⇒ v 2 = 16 V 5 20 10 v 2 16 ia = = = 0.8 A and v 3 = 40 i a = 40 ( 0.8 ) = 32 V 20 20 ⎛ v3 − v2 ⎞ ⎛ 32 − 16 ⎞ v3 ⎜ ⎟ = 32 ⎜ ⎟ = 51.2 W ⎝ 10 ⎠ ⎝ 10 ⎠

Figure P4.4-7 P4.4-7 The encircled numbers in the circuit shown Figure 4.4-27 are node numbers. The corresponding node voltages are:

v1 = 9.74 V and v2 = 6.09 V Determine the values of the gains of the dependent sources, r and g. Solution:

Using Ohm’s law, i b =

Then

v1 8

=

9.74 = 1.2175 A . Using KVL, the voltage across the CCVS is 8 r i b = v1 − v 2 = 9.74 − 6.09 = 3.65 V r=

r ib ib

=

3.65 = 2.9979 V/A 1.2175

Using KVL, v b = 12 − v1 = 12 − 9.74 = 2.26 V . Apply KCL to the supernode corresponding to the CCVS to get 12 − v1 v1 v 2 12 − 9.74 9.74 6.09 = + + g vb ⇒ = + + g v b ⇒ g v b = −1.6963 A 8 8 8 8 8 8 g v b −1.6963 g= = = −0.7506 A/V Then vb 2.26

P 4.4-8 Determine the value of the power supplied by the dependent source in Figure P 4.4-8.

Figure P 4.4-8 Solution:

Label the node voltages. First, v2 = 10 V due to the independent voltage source. Next, express the controlling current of the dependent source in terms of the node voltages: ia =

v3 − v 2 16

=

v 3 − 10 16

Now the controlled voltage of the dependent source can be expressed as ⎛ v 3 − 10 ⎞ v1 − v 3 = 8 i a = 8 ⎜ ⎟ ⎝ 16 ⎠



v1 =

3 v3 − 5 2

Apply KCL to the supernode corresponding to the dependent source to get v1 − v 2

+

4

v1 12

+

v3 − v 2 16

+

v3 8

=0

Multiplying by 48 and using v2 = 10 V gives 16v1 + 9v 3 = 150 Substituting the earlier expression for v1 ⎛3 ⎞ 16 ⎜ v 3 − 5 ⎟ + 9v 3 = 150 ⎝2 ⎠



v 3 = 6.970 V

Then v1 = 5.455 V and ia = -0.1894 A. Applying KCL at node 2 gives v1 12 So

= ib +

10 − v1 4



12 i b = −3 + 4 v1 = −30 + 4 ( 5.455 )

i b = −0.6817 A.

Finally, the power supplied by the dependent source is

p = ( 8 i a ) i b = 8 ( −0.1894 ) ( −0.6817 ) = 1.033 W (checked: LNAP 5/24/04)

P 4.4-9 The node voltages in the circuit shown in Figure P 4.4-9 are

v1 = 4 V, v2 = 0 V, and v3 = –6 V Determine the values of the resistance, R, and of the gain, b, of the CCCS.

Figure P 4.4-9 Solution:

Apply KCL at node 2: i a + bi a = i b = but ia =

v3 − v 2 20

v 2 − v1 40

=

=

−6 − ( 0 ) = −0.3 A 20

0−4 = −0.1 40

so

(1 + b )( −0.1) = ( −0.3)



b=2

A A

Next apply KCL to the supernode corresponding to the voltage source. v1 10

+ 2 ia +

v3 R

=0



4 −6 + 2 ( −0.1) + =0 10 R



R=

6 = 30 Ω .2 (checked: LNAP 6/9/04)

P 4.4-10 The value of the node voltage at node b in the circuit shown in Figure P 4.4-10 is vb = 18 V. (a)

Determine the value of A, the gain of the dependent source.

(b)

Determine the power supplied by the dependent source. Figure P 4.4-10

Solution: (a) Express the controlling voltage of the dependent source in terms of the node voltages:

va = 9 − vb Apply KCL at node b to get 9 − vb 100

= A (9 − v b ) +

vb 200



A=

18 − 3v b

200 ( 9 − v b )

= 0.02

(b) The power supplied by the dependent source is

− ( Av a ) v b = − ( 0.02 ( 9 − 18 ) ) (18 ) = 3.24 W (checked: LNAP 6/06/04)

P 4.4-11 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.4-11.

Figure P 4.4-11 Solution: This circuit contains two ungrounded voltage sources, both incident to node x. In such a circuit it is necessary to merge the supernodes corresponding to the two ungrounded voltage sources into a single supernode. That single supernode separates the two voltage sources and their nodes from the rest of the circuit. It consists of the two resistors and the current source. Apply KCL to this supernode to get v x − 20 v x + +4=0 ⇒ v x = 10 V . 2 10

The power supplied by the dependent source is

( 0.1 v ) ( −30 ) = −30 W . x

(checked: LNAP 6/6/04)

P 4.4-12 Determine values of the node voltages, v1, v2, v3, v4, and v5, in the circuit shown in Figure P 4.4-12.

Figure P 4.4-12. Solution:

Express the voltages of the independent voltage sources in terms of the node voltages v1 − v 2 = 16 and v 4 − v 5 = 8 Express the controlling current of the dependent source in terms of the node voltages ix =

v3 6

Express the controlled voltage of the dependent source in terms of the node voltages ⎛ v3 ⎞ v 2 − v 4 = 4i x = 4 ⎜ ⎟ ⎝6⎠



− 6v 2 + 4v 3 + 6v 4 = 0

Apply KCL to the supernode to get v1 − v 3 2

+

v 4 − v3 3

+

v5 8

=1



12v1 − 20v 3 + 8v 4 + 3v 5 = 24

Apply KCL at node 3 to get v 3 − v1 2

+

v3 6

+

v3 − v 4 3

=0



− 3v1 + 6v 2 − 2v 4 = 0

Solving, e.g. using MATLAB, gives 0 0 ⎤ ⎡ v1 ⎤ ⎡16 ⎤ ⎡ 1 −1 0 ⎢ ⎥ ⎢0 0 0 1 −1⎥⎥ ⎢v 2 ⎥ ⎢⎢ 8 ⎥⎥ ⎢ ⎢ 0 −6 4 6 0 ⎥ ⎢v 3 ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 12 0 20 8 3 − ⎢ ⎥ ⎢v 4 ⎥ ⎢ 24 ⎥ ⎢⎣ −3 0 6 −2 0 ⎥⎦ ⎢⎣ v 5 ⎥⎦ ⎢⎣ 0 ⎥⎦



⎡ v1 ⎤ ⎡ 24 ⎤ ⎢v ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ 8 ⎥ ⎢ v 3 ⎥ = ⎢12 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢v 4 ⎥ ⎢ 0 ⎥ ⎢ v 5 ⎥ ⎢⎣ −8⎥⎦ ⎣ ⎦

(checked: LNAP 6/13/04)

P 4.4-13 Determine values of the node voltages, v1, v2, v3, v4, and v5, in the circuit shown in Figure P 4.4-13.

Figure P 4.4-13 Solution: Express the voltage source voltages in terms of the node voltages:

v1 − v 2 = 8 and v 4 − v 3 = 16 Express the controlling current of the dependent source in terms of the node voltages: ix =

v 2 − v3 10

+

v1 − v 3 5

= 0.2v1 + 0.1v 2 − 0.3v 3

Express the controlled voltage of the dependent source in terms of the node voltages: v 5 = 4i x = 0.8v1 = 0.4v 2 − 1.2v 3



0.8v1 + 0.4v 2 − 1.2v 3 − v 5 = 0

Apply KVL to the supernodes v1 − v 5

+

2

v2 − v4

v4 − v2 4

4 +

v3 8

+

+

v 2 − v3 10

v3 − v 2 10

+

+

v1 − v 3 5

v 3 − v1 5

=0

=2

⇒ ⇒

14v1 + 7v 2 − 6v 3 − 5v 4 − 10v 5 = 0 − 8v1 − 14v 2 + 17v 3 + 10v 4 = 80

Solving, e.g. using MATLAB, gives 0 0 0 ⎤ ⎡ v1 ⎤ ⎡ 8 ⎤ −1 ⎡1 ⎢ ⎥ ⎢0 0 1 0 ⎥⎥ ⎢v 2 ⎥ ⎢⎢16 ⎥⎥ −1 ⎢ ⎢ 0.8 0.4 −1.2 0 −1 ⎥ ⎢ v 3 ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 7 −6 −5 −10 ⎥ ⎢v 4 ⎥ ⎢ 0 ⎥ ⎢ 14 ⎢⎣ −8 −14 17 10 0 ⎥⎦ ⎢⎣ v 5 ⎥⎦ ⎢⎣80 ⎥⎦



⎡ v1 ⎤ ⎡11.32 ⎤ ⎢v ⎥ ⎢ ⎥ ⎢ 2 ⎥ ⎢ 3.32 ⎥ ⎢ v 3 ⎥ = ⎢ 2.11 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢v 4 ⎥ ⎢18.11⎥ ⎢ v 5 ⎥ ⎢⎣ 7.85 ⎥⎦ ⎣ ⎦

(checked: LNAP 6/13/04)

P 4.4-14 The voltages v1, v2, v3, and v4 are the node voltages corresponding to nodes 1, 2, 3, and 4 in Figure P 4.4-14. Determine the values of these node voltages.

Figure P 4.4-14 Solution:

Express the controlling voltage and current of the dependent sources in terms of the node voltages: v3 − v4 v a = v 4 and i b = R2 Express the voltage source voltages in terms of the node voltages: v1 = V s and v 2 − v 3 = Av a = Av 4 Apply KCL to the supernode corresponding to the dependent voltage source v 2 − v1 R1

+

v3 − v4 R2

= Is

⇒ − R 2 v1 + R 2 v 2 + R1 v 3 − R1 v 4 = R1 R 2 I s

Apply KCL at node 4: B

v3 − v4 R2

+

v3 − v4 R2

=

v4 R3





( B + 1) v 3 − ⎜⎜ B + 1 + ⎝

R2 ⎞ ⎟ v4 = 0 R 3 ⎟⎠

Organizing these equations into matrix form: ⎡ 1 ⎢ 0 ⎢ ⎢− R2 ⎢ ⎢ ⎢ 0 ⎢⎣ With the given values:

0 1 R2 0

⎤ ⎥ ⎡ v1 ⎤ ⎡ V s ⎤ ⎥ ⎢v ⎥ ⎢ ⎥ ⎥ ⎢ 2⎥ = ⎢ 0 ⎥ ⎥ ⎢v 3 ⎥ ⎢ R R I ⎥ ⎛ R 2 ⎞⎥ ⎢ ⎥ ⎢ 1 2 s ⎥ B +1 − ⎜ B +1+ ⎟⎟ ⎥ ⎣⎢v 4 ⎦⎥ ⎣ 0 ⎦ ⎜ R 3 ⎝ ⎠ ⎥⎦ 0 −1 R1

0 −A − R1

⎡ v1 ⎤ ⎡ 25 ⎤ 0 0 0 ⎤ ⎡ v1 ⎤ ⎡ 25 ⎤ ⎡ 1 ⎢v ⎥ ⎢ ⎢v ⎥ ⎢ ⎢ 0 ⎥ ⎥ −5 ⎥ ⎢ 2 ⎥ ⎢ 0 ⎥ 1 −1 44.4 ⎥⎥ 2⎥ ⎢ ⎢ ⎢ = ⇒ = ⎢ v 3 ⎥ ⎢ 8.4 ⎥ ⎢ −20 20 10 −10 ⎥ ⎢ v 3 ⎥ ⎢ 400 ⎥ ⎢ ⎥ ⎢ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎥ 0 4 −4.667 ⎦ ⎢⎣v 4 ⎥⎦ ⎣ 0 ⎦ ⎢⎣ v 4 ⎥⎦ ⎣ 7.2 ⎦ ⎣ 0

(Checked using LNAP 9/29/04)

P 4.4-15 The voltages v1, v2, v3, and v4 in Figure P 4.4-15 are the node voltages corresponding to nodes 1, 2, 3, and 4. The values of these voltages are

v1 = 10 V, v2 = 75 V, v3 = –15 V, and v4 = 22.5 V Determine the values of the gains of the dependent sources, A and B, and of the resistance R1.

Figure P 4.4-15 Solution: Express the controlling voltage and current of the dependent sources in terms of the node voltages: v a = v 4 = 22.5 V ib =

and

v3 − v4 R2

=

−15 − 22.5 = −0.75 50

Express the dependent voltage source voltage in terms of the node voltages: v 2 − v 3 = Av a = Av 4 A=

so

v 2 − v3 v4

=

75 − ( −15 ) = 4 V/V 22.5

Apply KCL to the supernode corresponding to the dependent voltage source v 2 − v1 R1

+

v3 − v4 R2

= Is



75 − 10 −15 − 22.5 + = 2.5 ⇒ R1 = 20 Ω R1 50

Apply KCL at node 4: v3 − v 4 R2

=

v4 R3

+B

v3 − v 4 R2



−15 − 22.5 22.5 −15 − 22.5 = +B ⇒ B = 2.5 A/A 50 20 50

(Checked using LNAP 9/29/04)

P 4.4-16 The voltages v1, v2, and v3 in Figure P 4.4-16 are the node voltages corresponding to nodes 1, 2, and 3. The values of these voltages are

v1 = 12 V, v2 = 21 V, and v3 = –3 V (a)

Determine the values of the resistances R1 and R2.

(b)

Determine the power supplied by each source. Figure P 4.4-16

Solution:

v 2 − v1

R1 =

(b)

The power supplied by the voltage source is 12 ( 0.5 + 1.25 − 2 ) = −3 W . The power supplied by

2 − 0.5

=

v2 21 − 12 −3 = 6 Ω and R 2 = = =4Ω 1.5 1.25 − 2 −0.75

(a)

the 1.25-A current source is 1.25 ( −3 − 12 ) = −18.75 W . The power supplied by the 0.5-A current source is −0.5 ( 21) = −10.5 W . The power supplied by the 2-A current source is 2 ( 21 − ( −3) ) = 48 W .

P 4.4-17 The voltages v1, v2, and v3 in Figure P 4.4-17 are the node voltages corresponding to nodes 1, 2, and 3. The values of these voltages are

v1 = 12 V, v2 = 9.6 V,

and v3= –1.33 V

(a)

Determine the values of the resistances R1 and R2.

(b)

Determine the power supplied by each source. Figure P 4.4-18

Solution:

12 − ( −1.33) = 1.666 A 8 9.6 i2 = = 2.4 A 4 v 2 − v1 9.6 − 12 = = 6 Ω and R1 = 2 − i1 2 − 2.4

i1 = and (a)

R2 =

v3 i1 − 2

=

−1.33 = 3.98  4 Ω 1.666 − 2

(b) The power supplied by the voltage source is 12 ( 2.4 + 1.66 − 2 ) = 24.7 W . The power supplied by

the current source is 2 ( 9.6 − ( −1.33) ) = 21.9 W .

(Checked using LNAP 10/2/04)

P4.4-18 The voltages v 2 , v 3 and v 4 for the circuit shown in Figure P4.4-18 are:

v 2 = 16 V, v 3 = 8 V and v 4 = 6 V Determine the values of the following: (a) The gain, A, of the VCVS (b) The resistance R 5 (c) The currents i b and i c (d) The power received by resistor R 4

Figure P4.4-18 Solution: Given the node voltages

v 2 = 16 V, v 3 = 8 V and v 4 = 6 V A=

Av a va

=

⎛ v3 − v4 ⎞ R5 ⎜ ⎟ = v4 ⎝ 15 ⎠

ib =

16 − 8 V =4 8−6 V

⇒ R5 =

15 ( 6 ) = 45 Ω , 8−6

40 − 24 40 − 16 16 = 2 A and i c = − = 0.6667 A 12 12 12 p4 =

va2 15

=

22 = 0.2667 W 15

P4.4-19 Determine the values of the node voltages v1 and v 2 for the circuit shown in Figure P4.4-19.

Figure P4.4-19 P4.4-19 The node equations are

28 − v1 4 and

+ 3 v1 =

v1 − v 2 5

v1 − v 2 5

= 3 v1 +

v2 6

⇒ 5 ( 28 − v1 ) + 20 ( 3 v1 ) = 4 ( v1 − v 2 ) ⇒ 140 = −51 v1 − 4 v 2 + 4 v 3 = 3 v1 +

v2 6

+ 4 ( v1 − v 2 ) ⇒ 0 = 204 v1 − 109 v 2

Using MATLAB to solve these equations:

Consequently

v1 = −2.3937 V and v 2 = −4.4800 V

P4.4-20 The encircled numbers in Figure P4.4-20 are node numbers. Determine the values of v1 , v 2 and v 3 , the

node voltages corresponding to nodes 1, 2 and 3.

Figure P4.4-20 Solution: Apply KCL to the supernode corresponding to the horizontal voltage source to get

v1 10

=

va 2

=

v3 − v 2 2

=

v 3 − ( v1 + 10 ) 2

(

⇒ v1 = 5 v 3 − ( v1 + 10 )

)

⇒ 50 = −6 v1 + 5 v 3

Looking at the dependent source we notice that

(

v 3 = 5 v a = 5 ( v 3 − v 2 ) = 5 v 3 − ( v1 + 10 )

)

Using MATLAB to solve these equations:

Consequently Then

v1 = −50 V and v 3 = −50 V v 2 = v1 + 10 = −40 V

⇒ 50 = −5 v1 + 4 v 3

P4.4-21 Determine the values of the node voltages v1 , v 2 and v 3 for the circuit shown in Figure P4.4-21.

Figure P4.4-21 Solution:

Represent the controlling current of the dependent source in terms of the node voltages: i a = Represent the controlled voltage of the dependent source in terms of the node voltages: 4 i a = v1 − v 2

⎛ v3 ⎞ ⇒ = 4 ⎜ ⎟ = v1 − v 2 ⎝ 2⎠

⇒ 0 = v1 − v 2 − 2 v 3

Apply KCL to the supernode corresponding to the dependent voltage source: 12 − v1 2

=

v2 2

+

v 2 − v3 2

⇒ 12 − v1 = v 2 + v 2 − v 3 ⇒ 12 = v1 + 2 v 2 − v 3

Apply KCL to top node of the current source: v 2 − v3 2

+1 =

v3 2

⇒ v 2 − v 3 + 2 = v 3 ⇒ 2 = −v 2 + 2 v 3

Solving these equations using MATLAB gives v1 = 8.2857 V , v 2 = 3.1459 V and v 3 = 2.5714 V

v3 2

P4.4-22 Determine the values of the node voltages v1 , v 2 and v 3 for the circuit shown in Figure P4.4-22.

Figure P4.4-22 Solution: The node equations are: 12 − v1 2 4 ia =

v2 2

⎛ v3 ⎞ = 4 i a = 4 ⎜ ⎟ ⇒ 12 − v1 = 4 v 3 ⇒ 12 = v1 + 4 v 3 ⎝ 2⎠

+

v 2 − v3

v 2 − v3 2

⎛ v3 ⎞ v 2 v 2 − v3 ⇒ 4⎜ ⎟ = + 2 ⎝ 2⎠ 2

2

+1 =

v3 2

⇒ v 2 − v 3 + 2 = v 3 ⇒ 2 = −v 2 + 2 v 3

Solving these equations using MATLAB gives v1 = 28 V , v 2 = −10 V and v 3 = −4 V

⇒ 0 = 2 v 2 − 5 v3

Section 4-5 Mesh Current Analysis with Independent Voltage Sources P 4.5-1 Determine the mesh currents, i1, i2, and i3, for the circuit shown in Figure P 4.5-1. Answers: i1 = 3 A, i2 = 2 A, and i3 = 4 A

Figure P 4.5-1 Solution: The mesh equations are

2 i1 + 9 (i1 − i 3 ) + 3(i1 − i 2 ) = 0 15 − 3 (i1 − i 2 ) + 6 (i 2 − i 3 ) = 0

−6 (i 2 − i 3 ) − 9 (i1 − i 3 ) − 21 = 0 or 14 i1 − 3 i 2 − 9 i 3 = 0 −3 i1 + 9 i 2 − 6 i 3 = −15 −9 i1 − 6 i 2 + 15 i 3 = 21 so i1 = 3 A, i2 = 2 A and i3 = 4 A. (checked using LNAP 8/14/02)

P 4.5-2 The values of the mesh currents in the circuit shown in Figure P 4.5-2 are

i1 = 2 A, i2 = 3 A, and i3 = 4 A. Determine the values of the resistance R and of the voltages v1 and v2 of the voltage sources. Answers: R = 12 Ω, v1 = –4 V, and v2 = –28 V

Figure P 4.5-2 Solution: The mesh equations are: Top mesh: so

Bottom, right mesh: so Bottom left mesh so

4 (2 − 3) + R(2) + 10 (2 − 4) = 0 R = 12 Ω. 8 (4 − 3) + 10 (4 − 2) + v 2 = 0 v2 = −28 V. −v1 + 4 (3 − 2) + 8 (3 − 4) = 0

v1 = −4 V.

(checked using LNAP 8/14/02)

P 4.5-3 The currents i1 and i2 in Figure P 4.5-3 are the mesh currents. Determine the value of the resistance R required to cause va = –6 V. Answer: R = 4 Ω

Figure P 4.5-3 Solution:

Ohm’s Law: i 2 =

−6 = −0.75 A 8

KVL for loop 1: R i1 + 4 ( i1 − i 2 ) + 3 + 18 = 0

KVL for loop 2 + (−6) − 3 − 4 ( i1 − i 2 ) = 0 ⇒ − 9 − 4 ( i1 − ( −0.75 ) ) = 0 ⇒ i1 = −3 A

R ( −3) + 4 ( −3 − ( −0.75 ) ) + 21 = 0 ⇒ R = 4 Ω

(checked using LNAP 8/14/02)

P 4.5-4 Determine the mesh currents, ia and

ib, in the circuit shown in Figure P 4.5-4.

Figure P 4.5-4 Solution:

KVL loop 1:

KVL loop 2: Solving these equations:

25 ia − 2 + 250 ia + 75 ia + 4 + 100 (ia − ib ) = 0 450 ia −100 ib = −2 −100(ia − ib ) − 4 + 100 ib + 100 ib + 8 + 200 ib = 0 −100 ia + 500 ib = − 4 ia = − 6.5 mA , ib = − 9.3 mA (checked using LNAP 8/14/02)

Find the current i for the circuit of Figure P 4.5-5.

P 4.5-5 Hint:

A short circuit can be treated as a 0-V voltage source.

Figure P 4.5-5 Solution:

Mesh Equations: mesh 1 : 2i1 + 2 (i1 − i2 ) + 10 = 0 mesh 2 : 2(i2 − i1 ) + 4 (i2 − i3 ) = 0 mesh 3 : − 10 + 4 (i3 − i2 ) + 6 i3 = 0 Solving: 5 i = i2 ⇒ i = − = −0.294 A 17

(checked using LNAP 8/14/02)

P 4.5-6 Simplify the circuit shown in Figure P 4.5-6 by replacing series and parallel resistors by equivalent resistors. Next, analyze the simplified circuit by writing and solving mesh equations. (a) Determine the power supplied by each source. (b) Determine the power absorbed by the 30-Ω resistor.

Figure P 4.5-6 Solution: Replace series and parallel resistors with equivalent resistors:

60 Ω & 300 Ω = 50 Ω , 40 Ω + 60 Ω = 100 Ω and 100 Ω + 30 Ω + ( 80 Ω & 560 Ω ) = 200 Ω

so the simplified circuit is

The mesh equations are

200 i1 + 50 ( i1 − i 2 ) − 12 = 0 100 i 2 + 8 − 50 ( i1 − i 2 ) = 0

or ⎡ 250 −50 ⎤ ⎡ i1 ⎤ ⎡12 ⎤ ⎢ −50 150 ⎥ ⎢i ⎥ = ⎢ −8⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦



⎡ i1 ⎤ ⎡ 0.04 ⎤ ⎢i ⎥ = ⎢ ⎥ ⎣ 2 ⎦ ⎣ −0.04 ⎦

The power supplied by the 12 V source is 12 i1 = 12 ( 0.04 ) = 0.48 W . The power supplied by the 8 V source is −8i 2 = −8 ( −0.04 ) = 0.32 W . The power absorbed by the 30 Ω resistor is i12 ( 30 ) = ( 0.04 ) ( 30 ) = 0.048 W . 2

(checked: LNAP 5/31/04)

Section 4-6 Mesh Current Analysis with Voltage and Current Sources

P 4.6-1 Find ib for the circuit shown in Figure P 4.6-1. Answer: ib = 0.6 A Figure P 4.6-1 Solution:

1 A 2 mesh 2: 75 i2 + 10 + 25 i2 = 0 mesh 1: i1 =

⇒ i2 = − 0.1 A

ib = i1 − i2 = 0.6 A (checked using LNAP 8/14/02)

P 4.6-2 Find vc for the circuit shown in Figure P 4.6-2. Answer: vc = 15 V

Figure P 4.6-2 Solution: Mesh currents: mesh a: ia = − 0.25 A mesh b: ib = − 0.4 A Ohm’s Law: vc = 100(ia − ib ) = 100(0.15) =15 V

(checked using LNAP 8/14/02)

P 4.6-3 Find v2 for the circuit shown in Figure P 4.6-3. Answer: v2 = 2 V

Figure P 4.6-3 Solution:

Express the current source current as a function of the mesh currents: i1 − i2 = − 0.5 ⇒ i1 = i2 − 0.5 Apply KVL to the supermesh: 30 i1 + 20 i2 + 10 = 0 ⇒ 30 (i2 − 0.5) + 20i2 = − 10 50 i2 − 15 = − 10 ⇒ i2 =

5 = .1 A 50

i1 =−.4 A and v2 = 20 i2 = 2 V (checked using LNAP 8/14/02)

Find vc for the circuit shown in

P 4.6-4

Figure P 4.6-4.

Figure P 4.6-4 Solution:

Express the current source current in terms of the mesh currents: ib = ia − 0.02 Apply KVL to the supermesh: 250 ia + 100 (ia − 0.02) + 9 = 0 ∴ ia = − .02 A = − 20 mA vc = 100(ia − 0.02) = −4 V (checked using LNAP 8/14/02)

P 4.6-5

Determine the value of the voltage measured by the voltmeter in Figure P 4.6-5.

Answer: 8 V

Figure P 4.6-5 Solution: Label the mesh currents:

Express the current source current in terms of the mesh currents:

i 3 − i 1 = 2 ⇒ i1 = i 3 − 2 Supermesh: Lower, left mesh:

6 i1 + 3 i 3 − 5 ( i 2 − i 3 ) − 8 = 0 ⇒ 6 i1 − 5 i 2 + 8 i 3 = 8

−12 + 8 + 5 ( i 2 − i 3 ) = 0 ⇒ 5 i 2 = 4 + 5 i 3

Eliminating i1 and i2 from the supermesh equation: 6 ( i 3 − 2 ) − ( 4 + 5 i 3 ) + 8 i 3 = 8 ⇒ 9 i 3 = 24 ⎛ 24 ⎞ The voltage measured by the meter is: 3 i 3 = 3 ⎜ ⎟ = 8 V ⎝ 9 ⎠ (checked using LNAP 8/14/02)

Determine the value of the current measured by the ammeter in Figure P 4.6-6.

P 4.6-6 Hint:

Write and solve a single mesh equation.

Answer: –5/6 A

Figure P 4.6-6 Solution:

Mesh equation for right mesh: 4 ( i − 2 ) + 2 i + 6 ( i + 3) = 0 ⇒ 12 i − 8 + 18 = 0 ⇒ i = −

10 5 A=− A 12 6

(checked using LNAP 8/14/02)

Figure P4.6-7 P4.6-7 The mesh currents are labeled in the circuit shown Figure 4.6-7. The value of these mesh currents are: i1 = −1.1014 A, i 2 = 0.8986 A and i 3 = −0.2899 A a.) Determine the values of the resistances R1 and R 3 . b.) Detemine the value of the current sourc current. c.) Determine the value of the power supplied by the 12 V voltage source. Solution: Label the resistor currents and the current source currrents in terms of the mesh currents:

a.) Apply KVL to the supermesh corresponding to the current source to get R1 i1 + 12 + 24 ( i 2 − i 3 ) − 24 = 0 ⇒ R1 =

12 − 24 ( i 2 − i 3 ) i1

=

12 − 24 ( 0.8986 − ( −0.2899 ) ) −1.1014

= 15 Ω

Apply KVL to the rightmost mesh to get R 3 i 3 + 32 − 24 ( i 2 − i 3 ) = 0 ⇒ R 3 =

b.)

−32 + 24 ( i 2 − i 3 ) i3

=

−32 + 24 ( 0.8986 − ( −0.2899 ) ) −0.2899

= 12 Ω

I s = i 2 − i1 = 0.8986 − ( −1.1014 ) = 2 A

c.) Noticing that 12 V and i 2 adhere to the passive convention, the power supplied by the 12 V voltage

source is

−12 i 2 = −12 ( 0.8986 ) = −10.783 W .

P 4.6-8

Determine values of the mesh currents, i1,

i2, and i3, in the circuit shown in Figure P 4.6-8.

Figure P 4.6-8 Solution: Use units of V, mA and kΩ. Express the currents to the supermesh to get

i1 − i 3 = 2 Apply KVL to the supermesh to get 4 ( i 3 − i 3 ) + (1) i 3 − 3 + (1) ( i1 − i 2 ) = 0



i1 − 5 i 2 + 5 i 3 = 3

Apply KVL to mesh 2 to get 2i 2 + 4 ( i 2 − i 3 ) + (1) ( i 2 − i1 ) = 0

( −1) i1 + 7i 2 − 4i 3 = 0



Solving, e.g. using MATLAB, gives ⎡ 1 0 −1⎤ ⎡ i1 ⎤ ⎡ 2 ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ 1 −5 5 ⎥ ⎢i 2 ⎥ = ⎢ 3 ⎥ ⎢⎣ −1 7 −4 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 0 ⎥⎦



⎡ i 1 ⎤ ⎡ 3⎤ ⎢ ⎥ ⎢ ⎥ ⎢i 2 ⎥ = ⎢1⎥ ⎢ i 3 ⎥ ⎢⎣1⎥⎦ ⎣ ⎦ (checked: LNAP 6/21/04)

Figure P4.6-9 P4.6-9 The mesh currents are labeled in the circuit shown Figure 4.6-9. Determine the value of the mesh currents i1 and i 2 . Solution: Determine the value of the mesh currents i1 and i 2 .

i 2 = i1 + 4 A , 8 i1 + 12 i1 + 5 i 2 = 0 ⇒ 8 i1 + 12 i1 + 5 ( 4 + i1 ) = 0 ⇒ i1 = −0.8 A i 2 = i1 + 4 = −0.8 + 4 = 3.2 A

and

P 4.6-10 The mesh currents in the circuit shown in Figure P 4.6-10 are

i1 = –2.2213 A, i2 = 0.7787 A, and i3 = 0.0770 A (a) Determine the values of the resistances R1 and R3. (b) Determine the value of the power supplied by the current source.

Figure P 4.6-10 Solution: (a)

50 ( i 3 − i 2 ) + R 3i 3 + 32 = 0 ⇒ 50 ( 0.0770 − 0.7787 ) + R 3 ( 0.0770 ) + 32 = 0 ⇒ R 3 = 40 Ω

i1 R1 + 20i 2 + 50 ( i 2 − i 3 ) − 24 = 0 ⇒

R1 ( −2.2213) + 20 ( 0.7787 ) + 50 ( 0.7787 − 0.0770 ) = 24 ⇒

(b)

R1 = 12 Ω

I s = i 2 − i1 = 0.7787 − ( −2.2213) = 3 A

The power supplied by the current source is p = I s ( 24 − R1 i1 ) = 3 ( 24 − 12 ( −2.2213) ) = 152 W (checked: LNAP 6/19/04)

P 4.6-11

Determine the value of the voltage measured by the voltmeter in Figure P 4.6-11.

Hint: Apply KVL to a supermesh to determine the current in the 2-Ω resistor. Answer: 4/3 V

Figure P 4.6-11 Solution:

3 3 = i1 − i 2 ⇒ i1 = + i 2 . 4 4 ⎛3 ⎞ Apply KVL to the supermesh: −9 + 4i1 + 3 i 2 + 2 i 2 = 0 ⇒ 4 ⎜ + i 2 ⎟ + 5 i 2 = 9 ⇒ 9 i 2 = 6 ⎝4 ⎠ 2 4 so i 2 = A and the voltmeter reading is 2 i 2 = V 3 3 Express the current source current in terms of the mesh currents:

P 4.6-12 Hint:

Determine the value of the current measured by the ammeter in Figure P 4.6-12.

Apply KVL to a supermesh.

Answer: –0.333 A

Figure P 4.6-12 Solution:

Express the current source current in terms of the mesh currents: 3 = i1 − i 2

⇒ i1 = 3 + i 2 .

Apply KVL to the supermesh: −15 + 6 i1 + 3 i 2 = 0 ⇒ 6 ( 3 + i 2 ) + 3 i 2 = 15 ⇒ 9 i 2 = −3 Finally, i 2 = −

1 A is the current measured by the ammeter. 3

Figure P4.6-13 P4.6-13 Determine the values of the mesh currents i1, i2 and i3 and the output voltage vo in the circuit shown Figure 4.6-13. Solution: Notice that the current source are each in a single mesh. Consequently, i1 = 2.4 A and

i2 = 1.2 A. Label the resistor currents in terms of the mesh currents:

Apply KVL to mesh 3 to get 12 i 3 + 15 − 16 (1.2 − i 3 ) − 18 ( 2.4 − i 3 ) = 0 ⇒ 46 i 3 = 47.4 ⇒ i 3 = 1.0304 A Apply KVL to the rightmost mesh to get v o − 15 − 12 i 3 = 0 ⇒ v o = 15 + 12 (1.0304 ) = 27.3648 V

Figure P4.6-14 P4.6-14 Determine the values of the power supplied by the sources in the circuit shown Figure P4.6-14. Solution: First, label the mesh currents, taking advantage of the current sources. Next, express the resistor currents in terms of the mesh currents:

Apply KVL to the middle mesh:

15 i + 25 ( i + 3) + 10 ( i − 5 ) = 0 ⇒ i = −

⎛ 1 ⎞ The 5 A current source supplies 5 (10 )( i − 5 ) = 5 (10 ) ⎜ − − 5 ⎟ = 275 W ⎝ 2 ⎠ ⎛ 1 ⎞ The 3 A current source supplies 3 ( 25 )( i + 3) = 3 ( 25 ) ⎜ − + 3 ⎟ = 187.5 W ⎝ 2 ⎠

1 A 2

Figure P4.6-15 P4.6-15 Determine the values of the resistance R and of the power supplied by the 6 A current source in the circuit shown Figure P4.6-15. Solution: First, label the mesh currents:

Notice that

i1 = −2.5 A, i 2 = 1 A and 6 = i 3 + 2.5 ⇒ i 3 = 3.5 A

Next, express the resistor currents in terms of the mesh currents:

Apply KVL to the bottom, left mesh: Apply KVL to the right mesh The 6 A current source supplies

4 ( 3.5 ) − 10 ( 2.5 ) + R (1) = 0 ⇒ R = 11 Ω

3.5 ( 5 ) + 2.5 (10 ) − v = 0 ⇒ v = 42.5 V 6 v = 6 ( 42.5 ) = 255 W

Section 4-7 Mesh Current Analysis with Dependent Sources P 4.7-1 Find v2 for the circuit shown in Figure P 4.7-1. Answer: v2 = 10 V

Figure P 4.7-1 Solution: Express the controlling voltage of the dependent source as a function of the mesh current v2 = 50 i1

Apply KVL to the right mesh: −100 (0.04(50i1 ) − i1 ) + 50i1 + 10 = 0 ⇒ i1 = 0.2 A v2 = 50 i1 = 10 V (checked using LNAP 8/14/02)

P4.7-2 Determine the values of the power supplied by the voltage source and by the CCCS in the circuit shown Figure P4.7-2

Figure P4.7-2 Solution: First, label the mesh currents, taking advantage of the current sources. Next, express the resistor currents in terms of the mesh currents:

Apply KVL to the left mesh:

4000 i a + 2000 ( 6 i a ) − 2 = 0 ⇒ i a =

The 2 A voltage source supplies

The CCCS supplies

2 i a = 2 ( 0.125 × 10−3 ) = 0.25 mW

( 5 i a ) ⎡⎣( 2000 ) ( 6 i a )⎤⎦ = ( 60 ×10 )( 0.125 ×10 ) 3

−3 2

1 = 0.125 mA 8

= 0.9375 × 10−3 = 0.9375 mW

P 4.7-3 Find vo for the circuit shown in Figure P 4.7-3. Answer: vo = 2.5 V

Figure P 4.7-3 Solution: Express the controlling current of the dependent source as a function of the mesh current:

ib = .06 − ia Apply KVL to the right mesh:

−100 (0.06 − i a ) + 50 (0.06 − i a ) + 250 i a = 0 ⇒

ia = 10 mA

vo = 50 i b = 50 (0.06 − 0.01) = 2.5 V

Finally:

(checked using LNAP 8/14/02)

P 4.7-4 Determine the mesh current ia for the circuit shown in Figure P 4.7-4. Answer: ia = –24 mA

Figure P 4.7-4 Solution: Express the controlling voltage of the dependent source as a function of the mesh current:

vb = 100 (.006 − ia ) Apply KVL to the right mesh: −100 (.006 − ia ) + 3 [100(.006 − ia )] + 250 ia = 0 ⇒ ia = −24 mA (checked using LNAP 8/14/02)

P 4.7-5 Although scientists continue to debate exactly why and how it works, the process of utilizing electricity to aid in the repair and growth of bones—which has been used mainly with fractures—may soon be extended to an array of other problems, ranging from osteoporosis and osteoarthritis to spinal fusions and skin ulcers. An electric current is applied to bone fractures that have not healed in the normal period of time. The process seeks to imitate natural electrical forces within the body. It takes only a small amount of electric stimulation to accelerate bone recovery. The direct current method uses an electrode that is implanted at the bone. This method has a success rate approaching 80 percent. The implant is shown in Figure P 4.7-5a and the circuit model is shown in Figure P 4.7-5b. Find the energy delivered to the cathode during a 24-hour period. The cathode is represented by the dependent voltage source and the 100-kΩ resistor.

Figure P 4.7-5 Solution:

Apply KVL to left mesh : − 3 + 10 × 103 i1 + 20 × 103 ( i1 − i2 ) = 0 ⇒ 30 × 103 i1 − 20 × 103 i2 = 3 Apply KVL to right mesh : 5 ×103 i1 + 100 ×103 i2 + 20 × 103 ( i2 − i1 ) = 0 ⇒ i1 = 8i2 Solving (1) & ( 2 ) simultaneously

⇒ i1 =

6 3 mA, i2 = mA 55 220

(1) ( 2)

Power delivered to cathode =

( 5 i1 ) ( i2 ) + 100 ( i2 )2

( 55)( 3 220) + 100 ( 3 220) = 0.026 mW ( 2.6 ×10−5 W ) ( 24 hr ) (3600 s hr ) = 2.25 J 2

= 5 6

∴ Energy in 24 hr. =

Figure P4.7-6 P4.7-6 Determine the value of the power supplied by the VCCS in the circuit shown Figure P4.7-6. Solution: First, label the mesh currents.

Next, express the controlling voltage of the VCCS in terms of the mesh currents: v a = 20 i 2 i1 = 2 A and i 3 =

va

= 10 i 2 2 Next, express the resistor currents in terms of the mesh currents: Notice that

Apply KVL to the middle mesh: Consequently The VCCS supplies

20 i 2 + 2 ( i 2 − 10 i 2 ) + 8 ( i 2 − 2 ) = 0 ⇒ i 2 = 1.6 A

v a = 20 i 2 = 20 (1.6 ) = 32 V and i 3 =

va 2

=

32 = 16 A 2

⎡ 2 ( i 3 − i 2 ) ⎤ = 32 ( 2 )(16 − 1.6 ) = 460.8 W ⎦ 2 2 ⎣

va

P 4.7-7

The currents i1, i2 and i3 are the mesh

currents of the circuit shown in Figure P 4.7-7. Determine the values of i1, i2, and i3.

Figure P 4.7-7 Solution: Express va and ib, the controlling voltage and current of the dependent sources, in terms of the mesh currents v a = 5 ( i 2 − i 3 ) and i b = −i 2

Next express 20 ib and 3 va, the controlled voltages of the dependent sources, in terms of the mesh currents 20 i b = −20 i 2 and 3 v a = 15 ( i 2 − i 3 ) Apply KVL to the meshes

−15 ( i 2 − i 3 ) + ( −20 i 2 ) + 10 i1 = 0 − ( −20 i 2 ) + 5 ( i 2 − i 3 ) + 20 i 2 = 0 10 − 5 ( i 2 − i 3 ) + 15 ( i 2 − i 3 ) = 0

These equations can be written in matrix form ⎡10 −35 15 ⎤ ⎡ i1 ⎤ ⎡ 0 ⎤ ⎢ 0 45 −5 ⎥ ⎢i ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ 0 10 −10 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ −10 ⎥⎦ Solving, e.g. using MATLAB, gives i1 = −1.25 A, i 2 = +0.125 A, and i 3 = +1.125 A (checked: MATLAB & LNAP 5/19/04)

P 4.7-8

Determine the value of the power supplied by

the dependent source in Figure P 4.7-8.

Figure P 4.7-8 Solution: Label the mesh currents:

Express ia, the controlling current of the CCCS, in terms of the mesh currents i a = i 3 − i1 Express 2 ia, the controlled current of the CCCS, in terms of the mesh currents: i 1 − i 2 = 2 i a = 2 ( i 3 − i 1 ) ⇒ 3 i1 − i 2 − 2 i 3 = 0 Apply KVL to the supermesh corresponding to the CCCS: 80 ( i1 − i 3 ) + 40 ( i 2 − i 3 ) + 60 i 2 + 20 i1 = 0



100i1 + 100i 2 − 120i 3 = 0

Apply KVL to mesh 3 10 + 40 ( i 3 − i 2 ) + 80 ( i 3 − i1 ) = 0



-80 i1 − 40 i 2 + 120 i 3 = −10

These three equations can be written in matrix form −1 −2 ⎤ ⎡ i1 ⎤ ⎡ 0 ⎤ ⎡ 3 ⎢100 100 −120 ⎥ ⎢i ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −80 −40 120 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ −10 ⎥⎦ Solving, e.g. using MATLAB, gives i1 = −0.2 A, i 2 = −0.1 A and i 3 = −0.25 A Apply KVL to mesh 2 to get v b + 40 ( i 2 − i 3 ) + 60 i 2 = 0 ⇒ v b = −40 ( −0.1 − ( −0.25 ) ) − 60 ( −0.1) = 0 V So the power supplied by the dependent source is p = v b ( 2i a ) = 0 W . (checked: LNAP 6/7/04)

P 4.7-9

Determine the value of the resistance R

in the circuit shown in Figure P 4.7-9.

Figure P 4.7-9 Solution: Notice that i b and 0.5 mA are the mesh currents. Apply KCL at the top node of the dependent source to

get 1 mA 6 Apply KVL to the supermesh corresponding to the dependent source to get i b + 0.5 × 10−3 = 4 i b

⇒ ib =

(

)

−5000 i b + (10000 + R ) 0.5 × 10−3 − 25 = 0

(

)

⎛1 ⎞ −5000 ⎜ × 10−3 ⎟ + (10000 + R ) 0.5 × 10−3 = 25 ⎝6 ⎠ 125 6 R= = 41.67 kΩ 0.5 ×10−3

P 4.7-10 The circuit shown in Figure P 4.7-10 is the small signal model of an amplifier. The input to the amplifier is the voltage source voltage, vs. The output of the amplifier is the voltage vo. (a) The ratio of the output to the input, vo/vs, is called the gain of the amplifier. Determine the gain of the amplifier. (b) The ratio of the current of the input source to the input voltage, ib/vs, is called the input resistance of the amplifier. Determine the input resistance.

(checked: LNAP 6/21/04)

Figure P 4.7-10

Solution: The controlling and controlled currents of the CCCS, i b and 40 i b, are the mesh currents. Apply KVL to the left mesh to get 1000 i b + 2000 i b + 300 ( i b + 40i b ) − v s = 0 ⇒ 15300i b = v s

The output is given by (a) The gain is (b) The input resistance is

v o = −3000 ( 40 i b ) = −120000 i b vo vs

=−

120000 = −7.84 V/V 15300

vs ib

= 15300 Ω

(checked: LNAP 5/24/04)

P 4.7-11

Determine the values of the mesh currents of

the circuit shown in Figure P 4.7-11.

Figure P 4.7-11 Solution: Label the mesh currents.

Express ix in terms of the mesh currents: i x = i1 Express 4ix in terms of the mesh currents: 4i x = i 3 Express the current source current in terms of the mesh currents to get: 0.5 = i1 − i 2



i 2 = i x − 0.5

Apply KVL to supermesh corresponding to the current source to get 5i1 + 20 ( i1 − i 3 ) + 10 ( i 2 − i 3 ) + 25i 2 = 0 Substituting gives

5i x + 20 ( −3i x ) + 10 ( i x − 0.5 − 4i x ) + 25 ( i x − 0.5 ) = 0



ix = −

35 = −0.29167 120

So the mesh currents are i1 = i x = −0.29167 A i 2 = i x − 0.5 = −0.79167 A i 3 = 4i x = −1.1667 A

(checked: LNAP 6/21/04)

P 4.7-12

The currents i1, i2, and i3 are the

mesh currents corresponding to meshes 1, 2, and 3 in Figure P 4.7-12. Determine the values of these mesh currents.

Figure P 4.7-12 Solution:

Express the controlling voltage and current of the dependent sources in terms of the mesh currents: v a = R 3 ( i1 − i 2 ) and i b = i 3 − i 2 Express the current source currents in terms of the mesh currents: i 2 = − I s and i1 − i 3 = B i b = B ( i 3 − i 2 ) Consequently i1 − ( B + 1) i 3 = B I s

Apply KVL to the supermesh corresponding to the dependent current source R1 i 3 + A R 3 ( i1 − i 2 ) + R 2 ( i 3 − i 2 ) + R 3 ( i1 − i 2 ) − V s = 0 or

( A + 1) R 3 i1 − ( R 2 + ( A + 1) R3 ) i 2 + ( R1 + R 2 ) i 3 = Vs

Organizing these equations into matrix form: ⎡ 0 ⎢ 1 ⎢ ⎢ ⎢⎣( A + 1) R 3

1 0 − ( R 2 + ( A + 1) R 3 )

0 ⎤ ⎡ i1 ⎤ ⎡ − I s ⎤ ⎥⎢ ⎥ ⎢ ⎥ − ( B + 1) ⎥ ⎢i 2 ⎥ = ⎢ B I s ⎥ ⎥ R1 + R 2 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣ V s ⎥⎦

With the given values: ⎡ i1 ⎤ ⎡ −0.8276 ⎤ 1 0 ⎤ ⎡ i1 ⎤ ⎡ −2 ⎤ ⎡0 ⎢ ⎥ ⎢1 ⎥ i = ⎢ 6 ⎥ ⇒ ⎢i ⎥ = ⎢ −2 ⎥ A − 0 4 ⎢ 2⎥ ⎢ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎥ ⎢ i 3 ⎥ ⎢⎣ −1.7069 ⎥⎦ ⎢⎣ 60 −80 50 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 25 ⎥⎦ ⎣ ⎦ (Checked using LNAP 9/29/04)

P 4.7-13 The currents i1, i2, and i3 are the mesh currents corresponding to meshes 1, 2, and 3 in Figure P 4.7-13. The values of these currents are

i1 = –1.375 A, i2 = –2.5 A, and i3 = –3.25 A Determine the values of the gains of the dependent source, A and B.

Figure P 4.7-13 Solution: Express the controlling voltage and current of the dependent sources in terms of the mesh currents: v a = 20 ( i1 − i 2 ) = 20 ( −1.375 − ( −2.5 ) ) = 22.5

and

i b = i 3 − i 2 = −3.25 − ( −2.5 ) = −0.75 A

Express the current source currents in terms of the mesh currents: i 2 = −2.5 A and i 3 − i1 = B i b

⇒ − 1.375 − ( −2.5 ) = B ( −0.75 ) ⇒ B = 2.5 A/A

Apply KVL to the supermesh corresponding to the dependent current source 0 = 20 i 3 + Av a + 50 i b + v a − 10 = 20 ( −3.25 ) + A ( 22.5 ) + 50 ( −0.75 ) + 22.5 − 10 ⇒

A = 4 V/V

(Checked using LNAP 9/29/04)

P 4.7-14 Determine the current i in the circuit shown in Figure P 4.7-14. Answer: i = 3 A

Figure P 4.7-14 Solution:

Label the node voltages as shown. The controlling currents of the CCCS is expressed as i = The node equations are and

12 =

va 28

+

va − vb 4

va − vb 4 +

va 14

=

+

va 28

.

va 14

vb 8

Solving the node equations gives v a = 84 V and v b = 72 V . Then i =

va 28

=

84 =3A . 28

(checked using LNAP 6/16/05)

P4.7-15 Determine the values of the mesh currents i1 and i2 for the circuit shown in Figure P4.7-15.

Figure P4.7-15 Solution: Expressing the dependent source currents in terms of the mesh currents we get:

i1 = 4 i a = 4 ( i 2 + 1) ⇒ 4 = i1 − 4 i 2 Apply KVL to mesh 2 to get 2 i 2 + 2 ( i 2 + 1) − 2 ( i1 − i 2 ) = 0 ⇒ − 2 = −2 i1 + 6 i 2

Solving these equations using MATLAB we get i1 = −8 A and i2 = −3 A

P4.7-16 Determine the values of the mesh currents i1 and i2 for the circuit shown in Figure P4.7-16 .

Figure P4.7-16

Apply KVL to mesh 1 to get 2 i1 + 4 i a + 2 ( i1 − i 2 ) − 12 = 0 ⇒ 2 i1 + 4 ( i 2 + 1) + 2 ( i1 − i 2 ) − 12 = 0 ⇒ 8 = 4 i1 + 2 i 2 Apply KVL to mesh 2 to get 2 i 2 + 2 ( i 2 + 1) − 2 ( i1 − i 2 ) = 0 ⇒ − 2 = −2 i1 + 6 i 2

Solving these equations using MATLAB we get i1 = 1.8571 A and i2 = 0.2857 A

Section 4.8 The Node Voltage Method and Mesh Current Method Compared P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, vs and is, and one output vo. The output is related to the inputs by the equation vo = ais + bvs where a and b are constants to be determined. Determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations. Figure P 4.8-2 Solution: (a)

Apply KVL to meshes 1 and 2:

32i1 − v s + 96 ( i1 − i s ) = 0

v s + 30i 2 + 120 ( i 2 − i s ) = 0 150i 2 = +120i s − v s vs 4 i2 = is − 5 150 1 v o = 30i 2 = 24i s − v s 5 So a = 24 and b = -.02. (b) Apply KCL to the supernode corresponding to the voltage source to get va − (vs + vo ) 96 So is =

+

vs + vo 120

va − vo 32 +

vo 30

=

=

vs + vo 120

vs 120

+

+

vo 30

vo 24

Then 1 v o = 24i s − v s 5 So a = 24 and b = -0.2. (checked: LNAP 5/24/04)

P 4.8-3 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.8-3 by writing and solving (a) node equations and (b) mesh equations. Figure P 4.8-3 Solution: (a) Label the reference node and node voltages.

v b = 120 V due to the voltage source. Apply KCL at the node between the resistors to get vb − va 50

=

va 10



v a = 20 V

and the power supplied by the dependent source is p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W

(b) Label the mesh currents. Express the controlling voltage of the dependent source in terms of the mesh current to get v a = 10 ( i 2 − i1 )

Express the controlled current of the dependent source in terms of the mesh currents to get −i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1



i1 = 2i 2

Apply KVL to the bottom mesh to get 50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0

⇒ ⇒

i 2 − i1 = 2

So

i 2 − 2i 2 = 2

i 2 = −2 A



i 1 = −4 A

Then

v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A

The power supplied by the dependent source is p = 120 ( i a ) = 120 ( 4 ) 480 W (checked: LNAP 6/21/04)

Section 4.8 The Node Voltage Method and Mesh Current Method Compared P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, vs and is, and one output vo. The output is related to the inputs by the equation vo = ais + bvs where a and b are constants to be determined. Determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations. Figure P 4.8-2 Solution: (a)

Apply KVL to meshes 1 and 2:

32i1 − v s + 96 ( i1 − i s ) = 0

v s + 30i 2 + 120 ( i 2 − i s ) = 0 150i 2 = +120i s − v s vs 4 i2 = is − 5 150 1 v o = 30i 2 = 24i s − v s 5 So a = 24 and b = -.02. (b) Apply KCL to the supernode corresponding to the voltage source to get va − (vs + vo ) 96 So is =

+

vs + vo 120

va − vo 32 +

vo 30

=

=

vs + vo 120

vs 120

+

+

vo 30

vo 24

Then 1 v o = 24i s − v s 5 So a = 24 and b = -0.2. (checked: LNAP 5/24/04)

P 4.8-3 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.8-3 by writing and solving (a) node equations and (b) mesh equations. Figure P 4.8-3 Solution: (a) Label the reference node and node voltages.

v b = 120 V due to the voltage source. Apply KCL at the node between the resistors to get vb − va 50

=

va 10



v a = 20 V

and the power supplied by the dependent source is p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W

(b) Label the mesh currents. Express the controlling voltage of the dependent source in terms of the mesh current to get v a = 10 ( i 2 − i1 )

Express the controlled current of the dependent source in terms of the mesh currents to get −i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1



i1 = 2i 2

Apply KVL to the bottom mesh to get 50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0

⇒ ⇒

i 2 − i1 = 2

So

i 2 − 2i 2 = 2

i 2 = −2 A



i 1 = −4 A

Then

v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A

The power supplied by the dependent source is p = 120 ( i a ) = 120 ( 4 ) 480 W (checked: LNAP 6/21/04)

Section 4.8 The Node Voltage Method and Mesh Current Method Compared P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, vs and is, and one output vo. The output is related to the inputs by the equation vo = ais + bvs where a and b are constants to be determined. Determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations. Figure P 4.8-2 Solution: (a)

Apply KVL to meshes 1 and 2:

32i1 − v s + 96 ( i1 − i s ) = 0

v s + 30i 2 + 120 ( i 2 − i s ) = 0 150i 2 = +120i s − v s vs 4 i2 = is − 5 150 1 v o = 30i 2 = 24i s − v s 5 So a = 24 and b = -.02. (b) Apply KCL to the supernode corresponding to the voltage source to get va − (vs + vo ) 96 So is =

+

vs + vo 120

va − vo 32 +

vo 30

=

=

vs + vo 120

vs 120

+

+

vo 30

vo 24

Then 1 v o = 24i s − v s 5 So a = 24 and b = -0.2. (checked: LNAP 5/24/04)

P 4.8-3 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.8-3 by writing and solving (a) node equations and (b) mesh equations. Figure P 4.8-3 Solution: (a) Label the reference node and node voltages.

v b = 120 V due to the voltage source. Apply KCL at the node between the resistors to get vb − va 50

=

va 10



v a = 20 V

and the power supplied by the dependent source is p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W

(b) Label the mesh currents. Express the controlling voltage of the dependent source in terms of the mesh current to get v a = 10 ( i 2 − i1 )

Express the controlled current of the dependent source in terms of the mesh currents to get −i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1



i1 = 2i 2

Apply KVL to the bottom mesh to get 50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0

⇒ ⇒

i 2 − i1 = 2

So

i 2 − 2i 2 = 2

i 2 = −2 A



i 1 = −4 A

Then

v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A

The power supplied by the dependent source is p = 120 ( i a ) = 120 ( 4 ) 480 W (checked: LNAP 6/21/04)

Section 4.8 The Node Voltage Method and Mesh Current Method Compared P 4.8-2 The circuit shown in Figure P 4.8-2 has two inputs, vs and is, and one output vo. The output is related to the inputs by the equation vo = ais + bvs where a and b are constants to be determined. Determine the values a and b by (a) writing and solving mesh equations and (b) writing and solving node equations. Figure P 4.8-2 Solution: (a)

Apply KVL to meshes 1 and 2:

32i1 − v s + 96 ( i1 − i s ) = 0

v s + 30i 2 + 120 ( i 2 − i s ) = 0 150i 2 = +120i s − v s vs 4 i2 = is − 5 150 1 v o = 30i 2 = 24i s − v s 5 So a = 24 and b = -.02. (b) Apply KCL to the supernode corresponding to the voltage source to get va − (vs + vo ) 96 So is =

+

vs + vo 120

va − vo 32 +

vo 30

=

=

vs + vo 120

vs 120

+

+

vo 30

vo 24

Then 1 v o = 24i s − v s 5 So a = 24 and b = -0.2. (checked: LNAP 5/24/04)

P 4.8-3 Determine the power supplied by the dependent source in the circuit shown in Figure P 4.8-3 by writing and solving (a) node equations and (b) mesh equations. Figure P 4.8-3 Solution: (a) Label the reference node and node voltages.

v b = 120 V due to the voltage source. Apply KCL at the node between the resistors to get vb − va 50

=

va 10



v a = 20 V

and the power supplied by the dependent source is p = v bi a = (120 ) ⎡⎣0.2 ( 20 ) ⎤⎦ = 480 W

(b) Label the mesh currents. Express the controlling voltage of the dependent source in terms of the mesh current to get v a = 10 ( i 2 − i1 )

Express the controlled current of the dependent source in terms of the mesh currents to get −i1 = i a = 0.2 ⎡⎣10 ( i 2 − i1 ) ⎤⎦ = 2i 2 − 2i1



i1 = 2i 2

Apply KVL to the bottom mesh to get 50 ( i 2 − i1 ) + 10 ( i 2 − i1 ) − 120 = 0

⇒ ⇒

i 2 − i1 = 2

So

i 2 − 2i 2 = 2

i 2 = −2 A



i 1 = −4 A

Then

v a = 10 ( −2 − ( −4 ) ) = 20 V and i a = 0.2 ( 20 ) = 4 A

The power supplied by the dependent source is p = 120 ( i a ) = 120 ( 4 ) 480 W (checked: LNAP 6/21/04)

Section 4.9 Circuit Analysis Using MATLAB

Figure P4.9-1 P4.9-1. The encircled numbers in the circuit shown Figure P4.9-1 are node numbers. Determine the values of the corresponding node voltages, v1, v2 and v3. Solution: First, express the resistor currents in terms of the node voltages:

Apply KCL at node 1 to get 5 = Apply KCL at node 2 to get Apply KCL at node 3 to get

In matrix form:

Solving using MATLAB:

v1 − v 2 5

v1 − v 2 5 v2 − v3 4

= +

+

v2 10

v1 − v 3 2 +

v2 − v3

v1 − v 3 2

4

⇒ 0.7 v1 − 0.2 v 2 − 0.5 v 3 = 5 ⇒ − 0.2 v1 + 0.55 v 2 − 0.25 v 3 = 0

= 3 ⇒ − 0.5 v1 − 0.25 v 2 + 0.75 v 3 = −3

⎡ 0.7 −0.2 −0.5 ⎤ ⎡ v1 ⎤ ⎡ 5 ⎤ ⎢ −0.2 0.55 −0.25⎥ ⎢v ⎥ = ⎢ 0 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ −0.5 −0.25 0.75 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ −3⎥⎦

v1 = 28.1818 V, v 2 = 20 V and v 3 = 21.4545

Figure P4.9-2 P4.9-2. Determine the values of the node voltages, v1 and v2, in the circuit shown Figure P4.9-2. Solution: First, express the resistor currents in terms of the node voltages:

Apply KCL at node 1 to get Apply KCL at node 2 to get In matrix form: Solving using MATLAB:

15 − v1 25 v1 − v 2 50

= =

v1 − v 2 50 v2 10

+

+

v1 + 8 20

v2 + 8 40

⇒ 0.11v1 − 0.02 v 2 = 0.2

⇒ − 0.02 v1 + 0.145 v 2 = −0.2

⎡ 0.11 −0.02 ⎤ ⎡ v1 ⎤ ⎡ 0.2 ⎤ ⎢ −0.02 0.145 ⎥ ⎢v ⎥ = ⎢ −0.2 ⎥ ⎣ ⎦ ⎣ 2⎦ ⎣ ⎦

v1 = 1.6077 V and v 2 = −1.1576 V

Figure P4.9-3 P4.9-3. Determine the values of the node voltages, v1, v2 and v3 in the circuit shown in Figure P4.9-3. Solution: First, express the resistor currents in terms of the node voltages:

Apply KCL at node 1 to get Apply KCL at node 2 to get Apply KCL at node 3 to get

In matrix form:

Solving using MATLAB:

−15 − v1 25 v1 − v 2 20 v1 − v 2 50

= +

=

v1 − v 2 20

v2 25

+

v1 − v 3 50

v2 − v3

v 2 − v3 40

+

40 =

⇒ − 0.05 v1 + 0.115 v 2 − 0.025 v 3 = 0

v 3 − 10 10

⇒ 0.11v1 − 0.05 v 2 − 0.02 v 3 = −0.6

⇒ − 0.02 v1 − 0.025 v 2 + 0.145 v 3 = 1

⎡ 0.11 −0.05 −0.02 ⎤ ⎡ v1 ⎤ ⎡ −0.6 ⎤ ⎢ −0.05 0.115 −0.025⎥ ⎢v ⎥ = ⎢ 0 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ −0.02 −0.025 0.145 ⎥⎦ ⎢⎣ v 3 ⎥⎦ ⎢⎣ 1 ⎥⎦ v1 = 1.6077 V and v 2 = −1.1576 V

P4.9-4 Determine the node voltages, v1 and v2, for the circuit shown in Figure P4.9-4.

Figure P.4.9-4 Solution: Emphasize and label the nodes:

Notice the 24 V source connected between node 3 and the reference node. Consequently v 3 = 24 V Apply KCL at node 1 to get v1 8 v1

+

v1 − v 2 25

+2=0

is the current directed downward in the 8 Ω resistor and

v1 − v 2

is the current 8 25 directed from left to right in the 25 Ω resistor. We will simplify this equation by doing two things: In this equation

1. Multiplying each side by 8 × 25 = 200 to eliminate fractions. 2. Move the terms that don’t involve node voltages to the right side of the equation. The result is 33 v1 − 8 v 2 = −400 Next, apply KCL at node 2 to get v2 9

+

v 2 − 24 14

=

v1 − v 2 25

In this equation

v2 9

is the current directed downward in the 9 Ω resistor, v1 − v 2

v 2 − 24 14

is the current directed

is the current directed from left to right in the 25 Ω 25 resistor. We will simplify this equation by doing two things:

from left to right in the 14 Ω resistor and

1. Multiplying each side by 8 × 25 × 14 = 2800 to eliminate fractions. 2. Move the terms that involve node voltages to the left side of the equation and move the terms that don’t involve node voltages to the right side of the equation. The result is − ( 9 ×14 ) v1 + ( 9 × 14 + 25 × 14 + 25 × 9 ) v 2 = 24 × 25 × 9 ⇒ − 126 v1 + 701v 2 = 5400

The simultaneous equations can be written in matrix form 33 v1 − 8 v 2 = −400

−8 ⎤ ⎡ v1 ⎤ ⎡ −400 ⎤ ⎡ 33 ⇒ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ −126 v1 + 701 v 2 = 5400 ⎣ −126 701⎦ ⎣ v 2 ⎦ ⎣ 5400 ⎦ We can use MATLAB to solve the matrix equation:

Then

⎡ v1 ⎤ ⎡ −10.7209 ⎤ ⎢v ⎥ = ⎢ ⎥ ⎣ 2 ⎦ ⎣ 5.7763 ⎦

That is, the node voltages are v1 = −10.7209 V and v 2 = 5.7763 V .

P4.9-5 Determine the mesh currents, i1 and i2, for the circuit shown in Figure P4.9-5.

Figure P4.9.5 Solution: Label the label the mesh currents. Then, label the element currents in terms of the mesh currents:

Notice that the 2 A source on the outside of the circuit is in mesh 3 and that the currents 2 A and i 3 have the same direction. Consequently i3 = 2 A Apply KVL to mesh 1 to get 25 ( i1 − i 3 ) + 9 ( i1 − i 2 ) + 8 i1 = 0 In this equation 25 ( i1 − i 3 ) is the voltage across the 25 Ω resistor (+ on the left), 9 ( i1 − i 2 ) is the voltage across the 9 Ω resistor (+ on top) and 8i1 is the voltage across the 8 Ω resistor (+ on bottom). Substituting i 3 = 2 A and doing a little algebra gives 42 i1 − 9 i 2 = 50 Next, apply KVL to mesh 2 to get 14 ( i 2 − i 3 ) + 24 − 9 ( i1 − i 2 ) = 0

In this equation 14 ( i 2 − i 3 ) is the voltage across the 14 Ω resistor (+ on the left), 24 is the voltage source voltage and 9 ( i1 − i 2 ) is the voltage across the 9 Ω resistor (+ on top). Substituting i 3 = 2 A and doing

a little algebra gives −9 i1 + 23 i 2 = −24 + 14 ( 2 ) = 4

The simultaneous equations can be written in matrix form 42 i1 − 9 i 2 = 50

⎡ 42 −9 ⎤ ⎡ i1 ⎤ ⎡50 ⎤ ⇒ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ −9 i1 + 23 i 2 = 4 ⎣ −9 23 ⎦ ⎣i 2 ⎦ ⎣ 4 ⎦

We can use MATLAB to solve the matrix equation:

Then

⎡ i1 ⎤ ⎡1.3401 ⎤ ⎢ ⎥=⎢ ⎥ ⎣i 2 ⎦ ⎣0.6983⎦

That is, the mesh currents are i1 = 1.3401 A and i 2 = 0.6983 A .

P4.9-6 Represent the circuit shown in Figure P4.9-6 by the matrix equation ⎡ a 11 ⎢ ⎣ a 21

a 12 ⎤ ⎡ v1 ⎤ ⎡ −40 ⎤ = a 22 ⎦⎥ ⎣⎢ v 2 ⎦⎥ ⎢⎣ −228⎥⎦

Determine the values of the coefficients a11 , a12 , a 21 and a 22 .

Figure P4.9-6 Solution: Emphasize and label the nodes:

Noticing the 10 V source connected between node 3 and the reference node, we determine that node voltage at node 3 is v 3 = −10 V Apply KCL at node 1 to get In this equation

v1

v1 10

+ 0.4 +

v1 − v 2 10

=0

is the current directed downward in the vertical 10 Ω resistor and

v1 − v 2

is the 10 10 current directed from left to right in the horizontal 10 Ω resistor. We will simplify this equation by doing two things: 1. Multiplying each side by 10 to eliminate fractions. 2. Move the terms that don’t involve node voltages to the right side of the equation.

2 v1 − v 2 = −4

The result is Next, apply KCL at node 2 to get v2 19 In this equation

v2 19

+

v 2 − ( −10 ) 22

=

v1 − v 2 10

+ 0.4

is the current directed downward in the 19 Ω resistor,

directed from left to right in the 22 Ω resistor and

v 2 − ( −10 )

v1 − v 2

22

is the current

is the current directed from left to right in the 10 horizontal 10 Ω resistor. We will simplify this equation by doing two things: 1. Multiplying each side by 19 × 22 × 10 = 4180 to eliminate fractions. 2. Move the terms that involve node voltages to the left side of the equation and move the terms that don’t involve node voltages to the right side of the equation. The result is

− (19 × 22 ) v1 + (19 × 10 + 22 ×10 + 19 × 22 ) v 2 = −10 × 10 × 19 + 0.4 × 19 × 22 × 10 ⇒ − 418 v1 + 828 v 2 = −228 Comparing our equations to the given equations, we see that we need to multiply both sides of our first equation by 10. Then 20 v1 − 10 v 2 = −40 ⎡ 20 −10 ⎤ ⎡ v1 ⎤ ⎡ −40 ⎤ ⇒ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ −418 v1 + 828 v 2 = −228 ⎣ −418 828 ⎦ ⎣v 2 ⎦ ⎣ −228⎦ Comparing coefficients gives a 11 = 20 , a 12 = −10 , a 21 = −418 and a 22 = 828 .

P4.9-7 Represent the circuit shown in Figure P4.9-7 by the matrix equation ⎡ a 11 ⎢ ⎣ a 21

a12 ⎤ ⎡ i1 ⎤ ⎡ 4 ⎤ = a 22 ⎦⎥ ⎣⎢i 2 ⎦⎥ ⎢⎣10 ⎥⎦

Determine the values of the coefficients a11 , a12 , a 21 and a 22 .

Figure P4.9-7

Solution: Label the label the mesh currents. Then, label the element currents in terms of the mesh currents:

Notice that the 0.4 A source on the inside of the circuit is in both mesh 1 and mesh 3. Mesh current i1 is directed in the same way as current source current but the mesh current i 3 is directed opposite to the current source current. Consequently i1 − i 3 = 0.4 A The current source is in both mesh 1 and mesh 3 so we apply KVL to the supermesh corresponding to the current source (i.e. the perimeter of meshes 1 and 3). The result is 10 i 3 + 19 ( i1 − i 2 ) + 10 i1 = 0

In this equation 10i 3 is the voltage across the horizontal 10 Ω resistor (+ on the left), 19 ( i1 − i 2 ) is the voltage across the 19 Ω resistor (+ on top) and 10i1 is the voltage across the vertical 10 Ω resistor (+ on bottom). Substituting i 3 = i1 − 0.4 and doing a little algebra gives 39 i1 − 19 i 2 = 4 Next, apply KVL to mesh 2 to get 22 i 2 − 10 − 19 ( i1 − i 2 ) = 0 In this equation 22i 2 is the voltage across the 22 Ω resistor (+ on the left), 10 is the voltage source

voltage and 19 ( i1 − i 2 ) is the voltage across the 19 Ω resistor (+ on top). Doing a little algebra gives −19 i1 + 41i 2 = 10 To summarize, the circuit is represented by the simultaneous equations: 39 i1 − 19 i 2 = 4

⎡ 39 −19 ⎤ ⎡ i1 ⎤ ⎡ 4 ⎤ ⇒ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ −19 i1 + 41i 2 = 10 ⎣ −19 41 ⎦ ⎣i 2 ⎦ ⎣10 ⎦

Comparing these equations to the given equations shows a 11 = 39 , a 12 = −19 , a 21 = −19 and a 22 = 41 .

P4.9-8 Determine the values of the power supplied by each of the sources for the circuit shown in Figure P4.9-8.

Figure P.4.9-8 Solution: First, label the mesh currents and then label the element currents:

Notice the 2.4 A source in both mesh 2 and mesh 3. We have i 3 − i 2 = 2.4 A Apply KVL to mesh 1 to get 40 i1 − 5 ( i 3 − i1 ) − 5 ( i 2 − i1 ) = 0 ⇒ 50 i1 − 5 i 2 − 5 i 3 = 0 Identify the supermesh corresponding to the 2.4 A current source:

Apply KVL to the supermesh to get 5 ( i 2 − i1 ) + 5 ( i 3 − i1 ) + 24 + 40 i 2 = 0 ⇒ − 10 i1 + 45 i 2 + 5 i 3 = −24 Writing the mesh equations in matrix form gives

⎡ 0 −1 1 ⎤ ⎡ i1 ⎤ ⎡ 2.4 ⎤ ⎢ 50 −5 −5⎥ ⎢i ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −10 45 5 ⎥⎦ ⎢⎣i 3 ⎥⎦ ⎢⎣ −24 ⎥⎦ Solving using MATLAB:

That is, the mesh currents are i1 = 0.1 A, i 2 = −0.7 A and i 3 = 1.7 A. The 24 V source supplies

−24 i 3 = ( −24 )(1.7 ) = −40.8 W

The power supplied by the current source depends on vs, the voltage across the current source. Apply KVL to mesh 3 to get 5 ( i 3 − i1 ) + 24 − v s = 0 ⇒ v s = 5 (1.7 − 0.1) + 24 = 32 V The current source supplies

2.4 vs = 2.4 ( 32 ) = 76.8 W

Figure P4.9-9 P4.9-9 The mesh currents are labeled in the circuit shown Figure 4.9-9. Determine the value of the mesh currents i1 and i 2 . Solution: Determine the value of the mesh currents i1 and i 2 .

Replace series resistors with an equivalent resistor and series voltage sources with and equivalent voltage source to get

Apply KVL to mesh 1

16 i1 + 8 ( i1 − i 2 ) − 9 = 0 ⇒ 24 i1 − 8 i 2 = 9

Apply KVL to mesh 2

4 i 2 + 5 i1 − 8 ( i1 − i 2 ) = 0 ⇒ − 3 i1 + 12 i 2 = 0

In matrix form Solving using MATLAB

⎡ 24 −8⎤ ⎡ i1 ⎤ ⎡9 ⎤ ⎢ −3 12 ⎥ ⎢i ⎥ = ⎢ 0 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦

So the mesh currents are

i1 = 0.4091 A and i 2 = 0.1023 A

Figure P4.9-10 P4.9-10 The encircled numbers in the circuit shown Figure P4.9-10 are node numbers. Determine the values of the corresponding node voltages, v1 and v2. Solution: Determine the value of the node voltages, v1 and v2.

Replace parallel resistors with an equivalent resistor and parallel sources with and equivalent current source to get

Apply KCL at node 1 Apply KCL at node 2

In matrix form

Solving using MATLAB

2.5 =

v1 4.44

v1 − v 2 2 1 ⎡ 1 ⎢ 4.44 + 2 ⎢ ⎢ 1 + 1.5 ⎣⎢ 2

+

v1 − v 2 2

+ 1.5 v1 =

=0

v2 4

1 ⎤ 2 ⎥ ⎡ v1 ⎤ ⎡ 2.5⎤ ⎥⎢ ⎥ = 1 1 ⎥ ⎣v 2 ⎦ ⎢⎣ 0 ⎥⎦ − − 2 4 ⎦⎥ −

So the node voltages are

v1 = −4.1111 V and v 2 = −10.9630 V

Section 4.10 How Can We Check … ? P 4.10-1 Computer analysis of the circuit shown in Figure P 4.10-1 indicates that the node voltages are va = 5.2 V, vb = –4.8 V, and vc = 3.0 V. Is this analysis correct? Hint: Use the node voltages to calculate all the element currents. Check to see that KCL is satisfied at each node.

Figure P 4.10-1 Solution: Apply KCL at node b:

vb − va v −v 1 − + b c = 0 4 2 5 −4.8 − 5.2 1 − 4.8 − 3.0 − + ≠0 4 2 5

The given voltages do not satisfy the KCL equation at node b. They are not correct.

P 4.10-2 An old lab report asserts that the node voltages of the circuit of Figure P 4.10-2 are

va = 4 V, vb = 20 V, and vc = 12 V. Are these correct?

Solution:

Apply KCL at node a:

v ⎛v −v ⎞ −⎜ b a ⎟ − 2 + a = 0 2 ⎝ 4 ⎠ 4 ⎛ 20 − 4 ⎞ −⎜ = −4≠ 0 ⎟−2+ 2 ⎝ 4 ⎠

The given voltages do not satisfy the KCL equation at node a. They are not correct.

P 4.10-3

Your lab partner forgot to record the values of R1,

R2, and R3. He thinks that two of the resistors in Figure P 4.10-3 had values of 10 kΩ and that the other had a value of 5 kΩ. Is this possible? Which resistor is the 5-kΩ resistor?

Figure P 4.10-3 Solution: ⎛ 12 − 7.5 ⎞ 7.5 7.5 − 6 −⎜ + =0 ⎟+ R2 ⎝ R1 ⎠ R3

Writing a node equation:

4.5 7.5 1.5 + + =0 R1 R3 R2 There are only three cases to consider. Suppose R1 = 5 kΩ and R 2 = R 3 = 10 kΩ. Then −

So



4.5 7.5 1.5 −0.9 + 0.75 + 0.15 + + = = 0 R1 R3 R2 1000

This choice of resistance values corresponds to branch currents that satisfy KCL. Therefore, it is indeed possible that two of the resistances are 10 kΩ and the other resistance is 5 kΩ. The 5 kΩ is R1.

P 4.10-4 Computer analysis of the circuit shown in Figure P 4.10-4 indicates that the mesh currents are

i1 = 2 A, i2 = 4 A, and i3 = 3 A. Verify that this analysis is correct. Hint: Use the mesh currents to calculate the element voltages. Verify that KVL is satisfied for each mesh. Figure P 4.10-4

Solution: Applying KVL to each mesh:

Top mesh: Bottom right mesh

10 (2 − 4) + 12(2) + 4 (2 − 3) = 0 8 (3 − 4) + 4 (3 − 2) + 4 = 0

Bottom, left mesh: 28 + 10 (4 − 2) + 8 (4 − 3) ≠ 0 (Perhaps the polarity of the 28 V source was entered incorrectly.) KVL is not satified for the bottom, left mesh so the computer analysis is not correct.

PSpice Problems SP 4-1

Use PSpice to determine the node voltages of the circuit shown in Figure SP 4-1

Figure SP 4-1 Solution: The PSpice schematic after running a “Bias Point” simulation:

SP 4-2

Use PSpice to determine the mesh currents of the circuit shown in Figure SP 4-2.

Figure SP 4-2 Solution: The PSpice schematic after running a “Bias Point” simulation:

From the PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2

-3.000E+00 -2.250E+00 V_V3

-7.500E-01

The voltage source labeled V3 is a short circuit used to measure the mesh current. The mesh currents are i1 = −3 A (the current in the voltage source labeled V1) and i2 = −0.75 A (the current in the voltage source labeled V3).

SP 4-3

The voltages va, vb, vc, and vd in Figure SP 4-3 are the node voltages corresponding to

nodes a, b, c and d. The current i is the current in a short circuit connected between nodes b and c. Use PSpice to determine the values of va, vb, vc, and vd and of i. Solution: The PSpice schematic after running a “Bias Point” simulation:

The PSpice output file: **** INCLUDING sp4_2-SCHEMATIC1.net **** * source SP4_2 V_V4 0 N01588 12Vdc R_R4 N01588 N01565 4k V_V5 N01542 N01565 0Vdc R_R5 0 N01516 4k V_V6 N01542 N01516 8Vdc I_I1 0 N01565 DC 2mAdc I_I2 0 N01542 DC 1mAdc VOLTAGE SOURCE CURRENTS NAME CURRENT V_V4 V_V5 V_V6

-4.000E-03 2.000E-03 -1.000E-03

From the PSpice schematic: va = −12 V, vb = vc = 4 V, vd = −4 V. From the output file: i = 2 mA.

SP 4-4

Determine the current, i, shown in Figure SP 4-4.

Answer: i = 0.56 A

Figure SP 4-4 Solution: The PSpice schematic after running a “Bias Point” simulation:

The PSpice output file: VOLTAGE SOURCE CURRENTS NAME CURRENT V_V7 V_V8

-5.613E-01 -6.008E-01

The current of the voltage source labeled V7 is also the current of the 2 Ω resistor at the top of the circuit. However this current is directed from right to left in the 2 Ω resistor while the current i is directed from left to right. Consequently, i = +5.613 A.

Design Problems DP 4-1 An electronic instrument incorporates a 15-V power supply. A digital display is added that requires a 5-V power supply. Unfortunately, the project is over budget and you are instructed to use the existing power supply. Using a voltage divider, as shown in Figure DP 4-1, you are able to obtain 5 V. The specification sheet for the digital display shows that the display will operate properly over a supply voltage range of 4.8 V to 5.4 V. Furthermore, the display will draw 300 mA (I) when the display is active and 100 mA when quiescent (no activity). (a)

Select values of R1 and R2 so that the display will be supplied with 4.8 V to 5.4 V under all conditions of current I.

(b)

Calculate the maximum power dissipated by each resistor, R1 and R2, and the maximum current drawn from the 15-V supply.

(c)

Is the use of the voltage divider a good engineering solution? If not, why? What problems might arise?

Figure DP 4-1 Solution: Model the circuit as:

(a) We need to keep v2 across R2 in the range 4.8 ≤ v2 ≤ 5.4 display is active ⎧0.3 A For I = ⎨ ⎩ 0.1 A display is not active

KCL at a:

v2 − 15 v2 + +I =0 R1 R2

Assumed that maximum I results in minimum v2 and visa-versa. Then ⎧4.8 V v2 = ⎨ ⎩5.4 V

when I = 0.3 A when I = 0.1 A

Substitute these corresponding values of v2 and I into the KCL equation and solve for the resistances 4.8 − 15 4.8 + + 0.3 = 0 R1 R2 5.4 − 15 5.4 + + 0.1 = 0 R1 R2 ⇒ R1 = 7.89 Ω, R2 = 4.83 Ω

(b)

15 − 4.8 = 1.292 A ⇒ PR = (1.292)2 (7.89) = 13.17 W 1max 1max 7.89 5.4 )2 ( 5.4 IR = = 1.118 A ⇒ PR = = 6.03 W 2max 2 max 4.83 4.83 maximum supply current = I R = 1.292 A IR

=

1max

(c) No; if the supply voltage (15V) were to rise or drop, the voltage at the display would drop below 4.8V or rise above 5.4V.

The power dissipated in the resistors is excessive. Most of the power from the supply is dissipated in the resistors, not the display.

DP 4-2 For the circuit shown in Figure DP 4-2, it is desired to set the voltage at node a equal to

0 V in order to control an electric motor. Select voltages v1 and v2 in order to achieve va = 0 V when v1 and v2 are less than 20 V and greater than zero and R = 2 Ω.

Figure DP 4-2 Solution: Express the voltage of the 8 V source in terms of its node voltages to get vb − va = 8 . Apply KCL to the supernode corresponding to the 8 V source: va − v1 R

+

va vb vb − ( −v 2 ) + + = 0 ⇒ 2 va − v1 + 2 vb + v 2 = 0 R R R ⇒ 2 va − v1 + 2 ( va + 8 ) + v 2 = 0 ⇒ 4 va − v1 + v 2 + 16 = 0 ⇒ va =

Next set va = 0 to get 0=

v1 − v 2

For example, v1 = 18 V and v2 = 2 V.

4

v1 − v 2 4

− 4 ⇒ v1 − v 2 = 16 V

−4

DP 4-3 A wiring circuit for a special lamp in a

home is shown in Figure DP 4-3. The lamp has a resistance of 2 Ω, and the designer selects R = 100 Ω. The lamp will light when I ≥ 50 mA but will Figure DP 4-3

burn out when I > 75 mA. (a)

Determine the current in the lamp and determine if it will light for R = 100 Ω.

(b)

Select R so that the lamp will light but will not burn out if R changes by ± 10 percent because of temperature changes in the home.

Solution: (a)

Apply KCL to left mesh:

−5 + 50 i1 + 300 (i1 − I ) = 0

Apply KCL to right mesh:

( R + 2) I + 300 ( I − i1 ) = 0

150 1570 + 35 R We desire 50 mA ≤ I ≤ 75 mA so if R = 100 Ω, then I = 29.59 mA ⇒ l amp so the lamp will not light.

Solving for I:

I=

(b) From the equation for I, we see that decreasing R increases I:

try R = 50 Ω ⇒ I = 45 mA (won't light) try R = 25Ω ⇒ I = 61 mA ⇒ will light Now check R±10% to see if the lamp will light and not burn out: −10% → 22.5Ω → I = 63.63 mA ⎫ lamp will ⎬ +10% → 27.5Ω → I = 59.23 mA ⎭ stay on

DP 4-4 In order to control a device using the circuit shown in Figure DP 4-4, it is necessary

that vab = 10 V. Select the resistors when it is required that all resistors be greater than 1 Ω and R3 + R4 = 20 Ω.

Figure DP 4-4

Solution:

R = R1 || R 2 || ( R 3 + R 4 )

Equivalent resistance:

R ( 25 ) 10 + R We require vab = 10 V. Apply the voltage division principle in the left circuit to get: Voltage division in the equivalent circuit: v1 =

10 =

(

)

R1 R 2 ( R3 + R4 ) R4 R4 v1 = × × 25 R3 + R4 R3 + R4 10 + R1 R 2 ( R3 + R4 )

(

)

This equation does not have a unique solution. Here’s one solution: choose R1 = R2 = 25 Ω and R3 + R4 = 20 Ω then 10 =

(12.5 20 ) × 25 ⇒ R = 18.4Ω R4 × 4 20 10 + (12.5 20 ) and R3 + R4 = 20 ⇒ R3 = 1.6 Ω

DP 4-5 The current i shown in the circuit of Figure DP 4-5 is used to measure the stress

between two sides of an earth fault line. Voltage v1 is obtained from one side of the fault, and v2 is obtained from the other side of the fault. Select the resistances R1, R2, and R3 so that the magnitude of the current i will remain in the range between 0.5 mA and 2 mA when v1 and v2 may each vary independently between +1 V and +2 V (1 V ≤ vn ≤ 2 V).

Solution:

Apply KCL to the left mesh:

(R

1

+ R 3 ) i1 − R 3 i2 − v1 = 0

Apply KCL to the left mesh:

(R

− R 3 i1 +

2

+ R 3 ) i2 + v2 = 0

Solving for the mesh currents using Cramer’s rule: − R3 ⎡v1 ⎤ ⎡( R1 + R 3 ) ⎢ ⎥ ⎢−R ⎢⎣ −v2 ( R 2 + R 3 ) ⎥⎦ 3 ⎣ and i2 = i1 = Δ Δ 2 where Δ = ( R1 + R 3 ) ( R 2 + R 3 ) − R 3

v1 ⎤ − v2 ⎦⎥

Try R1 = R2 = R3 = 1 kΩ = 1000 Ω. Then Δ = 3 MΩ. The mesh currents will be given by i

=

[ 2v1 − v2 ] 1000

and i 2 =

[ −2v2 + v1 ] 1000

3 × 10 3 ×10 Now check the extreme values of the source voltages: 1

6

6

if v1 = v2 = 1 V ⇒ i = 2

mA 3 if v1 = v2 = 2 V ⇒ i = 4 mA 3

⇒ i = i1 − i2 =

okay okay

v1 + v2 3000

Chapter 5 Circuit Theorems Exercises R

Exercise 5.2-1 Determine values of R and is so that the circuits shown in Figures E 5.2-1a,b are equivalent to each other due to a source transformation.

+ –

Answer: R = 10 Ω and is = 1.2 A

12 V

10 Ω

is

(a)

(b)

Figures E 5.2-1 Exercise 5.2-2 Determine values of R and is so that the circuits shown in Figures E 5.2-2a,b are equivalent to each other due to a source transformation.

R

– +

Hint: Notice that the polarity of the voltage source in Figure E 5.2-2a is not the same as in Figure E 5.2-1a.

12 V

10 Ω

is

(b)

(a)

Figures E 5.2-2

Answer: R = 10 Ω and is = –1.2 A 8Ω

Exercise 5.2-3 Determine values of R and vs so that the circuits shown in Figures E 5.2-3a,b are equivalent to each other due to a source transformation.

+ –

Answer: R = 8 Ω and vs = 24 V

vs

R

3A

(a)

(b)

Figure E 5.2-3 Exercise 5.2-4 Determine values of R and vs so that the circuits shown in Figures E 5.2-4a,b are equivalent to each other due to a source transformation. Hint: Notice that the reference direction of the current source in Figure E 5.2-4b is not the same as in Figure E 5.2-3b. Answer: R = 8 Ω and vs = –24 V



+ –

vs

R

3A

(b)

(a)

Figure E 5.2-4



Exercise 5.4-1 Determine values of Rt and voc that cause the circuit shown in Figure E 5.4-1b to be the Thévenin equivalent circuit of the circuit in Figure E 5.4-1a.

+ –



+ –



3V

Answer: Rt = 8 Ω and voc = 2 V

b

(b)

Figure E 5.2-1

a

voc

b

(a)

Solution:

Rt

a



Exercise 5.4-2 Determine values of Rt and voc that cause the circuit shown in Figure E 5.4-2b to be the Thévenin equivalent circuit of the circuit in Figure E 5.4-2a.

12 V

3Ω + –

+ –

ia

Rt

a + –

2ia

voc

b

Answer: Rt = 3 Ω and voc = –6 V

(a)

Solution:

ia =

2 i a − 12

⇒ i a = −3 A 6 voc = 2 i a = −6 V

12 + 6 i a = 2 i a 3 i sc = 2 i a

Rt =

−6 =3Ω −2

b

(b)

Figure E 5.2-2

⇒ i a = −3 A

⇒ i sc =

2 ( − 3 ) = −2 A 3

a



Exercise 5.5-1 Determine values of Rt and isc that cause the circuit shown in Figure E 5.5-1b to be the Norton equivalent circuit of the circuit in Figure E 5.5-1a. Answer: Rt = 8 Ω and isc = 0.25 A

+ –



a



3V

isc

Rt

b

(a)

b

(b)

Figure E 5.5-1 Solution:

a



Exercise 5.6-1 Find the maximum power that can be delivered to RL for the circuit of Figure E 5.6-1 using a Thévenin equivalent circuit. Answer: 9 W when RL = 4 Ω

18 V

+ –





Figure E 5.6-1

Solution:

voc =

6 (18) = 12 V 6+3

Rt = 2 +

( 3)( 6 ) = 4 Ω 3+ 6

For maximum power, we require

R L = Rt = 4 Ω Then 2

pmax =

voc 122 = =9 W 4 Rt 4 ( 4 )

RL

Section 5-2: Source Transformations P 5.2-1 The circuit shown in Figure P 5.2-1a has been divided into two parts. The circuit shown in Figure P 5.2-1b was obtained by simplifying the part to the right of the terminals using source transformations. The part of the circuit to the left of the terminals was not changed.



i

2V



– +

+ 9V

+ –

v ia



0.5 A



(a)

(a)

Determine the values of Rt and vt in Figure P 5.2-1b.

(b)

+ Determine the values of the + v vt current i and the voltage v in 9 V +– – Figure P 5.2-1b. The circuit in ia – Figure P 5.2-1b is equivalent to the circuit in Figure P 5.2-1a. (b) Consequently, the current i and Figure P 5.2-1 the voltage v in Figure P 5.2-1a have the same values as do the current i and the voltage v in Figure P 5.2-1b.

(c)



Determine the value of the current ia in Figure P 5.2-1a.

Solution: (a)

i

Rt



∴ Rt = 2 Ω

(b)

−9 − 4i − 2i + (−0.5) = 0 −9 + (−0.5) i = = −1.58 A 4+2 v = 9 + 4 i = 9 + 4(−1.58) = 2.67 V

(c)

ia = i = − 1.58 A

vt = − 0.5 V

(checked using LNAP 8/15/02)



P 5.2-2 Consider the circuit of Figure P 5.2-2. Find ia by simplifying the circuit (using source transformations) to a single-loop circuit so that you need to write only one KVL equation to find ia.



10 V

+ –

ia



2A



Figure P 5.2-2 Solution:

Finally, apply KVL:

−10 + 3 ia + 4 ia −

16 =0 3

∴ ia = 2.19 A (checked using LNAP 8/15/02)

3A

P 5.2-3 Find vo using source transformations if i = 5/2 A in the circuit shown in Figure P 5.2-3. Hint: Reduce the circuit to a single mesh that contains the voltage source labeled vo.

6Ω 8V



10 Ω

+–

2A

16 Ω

Answer: vo = 28 V

12 Ω

20 Ω

v0

7Ω i

+–

Figure P 5.2-3 Solution:

Source transformation at left; equivalent resistor for parallel 6 and 3 Ω resistors:

Equivalents for series resistors, series voltage source at left; series resistors, then source transformation at top:

Source transformation at left; series resistors at right:

Parallel resistors, then source transformation at left:

Finally, apply KVL to loop − 6 + i (9 + 19) − 36 − vo = 0 i = 5 / 2 ⇒ vo = −42 + 28 (5 / 2) = 28 V (checked using LNAP 8/15/02)

P 5.2-4 Determine the value of the current ia in the circuit shown in Figure P 5.2-4.

6 kΩ

10 V

4 kΩ

4 kΩ

+–

ia + –

12 V

4 kΩ

3 kΩ

– +

6V

Figure P 5.2-4 Solution:

− 4 − 2000 ia − 4000 ia + 10 − 2000 ia − 3 = 0 ∴ ia = 375 μ A (checked using LNAP 8/15/02)

4A

P 5.2-5 Use source transformations to find the current ia in the circuit shown in Figure P 5.2-5.



6V – +

Answer: ia = 1 A 12 V

+ –

1A

ia



Figure P 5.2-5. Solution:

−12 − 6 ia + 24 − 3 ia − 3 = 0 ⇒ ia = 1 A (checked using LNAP 8/15/02)

P 5.2-6 Use source transformations to find the value of the voltage va in Figure P 5.2-6. Answer: va = 7 V

8V

100 Ω

+ – + –

10 V

+ va –

100 Ω

100 Ω

30 mA

Figure P 5.2-6 Solution: A source transformation on the right side of the circuit, followed by replacing series resistors with an equivalent resistor:

Source transformations on both the right side and the left side of the circuit:

Replacing parallel resistors with an equivalent resistor and also replacing parallel current sources with an equivalent current source:

Finally,

va =

50 (100 ) 100 ( 0.21) = ( 0.21) = 7 V 50 + 100 3 (checked using LNAP 8/15/02)

P5.2-7

Figure P5.2-7

The equivalent circuit in Figure P5.2-7 is obtained from the original circuit using source transformations and equivalent resistances. (The lower case letters a and b identify the nodes of the capacitor in both the original and equivalent circuits.) Determine the values of Ra, Va, Rb and I b in the equivalent circuit. Solution

Performing a source transformation at each end of the circuit yields

Thenx

where V a = 2.2 (10 ) − 32 = −10 V , R a = 18 + 10 = 28 Ω , R b = 18 || 9 = 6 Ω and I b = 2.5 +

36 18

= 4.5 A

P 5.2-8 The circuit shown in Figure P 5.2-8 contains an unspecified resistance R. (a) Determine the value of the current i when R = 4 Ω. (b) Determine the value of the voltage v when R = 8 Ω. (c) Determine the value of R that will cause i = 1 A. (d) Determine the value of R that will cause v = 16 V. i

R +

24 Ω

+ –

12 V

v



18 Ω

2A

24 Ω

Figure P 5.2-8 Solution:

Replace series and parallel resistors by an equivalent resistor. 18 & (12 + 24 ) = 12 Ω

Do a source transformation, then replace series voltage sources by an equivalent voltage source.

12 Ω

Do two more source transformations Now current division gives 24 ⎛ 8 ⎞ i=⎜ ⎟3 = 8+ R ⎝8+ R ⎠ Then Ohm’s Law gives 24 R v = Ri = 8+ R

(a )

i=

24 =2A 8+ 4

(b) v =

24 ( 8 ) = 12 V 8+8

(c) 1 =

24 8+ R

(d) 16 =

24 R 8+ R





R = 16 Ω

R = 16 Ω (checked: LNAP 6/9/04)

15 Ω

P 5.2-9 Determine the value of the power supplied by the current source in the circuit shown in Figure P 5.2-9.

+

24 V –

2A

25 Ω

24 Ω –

32 V +

Figure P 5.2-9 Solution: Use source transformations and equivalent resistances to reduce the circuit as follows

The power supplied by the current source is given by

p = ⎡⎣ 23.1 + 2 (10.3125 ) ⎤⎦ 2 = 87.45 W

12 Ω

Section 5-3 Superposition P5.3-1 The inputs to the circuit shown in Figure P5.3-1 are the voltage source voltages v1 and v2.. The output of the circuit is the voltage vo. The output is related to the inputs by vo = a v1 + b v2 where a and b are constants. Determine the values of a and b.

Figure P5.3-1 Solution: Let vo1 = a v1 be the output when v2 = 0. In this case, the right voltage source acts like a short circuit so we have the circuit show to the right. Then v o1 =

20 || 5 4 1 1 v1 = v1 = v1 ⇒ a = 20 + ( 20 || 5 ) 20 + 4 6 6

Let vo2 = b v2 be the output when v1 = 0. In this case, the left voltage source acts like a short circuit so we have the circuit show to the right. Then v o2 =

20 || 20 10 2 v2 = v2 = v2 5 + ( 20 || 20 ) 5 + 10 3

⇒ b=

2 3

P5.3-2 A particular linear circuit has two inputs, v1 and v2, and one output, vo. Three measurements are made. The first measurement shows that the output is vo = 4 V when the inputs are v1 = 2 V and v2 = 0. The second measurement shows that the output is vo = 10 V when the inputs are v1 = 0 and v2 = −2.5 V. In the third measurement the inputs are v1 = 3 V and v2 = 3 V. What is the value of the output in the third measurement? Solution: The output of a linear circuit is a linear combination of the inputs:

v o = a1v1 + a 2 v 2 From the first two measurements we have: ⎛ a1 ⎞ ⎛ 2 ⎞ 0 ⎞ ⎛ a1 ⎞ ⎛ 4 ⎞ ⎛2 ⎜ ⎟=⎜ ⎟⎜ a ⎟ ⇒ ⎜ a ⎟ = ⎜ ⎟ ⎝10 ⎠ ⎝ 0 −2.5 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ −4 ⎠

Now the output of the third measurement can be determine to be v o = a 1 ( 3) + a 2 ( 3) = ( 2 )( 3) + ( −4 )( 3) = −6 V

P5.3-3 The circuit shown in Figure P5.3-3 has two inputs, vs and is, and one output io. The output is related to the inputs by the equation io = a is + b vs Given the following two facts: The output is io = 0.45 A when the inputs are is = 0.25 A and vs = 15 V.

and

The output is io = 0.30 A when the inputs are is = 0.50 A and vs = 0 V. Determine the values of the constants a and b and the values of the resistances are R 1 and R 2. Answers: a = 0.6 A/A, b = 0.02 A/V, R 1 = 30 Ω and R 2 = 20 Ω.

Figure P5.3-3 Solution: From the 1st fact:

0.45 = a ( 0.25 ) + b (15 )

From the 2nd fact:

0.30 = a ( 0.50 ) + b ( 0 ) ⇒ a =

0.30 = 0.60 0.50

0.45 = ( 0.60 )( 0.25 ) + b (15 ) ⇒ b =

Substituting gives

Next, consider the circuit: a i s = i o1 = i o so

0.60 =

R1 + R 2

b v s = i o2 = i o

and so

R1

0.02 =

1 R1 + R 2

v s =0

⎛ R1 ⎞ i =⎜ ⎜ R1 + R 2 ⎟⎟ s ⎝ ⎠

⇒ 2 R1 = 3 R 2

i s =0

=

vs R1 + R 2

⇒ R1 + R 2 =

1 = 50 Ω 0.02

Solving these equations gives R 1 = 30 Ω and R 2 = 20 Ω.

0.45 − ( 0.60 )( 0.25 ) = 0.02 15

P 5.3-4 Use superposition to find the value of 10 Ω the voltage v in Figure P 5.3-4.

9A 20 Ω +

v

6A

15 Ω



Figure P 5.3-4 Solution: Consider 6 A source only (open 9 A source)

Use current division:

v1 ⎡ 15 ⎤ = 6 ⎢ ⇒ v1 = 40 V 20 ⎣15 + 30 ⎥⎦

Consider 9 A source only (open 6 A source)

Use current division: v2 ⎡ 10 ⎤ = 9 ⎢ ⇒ v2 = 40 V 20 ⎣10 + 35 ⎥⎦

∴ v = v1 + v2 = 40 + 40 = 80 V

(checked using LNAP 8/15/02)

Figure P5.3-5 P5.3-5 Determine v(t), the voltage across the vertical resistor in the circuit in Figure P5.3-5. Solution; We’ll use superposition. Let v1(t) the be the part of v(t) due to the voltage source acting alone. Similarly, let v2(t) the be the part of v(t) due to the voltage source acting alone. We can use these circuits to calculate v1(t) and v2(t).

Notice that v1(t) is the voltage across parallel resistors. Using equivalent resistance, we calculate 40||10 = 8 Ω. Next, using voltage division we calculate 8 v1 ( t ) = (12 ) = 2 V 8 + 40 Similarly v2(t) is the voltage across parallel resistors Using equivalent resistance we first determine 40||40 = 20 Ω and then calculate

20 (12 cos ( 5t ) ) = 8cos ( 5t ) V 10 + 20 v ( t ) = v1 ( t ) + v 2 ( t ) = 2 + 8cos ( 5 t ) V

v 2 (t ) = Using superposition

15 mA

P 5.3-6 Use superposition to find the value of the current i in Figure P 5.3-6. 4 kΩ

Answer: i = 3.5 mA

15 V +–

2 kΩ

30 mA

12 kΩ

6 kΩ i

Figure P5.3-6 Solution: Consider 30 mA source only (open 15 mA and short 15 V sources). Let i1 be the part of i due to the 30 mA current source.

⎛ 2 ⎞ ⎛ 6 ⎞ ia = 30 ⎜ ⎟ = 6 mA ⇒ i1 = ia ⎜ ⎟ = 2 mA ⎝ 2+8⎠ ⎝ 6 + 12 ⎠ Consider 15 mA source only (open 30 mA source and short 15 V source) Let i2 be the part of i due to the 15 mA current source.

⎛ 4 ⎞ ib = 15 ⎜ ⎟ = 6 mA ⇒ ⎝ 4+6⎠

⎛ 6 ⎞ i2 = ib ⎜ ⎟ = 2 mA ⎝ 6 + 12 ⎠

Consider 15 V source only (open both current sources). Let i3 be the part of i due to the 15 V voltage source.

⎛ 6 || 6 ⎞ ⎛ 3 ⎞ i3 = − 2.5 ⎜⎜ ⎟⎟ = − 10 ⎜ ⎟ = −0.5 mA ⎝ 3 + 12 ⎠ ⎝ ( 6 || 6 ) + 12 ⎠ Finally,

i = i1 + i2 + i3 = 2 + 2 − 0.5 = 3.5 mA

(checked using LNAP 8/15/02)

Figure P5.3-7 P5.3-7 Determine v(t), the voltage across the 40 Ω resistor in the circuit in Figure P5.3-7. Solution: We’ll use superposition. Let v1(t) the be the part of v(t) due to the voltage source acting alone. Similarly, let v2(t) the be the part of v(t) due to the voltage source acting alone. We can use these circuits to calculate v1(t) and v2(t).

Using voltage division we calculate v1 ( t ) = −

40 (12 + 15cos (8 t ) ) = −9.6 − 12 cos (8t ) 10 + 40

Using equivalent resistance we first determine 10||40 = 8 Ω and then calculate v 2 ( t ) = 8 (1 + sin ( 5t ) ) = 8 + 8sin ( 5t )

Using superposition

v ( t ) = v1 ( t ) + v 2 ( t ) = −1.6 + 8sin ( 5 t ) − 12 cos ( 8 t ) V

ix

P 5.3-8 Use superposition to find the value of the current ix in Figure P 5.3-8. Answer: i = 3.5 mA

+ –





8V

2A

+ –

3ix

Figure P5.3-8 Solution: Consider 8 V source only (open the 2 A source)

Let i1 be the part of ix due to the 8 V voltage source. Apply KVL to the supermesh:

6 ( i1 ) + 3 ( i 1 ) + 3 ( i 1 ) − 8 = 0 8 2 = A 12 3 Let i2 be the part of ix due to the 2 A current source. i1 =

Consider 2 A source only (short the 8 V source)

Apply KVL to the supermesh:

6 ( i 2 ) + 3 ( i 2 + 2 ) + 3 i2 = 0 i2 = Finally,

i x = i1 + i 2 =

2 1 1 − = A 3 2 6

−6 1 =− A 12 2

P 5.3-9 The input to the circuit shown in Figure P 5.3-9 is the voltage source voltage, vs. The output is the voltage vo. The current source current, ia, is used to adjust the relationship between the input and output. Design the circuit so that input and output are related by the equation vo = 2vs + 9. Hint: Determine the required values of A and ia. A ix



+ –

ix

+ –

vs

12 Ω

12 Ω

ia

+ vo −

Figure P 5.3-9 Soluton: ix =

vs − va R1

va − vo = A ix = A va =

vs − va R1

R1 v o + A v s R1 + A

Apply KCL to the supernode corresponding to the CCVS to get va − vs R1 R1 + R 2 R1 R 2

+

va R2

va −

+ ia +

vs R1

vo R3

+ ia +

=0

vo R3

=0

R1 + R 2 ⎛ R 1 v o + A v s ⎞ v s vo + ia + =0 ⎜ ⎟− R1 R 2 ⎜⎝ R1 + A ⎟⎠ R1 R3

(

)

⎛ R +R ⎛ R1 + R 2 A 1 ⎞⎟ 1 ⎞ 1 2 ⎜ + vo + ⎜ − ⎟ vs + ia = 0 ⎜ R 2 R1 + A R 3 ⎟ ⎜ R1 R 2 R1 + A R1 ⎟ ⎝ ⎠ ⎝ ⎠

(

(

)

) ( R 2 R 3 ( R1 + A )

R 3 R1 + R 2 + R 2 R 1 + A

)v

(

o+

)

A − R2

(

R 2 R1 + A

)

vs + ia = 0

vo =

(

R 2 R 3 ( R1 + A ) ) vs − ia R 3 ( R1 + R 2 ) + R 2 ( R 1 + A ) R 3 ( R 1 + R 2 ) + R 2 ( R1 + A ) R3 R 2 − A

When R1 = 6 Ω, R 2 = 12 Ω and R 3 = 12 Ω vo =

12 ( 6 + A ) 12 − A vs − ia 24 + A 24 + A

Comparing this equation to v o = 2 v s + 9 , we requires 12 − A =2 ⇔ 24 + A

A = −12

V A

Then 2 v s + 9 = v o = 2v s + 6i a so we require 9 = 6ia

⇒ i a = 1.5 A

(checked: LNAP 6/22/04)

P 5.3-10 The circuit shown in Figure P 5.3-10 has three inputs: v1, v2, and i3. The output of the circuit is vo. The output is related to the inputs by

vo = av1 + bv2 + ci3 where a, b, and c are constants. Determine the values of a, b, and c. v2 +





+ + –

v1

40 Ω

10 Ω

vo

i3



Figure P 5.3-10 Solution:

vo1 =

vo2 = −

40 ||10 1 1 v1 = v1 ⇒ a = 8 + 40 ||10 2 2

10 3 v1 = − v 2 8 || 40 + 10 5

⇒ b=−

3 5

vo3 = ( 8 ||10 || 40 ) i 3 = 4 i 3 ⇒ c = 4

(checked: LNAP 6/22/04)

P 5.3-11 Determine the voltage vo(t) for the circuit shown in Figure P 5.3-11.



+

12 cos 2t V 4 ix

10 Ω

+ 40 Ω

+ –

2 V 10 Ω

ix



vo(t) –

Figure P 5.3-11 Solution: Using superposition: v x = 10 i x

and

v x − 12 cos 2t 40

+

vx 10

+

vx 10

= 4ix

so 10 i x − 12 cos 2t 40

= 2ix

⇒ ix = −

12 cos 2t 70

Finally, v o1 = −5 4 i x = 3.429 cos 2t V

( )

v x = 10 i x

and vx 40

+

vx − 2 10

+

vx 10

= 4ix

so −0.2 = 1.75 i x

Finally,

⇒ i x = −0.11429 A

( )

v o1 = −5 4 i x = 2.286 V v o = v o1 + v o2 = 3.429 cos 2t + 2.286 V

(checked: LNAP 6/22/04)

P 5.3-12 Determine the value of the voltage vo in the circuit shown in Figure P 5.3-12.

96 Ω

32 Ω 20 V +

0.3 A



120 Ω

30 Ω

+ vo –

Figure P 5.3-12 Solution: Using superposition:

v o1 = 24 ( 0.3) = 7.2 V

v o2 = −

30 20 = −4 V 120 + 30

v o = v o1 + v o 2 = 3.2 V

(checked: LNAP 5/24/04)

P 5.3-13 Determine the value of the voltage vo in the circuit shown in Figure P 5.3-13.



a

vo

+

b

2Ω i1

R







i2

Figure P 5.3-13 Solution: Using superposition

⎛ R || 4 ⎞ ⎛ ⎞ 4 v o = −2 ⎜⎜ ⎟⎟ i1 + 2 ⎜⎜ ⎟⎟ i 2 6 + R || 4 2 + R || 4 + 4 ( ) ( ) ⎝ ⎠ ⎝ ⎠ Comparing to v o = −0.5 i1 + 4 , we require ⎛ R || 4 ⎞ −2 ⎜⎜ ⎟⎟ = −0.5 ⇒ 4 ( R || 4 ) = 6 + ( R || 4 ) ⇒ R || 4 = 2 ⇒ R = 4 Ω ⎝ 6 + ( R || 4 ) ⎠ and ⎛ ⎞ ⎛ ⎞ 4 4 2 ⎜⎜ ⎟⎟ i 2 = 4 ⇒ 2 ⎜⎜ ⎟⎟ i 2 = 4 ⇒ i 2 = 4 A ⎝ 2 + ( R || 4 ) + 4 ⎠ ⎝ 2 + ( 4 || 4 ) + 4 ⎠ (checked LNAP 6/12/04)

P 5.3-14 Determine values of the current, ia, and the resistance, R, for the circuit shown in Figure P 5.3-14.

+

8V –

ia 5 kΩ

20 kΩ 7 mA 4 kΩ

Figure P 5.3-14 Solution: Use units of mA, kΩ and V.

4 + (5||20) = 8 kΩ (a) Using superposition 8 ⎛ 8 ⎞ 2=⎜ ⇒ 2 ( R + 8 ) = 48 ⇒ R = 16 kΩ ⎟7− R +8 ⎝ R +8⎠ (b) Using superposition again 8 ⎤ 4⎛ 2 1⎞ ⎛ 5 ⎞ ⎡⎛ 16 ⎞ ia = ⎜ = ⎜ × 7 + ⎟ = 4 mA ⎟ ⎢⎜ ⎟7+ ⎥ 8 + 16 ⎦ 5 ⎝ 3 3⎠ ⎝ 5 + 20 ⎠ ⎣⎝ 8 + 16 ⎠

R

2 mA

20 Ω

P 5.3-15 The circuit shown in Figure P 5.3-15 has three inputs: v1, i2, and v3. The output of the circuit is the v1 current io. The output of the circuit is related to the inputs by

+ –

i2

io

12 Ω

i1 = avo + bv2 + ci3 where a, b, and c are constants. Determine the values of a, b, and c.

40 Ω v3 –

+

10 Ω

Figure P 5.3-15 Solution:

⎞ ⎞ ⎛ v1 10 ⎞ ⎛ 10 ⎞ ⎛ 20 ⎛ + − io = ⎜ − ⎜ ⎟ i2 ⎜ ⎟ ⎟ ⎜ ⎟ ⎝ 10 + 40 ⎠ ⎜⎝ 20 + 12 + ( 40 & 10 ) ⎟⎠ ⎝ 10 + 40 ⎠ ⎜⎝ 20 + ⎡⎣12 + ( 40 & 10 ) ⎤⎦ ⎟⎠ ⎞ ⎛ ⎞⎛ v3 20 + 12 + ⎜⎜ − ⎟ ⎟⎟ ⎜⎜ ⎟ ⎝ 40 + ( 20 + 12 ) ⎠ ⎝ 10 + ⎡⎣ 40 & ( 20 + 12 ) ⎤⎦ ⎠ ⎛ 1 ⎞ ⎛ 1⎞ ⎛ 1 ⎞ io = ⎜ − ⎟ v1 + ⎜ − ⎟ i 2 + ⎜ − ⎟ v3 ⎝ 200 ⎠ ⎝ 10 ⎠ ⎝ 62.5 ⎠ So

a = −0.05, b = −0.1 and c = −0.016 (checked: LNAP 6/19/04)

25 V

P 5.3-16 Using the superposition principle, find the value of the current measured by the ammeter in Figure P 5.3-16a.

5A

Hint: Figure P 5.3-16b shows the circuit after the ideal ammeter has been replaced by the equivalent short circuit and a label has been added to indicate the current measured by the ammeter, im. Answer:

25 3 im = − 5 = 5−3 = 2 A 3+ 2 2+3

Ammeter

–+

3Ω 2Ω

(a) 25 V –+



5A



(b)

Figure P 5.3-16

Solution:

im =

25 3 − ( 5) = 5 − 3 = 2 A 3+ 2 2+3

im

Section 5-4: Thèvenin’s Theorem P 5.4-1 Determine values of Rt and voc that cause the circuit shown in Figure P 5.4-1b to be the Thévenin equivalent circuit of the circuit in Figure P 5.4-1a. Hint: Use source transformations and equivalent resistances to reduce the circuit in Figure P 5.4-1a until it is the circuit in Figure P 5.4-1b. Answer: Rt = 5 Ω and voc = 2 V 3Ω

+ – 12 V





Rt

a + –

3A

voc

b

(a)

a

b

(b)

Figure P 5.4-1 Solution:

(checked using LNAP 8/15/02)

P 5.4-2 The circuit shown in Figure P 5.4-2b is the Thévenin equivalent circuit of the circuit shown in Figure P 5.4-2a. Find the value of the open-circuit voltage, voc, and Thévenin resistance, Rt. Answer: voc = –12 V and Rt = 16 Ω 10 Ω

15 V



40 Ω

+ –

(a)

Rt

+ –

voc

(b)

Figure P 5.4-2 Solution: The circuit from Figure P5.4-2a can be reduced to its Thevenin equivalent circuit in four steps:

(a)

(b)

(c)

(d)

Comparing (d) to Figure P5.4-2b shows that the Thevenin resistance is Rt = 16 Ω and the open circuit voltage, voc = −12 V.

P 5.4-3 The circuit shown in Figure P 5.4-3b is the Thévenin equivalent circuit of the circuit shown in Figure P 5.4-3a. Find the value of the open-circuit voltage, voc, and Thévenin resistance, Rt. Answer: voc = 2 V and Rt = 4 Ω 12 V

Rt

–+

1A





+ –

voc



(a)

(b)

Figure P 5.4-3 Solution: The circuit from Figure P5.4-3a can be reduced to its Thevenin equivalent circuit in five steps:

(a)

(c) (b)

(d)

(e) Comparing (e) to Figure P5.4-3b shows that the Thevenin resistance is Rt = 4 Ω and the open circuit voltage, voc = 2 V. (checked using LNAP 8/15/02)

12 Ω

P 5.4-4 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-4.

6Ω + –

10 Ω

a



18 V

b

Figure P 5.4-4 Find Rt:

Rt = Write mesh equations to find voc:

12 (10 + 2 ) =6Ω 12 + (10 + 2 ) Mesh equations: 12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0 6 ( i2 − i1 ) + 3 i 2 − 18 = 0 28 i1 = 6 i 2 9 i 2 − 6 i1 = 18

36 i1 = 18 ⇒ i1 = i2 = Finally,

1 A 2

14 ⎛ 1 ⎞ 7 ⎜ ⎟= A 3 ⎝2⎠ 3

⎛7⎞ ⎛1⎞ voc = 3 i 2 + 10 i1 = 3 ⎜ ⎟ + 10 ⎜ ⎟ = 12 V ⎝ 3⎠ ⎝ 2⎠ (checked using LNAP 8/15/02)

0.75va

P 5.4-5 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-5.



Answer: voc = –2 V and Rt = –8/3 Ω + –

6V

a – va +

4Ω b

Figure P 5.4-4

Solution: Find voc:

Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: va = −voc Apply KCL at node a:

⎛ 6 − voc ⎞ voc ⎛ 3 ⎞ −⎜ + ⎜ − voc ⎟ = 0 ⎟+ ⎝ 8 ⎠ 4 ⎝ 4 ⎠ −6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V Find Rt: We’ll find isc and use it to calculate Rt. Notice that the short circuit forces Apply KCL at node a:

va = 0

⎛ 6−0⎞ 0 ⎛ 3 ⎞ −⎜ ⎟ + + ⎜ − 0 ⎟ + i sc = 0 ⎝ 8 ⎠ 4 ⎝ 4 ⎠ i sc = Rt =

6 3 = A 8 4

voc −2 8 = =− Ω 3 i sc 3 4

(checked using LNAP 8/15/02)





P 5.4-6 Find the Thévenin equivalent circuit for the circuit shown in Figure P 5.4-6.

2va

+ –



+ va –

a

3A b

Figure P 5.4-6 Solution: Find voc:

2 va − va va = + 3 + 0 ⇒ va = 18 V 3 6 The voltage across the right-hand 3 Ω resistor is zero so: va = voc = 18 V Apply KCL at the top, middle node:

Find isc:

2 va − va va v = + 3 + a ⇒ va = −18 V 3 6 3 v −18 Apply Ohm’s law to the right-hand 3 Ω resistor : = −6 V i sc = a = 3 3 v 18 R t = oc = = −3 Ω Finally: i sc −6 Apply KCL at the top, middle node:

(checked using LNAP 8/15/02)

P5.4-7 The equivalent circuit in Figure P5.4-7 is obtained by replacing part of the original circuit by its Thevenin equivalent circuit. The values of the parameters of the Thevenin equivalent circuit are v oc = 15 V and R t = 60 Ω

Determine the following: a.) The values of V s and R a . (Three resistors in the original circuit have equal resistance, Ra.) b.) The value of R b required to cause i = 0.2 A. c.) The value of R b required to cause v = 12 V.

Figure P5.4-7 Solution: a.) From

We see that v oc =

b.) i = c.) v =

v oc Rt + Rb Rb Rt + Rb

Vs 2 and R t = R a . With the given values of voc and Rt we calculate 5 5 V 2 15 = s ⇒ Vs = 75 V and 60 = R a ⇒ R a = 150 Ω . 5 5

⇒ 0.2 = v oc

15 60 + R b

⇒ 12 =

⇒ R b = 15 Ω

15 R b 60 + R b

⇒ R b = 240 Ω

P 5.4-8 A resistor, R, was connected to a circuit box as shown in Figure P 5.4-8. The voltage, v, was measured. The resistance was changed, and the voltage was measured again. The results are shown in the table. Determine the Thévenin equivalent of the circuit within the box and predict the voltage, v, when R = 8 kΩ.

i + Circuit

v –

R

R

v

2 kΩ 4 kΩ

6V 2V

Figure P 5.4-8

Solution:

From the given data: 2000 ⎫ voc ⎪ R t + 2000 ⎪ ⎧ voc = 1.2 V ⎬ ⇒ ⎨ 4000 ⎩ R t = −1600 Ω voc ⎪ 2= R t + 4000 ⎪⎭

6=

When R = 8000 Ω, v=

R voc Rt + R

v=

8000 (1.2 ) = 1.5 V −1600 + 8000

P 5.4-9 A resistor, R, was connected to a circuit box as shown in Figure P 5.4-9. The current, i, was measured. The resistance was changed, and the current was measured again. The results are shown in the table.

(a) (b) Hint:

i R

i

2 kΩ 4 kΩ

4 mA 3 mA

+ Circuit

Specify the value of R required to cause i = 2 mA. Given that R > 0, determine the maximum possible value of the current i.

v

R



Figure P 5.4-9

Use the data in the table to represent the circuit by a Thévenin equivalent.

Solution:

From the given data: voc ⎫ R t + 2000 ⎪⎪ ⎧ voc = 24 V ⎬ ⇒ ⎨ voc ⎩ R t = 4000 Ω ⎪ 0.003 = R t + 4000 ⎪⎭

0.004 =

i=

voc Rt + R

(a) When i = 0.002 A: 24 ⇒ R = 8000 Ω 0.002 = 4000 + R (b) Maximum i occurs when R = 0: 24 = 0.006 = 6 mA ⇒ i ≤ 6 mA 4000

P 5.4-10 For the circuit of Figure P 5.4-10, specify the resistance R that will cause current ib to be 2 mA. The current ia has units of amps. Hint: Find the Thévenin equivalent circuit of the circuit connected to R. 2000ia 6 kΩ + –

12 V

+ –

1 kΩ

ia

ib

R

Figure P 5.4-10 Solution: −12 + 6000 ia + 2000 ia + 1000 ia = 0 ia = 4 3000 A voc = 1000 ia =

4 V 3

ia = 0 due to the short circuit −12 + 6000 isc = 0 ⇒ isc = 2 mA 4 voc Rt = = 3 = 667 Ω isc .002

4 3 ib = 667 + R ib = 0.002 A requires

4 3 − 667 = 0 R = 0.002 (checked using LNAP 8/15/02)

4i a

P 5.4-11 For the circuit of Figure P 5.4-11, specify the value of the resistance RL that will cause current iL to be –2 A.

+ –

10 A



iL

RL

i

Answer: RL = 12 Ω

b

Figure P 5.4-11 Solution:

10 = i + 0 ⇒ i = 10 A voc + 4 i − 2 i = 0 ⇒ voc = −2 i = −20 V

i + i sc = 10 ⇒ i = 10 − i sc

4 i + 0 − 2 i = 0 ⇒ i = 0 ⇒ i sc = 10 A Rt =

−2 = iL =

voc −20 = = −2 Ω isc 10

−20 ⇒ RL = 12 Ω RL − 2

(checked using LNAP 8/15/02)

P 5.4-12 The circuit shown in Figure P 5.4-12 contains an adjustable resistor. The resistance R can be set to any value in the range 0 ≤ R ≤ 100 kΩ. (a) Determine the maximum value of the current ia that can be obtained by adjusting R. Determine the corresponding value of R. (b) Determine the maximum value of the voltage va that can be obtained by adjusting R. Determine the corresponding value of R. (c) Determine the maximum value of the power supplied to the adjustable resistor that can be obtained by adjusting R. Determine the corresponding value of R. ia + + –

12 kΩ

R

12 V

va

− 2 mA

18 kΩ 24 kΩ

Figure P 5.4-12 Solution: Replace the part of the circuit that is connected to the variable resistor by its Thevenin equivalent circuit:

18 kΩ || (12 kΩ + 24 kΩ ) = 18 kΩ || 36 kΩ = 12 kΩ

ia =

R 36 36 and v a = R + 12000 R + 12000 2

36 ⎛ ⎞ p = ia va = ⎜ ⎟ R ⎝ R + 12000 ⎠

36 = 3 mA when R = 0 Ω (a short circuit). 0 + 12000 105 36 = 32.14 V when R is as large as possible, i.e. R = 100 kΩ. (b) v a = 5 10 + 12000 (c) Maximum power is delivered to the adjustable resistor when R = R t = 12 kΩ . Then (a) i a =

2

36 ⎛ ⎞ p = ia va = ⎜ ⎟ 12000 = 0.027 = 27 mW ⎝ 12000 + 12000 ⎠

(checked: LNAP 6/22/04)

P 5.4-13 The circuit shown in Figure P 5.4-13 consists of two parts, the source (to the left of the terminals) and the load. The load consists of a single adjustable resistor having resistance 0 ≤ RL ≤ 20 Ω. The resistance R is fixed, but unspecified. When RL = 4 Ω, the load current is measured to be io = 0.375 A. When RL = 8 Ω, the value of the load current is io = 0.300 A.

(a)

Determine the value of the load current when RL = 10 Ω.

(b)

Determine the value of R.

48 Ω io 24 V

+ –

R source

RL load

Figure P 5.4-13

Solution: Replace the source by it’s Thevenin equivalent circuit to get io =

v oc R t +R L

Using the given formation v oc ⎫ ⎪ R t + 4⎪ ⎬ v oc ⎪ 0.300 = R t + 8 ⎪⎭

0.375 =

So Rt =



0.375 ( R t + 4 ) = 0.300 ( R t + 8 )

( 0.300 ) 8 − ( 0.375) 4 = 12 Ω and v

0.075 6 (a) When R L = 10 Ω, i o = = 0.2727 A. 12 + 10 (b) 12 Ω = R t = 48 11R ⇒ R = 16 Ω .

oc

= 0.3 (12 + 8 ) = 6 V

(checked: LNAP 5/24/04)

P 5.4-14 The circuit shown in Figure P 5.4-14 contains an unspecified resistance, R. Determine the value of R in each of the following two ways.

(a) (b)

Write and solve mesh equations. Replace the part of the circuit connected to the resistor R by a Thévenin equivalent circuit. Analyze the resulting circuit.

40 Ω

20 Ω R + –

40 V

0.25 A 20 Ω

10 Ω

Figure P 5.4-14

Solution: (a)

i 3 − i 2 = 0.25 A Apply KVL to mesh 1 to get

20 ( i1 − i 2 ) + 20 ( i1 − i 3 ) − 40 = 0

Apply KVL to the supermesh corresponding to the unspecified resistance to get 40i 2 + 10i 3 − 20 ( i1 − i 3 ) − 20 ( i1 − i 2 ) = 0 Solving, for example using MATLAB, gives −1 1 ⎤ ⎡ i1 ⎤ ⎡0.25⎤ ⎡ 0 ⎢ 40 −20 −20 ⎥ ⎢i ⎥ = ⎢ 40 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ −40 60 30 ⎥⎦ ⎢⎣ i 3 ⎥⎦ ⎢⎣ 0 ⎥⎦



Apply KVL to mesh 2 to get 40i 2 + R ( i 2 − i 3 ) − 20 ( i1 − i 2 ) = 0



R=

⎡ i1 ⎤ ⎡1.875 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢i 2 ⎥ = ⎢0.750 ⎥ ⎢ i 3 ⎥ ⎢⎣1.000 ⎥⎦ ⎣ ⎦ 20 ( i1 − i 2 ) − 40i 2 i2 − i3

= 30 Ω

(b)

⎛ 20 ⎞ ⎛ 40 ⎞ v oc = ⎜ ⎟ 40 − ⎜ ⎟ 40 = −12 V ⎝ 20 + 20 ⎠ ⎝ 10 + 40 ⎠

R t = 18 Ω

0.25 =

12 18 + R



R = 30 Ω

(checked: LNAP 5/25/04)

a

96 Ω

io

32 Ω

20 V +

P 5.4-15 Consider the circuit shown in Figure P 5.4-15. Replace the part of the circuit to the left of terminals a–b by its Thévenin equivalent circuit. Determine the value of the current io.

32 Ω



+

120 Ω

30 Ω

vo – b

Figure P 5.4-15 Solution: Find the Thevenin equivalent circuit for the part of the circuit to the left of the terminals a-b.

Using voltage division twice

v oc =

32 30 20 − 20 = 5 − 4 = 1 V 32 + 96 120 + 30

R t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48 Ω

Replacing the part of the circuit to the left of terminals a-b by its Thevenin equivalent circuit gives

io =

1 = 0.0125 A = 12.5 mA 48 + 32 (checked: LNAP 5/24/04)

P 5.4-16 An ideal voltmeter is modeled as an open circuit. A more realistic model of a voltmeter is a large resistance. Figure P 5.4-16a shows a circuit with a voltmeter that measures the voltage vm. In Figure P 5.4-16b the voltmeter is replaced by the model of an ideal voltmeter, an open circuit. The voltmeter measures vmi, the ideal value of vm.

Determine the value of vmi.

(b)

Express the measurement error that occurs when Rm = 1000 Ω as a percentage of vmi.

(c)

10 Ω

+ –

25 V

50 Ω

vm –

(a) 200 Ω

10 Ω

+ –

25 V

50 Ω

vmi –

(b) 200 Ω

10 Ω +

+ –

25 V

50 Ω

Rm

vm –

(c)

Figure P 5.4-16

Determine the minimum value of Rm required to ensure that the measurement error is smaller than 2 percent of vmi.

Solution: Replace the circuit by its Thevenin equivalent circuit:

(a)

Voltmeter

+

+

As Rm → ∞, the voltmeter becomes an ideal voltmeter and vm → vmi. When Rm < ∞, the voltmeter is not ideal and vm > vmi. The difference between vm and vmi is a measurement error caused by the fact that the voltmeter is not ideal. (a)

200 Ω

⎛ Rm ⎞ vm = ⎜ 5 ⎜ R m + 50 ⎟⎟ ⎝ ⎠ v mi = lim v m = 5 V R m →∞

(b) When R m = 1000 Ω, v m = 4.763 V so % error =

5 − 4.762 × 100 = 4.76% 5

(c)

⎛ Rm ⎞ 5−⎜ 5 ⎜ R m + 50 ⎟⎟ ⎝ ⎠ 0.02 ≥ 5



Rm R m + 50

≥ 0.98



R m ≥ 2450 Ω (checked: LNAP 6/16/04)

P5.4-17 Given that 0 ≤ R ≤ ∞ in the circuit shown in Figure P5.4-17, consider these two observations:

Observation 1:

When R = 2 Ω then v R = 4 V and i R = 2 A.

Observation 1:

When R = 6 Ω then v R = 6 V and i R = 1 A.

Determine the following a) The maximum value of i R and the value of R that causes i R to be maximal. b) The maximum value of v R and the value of R that causes v R to be maximal. c) The maximum value of p R = i R v R and the value of R that causes p R to be maximal.

Figure P5.4-17 Solution: We can replace the part of the circuit to the left of the terminals by its Thevenin equivalent circuit:

Using voltage division v R = iR =

v oc R + Rt

R v oc and using Ohm’s law R + Rt

.

v oc R v oc = will be maximum when Rt R + Rt 1+ R R = ∞. The maximum value of v R will be v oc . Similarly,

By inspection , v R =

iR =

v oc R + Rt

will be maximum when R = 0. The maximum value

of i R will be

v oc Rt

= i sc .

The maximum power transfer theorem tells use that p R = i R v R will be maximum when R = R t . ⎛ v oc Then p R = i R v R = ⎜ ⎜ R + Rt ⎝

⎞⎛ R ⎞ ⎛ v oc v oc ⎟ = R ⎜ ⎟⎜ ⎟⎜ R + R t ⎟ ⎜ R + Rt ⎠⎝ ⎠ ⎝

2

⎞ ⎟⎟ . ⎠

Let’s substitute the given data into the equation i R = When R = 2 Ω we get 2 = 1=

v oc 6 + Rt

v oc 2 + Rt

v oc R + Rt

.

⇒ 4 + 2 R t = v oc . When R = 6 Ω we get

⇒ 6 + R t = v oc .

So 6 + R t = 4 + 2 R t

⇒ R t = 2 Ω and v oc = 4 + 2 R t = 8 V . Also i sc =

v oc Rt

=

8 = 4 A. 2

P5.4-18 Consider the circuit shown in Figure P5.4-18. Determine

a) The value of v R that occurs when R = 9 Ω. b) The value of R that causes v R = 5.4 V. c) The value of R that causes i R = 300 mA.

Figure P5.4-18 Solution: Reduce this circuit using source transformations and equivalent resistance:

9 ⎛ R ⎞ so the questions can be easily answered: Now v R = ⎜ ⎟ 9 and i R = R + 18 ⎝ R + 18 ⎠ a) When R = 9 Ω then v R = 3 V. b) When R = 27 Ω then v R = 5.4 V. c) When R = 12 Ω then i R = 300 mA.

P5.4-19 The circuit shown in Figure P5.4-19a can be reduced to the circuit shown in Figure P5.4-19b using source transformations and equivalent resistances. Determine the values of the source voltage v oc and the

(a)

resistance R.

(b) Figure P5.4-19 Solution

46 = R t = R + ( 42 || 84 ) = R + 28 ⇒ R = 18 Ω

v oc =

84 (18) = 12 V 42 + 84

P5.4-20 The equivalent circuit in Figure P5.4-20 is obtained by replacing part of the original circuit by its Thevenin equivalent circuit. The values of the parameters of the Thevenin equivalent circuit are v oc = 15 V and R t = 60 Ω

Determine the following: a.) The values of V s and R a . (Three resistors in the original circuit have equal resistance, Ra.) b.) The value of R b required to cause i = 0.2 A. c.) The value of R b required to cause v = 5 V.

Figure P5.4-20 Solution a.) From

We see that v oc =

b.) i = c.) v =

v oc Rt + Rb Rb Rt + Rb

Vs 3 and R t = R a . With the given values of voc and Rt we calculate 2 2 V 3 15 = s ⇒ Vs = 30 V and 60 = R a ⇒ R a = 40 Ω . 2 2 ⇒ 0.2 =

v oc

15 60 + R b

⇒ 5=

15 R b 60 + R b

⇒ R b = 15 Ω ⇒ R b = 30 Ω

Section 5-5: Norton’s Theorem P5.5-1 The part of the circuit shown in Figure P5.3-1a to the left of the terminals can be reduced to its Norton equivalent circuit using source transformations and equivalent resistance. The resulting Norton equivalent circuit, shown in Figure P5.3-1b, will be characterized by the parameters: i sc = 0.5 A and R t = 20 Ω

a) Determine the values of v s and R1 . b) Given that 0 ≤ R 2 ≤ ∞ , determine the maximum values of the voltage, v, and of the power, p = vi. Answers: v s = 37.5 V, R1 = 25 Ω, max v = 10 V and max p = 1.25 W

(a)

(b) Figure P5.5-1

Solution: Two source transformations reduce the circuit as follows:

Recognizing the parameters of the Norton equivalent circuit gives: 0.5 = i sc =

12.5 + v s 100

⇒ v s = 37.5 V and 20 = R t = 100 || R1 =

(

)

Next, the voltage across resistor R 2 is given by v = i sc R t || R 2 =

100 R1 100 + R1

R t R 2 i sc Rt + R2

=

⇒ R1 = 25 Ω

R t i sc so this Rt +1 R2

voltage is maximum when R 2 = ∞ and max v = R t i sc =10 V. The power p = vi will be

maximum when R 2 = R t = 20 Ω . Then v =

R t i sc

2

=

20 ( 0.5 ) 2

=5 V, i=

v 5 = = 0.25 A and R 2 20

p = v i = 5 ( 0.25 ) = 1.25 W.

P 5.5-2 Two black boxes are shown in Figure P 5.5-2. Box A contains the Thévenin equivalent of some linear circuit, and box B contains the Norton equivalent of the same circuit. With access to just the outsides of the boxes and their terminals, how can you determine which is which, using only one shorting wire?

Box A 1Ω

1V

Box B a

+ –

a

1A

b



b

Figure P 5.5-2

Solution: When the terminals of the boxes are open-circuited, no current flows in Box A, but the resistor in Box B dissipates 1 watt. Box B is therefore warmer than Box A. If you short the terminals of each box, the resistor in Box A will draw 1 amp and dissipate 1 watt. The resistor in Box B will be shorted, draw no current, and dissipate no power. Then Box A will warm up and Box B will cool off.

(a)

(b) Figure P5.5-3 P5.5-3 The circuit shown in Figure P5.5-3a can be reduced to the circuit shown in Figure P5.53b using source transformations and equivalent resistances. Determine the values of the source current i sc and the resistance R. Solution:

48 = R t = R || ( 80 + 160 ) = i sc =

240 R ⇒ R = 60 Ω R + 240

80 ( 4.8 ) = 1.6 A 80 + 160

P 5.5-4

Find the Norton equivalent circuit for the circuit shown in Figure P 5.5-4. 3Ω

5Ω a

4A



5A b

Figure P 5.5-4

Solution:

P 5.5-5 The circuit shown in Figure P 5.5-5b is the Norton equivalent circuit of the circuit shown in Figure P 5.5-5a. Find the value of the short-circuit current, isc, and Thévenin resistance, Rt. Answer: isc = 1.13 A and Rt = 7.57 Ω 3Ω

5Ω – +

+ –

10 V

2ia



isc

Rt

ia

(a)

(b)

Figure P 5.5-5 Solution: To determine the value of the short circuit current, isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 5.6-4a after adding the short circuit and labeling the short circuit current. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (a), mesh current i2 is equal to the current in the short circuit. Consequently, i2 = isc . The controlling current of the CCVS is expressed in terms of the mesh currents as

i a = i1 − i 2 = i1 − isc Apply KVL to mesh 1 to get

3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 7 i1 − 4 i 2 = 10 Apply KVL to mesh 2 to get 5 i 2 − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = 0 ⇒ i1 =

11 i2 6

Substituting into equation 1 gives ⎛ 11 ⎞ 7 ⎜ i 2 ⎟ − 4 i 2 = 10 ⇒ i 2 = 1.13 A ⇒ i sc = 1.13 A ⎝6 ⎠

Figure (a) Calculating the short circuit current, isc, using mesh equations.

(1)

To determine the value of the Thevenin resistance, Rt, first replace the 10 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source across the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v Rt = T iT In Figure (b), the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (b), mesh current i2 is equal to the negative of the current source current. Consequently, i2 = i T . The controlling current of the CCVS is expressed in terms of the mesh currents as i a = i1 − i 2 = i1 + i T

Apply KVL to mesh 1 to get 3 i 1 − 2 ( i 1 − i 2 ) + 6 ( i1 − i 2 ) = 0 ⇒ 7 i 1 − 4 i 2 = 0 ⇒ i 1 =

4 i2 7

(2)

Apply KVL to mesh 2 to get 5 i 2 + vT − 6 ( i1 − i 2 ) = 0 ⇒ − 6 i1 + 11 i 2 = −vT Substituting for i1 using equation 2 gives ⎛4 ⎞ −6 ⎜ i 2 ⎟ + 11 i 2 = −vT ⎝7 ⎠ Finally,

Rt =

⇒ 7.57 i 2 = −vT

vT −vT −vT = = = 7.57 Ω iT i2 −iT

Figure (b) Calculating the Thevenin resistance, R t =

vT , using mesh equations. iT

To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure 4.6-4a after adding the open circuit and labeling the

open circuit voltage. Also, the meshes have been identified and labeled in anticipation of writing mesh equations. Let i1 and i2 denote the mesh currents in meshes 1 and 2, respectively. In Figure (c), mesh current i2 is equal to the current in the open circuit. Consequently, i2 = 0 A . The controlling current of the CCVS is expressed in terms of the mesh currents as

i a = i1 − i 2 = i1 − 0 = i1 Apply KVL to mesh 1 to get 3 i1 − 2 ( i1 − i 2 ) + 6 ( i1 − i 2 ) − 10 = 0 ⇒ 3 i1 − 2 ( i1 − 0 ) + 6 ( i1 − 0 ) − 10 = 0

⇒ i1 =

10 = 1.43 A 7

Apply KVL to mesh 2 to get 5 i 2 + voc − 6 ( i1 − i 2 ) = 0 ⇒ voc = 6 ( i1 ) = 6 (1.43) = 8.58 V

Figure (c) Calculating the open circuit voltage, voc, using mesh equations.

As a check, notice that R t isc = ( 7.57 )(1.13) = 8.55 ≈ voc (checked using LNAP 8/16/02)



P 5.5-6 The circuit shown in Figure P 5.5-6b is the Norton equivalent circuit of the circuit shown in Figure P 5.5-6a. Find the value of the short-circuit current, isc, and Thévenin resistance, Rt.

6Ω +

– +

24 V

va

1.33va

isc

Rt



Answer: isc = –24 A and Rt = –3 Ω

(a)

(b)

Figure P 5.5-6 Solution: To determine the value of the short circuit current, Isc, we connect a short circuit across the terminals of the circuit and then calculate the value of the current in that short circuit. Figure (a) shows the circuit from Figure 4.6-5a after adding the short circuit and labeling the short circuit current. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (a), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The voltage at node 3 is equal to the voltage across a short, v3 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across a short, i.e. v3 = 0 . Apply KCL at node 2 to get

v1 − v 2 3

=

v 2 − v3 6

⇒ 2 v1 + v 3 = 3 v 2

⇒ − 48 = 3 v a

⇒ v a = −16 V

Apply KCL at node 3 to get

v 2 − v3 6

+

4 v 2 = isc 3



9 v a = isc 6

⇒ isc =

9 ( −16 ) = −24 A 6

Figure (a) Calculating the short circuit current, Isc, using mesh equations.

To determine the value of the Thevenin resistance, Rth, first replace the 24 V voltage source by a 0 V voltage source, i.e. a short circuit. Next, connect a current source circuit across

the terminals of the circuit and then label the voltage across that current source as shown in Figure (b). The Thevenin resistance will be calculated from the current and voltage of the current source as v R th = T iT Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. In Figure (b), node voltage v1 is equal to the across a short circuit, i.e. v1 = 0 . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the voltage across the current source, i.e. v3 = vT . Apply KCL at node 2 to get

v1 − v 2 3

=

v2 − v3 6

⇒ 2 v1 + v 3 = 3 v 2

⇒ vT = 3 v a

Apply KCL at node 3 to get v2 − v3 6

+

4 v 2 + iT = 0 ⇒ 9 v 2 − v3 + 6 iT = 0 3 ⇒ 9 v a − vT + 6 iT = 0 ⇒ 3 v T − vT + 6 iT = 0 ⇒ 2 vT = −6 iT

Finally, Rt =

vT = −3 Ω iT

vT , using mesh equations. iT To determine the value of the open circuit voltage, voc, we connect an open circuit across the terminals of the circuit and then calculate the value of the voltage across that open circuit. Figure (c) shows the circuit from Figure P 4.6-5a after adding the open circuit and labeling the open circuit voltage. Also, the nodes have been identified and labeled in anticipation of writing node equations. Let v1, v2 and v3 denote the node voltages at nodes 1, 2 and 3, respectively. Figure (b) Calculating the Thevenin resistance, R th =

In Figure (c), node voltage v1 is equal to the negative of the voltage source voltage. Consequently, v1 = −24 V . The controlling voltage of the VCCS, va, is equal to the node voltage at node 2, i.e. va = v2 . The voltage at node 3 is equal to the open circuit voltage, i.e. v3 = voc . Apply KCL at node 2 to get v1 − v 2

=

v 2 − v3

3 Apply KCL at node 3 to get

6

v2 − v3

⇒ 2 v1 + v 3 = 3 v 2

⇒ − 48 + v oc = 3 v a

4 v 2 = 0 ⇒ 9 v 2 − v 3 = 0 ⇒ 9 v a = v oc 6 3 Combining these equations gives +

3 ( −48 + voc ) = 9 v a = voc

⇒ voc = 72 V

Figure (c) Calculating the open circuit voltage, voc, using node equations.

As a check, notice that

R th I sc = ( −3)( −24 ) = 72 = Voc

(checked using LNAP 8/16/02)

P 5.5-7 Determine the value of the resistance R in the circuit shown in Figure P 5.5-7 by each of the following methods:

(a)

(b)

5 kΩ

10 kΩ

ib

25 V +–

Replace the part of the circuit to the left of terminals a–b by its the Norton equivalent circuit. Use current division to determine the value of R.

4 ib

a

R

0.5 mA

b

Figure P 5.5-7

Analyze the circuit shown Figure P 5.5-6 using mesh equations. Solve the mesh equations to determine the value of R.

Solution: (a) Replace the part of the circuit that is connected to the left of terminals a-b by its Norton equivalent circuit:

Apply KCL at the top node of the dependent source to see that i b = 0 A . Then

( )

v oc = 25 + 5000 i b = 25 V Apply KVL to the supermesh corresponding to the dependent source to get

( )

−5000 i b + 10000 3 i b − 25 = 0 ⇒ i b = 1 mA Apply KCL to get i sc = 3 i b = 3 mA Then Rt =

v oc i sc

= 8.33 kΩ

Current division gives 0.5 =

8333 3 ⇒ R = 41.67 kΩ R + 8333

(b) Notice that i b and 0.5 mA are the mesh currents. Apply KCL at the top node of the dependent source to get 1 i b + 0.5 × 10−3 = 4 i b ⇒ i b = mA 6 Apply KVL to the supermesh corresponding to

the dependent source to get

(

)

−5000 i b + (10000 + R ) 0.5 × 10−3 − 25 = 0

(

)

⎛1 ⎞ −5000 ⎜ × 10−3 ⎟ + (10000 + R ) 0.5 × 10−3 = 25 ⎝6 ⎠ 125 6 R= = 41.67 kΩ 0.5 ×10−3

P5.5-8 Find the Norton equivalent circuit of this circuit:

Solution Simplify the circuit using a source transformation:

Identify the open circuit voltage and short circuit current. Apply KVL to the mesh to get:

(10 + 2 + 3) i x − 15 = 0

⇒ ix = 1 A

Then v oc = 3 i x = 3 V

Express the controlling current of the CCVS in terms of the mesh currents: i x = i1 − i sc The mesh equations are 10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15 and

i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 =

4 i sc 3

so ⎛4 ⎞ 15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A ⎝3 ⎠ The Thevenin resistance is Rt =

3 =3Ω 1

Finally, the Norton equivalent circuit is

(checked: LNAP 6/21/04)

P5.5-9 Find the Norton equivalent circuit of this circuit:

Solution Identify the open circuit voltage and short circuit current. ⎛1⎞ v1 = ⎜ ⎟ 3 = 1 V ⎝ 3⎠

Then v oc = v1 − 4 ( 2.5 v 1 ) = −9 V

⎛1 ⎞ v1 = 3 ⎜ − i sc ⎟ = 1 − 3 i sc ⎝3 ⎠ 4 ( 2.5 v1 + i sc ) + 5 i sc − v1 = 0 ⇒ 9 v1 + 9 i sc = 0 9 (1 − 3 i sc ) + 9 i sc = 0 ⇒ i sc = The Thevenin resistance is Rt =

1 A 2

−9 = −18 Ω 0.5

Finally, the Norton equivalent circuit is

(checked: LNAP 6/21/04)

P 5.5-10 An ideal ammeter is modeled as a short circuit. A more realistic model of an ammeter is a small resistance. Figure P 5.5-10a shows a circuit with an ammeter that measures the current im. In Figure P 5.5-10b the ammeter is replaced by the model of an ideal ammeter, a short circuit. The ammeter measures imi, the ideal value of im.

As Rm → 0, the ammeter becomes an ideal ammeter and im → imi. When Rm > 0, the ammeter is not ideal and im < imi. The difference between im and imi is a measurement error caused by the fact that the ammeter is not ideal. (a)

Determine the value of imi.

(b)

Express the measurement error that occurs when Rm = 20 Ω as a percentage of imi.

(c)

im

4 kΩ

3 mA

Ammeter

4 kΩ

2 kΩ

(a) imi 4 kΩ

3 mA

4 kΩ

2 kΩ

(b) im 4 kΩ

3 mA

Determine the maximum value of Rm required to ensure that the measurement error is smaller than 2 percent of imi.

4 kΩ

2 kΩ

(c)

Figure P 5.5-10

Solution: Replace the circuit by its Norton equivalent circuit:

⎛ 1600 ⎞ im = ⎜ 1.5 × 10−3 ) ⎜ 1600 + R m ⎟⎟ ( ⎝ ⎠ (a)

Rm

i mi = lim

R m →0

i m = 1.5 mA

(b) When Rm = 20 Ω then i m = 1.48 mA so % error =

1.5 − 1.48 × 100 = 1.23% 1.5

(c) ⎛ 1600 ⎞ 0.015 − ⎜ 0.015 ) ⎜ 1600 + R m ⎟⎟ ( ⎝ ⎠ 0.02 ≥ 0.015



1600 ≥ 0.98 ⇒ R m ≤ 32.65 Ω 1600 + R m (checked: LNAP 6/18/04)





P 5.5-11 Determine values of Rt and isc that cause the circuit shown in Figure P 5.5-11b to be the Norton equivalent circuit of the circuit in Figure P 5.5-11a.

+ –

+ –

12 V ia

a

a

isc

2ia

Rt

b

Answer: Rt = 3 Ω and isc = –2 A

b

(a)

(b)

Figure P 5.5-11 Solution:

ia =

2 i a − 12

⇒ i a = −3 A 6 voc = 2 i a = −6 V

12 + 6 i a = 2 i a 3 i sc = 2 i a

Rt =

−6 =3Ω −2

⇒ i a = −3 A

⇒ i sc =

2 ( − 3 ) = −2 A 3

12 Ω

P 5.5-12 Use Norton’s theorem to formulate a general expression for the current i in terms of the variable resistance R shown in Figure P 5.5-12.

8 Ω a

30 V

+ –

i

R b

Answer: i = 20/(8 + R) A

Figure P 5.5-12

Solution:

12 × 24 12 × 24 = = 8Ω 12 + 24 36 24 voc = ( 30 ) = 20 V 12 + 24 Rt =

i=

20 8+ R

16 Ω

Section 5-6: Maximum Power Transfer P 5.6-1 The circuit model for a photovoltaic cell is given in Figure P 5.6-1 (Edelson, 1992). The current is is proportional to the solar insolation (kW/m2). (a) (b)

Find the load resistance, RL, for maximum power transfer. Find the maximum power transferred when is = 1 A.



ia

R2 +

+ –

vs

+ –

4 ia

v

R



i

Figure P 5.6-1

Solution: (a) The value of the current in R 2 is 0 A so v oc = 4 i a . Then KVL gives

8 i a + 4 i a − V s = 0 ⇒ V s = 12 i a = 3 ( 4 i a ) = 3 ( v oc ) = 24 V

Next, KVL gives

8 i a + 4 i a − 24 = 0 ⇒ i a = 2 A and 4 i a = R 2 i sc

⇒ 4 ( 2) = R2 ( 2) ⇒ R2 = 4 Ω

(b) The power delivered to the resistor to the right of the terminals is maximized by setting R equal to the Thevenin resistance of the part of the circuit to the left of the terminals: R = Rt =

Then

p max =

v oc i sc

v oc 2 4 Rt

=

=

8 =4Ω 2

82 =4W 4 ( 4)



P 5.6-2 For the circuit in Figure P 5.6-2, (a) find R such that maximum power is dissipated in R and (b) calculate is the value of maximum power.

100 Ω

Answer: R = 60 Ω and Pmax = 54 mW Figure P 5.6-2 Solution:

a) For maximum power transfer, set RL equal to the Thevenin resistance: R L = R t = 100 + 1 = 101 Ω b) To calculate the maximum power, first replace the circuit connected to RL be its Thevenin equivalent circuit:

The voltage across RL is

vL =

Then

pmax

101 (100 ) = 50 V 101 + 101 2 vL 502 = = = 24.75 W R L 101

RL

150 Ω

P 5.6-3 For the circuit in Figure P 5.6-3, prove that for Rs variable and RL fixed, the power dissipated in RL is maximum when Rs = 0.

6V

+ –

100 Ω

R

+ –

2V

Figure P 5.6-3 Solution: Reduce the circuit using source transformations:

Then (a) maximum power will be dissipated in resistor R when: R = Rt = 60 Ω and (b) the value of that maximum power is P = i 2 ( R) = (0.03)2 (60) = 54 mW max R

Rs

P 5.6-4 Find the maximum power to the load RL if the maximum power transfer condition is met for the circuit of Figure P 5.6-4.

vs

+ –

RL

Answer: max pL = 0.75 W source network

load

Figure P 5.6-4 Solution: ⎡ RL ⎤ v L = vS ⎢ ⎥ ⎣⎢ R S + R L ⎦⎥ ∴ pL =

v L2 RL

=

v S2 R L ( RS + R L )2

By inspection, pL is max when you reduce RS to get the smallest denominator. ∴ set RS = 0



P 5.6-5 Determine the maximum power that can be absorbed by a resistor, R, connected to terminals a–b of the circuit shown in Figure P 5.6-5. Specify the required value of R.

20 Ω

a

10 Ω 20 A

120 Ω

50 Ω b

Figure P 5.6-5 Solution:

The required value of R is

R = Rt = 8 +

( 20 + 120 ) (10 + 50 ) = 50 Ω ( 20 + 120 ) + (10 + 50 )

30 ⎡ 170 voc = ⎢ ( 20 )⎤⎥ 10 − ⎡⎢ ( 20 )⎤⎥ 50 ⎣170 + 30 ⎦ ⎣170 + 30 ⎦ 170(20)(10) − 30(20)(50) 4000 = = = 20 V 200 200 The maximum power is given by 2 voc 202 pmax = = =2W 4 R t 4 ( 50 )

P 5.6-6 Figure P 5.6-6 shows a source Ro connected to a load through an amplifier. The load can safely receive + up to 15 W of power. Consider three + + Ava v 500 mV 100 kΩ RL a cases: – – – (a) A = 20 V/V and Ro = 10 Ω. Determine the value of RL that source amplifier load maximizes the power delivered Figure P 5.6-6 to the load and the corresponding maximum load power. (b) A = 20 V/V and RL = 8 Ω. Determine the value of Ro that maximizes the power delivered to the load and the corresponding maximum load power. (c) Ro = 10 Ω and RL = 8 Ω. Determine the value of A that maximizes the power delivered to the load and the corresponding maximum load power. Solution:

iL =

A vs Ro +RL

PL = i L R L = 2

A 2v s 2 R L

(R

o

+RL)

2

(a) R t =R o so R L =R o = 10 Ω maximizes the power delivered to the load. The corresponding load power is 2

⎛1⎞ 20 ⎜ ⎟ 10 ⎝2⎠ PL = = 2.5 W . 2 (10 + 10 ) 2

(b) Ro = 0 maximizes PL (The numerator of PL does not depend on Ro so PL can be maximized by making the denominator as small as possible.) The corresponding load power is

2

PL =

A 2v s 2 R L R L2

=

A 2v s 2 RL

⎛1⎞ 20 ⎜ ⎟ ⎝ 2 ⎠ = 12.5 W. = 8 2

(c) PL is proportional to A2 so the load power continues to increase as A increases. The load can safely receive 15 W. This limit corresponds to 2

⎛1⎞ A ⎜ ⎟ 8 ⎝2⎠ 15 = 2 (18) 2



A = 36

15 = 49.3 V. 8

(checked: LNAP 6/9/04)

P 5.6-7 The circuit in Figure P 5.6-7 contains a variable resistance, R, implemented using a potentiometer. The resistance of the variable resistor varies over the range 0 ≤ R ≤ 1000 Ω. The variable resistor can safely receive 1/4 W power. Determine the maximum power received by the variable resistor. Is the circuit safe? 180 Ω

+ –

10 V

R

120 Ω

150 Ω

470 Ω

+ –

20 V

Figure P 5.6-7 Solution: Replace the part of the circuit connected to the variable resistor by its Thevenin equivalent circuit. First, replace the left part of the circuit by its Thevenin equivalent: ⎛ 150 ⎞ v oc1 = ⎜ ⎟10 = 4.545 V ⎝ 150 + 180 ⎠ R t1 = 180 & 150 = 81.8 Ω

Next, replace the right part of the circuit by its Thevenin equivalent: ⎛ 470 ⎞ v oc2 = ⎜ ⎟ 20 = 15.932 V ⎝ 470 + 120 ⎠ R t2 = 120 & 470 = 95.6 Ω

Now, combine the two partial Thevenin equivalents: v oc = v oc1 − v oc2 = −10.387 V and R t = R t1 + R t2 = 177.4 Ω

So The power received by the adjustable resistor will be maximum when R = Rt = 177.4 Ω. The maximum power received by the adjustable

( −11.387 ) = 0.183 W . p= 4 (177.4 Ω ) 2

resistor will be

(checked LNAPDC 7/24/04)

Rt

P 5.6-8 For the circuit of Figure P 5.6-8, find the power delivered to the load when RL is fixed and Rt may be varied between 1 Ω and 5 Ω. Select Rt so that maximum power is delivered to RL.

10 V

Answer: 13.9 W

+ –

RL = 5 Ω

Figure P 5.6-8

Solution: ⎛ 10 ⎞ ⎡ R L ⎤ 100 R L p=iv=⎜ 10 ) ⎥ = ( ⎟ ⎢ 2 ⎜ Rt + R L ⎟ ⎢ Rt + R L ⎝ ⎠⎣ ⎦⎥ ( R t + R L )

The power increases as Rt decreases so choose Rt = 1 Ω. Then pmax = i v =

100 ( 5 )

(1 + 5)

2

= 13.9 W

5

P 5.6-9 A resistive circuit was connected to a variable resistor, and the power delivered to the resistor was measured as shown in Figure P 5.6-9. Determine the Thévenin equivalent circuit.

Power (W)

0

Answer: Rt = 20 Ω and voc = 20 V

10

20

30

R (ohms)

Figure P 5.6-9 Solution: From the plot, the maximum power is 5 W when R = 20 Ω. Therefore:

Rt = 20 Ω

and 2

pmax

v = oc 4 Rt

⇒ voc =

pmax 4 Rt = 5 ( 4 ) 20 = 20 V

40

Figure P5.6-10 P5.6-10 The part circuit shown in Figure P5.6-10a to left of the terminals can be reduced to its Norton equivalent circuit using source transformations and equivalent resistance. The resulting Norton equivalent circuit, shown in Figure P5.6-10b, will be characterized by the parameters:

i sc = 1.5 A and

R t = 80 Ω

(a) Determine the values of i s and R1 . (b) Given that 0 ≤ R 2 ≤ ∞ , determine the maximum value of p = vi, the power delivered to R2. Solution: Two source transformations reduce the circuit as follows:

(a) Recognizing the parameters of the Norton equivalent circuit gives: 1.5 = i sc = i s + 0.25 ⇒ i s = 1.25 A and 80 = R t = 100 || R1 =

100 R1 100 + R1

⇒ R1 = 400 Ω

(b) The maximum value of the power delivered to R2 occurs when R 2 = R t = 80 Ω . Then 2

1 ⎛1 ⎞ i = i sc = 0.75 A and p = ⎜ i sc ⎟ R t = ( 0.6252 ) 80 = 45 W 2 ⎝2 ⎠

P5.6-11. Given that 0 ≤ R ≤ ∞ in the circuit shown in Figure P5.6-12, determine (a) maximum value of i a , (b)

the maximum value of v a , and (c) the maximum value of pa = iava . Figure P5.6-11 Solution:

Replace the parallel combination of resistor R and the 8 Ω resistor by an equivalent resistance.

Using voltage division

va =

R eq 4 + R eq

(12 ) =

1 (12 ) 4 +1 R eq

Consequently, the maximum value of va corresponds to the is obtained by maximizing Req. The maximum of Req is obtained by maximizing R. Given that 0 ≤ R ≤ ∞, the maximum value of Req is 8 Ω and the maximum value of va is

v a max =

Using Ohm’s law

1 4 +1 8

(12 ) = 8 V

ia =

12 4 + R eq

Consequently, the maximum value of ia corresponds to the is obtained by minimizing Req. The minimum of Req is obtained by maximizing R. Given that 0 ≤ R ≤ ∞, the minimum value of Req is 0 Ω and the maximum value of ia is

i a max =

12 =3A 4+0

The maximum power theorem indicates that the maximum value of p a = i a v a occurs when Req = Rt. In this case, Rt = 4 Ω. We require Req = 4 Ω which is accomplished by making R = 8 Ω, an acceptable value since 0 ≤ 8 ≤ ∞. Then 2

2

⎛ 12 ⎞ ⎛ 12 ⎞ ⎜ ⎟ ⎜ ⎟ 2 2 pa = ⎝ ⎠ = ⎝ ⎠ = 9 W R eq 4

P5.6-12. Given that 0 ≤ R ≤ ∞ in the circuit shown in Figure P5.6-12, determine value of R that maximizes the power p a = i a v a and the corresponding maximum value

of p a . Figure P5.6-12 Solution:

Replace the combination of resistor R and the 20 Ω and 2 Ω resistors by an equivalent resistance.

The maximum power theorem indicates that the maximum value of p a = i a v a occurs when Req = Rt. In this case, Rt = 8 Ω. We require 8 = R eq =

20 ( R + 2 ) 20 R + 40 = 20 + ( R + 2 ) R + 22

8 ( R + 22 ) = 20 R + 40 ⇒ R =

8 ( 22 ) − 40 = 11.333 Ω 20 − 8

This isn’t a standard resistance value but it is an acceptable value for this problem since 0 ≤ 11.333 ≤ ∞. Then 2

2

⎛6⎞ ⎛6⎞ ⎜ ⎟ ⎜ ⎟ 2 2 p a = ⎝ ⎠ = ⎝ ⎠ = 1.125 W 8 R eq

Section 5.8 Using PSpice to Determine the Thevenin Equivalent Circuit

P5.8-1

a) Here are the results of simulating the circuit in PSpice. The numbers shown in white on a black background are the node voltages.

b) Add a resistor across the terminals of Circuit A. Then

v oc = v R i sc =

vR R

when

R≈∞

when

R≈0

1

Here are the PSpice simulation results:

v oc = 10 V

i sc = Rt =

1× 10−6 =1 A 1× 10−6

v oc i sc

=

10 = 10 Ω 1

c) Here is the result of simulation the circuit after replacing Circuit A by its Thevenin equivalent:

d) The node voltages of Circuit B are the same before and after replacing Circuit A by its Thevenin equivalent circuit.

2

Section 5-9 How Can We Check…? P 5.9-1 For the circuit of Figure P 5.9-1, the current i has been measured for three different values of R and is listed in the table. Are the data consistent? 1 kΩ

ix

R(Ω)

i(mA)

5000ix

5000 500 0

16.5 43.8 97.2

+ –

R

i

4 kΩ + –

10 V

4 kΩ

Figure P 5.9-1 Solution: Use the data in the first two lines of the table to determine voc and Rt: voc ⎫ R t + 0 ⎪⎪ ⎧voc = 39.9 V ⎬ ⇒ ⎨ voc ⎪ ⎩ R t = 410 Ω 0.0438 = R t + 500 ⎪⎭ 0.0972 =

Now check the third line of the table. When R= 5000 Ω: v 39.9 i = oc = = 7.37 mA R t + R 410 + 5000 which disagree with the data in the table. The data is not consistent.

i

P 5.9-2

R v –

+

Your lab partner built the circuit

shown in Figure P 5.9-2 and measured the

6 kΩ

current i and voltage v corresponding to several values of the resistance R. The results

12 V

+ –

2 mA

18 kΩ

are shown in the table in Figure P 5.9-2. Your lab partner says that RL = 8000 Ω is required to cause i = 1 mA. Do you agree? Justify your answer.

24 kΩ R

i

v

open 10 kΩ short

0 mA 0.857 mA 3 mA

12 V 8.57 V 0V

Figure P 5.9-2

Solution: Use the data in the table to determine voc and isc: voc = 12 V (line 1 of the table) isc = 3 mA so

Rt =

(line 3 of the table)

voc = 4 kΩ isc

Next, check line 2 of the table. When R = 10 kΩ: v 12 i = oc = = 0.857 mA 3 R t + R 10 (10 ) + 5 (103 )

To cause i = 1 mA requires

which agrees with the data in the table. v 12 0.001 = i = oc = ⇒ R = 8000 Ω R t + R 10 (103 ) + R

I agree with my lab partner’s claim that R = 8000 causes i = 1 mA.

12 kΩ

P 5.9-3

In preparation for lab, your lab 3R

partner determined the Thévenin equivalent of the circuit connected to RL in

2R

Figure P 5.9-3. She says that the Thévenin resistance is Rt = voltage is voc =

60 11

6 R 11

and the open-circuit

30 V

20 V

V. In lab, you built the

i

R

+ –

+ –

Load 10 V

+ –

circuit using R = 110 Ω and RL = 40 Ω and measured that i = 54.5 mA. Is this

Figure P 5.9-3

measurement consistent with the prelab calculations? Justify your answers. Solution:

1 1 1 1 11 = + + = R t R 2 R 3R 6 R

⇒ Rt =

6R 11

and ⎛ 23 ⎞ ⎛ 34 ⎞ ⎛ 65 ⎞ 180 voc = ⎜ ⎟ 30 + ⎜ ⎟ 20 + ⎜ ⎟ 10 = 11 ⎝ 3+ 2 3⎠ ⎝ 2+3 4⎠ ⎝ 1+ 6 5 ⎠

so the prelab calculation isn’t correct. But then 180 180 voc 11 i= = = 11 = 163 mA ≠ 54.5 mA 6 Rt + R (110 ) + 40 60 + 40 11 so the measurement does not agree with the corrected prelab calculation.

RL

P 5.9-4 Your lab partner claims that the current i in Figure P 5.9-4 will be no greater than 12.0 mA, regardless of the value of the resistance R. Do you agree? i

500 Ω

R 12 V

+ –

3 kΩ

6 kΩ

1500 Ω

Figure P 5.9-4

Solution: 6000 & 3000 & ( 500 + 1500 ) = 2000 & 2000 = 1000 Ω

i=

12 12 ≤ = 12 mA R + 1000 1000

How about that?! Your lab partner is right. (checked using LNAP 6/21/05)

P 5.9-5 Figure P 5.9-5 shows a circuit and some corresponding data. Two resistances, R1 and R, and the current source current are unspecified. The tabulated data provide values of the current, i, and voltage, v, corresponding to several values of the resistance R. (a) Consider replacing the part of the circuit connected to the resistor R by a Thévenin equivalent circuit. Use the data in rows 2 and 3 of the table to find the values of Rt and voc, the Thévenin resistance and the open-circuit voltage. (b) Use the results of part (a) to verify that the tabulated data are consistent. (c) Fill in the blanks in the table. (d) Determine the values of R1 and is.

12 V

+ –

24 Ω is

i + v

R



R, Ω

i, A

v, V

0 10 20 40 80

3 1.333 0.857 0.5 ?

0 13.33 17.14 ? 21.82

18 Ω

(b)

R1 12 Ω

(a)

Figure P5.9-5 Solution: (a)

KVL gives

v oc = ( R t + R ) i

from row 2

from row 3 So

(R

t

v oc = ( R t + 10 ) (1.333) v oc = ( R t + 20 ) ( 0.857 )

+ 10 ) (1.333) = ( R t + 20 ) ( 0.857 )

28 ( R t + 10 ) = 18 ( R t + 20 ) Solving gives 10 R t = 360 − 280 = 80 and



Rt = 8 Ω

v oc = ( 8 + 10 )(1.333) = 24 V

(b) i=

v oc Rt + R

=

24 R 24 R and v = v oc = 8+ R R + Rt R+8

When R = 0, i = 3 A, and v = 0 V. 1 When R = 40 Ω, i = A . 2 24 ( 80 ) 240 = = 21.82 . When R = 80 Ω, v = 88 11 These are the values given in the tabulated data so the data is consistent. 24 ( 40 ) (c) When R = 40 Ω, v = = 20 V . 48 24 = 0.2727 A . When R = 80 Ω, i = 88 (d) First 8 = R t = 24 & 18 & ( R1 + 12 ) ⇒ R1 = 24 Ω the, using superposition, 24 = v oc =

(

24

24 + 18 & ( R1 + 12 )

)

(

)

12 + 24 & 18 ( R1 + 12 ) i s = 8 + 8i s



is = 2 A

(checked using LNAP 6/21/05)

PSpice Problems SP 5-1 The circuit in Figure SP 5.1 has three inputs: v1, v2, and i3. The circuit has one output, vo. The equation vo = av1 + bv2 + ci3 expresses the output as a function of the inputs. The coefficients a, b, and c are real constants. (a) Use PSpice, and the principle of superposition, to determine the values of a, b, and c. (b) Suppose v1 = 10 V, v2 = 8 V, and we want the output to be vo = 7 V. What is the required value of i3? Hint:

The output is given by vo = a when v1 = 1 V, v2 = 0 V, and i3 = 0 A.

Answer: (a) vo = 0.3333v1 + 0.3333v2 + 33.33i3, 100 Ω

(b) i3 = 30 mA

v2 +–

+ –

v1

+ vo –

100 Ω

100 Ω

i3

Figure SP 5.1 Solution:

a = 0.3333

b = 0.3333

c =33.33 V/A

(a)

(b)

vo = 0.3333 v1 + 0.3333 v2 + 33.33 i 3

7 = 0.3333 (10 ) + 0.3333 ( 8 ) + 33.33 i 3

18 3 = 3 = 30 mA ⇒ i3 = 100 100 3 7−

SP 5-2 The pair of terminals a–b partitions the circuit in Figure SP 5.2 into two parts. Denote the node voltages at nodes 1 and 2 as v1 and v2. Use PSpice to demonstrate that performing a source transformation on the part of the circuit to the left of the terminal does not change anything to the right of the terminals. In particular, show that the current, io, and the node voltages, v1 and v2, have the same values after the source transformation as before the source transformation. 100 Ω

a

1

8V

2

+–

10 V

+ –

100 Ω

io

100 Ω

b

Figure SP 5.2 Solution: Before the source transformation:

VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2

-3.000E-02 -4.000E-02

After the source transformation:

VOLTAGE SOURCE CURRENTS NAME CURRENT V_V2

-4.000E-02

30 mA

0.75va

SP 5-3 Use PSpice to find the Thévenin equivalent circuit for the circuit shown in Figure SP 5.3.



Answer: voc = –2 V and Rt = –8/3 Ω + –

Solution:

6V

a – va +

4Ω b

Figure SP 5.3

voc = −2 V

VOLTAGE SOURCE CURRENTS NAME CURRENT V_V3 V_V4

-7.500E-01 7.500E-01 isc = 0.75 A

Rt = −2.66 Ω

SP 5-4 The circuit shown in Figure SP 5-4b is the Norton equivalent circuit of the circuit shown in Figure SP 5-4a. Find the value of the short-circuit current, isc, and Thévenin resistance, Rt. Answer: isc = 1.13 V and Rt = 7.57 Ω 5Ω

3Ω – + + –

10 V

2ia



isc

Rt

ia

(a)

(b)

Figure SP 5-4 Solution:

voc = 8.571 V

VOLTAGE SOURCE CURRENTS NAME

CURRENT

V_V5 -2.075E+00 V_V6 1.132E+00 X_H1.VH_H1 9.434E-01

isc = 1.132 A

Rt = 7.571 Ω

Design Problems R1

i +

DP 5-1 The circuit shown in Figure DP 51a has four unspecified circuit parameters: vs, R1, R2, and R3. To design this circuit, we must specify the values of these four parameters. The graph shown in Figure DP 5-1b describes a relationship between the current i and the voltage v. Specify values of vs, R1, R2, and R3 that cause the current i and the voltage v in Figure DP 5-1a to satisfy the relationship described by the graph in Figure DP 5-1b. First Hint: The equation representing the straight line in Figure DP 5-1b is v = –Rti + voc That is, the slope of the line is equal to –1 times the Thévenin resistance and the “vintercept” is equal to the open-circuit voltage. Second Hint: There is more than one correct answer to this problem. Try setting R1 = R2.

R3

+ –

R2

vs

v –

(a) v, V 6 4 2 –6

–4

–2

2

4

6

8

i, mA

–2 –4 –6 –8

(b)

Figure DP 5-1 Solution: The equation of representing the straight line in Figure DP 5-1b is v = − R t i + voc . That is, the slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0−5 open circuit voltage. Therefore: R t = − = 625 Ω and voc = 5 V. 0.008 − 0 Try R1 = R 2 = 1 kΩ . (R1 || R2 must be smaller than Rt = 625 Ω.) Then

5=

R2

1 vs = vs R1 + R 2 2

⇒ vs = 10 V

and

R1 R2 = R3 + 500 ⇒ R3 = 125 Ω R1 + R2 Now vs, R1, R2 and R3 have all been specified so the design is complete. 625 = R 3 +

DP 5-2 The circuit shown in Figure DP 5.2a has four unspecified circuit parameters: is, R1, R2, and R3. To design this circuit, we must specify the values of these four parameters. The graph shown in Figure DP 5-2b describes a relationship between the current i and the voltage v. Specify values of is, R1, R2, and R3 that cause the current i and the voltage v in Figure DP 5-2a to satisfy the relationship described by the graph in Figure DP 5-2b. First Hint: Calculate the open-circuit voltage, voc, and the Thévenin resistance, Rt, of the part of the circuit to the left of the terminals in Figure DP 5-2a. Second Hint: The equation representing the straight line in Figure DP 5-2b is v = –Rti + voc That is, the slope of the line is equal to –1 times the Thévenin resistance and the “v-intercept” is equal to the open-circuit voltage. Third Hint: There is more than one correct answer to this problem. Try setting both R3 and R1 + R2 equal to twice the slope of the graph in Figure DP 5-2b. v, V 6 4 R2

i

2

+ is

R1

R3

v

–6



–4

–2

2

4

6

8

i, mA

–2

(a)

–4 –6 –8

(b)

Figure DP 5-2 Solution: The equation of representing the straight line in Figure DP 5-2b is v = − R t i + voc . That is, the

slope of the line is equal to -1 times the Thevenin resistance and the "v - intercept" is equal to the 0 − ( −3 ) open circuit voltage. Therefore: R t = − = 500 Ω and voc = −3 V. −0.006 − 0 From the circuit we calculate R 3 ( R1 + R 2 ) R1 R 3 is and voc = − Rt = R1 + R 2 + R 3 R1 + R 2 + R 3 so

500 Ω =

R 3 ( R1 + R 2 ) R1 + R 2 + R 3

and −3 V = −

R1 R 3 R1 + R 2 + R 3

is

Try R 3 = 1kΩ and R1 + R 2 = 1kΩ . Then R t = 500 Ω and −3 = −

1000 R1

is = −

R1

i s ⇒ 6 = R1 i s 2000 2 This equation can be satisfied by taking R1 = 600 Ω and is = 10 mA. Finally, R2 = 1 kΩ - 400 Ω = 400 Ω. Now is, R1, R2 and R3 have all been specified so the design is complete.

R1

i

R3

+

DP 5-3 The circuit shown in Figure DP 5-3a has four unspecified circuit parameters: vs, R1, R2, and R3. To design this circuit, we must specify the values of these four parameters. The graph shown in Figure DP 5-3b describes a relationship between the current i and the voltage v. Is it possible to specify values of vs, R1, R2, and R3 that cause the current i and the voltage v in Figure DP 5-1a to satisfy the relationship described by the graph in Figure DP 5-3b? Justify your answer.

+ –

R2

vs

v –

(a) v, V 6 4 2 –6

–4

–2

2

4

6

8

i, mA

–2 –4 –6 –8

(b)

Figure DP 5-3

Solution: The slope of the graph is positive so the Thevenin resistance is negative. This would require R1 R 2 R3 + < 0 , which is not possible since R1, R2 and R3 will all be non-negative. R1 + R 2 Is it not possible to specify values of vs, R1, R2 and R3 that cause the current i and the voltage v in Figure DP 5-3a to satisfy the relationship described by the graph in Figure DP 5-3b.

DP 5-4 The circuit shown in Figure DP 5-4a has four unspecified circuit parameters: vs, R1, R2, and d, where d is the gain of the CCCS. To design this circuit, we must specify the values of these four parameters. The graph shown in Figure DP 5-4b describes a relationship between the current i and the voltage v. Specify values of vs, R1, R2, and d that cause the current i and the voltage v in Figure DP 5-4a to satisfy the relationship described by the graph in Figure DP 5-4b.

+ –

ia

i

R1

+

That is, the slope of the line is equal to –1 times the Thévenin resistance and the “v-intercept” is equal to the open-circuit voltage. Second Hint: There is more than one correct answer to this problem. Try setting R1 = R2.

v –

(a) v, V 6 4

First Hint: The equation representing the straight line in Figure DP 5-4b is

v = –Rti + voc

R2

dia

vs

2 –6

–4

–2

2

4

6

8

i, mA

–2 –4 –6 –8

(b)

Figure DP 5-4 Solution: The equation of representing the straight line in Figure DP 5-4b is v = − R t i + voc . That is, the

slope of the line is equal to the Thevenin impedance and the "v - intercept" is equal to the open −5 − 0 = −625 Ω and voc = −5 V. circuit voltage. Therefore: R t = − 0 − 0.008 The open circuit voltage, voc, the short circuit current, isc, and the Thevenin resistance, Rt, of this circuit are given by R 2 ( d + 1) voc = vs R1 + ( d + 1) R 2 , ( d + 1) v isc = s R1 and

Rt =

R1 R 2

R1 + ( d + 1) R 2

Let R1 = R2 = 1 kΩ. Then

−625 Ω = R t = and −5 =

( d + 1) vs d +2

1000 1000 ⇒ d= − 2 = −3.6 A/A −625 d +2 ⇒ vs =

−3.6 + 2 ( − 5) = −3.077 V −3.6 + 1

Now vs, R1, R2 and d have all been specified so the design is complete.

Chapter 6 The Operational Amplifier Exercises Exercise 6.6-1 Specify the values of R1 and R2 in Figure E 6.6-1 that are required to cause v3 to be related to v1 and v2 by the equation v3 = (4)v1 – ( 15 )v2. Answer: R1 = 10 kΩ and R2 = 2.5 kΩ

Figure E 6.6-1 Solution: ⎞ R2 R1 ⎞ ⎛ 10 × 103 ⎞ ⎛ ⎛ 10 ×103 ⎞ ⎛ v3 = ⎜ − v + + + v 1 1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 2 3 3 3 3 ⎟ 1 ⎝ 10 × 10 ⎠ ⎜⎝ R 2 + 10 × 10 ⎟⎠ ⎝ 10 × 10 ⎠ ⎝ 10 × 10 ⎠ ⎛ ⎞ R2 R1 ⎞ ⎛ = −⎜ v2 + 2 ⎜1 + v ⎟ 3 3 ⎟ 1 ⎜ ⎟ ⎝ 10 × 10 ⎠ ⎝ R 2 + 10 × 10 ⎠ ⎛1⎞ We require v3 = ( 4 ) v1 − ⎜ ⎟ v2 , so ⎝5⎠

R1 ⎞ ⎛ 4 = 2 ⎜1 + ⇒ R1 = 10 × 103 = 10 kΩ 3 ⎟ 10 × 10 ⎝ ⎠ and R2 1 = 5 R 2 + 10 × 103

⇒ R 2 + 10 × 103 = 5 R 2



R 2 = 2.5 kΩ

Exercise 6.6-2 Specify the values of R1 and R2 in Figure E 6.6-1 that are required to cause v3 to be related to v1 and v2 by the equation v3 = (6)v1 – ( 54 )v2. Answer: R1 = 20 kΩ and R2 = 40 kΩ

Figure E 6.6-1 Solution: As in Ex 6.7-1 ⎛ ⎞ R2 R1 ⎞ ⎛ v3 = − ⎜ v + 2 1 + v ⎟ ⎜ 2 3 ⎟ 1 ⎜ R 2 + 10 × 103 ⎟ ⎝ 10 × 10 ⎠ ⎝ ⎠ ⎛ 4⎞ We require v3 = ( 6 ) v1 − ⎜ ⎟ v2 , so ⎝5⎠

R1 ⎞ ⎛ 6 = 2 ⎜1 + ⇒ R1 = 20 ×103 = 20 kΩ 3 ⎟ ⎝ 10 ×10 ⎠ and R2 4 = 5 R 2 + 10 × 103

⇒ 4 R 2 + 40 × 103 = 5 R 2

⇒ R 2 = 40 kΩ

Exercise 6.7-1 The input offset voltage of a typical μA741 operational amplifier is 1 mV and the bias current is 80 nA. Suppose the operational amplifier in Figure 6.72a is a typical μA741. Show that the output offset voltage of the inverting amplifier will be at most 10 mV.

Figure 6.7-2a Solution:

Analysis of the circuit in Section 6.7 showed that output offset voltage = 6 vos + (50 × 103 ) ib1 For a μ A741 op amp, vos ≤ 1 mV and ib1 ≤ 80 nA so output offset voltage = 6 vos + (50 × 103 )ib1 ≤ 6 (10−3 )+(50.103 )(80×10−9 ) = 10 mV

Exercise 6.7-2 Suppose the 10-kΩ resistor in Figure 6.72a is changed to 2 kΩ and the 50-kΩ resistor is changed to 10 kΩ. (These changes will not change the gain of the inverting amplifier. It will still be – 5.) Show that the maximum output offset voltage is reduced to 35 mV. (Use ib = 500 nA and vos = 5 mV to calculate the maximum output offset voltage that could be caused by the μA741 amplifier.)

Figure 6.7-2a Solution:

vo = −

⎛ R ⎞ R2 vin + ⎜1 + 2 ⎟ vos + R2ib1 R1 ⎝ R1 ⎠

When R2 = 10 kΩ, R1 = 2 kΩ, vos ≤ 5 mV and

(

) (

output offset voltage ≤ 6 5 × 10−3 + 10 × 103

ib1 ≤ 500 nA then

) ( 500.10 ) ≤ 35×10 −9

−3

= 35 mV

Exercise 6.7-3 Suppose the μA741 operational amplifier in Figure 6.7-2a is replaced with a typical OPA101AM operational amplifier. Show that the output offset voltage of the inverting amplifier will be at most 0.6 mV.

Figure 6.7-2a Solution:

Analysis of this circuit in Section 6.7 showed that output offset voltage = 6 vos + ( 50 ×103 ) ib1 For a typical OPA1O1AM, vos = 0.1 mV and ib = 0.012 nA so

output offset voltage ≤ 6 ⎡⎣0.1×10−3 ⎤⎦ + ( 50 ×103 ) ⎡⎣0.012 ×10−9 ⎤⎦ ≤ 0.6 ×10−3 + 0.6 ×10−6 − 0.6 ×10−3 = 0.6 mV

Rf

Exercise 6.7-4

(a) (b)

Ra

Determine the voltage ratio vo/vs for the op amp circuit shown in Figure E 6.7-4 5

Calculate vo/vs for a practical op amp with A = 10 , Ro = 100 Ω, and Ri = 500 kΩ. The circuit resistors are Rs = 10 kΩ, Rf = 50 kΩ, and Ra = 25 kΩ.

– + + –

vs

Rs

Figure E 6.7-4 Solution:

Writing node equations v− − vs v− − vo v− + + =0 Ra Rb Ri + Rs ⎛ ⎞ R i v ⎟ vo − ⎜ − A − ⎜ ⎟ R +R i s ⎠ + vo − v− = 0 ⎝ R0 Rb Av =

R0 ( Ri + Rs ) + ARi R f vo = vs ( R f + R0 ) ( Ri + Rs ) + Ra ( R f + R0 + Ri + Rs ) − ARi Ra

For the given values, Av = −2.00006 V/V.

vo –

Answer: (b) vo/vs = – 2

After some algebra

+

Section 6-3: The Ideal Operational Amplifier 20 kΩ

P 6.3-1 Determine the value of voltage measured by the voltmeter in Figure P 6.3-1.



Voltmeter

+

Answer: – 4 V 20 kΩ 50 kΩ – +

4V

Figure P 6.3-1 Solution:

(checked using LNAP 8/16/02)

3 kΩ

P 6.3-2

4 kΩ

Find vo and io for the circuit of Figure P 6.3-2. 12 V

io

+ –

2 kΩ

– +

+ 1 kΩ

R

vo –

Figure P 6.3-2

Solution: Apply KVL to loop 1:

− 12 + 3000 i1 + 0 + 2000 i1 = 0 12 = 2.4 mA 5000 The currents into the inputs of an ideal op amp are zero so io = i1 = 2.4 mA ⇒ i1 =

i2 = − i1 = − 2.4 mA

va = i2 (1000 ) + 0 = −2.4 V Apply Ohm’s law to the 4 kΩ resistor vo = va − io ( 4000 )

= −2.4 − ( 2.4 ×10−3 ) ( 4000 ) = −12 V (checked using LNAP 8/16/02)

4 kΩ

8 kΩ

P 6.3-3 Find vo and io for the circuit of Figure P 6.3-3. Answer: vo = – 30 V and io = 3.5 mA

io –

12 V

+ –

+

2V

– +

+ 20 kΩ

vo –

Figure P 6.3-3 Solution: The voltages at the input nodes of an ideal op amp are equal so va = −2 V . Apply KCL at node a: vo − ( −2 ) 12 − ( −2 ) + = 0 ⇒ vo = −30 V 8000 4000 Apply Ohm’s law to the 8 kΩ resistor io =

−2 − vo = 3.5 mA 8000

(checked using LNAP 8/16/02)

10 kΩ

P 6.3-4

Find v and i for the circuit of Figure P 6.3-4. –

0.1 mA

+ v –

+ + –

5V

20 kΩ i

Figure P 6.3-4 Solution: The voltages at the input nodes of an ideal op amp are equal so v = 5 V . Apply KCL at the inverting input node of the op amp: ⎛ v −5 ⎞ −3 −⎜ a ⎟ − 0.1× 10 − 0 = 0 ⇒ va = 4 V ⎝ 10000 ⎠ Apply Ohm’s law to the 20 kΩ resistor va 1 i = = mA 20000 5 (checked using LNAP 8/16/02)

3 kΩ 4 kΩ

P 6.3-5 Find vo and io for the circuit of Figure P 6.3-5. Answer: vo = – 15 V and io = 7.5 mA



io

+

12 V

+ –

2 mA

6 kΩ

+ vo –

Figure P 6.3-5 Solution: The voltages at the input nodes of an ideal op amp are equal, so va = 0 V . Apply KCL at node a: ⎛ v − 0 ⎞ ⎛ 12 − 0 ⎞ −3 −⎜ o ⎟−⎜ ⎟ − 2 ⋅10 = 0 3000 4000 ⎝ ⎠ ⎝ ⎠ ⇒ vo = − 15 V Apply KCL at the output node of the op amp: v v io + o + o = 0 ⇒ io = 7.5 mA 6000 3000

(checked using LNAP 8/16/02)

6 kΩ

P 6.3-6 Determine the value of voltage measured by the voltmeter in Figure P 6.3-6. Answer: 7.5 V

+

Voltmeter



8 kΩ + –

2.5 V 6 kΩ 4 kΩ

Figure P 6.3-6

Soluton: The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal so va = 2.5 V . Apply Ohm’s law to the 4 kΩ resistor: v 2.5 ia = a = = 0.625 mA 4000 4000 Apply KCL at node a: ib = ia = 0.625 mA Apply KVL: vo = 8000 ib + 4000 ia

= (12 × 103 )( 0.625 × 10−3 ) = 7.5 V

(checked using LNAP 8/16/02)

P 6.3-7

Find vo and io for the circuit of Figure P 6.3-7. R1

R2

R4

io

– + –

R3 –

+

vs

+

+

R5

vo –

Figure P 6.3-7 Solution: Label the circuit to account for the properties of the ideal op amps:

Apply KCL at the inverting node of the left op amp to get: R2 ⎛ vs − 0 ⎞ ⎛ va − 0 ⎞ −⎜ vs ⎟⎟ + 0 = 0 ⇒ v a = − ⎟ − ⎜⎜ R1 ⎝ R1 ⎠ ⎝ R 2 ⎠ Apply KCL at the output node of the left op amp to get: io =

0 − va R2

+

0 − va R3

=−

R 2 + R3 R 2 R3

⎛ R 2 + R3 ⎞ va = ⎜ v ⎜ R1 R 3 ⎟⎟ s ⎝ ⎠

Apply KCL at the inverting node of the right op amp to get: ⎛ vo − 0 ⎞ ⎛ va − 0 ⎞ R4 R2 R4 −⎜ −⎜ + 0 = 0 ⇒ vo = − va = vs ⎟ ⎟ ⎜ R 4 ⎟ ⎜ R3 ⎟ R3 R1 R 3 ⎝ ⎠ ⎝ ⎠

6 kΩ +

P 6.3-8 Determine the current io for the circuit shown in Figure P 6.3-8.



8 kΩ

Answer: io = 2.5 mA

io

6 kΩ

+ –

4 kΩ

2V

4 kΩ

6 kΩ

8 kΩ – + + –

6 kΩ 5.8 V

Figure P 6.3-8 Solution: The node voltages have been labeled using: 1. The currents into the inputs of an ideal op amp are zero and the voltages at the input nodes of an ideal op amp are equal. 2. KCL 3. Ohm’s law then and

v0 = 11.8 − 1.8 = 10 V 10 io = = 2.5 mA 4000

(checked using LNAP 8/16/02)

P 6.3-9 Determine the current io for the circuit shown in Figure P 6.3-9. Answer: io = 2.5 mA

4 kΩ – +

18 V

4 kΩ



a

+

+

b

8 kΩ

8 kΩ

vo –

Figure P 6.3-9

Solution: Apply KCL at node a:

v a − ( −18 ) 4000

+

va 8000

+ 0 = 0 ⇒ v a = −12 V

The node voltages at the input nodes of ideal op amps are equal, so v b = v a . Using voltage division: vo =

8000 v = −8 V 4000 + 8000 b

(check using LNAP 8/16/02)

R1

P 6.3-10 Determine the current io for the circuit shown in Figure P 6.3-10.

is



Answer: io = 2.5 mA

+

+ R2

R3

vo



Figure P 6.3-10 Solution: Label the circuit as shown. The current in resistor R 3 is i s . Consequently:

v a = i s R3

Apply KCL at the top node of R 2 to get ⎛ R3 ⎞ va i= + i s = ⎜1 + i ⎜ R 2 ⎟⎟ s R2 ⎝ ⎠ Using Ohm’s law gives vo − va R1

⎛ R3 ⎞ = i = ⎜1 + i ⎜ R 2 ⎟⎟ s ⎝ ⎠

⎛ R1 R 3 ⎞ ⇒ v o = ⎜ R1 + R 3 + ⎟ is ⎜ R 2 ⎟⎠ ⎝

We require R1 + R 3 +

R1 R 3 R2

= 20

e.g. R1 = 5 kΩ and R 2 = R 3 = 10 kΩ . (checked: LNAP 6/2/04)

R1

P 6.3-11 Determine the voltage vo for the circuit shown in Figure P 6.3-11.

+

vo

– + –

vs

R2

Answer: vo = – 8 V R4

R3

Figure P 6.3-11

Solution: Label the circuit as shown. Apply KCL at the top node of R 2 to get

vs − va R1

=

⎛ R2 ⎞ + 0 ⇒ va = ⎜ ⎟⎟ v s ⎜ R2 ⎝ R1 + R 2 ⎠ va

Apply KCL at the inverting node of the op amp to get R 2 ( R3 + R 4 ) ⎛ R3 + R 4 ⎞ ⎛ R3 + R 4 ⎞ ⎛ R 2 ⎞ vo − va va = + 0 ⇒ vo = ⎜ va = ⎜ vs = v ⎟ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ R3 R4 ( R1 + R 2 ) R 4 s ⎝ R4 ⎠ ⎝ R 4 ⎠ ⎝ R1 + R 2 ⎠ We require

R 2 ( R3 + R 4 )

(R

1

+ R2 ) R4

=5

e.g. R1 = R 2 = 10 kΩ , R 3 = 90 kΩ and R 4 = 10 kΩ . (checked: LNAP 6/2/04)

5 kΩ

P 6.3-12 The circuit shown in Figure P 6.3-12 has one input, is, and one output, vo. Show that the output is proportional to the input. Design the

circuit so that the gain is

vo is

vs

+ –

20 kΩ



V = 20 . mA

+

1.5 V

+ –

+ 10 kΩ

vo –

Figure P 6.3-12

Solution: The node voltage at both input nodes of the op amp is 1.5 V. Apply KCL at the inverting input node of the op amp to get

v s − 1.5 5000

+

v o − 1.5 20000

= 0 ⇒ 4 ( v s − 1.5 ) + v o − 1.5 = 0 ⇒ v o = −4 ( v s − 1.5 ) + 1.5 = 0 ⇒ v o = −4 v s + 7.5

Comparing this equation to v o = m v s + b , we determine that m = −4 V/V and b = 75. V.

R

P 6.3-13 The circuit shown in Figure P 6.3-13 has one input, vs, and one output, vo. Show that the output is proportional to the input. Design the circuit so that vo = 5 vs.

0.2 mA

20 kΩ ioa

– +

25 kΩ 1.5 V

+ –

10 kΩ

+ vo –

Figure P 6.3-13 Solution: The output of this circuit is v o = 3.5 V.

(a) The current in the 25 kΩ resistor is 0 A because this current is also the input current of an ideal op amp. Consequently, the voltage at the input nodes of the op amp is 1.5 V. Apply KCL at the inverting input of the op amp to get 1.5 3.5 − 1.5 = 0.2 + = 0.3 R 20 so R=

1.5 = 5 kΩ 0.3

(b) The voltage source current is 0A so the voltage source supplies 0 W of power. The voltage across the current source is equal to the node voltage at the inverting input of the op amp, 1.5 V. Notice that this voltage and the given current source current do not adhere to the passive convention so (0.2)(1.5) = 0.3 mW is the power supplied by the current source. (c) Apply KCL at the output node of the op amp to get

i oa =

3.5 3.5 − 1.5 + = 0.45 A 10 20

The op amp supplies p oa = i oa × v o = ( 0.45 )( 3.5 ) = 1.575 W

P 6-3.14 Determine the values of the node voltages at nodes a, b, c, and d of the circuit shown in

Figure P 6-3.14. 25 kΩ 5 kΩ

a

20 kΩ c

– +

b

15 kΩ –

d

+

2.5 mA

5 kΩ

5 kΩ

10 kΩ

Figure P 6.3-14 Solution: Label the circuit as shown, noticing that some resistor currents are zero because these currents are also input currents of op amps:

Now

v a = −2.5 ( 5 ) = −12.5 V , v b = v a = −12.5 V, v c = 0 V

Apply KCL to get vb − 0 20

= 0+

0 − vd 15



vd = −

15 3 v b = − ( −12.5 ) = 9.375 V 20 4

P 6-3.15 Determine the values of the node voltages at nodes a, b, c, and d of the circuit shown in Figure P 6-3.15.

10 kΩ a

40 kΩ

20 kΩ –

b

10 kΩ c

+ – +

5V 40 kΩ



d

+

10 kΩ 60 kΩ

10 kΩ

40 kΩ

Figure P 6.3-15 Solution: Label the circuit as shown, noticing that some resistor currents are zero because these currents are also input currents of op amps:

vb ⎛ 20 ⎞ = 5 V and v d = v c = 5 V. Now v a = 0 V, v b = − ⎜ ⎟ ( −5 ) = 10 V, v c = 2 ⎝ 10 ⎠

Figure P6.3-16 P6.3-16 . Figure P6.3-16 shows four similar circuits. The outputs of the circuits are the voltages v1, v2, v3 and v4. Determine the values of these four outputs. Solution: 80 25 80 v1 = − ( 2.1) = −6.72 V , v 2 = − ( −2.1) = 0.65625 V , v 3 = − ( −2.1) = 6.72 V and 25 80 25 25 v 4 = − ( 2.1) = −0.65625 V 80

Figure P6.3-17 P6.3-17 . Figure P6.3-17 shows four similar circuits. The outputs of the circuits are the voltages v1, v2, v3 and v4. Determine the values of these four outputs. Solution ⎛ 80 ⎞ ⎛ 25 ⎞ ⎛ 80 ⎞ v1 = ⎜ 1 + ⎟ ( 2.1) = 8.82 V , v 2 = ⎜1 + ⎟ ( −2.1) = −2.75625 V , v 3 = ⎜ 1 + ⎟ ( −2.1) = −8.82 V a ⎝ 25 ⎠ ⎝ 80 ⎠ ⎝ 25 ⎠ ⎛ 25 ⎞ nd v 4 = ⎜1 + ⎟ ( 2.1) = 2.75625 V . ⎝ 80 ⎠

Section 6-4: Nodal Analysis of Circuits Containing Ideal Operational Amplifiers P 6.4-1 Determine the node voltages for the circuit shown in Figure P 6.4-1.

a

20 kΩ

b +

d –

Answer: va = 2 V, vb = – 0.25 V, vc = – 5 V, vd = – 2.5 V, and ve = – 0.25 V

9 kΩ 2V

+ –

c

40 kΩ

– +

5V

e 40 kΩ

1 kΩ

Figure P 6.4-1 Solution:

KCL at node b:

vb − 2 vb v +5 1 + + b = 0 ⇒ vb = − V 20000 40000 40000 4

The node voltages at the input nodes of an ideal op amp are equal so ve = vb = − KCL at node e:

ve v −v 10 V + e d = 0 ⇒ vd = 10 ve = − 1000 9000 4

1 V. 4

(checked using LNAP 8/16/02)

6 kΩ

P 6.4-2 Find vo and io for the circuit of Figure P 6.4-2. Answer: vo = – 4 V and io = 1.33 mA

6 kΩ

6 kΩ –

+ –

12 V

6 kΩ

io

+

+ 6 kΩ

vo –

Figure P 6.4-2 Solution: Apply KCL at node a: 0=

va − 12 v v −0 + a + a ⇒ va = 4 V 6000 6000 6000

Apply KCL at the inverting input of the op amp: ⎛v −0⎞ ⎛ 0 − vo ⎞ −⎜ a ⎟+0+⎜ ⎟ = 0 ⎝ 6000 ⎠ ⎝ 6000 ⎠ ⇒ vo = −va = −4 V

Apply KCL at the output of the op amp: vo ⎛ 0 − vo ⎞ io − ⎜ = 0 ⎟+ ⎝ 6000 ⎠ 6000 v ⇒ io = − o = 1.33 mA 3000

(checked using LNAP 8/16/02)

Figure P6.4-3 P6.4-3. Determine the values of the node voltages, va and vo, of the circuit shown in Figure P6.4-3.

Solution:

Writing node equations: va 2.25 2.25 + = 0 ⇒ v a = − 40 × 10 3 = −4.5 V 3 3 20 ×10 40 ×10 20 × 10 3 and va va va − vo ⎛ 8 8 8⎞ + + = 0 ⇒ v o = ⎜ + + ⎟ v a = 2 ( −4.5 ) = −9 V 3 3 3 40 ×10 10 ×10 8 × 10 ⎝ 40 10 8 ⎠

(

)

+

P 6.4-4 The output of the circuit shown in Figure P 6.4-4 is vo. The inputs are v1 and v2. Express the output as a function of the inputs and the resistor resistances.

– + –

v1

+ R1

vo

R2 R3

– + + –

v2

Figure P 6.4-4

Solution: Ohm’s law: i=

v1 − v2 R2

KVL: v0 = ( R1 + R2 + R3 ) i =

R1 + R2 + R3 ( v1 − v2 ) R2



R3

+

P 6.4-5 The outputs of the circuit shown in Figure P 6.4-5 are vo and io. The inputs are v1 and v2. Express the outputs as functions of the inputs and the resistor resistances.

R5

– + –

v1

R1

io –

R7

+

+ vo

R2 –

– + + –

R4

v2

Figure P 6.4-5 Solution:

⎛ R ⎞ v1 −va v1 −v2 R + + 0 = 0 ⇒ va = ⎜1+ 1 ⎟ v1 − 1 v2 R1 R7 R7 ⎝ R7 ⎠ ⎛ R ⎞ v2 − vb v −v R − 1 2 + 0 = 0 ⇒ vb = ⎜ 1+ 2 ⎟ v2 − 2 v1 R2 R7 R7 ⎝ R7 ⎠

R6

⎛ v −v ⎞ v −0 R6 − ⎜ b c ⎟ + c + 0 = 0 ⇒ vc = vb R4 + R6 ⎝ R4 ⎠ R6 ⎛ v −v ⎞ ⎛ v −v ⎞ −⎜ a c ⎟ + ⎜ c 0 ⎟ + 0 = 0 ⎝ R3 ⎠ ⎝ R5 ⎠

⇒ v0 = −

R5 R va + (1+ 5 )vc R3 R3

⎡ R R R (R +R ) ⎡R R (R +R ) R ⎤ R ⎤ R v0 = ⎢ 5 1 + 6 3 5 (1+ 2 ) ⎥ v2 − ⎢ 5 (1+ 1 ) + 6 3 5 2 ⎥ v1 R7 ⎦ R7 R3 ( R4 + R6 ) R7 ⎦ ⎣ R3 R7 R3 ( R4 + R6 ) ⎣ R3 v −v i0 = c 0 = " R5

10 kΩ

P 6.4-6 Determine the node voltages for the circuit shown in Figure P 6.4-6. Answer: va = – 0.75 V, vb = 0 V, and vc = – 0.9375 V

40 kΩ a – +

12 V

20 kΩ

40 kΩ

b

25 kΩ c –

15 kΩ

+

Figure P 6.4-6

Solution:

KCL at node b:

KCL at node a:

so

va vc 5 + = 0 ⇒ vc = − va 3 3 20 × 10 25 × 10 4 5 ⎛ ⎞ va − ⎜ − va ⎟ va − ( −12 ) va va + 0 ⎝ 4 ⎠ = 0 ⇒ v = − 12 V + + + a 3 3 3 40 ×10 40 × 10 20 × 10 10 × 103 13 vc = −

5 15 va = − 4 13

(checked using LNAP 6/21/05)

P 6.4-7 Find vo and io for the circuit shown in Figure P 6.4-7. Assume an ideal operational amplifier.

10 kΩ 30 kΩ

10 kΩ 6V +–

30 kΩ 30 kΩ

10 kΩ



io +

+

30 kΩ

vo –

Figure P 6.4-7

Solution:

Label the circuit to account for the properties of the ideal op amp and to identify the supernode corresponding to the voltage source. Apply KCL at the inverting input node of the op amp

⎛ ( va + 6 ) −0 ⎞ ⎛ v −0 ⎞ −⎜ a ⎟ = 0 ⎟+0−⎜ ⎝ 10000 ⎠ ⎝ 30000 ⎠ or va = − 1.5 V Apply KCL to the super node corresponding the voltage source: va − 0 va + 6− 0 va − v b ( va + 6 ) −v b + + + = 0 10000 30000 30000 10000 Multiply both sides by 30000 to get

3va + ( va + 6 ) + ( va − vb ) + 3 ⎡⎣( va + 6 )− v b ⎤⎦ = 0

Solving gives vb = 2 va + 6 = 3 V

Apply KCL at node b:

⎛ v −v ⎞ ⎛ ( v a + 6 )−v b ⎞ ⎟ = 0 −⎜ a b ⎟−⎜ 10000 30000 ⎝ 30000 ⎠ ⎜ 10000 ⎟ ⎝ ⎠ Multiply both sides by 30000 to get vb

+

v b −v o

3v b + ( v b − v o ) − ( v a −v b ) − 3 ⎡⎣( v a + 6 )− v b ⎤⎦ = 0 Solving gives v o = 8 v b − 4 va − 18 = 12 V Apply KCL at the output node of the op amp: io +

vo

30000

+

v o −v b

30000

= 0 ⇒ i o = −0.7 mA

10 kΩ

20 kΩ +

P 6.4-8 Find vo and io for the circuit shown in Figure P 6.4-8. Assume an ideal operational amplifier.



vo

+

– io

20 kΩ 10 kΩ

Figure P 6.4-8 Solution: Apply KVL to the bottom mesh:

−i0 (10000) − i0 (20000) + 5 = 0 ⇒ i0 =

1 mA 6

The node voltages at the input nodes of an ideal op amp are equal. Consequently va = 10000 i0 =

10 V 6

Apply KCL at node a: va v −v + a 0 = 0 ⇒ 10000 20000

v0 = 3va = 5 V

+ –

5V

P 6.4-9

Determine the node voltages for the circuit shown in Figure P 6.4-9.

Answer: va = – 12 V, vb = – 4 V, vc = – 4 V, vd = – 4 V, ve = – 3.2 V, vf = – 4.8 V, and vg = – 3.2 c

10 kΩ

d

20 kΩ

e

20 kΩ a

40 kΩ

20 kΩ



b

– + 12 V



g

+ +

f 40 kΩ

20 kΩ

40 kΩ

Figure P 6.4-9 Solution:

vb + 12 vb + = 0 ⇒ vb = −4 V 40000 20000 The node voltages at the input nodes of an ideal op amp are equal, so vc = vb = −4 V .

KCL at node b:

The input currents of an ideal op amp are zero, so vd = vc + 0 × 10 4 = −4 V . KCL at node g:

vg ⎛ v f − vg ⎞ 2 −⎜ + = 0 ⇒ vg = v f 3 ⎟ 3 3 ⎝ 20 ×10 ⎠ 40 × 10

The node voltages at the input nodes of an ideal op amp are equal, so ve = vg = 2 vd − v f vd − 3 v f vd − ve KCL at node d: 0 = + = + 20 ×103 20 ×103 20 ×103 20 ×103 vd − v f

Finally, ve = vg =

2 16 vf = − V. 3 5

2 vf . 3

6 24 ⇒ v f = vd = − V 5 5

R = Ro + ΔR

P 6.4-10 The circuit shown in Figure P 6.4-10 includes a simple strain gauge. The resistor R changes its value by ΔR when it is twisted or bent. Derive a relation for the voltage gain vo/vs and show that it is proportional to the fractional change in R, namely ΔR/Ro. Answer: vo =

Ro ΔR Ro + R1 Ro

R1 –

+ vs

+

+

R1

Ideal Ro





Figure P 6.4-10

Solution:

By voltage division (or by applying KCL at node a) R0 va = vs R1 + R0 Applying KCL at node b: vb − vs vb − v0 + = 0 R1 R0 +ΔR ⇒

vo

R0 +ΔR ( vb −vs )+ vb = v0 R1

The node voltages at the input nodes of an ideal op amp are equal so vb = va . ⎡⎛ R +ΔR ⎞ R0 ⎛ R +ΔR ⎤ R0 ⎞ ΔR ΔR v0 = ⎢⎜ 0 +1⎟ − 0 v s = ⎜ − vs ⎥ vs = − ⎟ R1 ⎦ R1 + R0 R1 + R0 ⎠ R0 ⎠ R1 + R0 ⎝ ⎣⎝ R1

+

P 6.4-11 Find vo for the circuit shown in Figure P 6.4-11. Assume an ideal operational amplifier.



io + – 1.5 V

20 kΩ 20 kΩ

8 kΩ 10 kΩ

+ vo –

Figure P 6.4-11

Solution: Node equations:

⎛ R2 ⎞ vs vs − va + = 0 ⇒ va = ⎜1 + ⎟ vs ⎜ R1 R2 R1 ⎟⎠ ⎝ and

vs − va va va − vo = + R2 R3 R4

so ⎛ R4 R4 ⎞ R4 vo = ⎜1 + + va − vs ⎟ ⎜ R 2 R3 ⎟ R 2 ⎝ ⎠ ⎛ R4 R4 ⎞ ⎛ R2 ⎞ ⎛ R4 R4 R2 R2 R4 ⎞ R4 vo = ⎜1 + 1+ vs − vs = ⎜ 1 + v + + + + ⎟ ⎜ ⎟ ⎜ R 2 R3 ⎟ ⎜ ⎜ R 3 R1 R1 R1 R 3 ⎟⎟ s R1 ⎟⎠ R2 ⎝ ⎠⎝ ⎝ ⎠

⎛ ( R1 + R 2 )( R 3 + R 4 ) + R 3 R 4 ⎞ ⎟ vs =⎜ ⎜ ⎟ R1 R 3 ⎝ ⎠

with the given values: ⎛ ( 20 + 20 )(10 + 8 ) + 10 × 8 ⎞ ⎛ 40 × 18 + 80 ⎞ vo = ⎜ ⎟ vs = ⎜ ⎟ vs = ( 4 ) vs 20 × 10 200 ⎝ ⎠ ⎝ ⎠

(checked: LNAP 5/24/04)

+

P 6.4-12 The circuit shown in Figure P 6.4-12 has one output, vo, and two inputs, v1 and v2. Show that

when

R1

vo



v1 +–

R3

R3 R6 = , the output is proportional to the R4 R5

R4

difference of the inputs, v1 – v2. Specify resistance values to cause vo = 5 (v1 – v2).

+

R2 v2



+ –

R5

R6

Figure P 6.4-12 Solution: Notice that the currents in resistance R1 and R2 are both zero, as shown. Consequently, the voltages at the noninverting inputs of the op amps are v1 and v2, as shown. The voltages at the inverting inputs of the ideal op amps are also v1 and v2, as shown.

Apply KCL at the top node of R6 to get va − v2 v2 = R5 R6

⎛ R5 + R 6 ⎞ ⇒ va = ⎜ v ⎜ R 6 ⎟⎟ 2 ⎝ ⎠

Apply KCL at the top node of R4 to get vo − v1 v1 − va = R3 R4

⎛ R3 ⎞ ⎛ R3 ⎞ ⇒ vo = ⎜ 1 + v1 − ⎜ ⎟ va ⎟ ⎜ R4 ⎟ ⎜ R4 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ R3 ⎞ ⎛ R3 ⎞ ⎛ R5 + R6 ⎞ vo = ⎜ 1 + v1 − ⎜ ⎟ ⎜ v ⎟ ⎜ R4 ⎟ ⎜ R 4 ⎟ ⎜ R 6 ⎟⎟ 2 ⎝ ⎠ ⎝ ⎠⎝ ⎠

When

R3 R4

=

R6 R5

⎛ R3 ⎞ ⎛ R3 ⎞ ⎛ R5 ⎞ ⎛ R3 ⎞ ⎛ R3 ⎞ ⎛ R 4 ⎞ vo = ⎜1 + v1 − ⎜ ⎟ ⎜1 + v2 = ⎜1 + v1 − ⎜ ⎟ ⎜ 1 + v ⎟ ⎟ ⎟ ⎜ R4 ⎟ ⎜ R4 ⎟ ⎜ R6 ⎟ ⎜ R4 ⎟ ⎜ R 4 ⎟ ⎜ R 3 ⎟⎟ 2 ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎛ R3 ⎞ v −v = ⎜1 + ⎜ R 4 ⎟⎟ ( 1 2 ) ⎝ ⎠

so vo is proportional to the difference of the inputs, v1 − v2, as required. R3 Next, to make vo = 5 ( v1 − v2 ) , we choose R3 and R4 so that 5 = 1 + , e.g. R3 = 40 kΩ and R4 = R4 10 kΩ. Then with

R3 R4

=

R6 R5

we have

R1 = 50 kΩ, R2 = 50 kΩ, R3 = 40 kΩ, R4 = 10 kΩ, R5 = 10 kΩ and R6 = 40 kΩ. (checked: LNAP 5/24/04)

R1

P 6.4-13 The circuit shown in Figure P 6.4-13 has one output, vo, and one input, vi. Show that the output is proportional to the input. Specify resistance values to cause vo = 20 vi.

vi



vo

+

R2

R3 R4 – +

Figure P 6.4-13 Solution: Write a node equation at the inverting input of the bottom op amp: vo R3

+

va R4

= 0 ⇒ va = −

R4 R3

vo

Write a node equation at the inverting input of the top op amp: R4 vo − vi va vi R3 R 2 R3 0= + = + ⇒ vo = vi R1 R 2 R 1 R2 R1 R 4 The output is proportional to the input and the constant of proportionality is vo = 20 vi so

R 2 R3 R1 R 4

R 2 R3 R1 R 4

. We require

= 20 . For example, R1 = R 4 = 10 kΩ, R 2 = 40 kΩ and R 3 = 50 kΩ .

R1

vs

R2



+ –

+

P 6.4-14 The circuit shown in Figure P 6.4-14 has one input, vs, and one output, vo. Show that the output is proportional to the input. Design the circuit so that vo = 20vs.

+

R4

R5

vo –

R3

Figure P 6.4-14 Solution:

Represent this circuit by node equations. vo − va R2

+

vo − vs R1

vo − va R4

+

=0

vo R5



=0

⎛ R4 ⎞ v a = ⎜1 + v ⎜ R 5 ⎟⎟ o ⎝ ⎠ ( R 1 + R 2 ) R 5 − R 1 ( R 4R 5 ) v o

⎛ R1 ⎞ ⎛ R1 ⎞⎛ R 4 ⎞ v s = ⎜1 + vo − ⎜ 1+ v = ⎟ ⎜ R2 ⎟ ⎜ R 2 ⎟⎟ ⎜⎜ R 5 ⎟⎟ o ⎝ ⎠ ⎝ ⎠⎝ ⎠ R 2R 5 20 = Then R 2 R 5 - R 1R 4 So

For example

R 2 v s = ( R 1 + R 2 ) v o − R 1v a



R 2R 5 ⇒

=

R 2R 5 R 2 R 5 − R 1R 4

vs

19 R 1R 4 = 20 R 2 R 5

R 1 = 19 kΩ, R 4 = 10 kΩ, R 2 = 20 kΩ, R 5 = 10 kΩ, R 3 = 10 kΩ. (checked: LNAP 5/24/04)

R

P 6.4-15 The circuit shown in Figure P 6.4-15 has one input, vs, and one output, vo. The circuit contains seven resistors having equal resistance, R. Express the gain of the circuit, vo/vs, in terms of the resistance R.

R

R

R

R

R – + –

vs

+

+

R

vo –

Figure P 6.4-15 Solution:

Writing node equations: vs

+

v1

=0 ⇒ v 1 = −v s R R v1 v1 v1 − v 2 + + =0 ⇒ v 2 = 3v1 = −3v s R R R v 2 − v1 v 2 v 2 − v o + + =0 ⇒ v o = 3v 2 − v1 = −8v s R R R vo = −8 , does not depend on R. The gain of this circuit, vs (checked: LNAP 6/21/04)

R3

P 6.4-16 The circuit shown in Figure P 6.4-16 has one input, vs, and one output, vo. Express the gain, vo/vs, in terms of the resistances R1, R2, R3, R4, and R5. Design the circuit so that vo = – 20 vs.

R1

R4

R2 –

+

+ + –

vs

R5

vo –

Figure P 6.4-16 Solution: Represent this circuit by node equations. vs − va R1

+

vo − va R3

=

va R2

and va R2

+

vo R4

= 0 ⇒ vo = −

R4 R2

va

So ⎛ 1 ⎛ R1 R 2 + R1 R 3 + R 2 R 3 ⎞ ⎛ R 2 ⎞ 1 1 ⎞ =⎜ + + va = ⎜ vo ⎟ ⎟ ⎟⎟ ⎜⎜ − ⎜ ⎟ R1 R 3 ⎜⎝ R1 R 2 R 2 ⎟⎠ R1 R 2 R 3 ⎝ ⎠ ⎝ R4 ⎠ ⎛ 1 R1 R 2 + R1 R 3 + R 2 R 3 ⎞ ⎛ R1 R 4 + R1 R 2 + R1 R 3 + R 2 R 3 ⎞ vs = −⎜ + vo = − ⎜ ⎟ ⎟⎟ v o ⎜ R3 ⎟ ⎜ R1 R R R R R R 1 3 4 1 3 4 ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ R3 R 4 vo = − ⎜ v ⎜ R1 R 4 + R1 R 2 + R1 R 3 + R 2 R 3 ⎟⎟ s ⎝ ⎠ R3 R 4 20 = We require R1 R 4 + R1 R 2 + R1 R 3 + R 2 R 3 vs

+

vo

Try R1 = R 2 = R and R 3 = R 4 = aR a2 20 = 3a + 1 2 a − 60a − 20 = 0

Then So

a= e.g.

+60 ± 3600 + 4 ( 80 )

= 60.332, − 0.332 2 R1 = R 2 = 10 kΩ and R 3 = R 4 = 603.32 kΩ (checked: LNAP 6/9/04)

R3

P 6.4-17 The circuit shown in Figure P 6.4-17 has one input, vs, and one output, vo. Express the gain of the circuit, vo/vs, in terms of the resistances R1, R2, R3, R4, R5, and R6. Design the circuit so that

vo = – 20 vs.

R2

R1 –

+

+ + –



vs R4

R5

+ R6

vo



Figure P 6.4-17 Solution:

Represent this circuit by node equations. vs R1

+

va R2

+

vo R3

=0

and va R4

+

va − vo R5

⎛ R4 ⎞ = 0 ⇒ va = ⎜ v ⎜ R 4 + R 5 ⎟⎟ o ⎝ ⎠ vs

So

R1

+

vo R3

=−

va R2

=−

R4

R 2 ( R 4 + R5 )

vo

⎛ 1 ⎞ R 2 ( R 4 + R5 ) + R 4 R3 R4 ⎟ vo = − = −⎜ + vo ⎜ R3 R 2 ( R 4 R5 ) ⎟ R1 R R R + R ( ) 2 3 4 5 ⎝ ⎠ vs

vo = −

We require Try

20 =

R 2 R3 ( R 4 + R5 )

R1 ( R 2 R 4 + R 2 R 5 + R 3 R 4 )

vs

R 2 R3 ( R 4 + R5 )

R1 ( R 2 R 4 + R 2 R 5 + R 3 R 4 )

R1 = R 4 = R 5 = R and R 2 = R 3 = aR

Then e.g.

20 =

2a 2 R 3 2 = a 3aR 3 3



a = 30

R1 = R 4 = R 5 = 10 kΩ and R 2 = R 3 = 300 kΩ (checked: LNAP 6/10/04)

R2

R1

P 6.4-18 The circuit shown in Figure P 6.4-18 has one input, vs, and one output, io. Express the gain of the circuit, io/vs, in terms of the resistances R1, R2, R3, and Ro. (This circuit contains a pair of resistors having resistance R1 and another pair having resistance R2.) Design the circuit so that io = 0.02 vs.

+ –



vs

+

R1

R2 – +

R3 Ro

io

Figure P 6.4-18 Solution: Label the node voltages as shown. Represent this circuit by node equations. vb − va R2 io +

vo − va R2

+

=

va R1

vo − vb R3

vs − va R1



=0

=0





va =

R1 R1 + R 2

vb

v b = R 3i o + v o

⎛ R1 + R2 ⎞ vo =⎜ va − ⎟ R1 ⎜⎝ R1 R 2 ⎟⎠ R2 vs

So vs

⎛ R1 + R 2 ⎞ ⎛ R1 ⎞ v o R3 =⎜ R 3i o + v o ) − = io ⎟ ⎜ ⎟ ( R1 ⎜⎝ R1 R 2 ⎟⎠ ⎜⎝ R1 + R 2 ⎟⎠ R2 R2 io R2 = v s R1 R 3

We require

R2 R1 R 3

= 0.02, e.g. R 2 = 8 kΩ, R1 = R 3 = 20 kΩ .

(checked: LNAP 6/21/04)

P 6.4-19 The circuit shown in Figure P 6.4-19 has one input, vs, and one output, vo. The circuit contains one unspecified resistance, R. (a) Express the gain of the circuit, vo/vs, in terms of the resistance R. (b) Determine the range of values of the gain that can be obtained by specifying a value for the resistance R. (c) Design the circuit so that vo = – 3 vs.

Figure P 6.4-19 Solution:

(a) Use units of volts, mA, and kΩ. Apply KCL at the inverting input of the left op amp to get vs 10

vo =

(b) (c) We require

+

va 50

+

50 ⎞ ⎛ = 0 ⇒ v a = − ⎜ 5v s + v o ⎟ R R ⎠ ⎝

vo

4 40 ⎛ 40 ⎞ v a = −4v s − v o ⇒ ⎜ 1 + ⎟ v o = − 4v s 5 R R⎠ ⎝ vo 4 4R =− =− 40 vs R + 40 1+ R vo 0≤R≤∞ ⇒ −4≤ ≤0 vs

−3 = −

4R R + 40



R = 120 kΩ (checked: LNAP 6/21/04)

P 6.4-20 The circuit shown in Figure P 6.4-20 has one input, vs, and one output, vo. The circuit contains one unspecified resistance, R. (a) Express the gain of the circuit, vo/vs, in terms of the resistance R. (b) Determine the range of values of the gain that can be obtained by specifying a value for the resistance R. (c) Design the circuit so that vo = – 5 vs. R 30 kΩ

10 kΩ

+ –

+ + –



vs

+

20 kΩ 10 kΩ

vo –

Figure P 6.4-20

Solution: (a) Use units of V, mA and kΩ. Apply KCL at the inverting input of the left op amp to get v1 10

+

va 30

+

30 ⎞ ⎛ = 0 ⇒ v a = − ⎜ 3v s + v o ⎟ R R ⎠ ⎝

vo

v o = 3v a = −9v s −

(c) We require

⎛ 90 ⎞ ⇒ ⎜ 1 + ⎟ v o = − 9v s R⎠ ⎝

9 9R =− 90 vs R + 90 1+ R vo 0≤R≤∞ ⇒ −9 ≤ ≤0 vs vo

(b)

90 vo R

−5 =

=−

−9 R R + 90



R = 112.5 kΩ (checked: LNAP 7/8/04)

P 6.4-21 The circuit shown in Figure P 6.4-21 has three inputs: v1, v2, and v3. The output of the circuit is vo. The output is related to the inputs by

vo = av1 + bv2 + cv3 where a, b, and c are constants. Determine the values of a, b, and c.

+ –

20 kΩ

20 kΩ

v1



40 kΩ

+

120 kΩ



20 kΩ

120 kΩ

+

+ –

+ –

v2

20 kΩ

+

vo –

20 kΩ

20 kΩ

30 kΩ

– +

v3

+ –

Figure P 6.4-21 Solution: Label the node voltages and identify some standard op amp circuits:

Either by recognizing the inverting amplifier or by writing and solving the node equation corresponding to node a, we obtain ⎛ 20 ⎞ v b = ⎜ − ⎟ v 1 = −v 1 ⎝ 20 ⎠ Either by recognizing the voltage divider and voltage follower or by writing and solving the node equation corresponding to node c, we obtain 1 ⎛ 20 ⎞ vd = ⎜ ⎟ v2 = v2 2 ⎝ 20 + 20 ⎠ Either by recognizing the noninverting amplifier or by writing and solving the node equation corresponding to node f, we obtain ⎛ 20 ⎞ v g = ⎜1 + ⎟ v 3 = 3 v 3 ⎝ 20 ⎠

Either by recognizing the summing amplifier or by writing and solving the node equation corresponding to node e, we obtain ⎡⎛ 120 ⎞ ⎛ 120 ⎞ ⎛ 120 ⎞ ⎤ v o = − ⎢⎜ ⎟ vb + ⎜ ⎟ vd + ⎜ ⎟ v g ⎥ = − ⎡⎣3 v b + v d + 4 v g ⎤⎦ ⎝ 120 ⎠ ⎝ 30 ⎠ ⎦ ⎣⎝ 40 ⎠ Substituting for v b , v d and v g gives ⎡ ⎤ ⎛1 ⎞ v o = − ⎢3 ( −v1 ) + ⎜ v 2 ⎟ + 4 ( 3 v 3 ) ⎥ = 3 v1 − 0.5 v 2 − 12 v 3 ⎝2 ⎠ ⎣ ⎦ so a = 3, b = −0.5 and c = −12 (checked: LNAP 6/21/04)

P 6.4-22 The circuit shown in Figure P 6.4-22 has two inputs: v1 and v2. The output of the circuit is vo. The output is related to the inputs by

vo = av1 + bv2 where a and b are constants. Determine the values of a and b. 20 kΩ

20 kΩ

40 kΩ

– +

+ –



v1

+

+ –

20 kΩ

20 kΩ

v2



+ vo

20 kΩ

+

20 kΩ

Figure P 6.4-22 Solution:

Label the node voltages as shown. Use units of V, mA and kΩ. v 3 = v1 and v 4 = −v 2



v5 − v3 40

+

v5 − v4 20

=0



v5 =

1 1 2 v 3 + 2v 2 ) = v 1 − v 2 ( 3 3 3

so a=−

1 2 and b = − 3 3 (checked: LNAP 6/21/04)

P6.4-23 The input to the circuit shown in Figure P6.4-23 is the voltage source voltage v s . The output is

the node voltage v o . The output is related to the input by the equation v o = k v s where k = is called the gain of the circuit. Determine the value of the gain k.

FigureP6.4-23

Solution: Label the node voltages as shown. Apply KCL at the inverting input node of the op amp to get

vs 30000

+

vs − va 80000

= 0 ⇒ 11v s = 3 v a

⇒ va =

11 vs 3

Apply KCL at the right node of the 80 kΩ resistor to get vs − va 80000

+

vo − va 20000

=

va 20000

⇒ vs − 9 va + 4 vo = 0 ⎛ 11 ⎞ ⇒ vs − 9 ⎜ vs ⎟ + 4 vo = 0 ⎝3 ⎠ ⇒ v s − 33 v s + 4 v o = 0 ⇒ 4 v o = 32 v s ⇒ vo = 8 vs

Comparing this equation to v o = k v s , we determine that k = 8 V/V.

vo vs

P6.4-24 The input to the circuit shown in Figure P6.4-23 is the voltage source voltage v s . The output is

the node voltage v o . The output is related to the input by the equation v o = m i s + b where m and b are constants. Determine the values of m and b.

FigureP6.4-24

Solution: Label the node voltages as shown. Apply KCL at the inverting input node of the op amp to get

is =

6 − va 50000

= 0 ⇒ v a = 6 − ( 50 × 10 3 ) i s

Apply KCL at the top node of the 25 kΩ resistor to get vo − va va is + = 25000 25000 Solving: ( 25 ×10 3 ) i s − 2 v a + v o = 0

( 25 ×10 ) i s − 2 ( 6 − ( 50 ×10 ) i s ) + v o = 0 (125 ×10 ) i s − 12 + v o = 0 v o = − (125 × 10 ) i s + 12 3

3

3

3

Comparing this equation to v o = m i s + b , we get m = − (125 ×10 3 ) V/A and b =12 V.

P6.4-25 The input to the circuit shown in Figure P6.4-25 is the node voltage v s . The output is the node

voltage v o . The output is related to the input by the equation v o = k v s where k = the gain of the circuit. Determine the value of the gain k.

FigureP6.4-25

Solution: Label the node voltages as shown. The node equations are

vs − va 5000 va − 0 5000

=

vo − vb 5000

=

va − 0 5000

0 − vb

vb 25000

Substituting v a = −

va − vo 50000

⇒ v b = −10 v a

50000 =

+

⇒ v o = 11v b

vo 110

into the first node equation gives:

vo vs

is called

vs +

vo



vo



vo

− vo 110 = 110 + 110 5000 5000 50000

vo ⎞ ⎛ ⎛ vo ⎞ vo ⇒ 10 ⎜ v s + − vo ⎟ = 10 ⎜ − ⎟− 110 ⎠ ⎝ ⎝ 110 ⎠ 110 131 ⇒ 10 v s = − vo 110 1100 v s = −8.397 v s ⇒ vo = − 131

Comparing this equation to v o = k v s , we determine that k = −8.397 V/V.

P6.4-26 The values of the node voltages v1, v2 and vo in Figure P6.4-26 are

v1 = 6.25 V, v2 = 3.75 V and vo = −15 V. Determine the value of the resistances R1, R2 and R3:

FigureP6.4-26 Solution: Label the node voltages as shown. Using units of Volts, kΩ and mAmps, the node equations are: 2.5 v1 − 2.5 v1 − v 2 v 2 = = , 20 R2 20 R1

And

v2 − 0 R3

+

vo − 0 20

=0

Using v1 = 6.25 V gives 2.5 6.25 − 2.5 = ⇒ R1 = 30 kΩ 20 R1 Using v1 = 6.25 V and v 2 = 3.75 V gives 6.25 − 3.75 3.75 = ⇒ R 2 = 30 kΩ 20 R2 Using v 2 = 3.75 V and v o = −15 V gives

3.75 − 0 −15 − 0 + = 0 ⇒ R 3 = 5 kΩ R3 20

P6.4-27 The input to the circuit shown in Figure P6.4-27 is the voltage source voltage, v i . The

output is the node voltage, v o .The output is related to the input by the equation v o = k v i where k =

vo vi

is called the gain of the circuit.

Determine the value of the gain k.

FigureP6.4-27 Solution: Label the node voltages as shown. The node voltages at the input nodes of an ideal op amp are equal so va = vb

Consequently i2 =

va − vb

=0 R The input currents of an ideal op amp are 0 A, so applying KCL at nodes a and b shows that i1 = 0 and i 3 = 0 . Consequently, the voltages across the corresponding resistor are 0 V. Finally va = v b = vc = vi

Write an node equation at the right node of the 24 kΩ resistor: 0=

vi 10

+

vi − vo 40

⇒ vo = 5 vi

Finally k=

vo vi

=5

Section 6-5: Design Using Operational Amplifier P 6.5-1 Design the operational amplifier circuit in Figure P 6.5-1 so that vout = r · iin where

r = 20

Operational amplifier circuit

iin

+ 20 kΩ

vout –

V mA

Figure P 6.5-1 Solution: Use the current-to-voltage converter, entry (g) in Figure 6.6-1.

P 6.5-2 Design the operational amplifier circuit in Figure P 6.5-2 so that iout = g · vin where

g =2

iout vin

+ –

Operational amplifier circuit

mA V

Figure P 6.5-2 Solution: Use the voltage –controlled current source, entry (i) in Figure 6.6-1.

5 kΩ

P 6.5-3 Design the operational amplifier circuit in Figure P 6.5-3 so that

v1

Operational amplifier circuit

+ –

vout = 5 · v1 + 2 · v2

+ 20 kΩ

v2

vout –

+ –

Figure P 6.5-3 Solution: Use the noninverting summing amplifier, entry (e) in Figure 6.6-1.

P 6.5-4

Design the operational amplifier circuit in Figure P 6.5-3 so that vout = 5 · (v1 – v2).

Solution: Use the difference amplifier, entry (f) in Figure 6.6-1.

P 6.5-5

Design the operational amplifier circuit in Figure P 6.5-3 so that vout = 5 · v1 – 2 · v2

Solution: Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1.

P 6.5-6 The voltage divider shown in Figure P 6.5-6 has a gain of vout −10 kΩ = =2 vin 5 kΩ + (−10 kΩ)

5 kΩ + vin

Design an operational amplifier circuit to implement the – 10-kΩ resistor.

Solution: Use the negative resistance converter, entry (h) in Figure 6.6-1.

+ –

–10 kΩ

vout –

Figure P 6.5-6

P 6.5-7 Design the operational amplifier circuit in Figure P 6.5-7 so that iin = 0 and vout = 3 · vin

iin + 5V

+ –

10 kΩ –5 V

+ –

Operational amplifier circuit

vin –

+ 20 kΩ

vout –

Figure P 6.5-7 Solution: Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces the current iin to be zero.

P 6.5-8 Design an operational amplifier circuit with output vo = 6 v1 + 2 v2, where v1 and v2 are input voltages. Solution:

Summing Amplifier: va = − ( 6 v1 + 2 v2 ) ⎫ ⎬ ⇒ vo = 6 v1 + 2 v2 Inverting Amplifier: vo = −va ⎭

8 kΩ

P 6.5-9 Determine the voltage vo for the circuit shown in Figure P 6.5-9. Hint: Use superposition. Answer: vo = (– 3)(3) + (4)(– 4) + (4)(8) = 7 V

4 kΩ

24 kΩ

– +

+

3V –

4V

– +

2 mA

+ 10 kΩ

vo –

Figure P 6.5-9 Solution:

Using superposition, vo = v1 + v2 + v3 = −9 − 16 + 32 = 7 V

P 6.5-10 For the op amp circuit shown in Figure P 6.5-10, find and list all the possible voltage gains that can be achieved by connecting the resistor terminals to either the input or the output voltage terminals.

6

12

12

– vs

+

+ –

Figure P 6.5-10 Solution: R1

6

12

24

6||12

6||24

R2 -vo/vs

12||12||24 0.8

6||12||24 0.286

6||12||12 0.125

12||24 2

12||12 1.25

12||24

6||12||12

6||12||24

12||12||24

6||12 0.5

24 8

12 3.5

6 1.25

R1 R2 -vo/vs

12||12 6||24 0.8

24

+ vo –

P 6.5-11 The circuit shown in Figure P 6.5-11 is called a Howland current source. It has one input, vin, and one output, iout. Show that when the resistances are chosen so that R2 R3 = R1 R4, the output is related to the input by the equation iout =

vin R1

R4

R3 – + R1

vin

+ –

R2

RL

iout

Figure P 6.5-11

Solution: Label the node voltages as shown. Apply KCL at the inverting input of the op amp to get va R3

+

va − vb R3

⎛ R3 + R 4 ⎞ = 0 ⇒ vb = ⎜ v ⎜ R 3 ⎟⎟ a ⎝ ⎠

Apply KCL at the noninverting input of the op amp to get v a − v in R1

+

va − vb R2

+ i out = 0

Solving gives ⎛ R1 + R 2 ⎞ v in v b ⎛ R1 + R 2 R 3 + R 4 ⎞ v in va ⎜ − − + i out = 0 ⇒ v a ⎜ − + i out = 0 ⎟ ⎟− ⎜ R1 R 2 ⎟ R1 R 2 ⎜ R1 R 2 R 2 R 3 ⎟⎠ R1 ⎝ ⎠ ⎝ When R 2 R 3 = R1 R 4 the quantity in parenthesis vanishes leaving

i out =

1 v in R1

P 6.5-12 The input to the circuit shown in Figure P 6.5-12a is the voltage vs. The output is the voltage vo. The voltage vb is used to adjust the relationship between the input and output. (a) Show that the output of this circuit is related to the input by the equation

vo = avs + b where a and b are constants that depend on R1, R2, R3, R4, and vb. Design the circuit so that its input and output have the relationship specified by the graph shown in Figure P 6.5-12b.

(b)

vo , V 8 + R1

6



+

4

R4 + –

R2

vs

2

vo + –

R3

vb

R5

–4 –6



–2

2

4

–2 –4

(a)

(b)

Figure P 6.5-12 Solution: (a) Label the node voltages as shown. The node equations are vs − va R1

+

vb − va R2

=

va R3

and

va R5

=

vo − va R4

⎛ R5 ⎞ ⇒ va = ⎜ v ⎜ R 4 + R 5 ⎟⎟ o ⎝ ⎠

Solving these equations gives vs ⎛ 1 v b ⎛ R1 R 2 + R 2 R 3 + R1 R 3 R5 ⎞ vb 1 1 ⎞ =⎜ + + =⎜ × va − vo − ⎟ ⎟ R1 ⎝⎜ R1 R 2 R 3 ⎟⎠ R 2 ⎜⎝ R1 R 2 R 3 R 4 + R 5 ⎟⎠ R2 So ⎛ R 2 R3 R 4 + R5 ⎞ R1 R 3 R 4 + R5 vo = ⎜ × vs + × × vb ⎟ ⎜ R1 R 2 + R 2 R 3 + R1 R 3 R 5 ⎟⎠ R1 R 2 + R 2 R 3 + R1 R 3 R5 ⎝ So

vs , V

⎛ R 2 R3 R 4 + R5 ⎞ R1 R 3 R 4 + R5 a=⎜ × v s and b = × × vb ⎟ ⎜ R1 R 2 + R 2 R 3 + R1 R 3 ⎟ R R R + R R + R R R 5 1 2 2 3 1 3 5 ⎝ ⎠ (b) The equation of the straight line is

5 vo = vs + 5 4 We require R 2 R3 R1 R 2 + R 2 R 3 + R1 R 3

×

R 4 + R5 R5

=

5 4

For example, let R1 = R 2 = R 3 = 10 kΩ, R 4 = 55 kΩ and R 5 = 20 kΩ . Next we require 5=

R1 R 3 R1 R 2 + R 2 R 3 + R1 R 3

×

R 4 + R5 R5

× vb =

5 vb 4

i.e. vb = 4 V

(checked: LNAP 6/20/04)

R3

R1

P 6.5-13 The input to the circuit shown in Figure P 6.5-13a is the voltage vs. The output is the voltage vo. The voltage vb is used to adjust the relationship between the input and output.

(a)

Show that the output of this circuit is related to the input by the equation

– + –

R2

vs + –

+

+ R4

vb

vo –

vo = avs + b

where a and b are constants that depend on R1, R2, R3, R4, and vb. (b)

(a)

Design the circuit so that its input and output have the relationship specified by the graph shown in Figure P 6.5-13b.

vo , V 8 6 4 2 –6

–4

2

–2

4 –2 –4

(b)

P 6.5-13 Solution: (a) Apply KCL at the inverting input of the op amp to get: ⎛ R3 ⎞ vs vb vo R3 + + = 0 ⇒ vo = ⎜ − ⎟ vs − vb ⎜ ⎟ R1 R 2 R 3 R2 ⎝ R1 ⎠ R3 R3 so a=− and b = − vb R1 R2

(b) The equation of the straight line is We require



R3 5 =− 2 R1

Next, we require

5=−

e.g. R 3v b R2

5 vo = − vs + 5 2

R1 = 20 kΩ and R 3 = 50 kΩ .

e.g.

R 2 = R 3 = 50 kΩ and v b = −5 V .

vs, V

P6.5-14 The input to the circuit shown in Figure P6.5-14 is the voltage source voltage v s . The

output is the node voltage v o . the output is related to the input by the equation v o = m v s + b where m and b are constants. (a) Specify values of R 3 and v a that cause the output to be related to the input by the equation v o = 4 v s + 7 . (b) Determine the values of m and b when

R 3 = 20 kΩ and v a = 2.5 V .

Figure P6.5-14 Solution: Label the node voltages:

⎛ 30 ⎞ Recognizing an inverting amplifier and a noninverting amplifier we write v b = ⎜ − ⎟ v s . ⎝ 10 ⎠ Applying KCL at the inverting input node of the right op amp gives ⎛ 20 ⎞ ⎛ 20 ⎞ ⎛ 20 ⎞ ⎛ 20 ⎞ ⎛ 30 ⎞ ⇒ v o = ⎜1 + v a − ⎜ ⎟ v b = ⎜1 + va − ⎜ ⎟ ⎜ − ⎟ vs ⎟ ⎟ ⎜ R3 ⎟ ⎜ R3 ⎟ ⎜ R3 ⎟ ⎜ R 3 ⎟ ⎝ 10 ⎠ R3 20 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 20 ⎞ ⎛ 60 ⎞ ⇒ v o = ⎜1 + va + ⎜ ⎟ vs ⎟ ⎜ R3 ⎟ ⎜ R3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ 60 ⎞ ⎛ 20 ⎞ (a) We require ⎜ ⎟ = 4 and ⎜1 + v = 7 V so R 3 = 15 kΩ and then v a = 3 V. ⎜ R3 ⎟ ⎜ R 3 ⎟⎟ a ⎝ ⎠ ⎝ ⎠ ⎛ 60 ⎞ ⎛ 60 ⎞ ⎛ 20 ⎞ V ⎛ 20 ⎞ and b = ⎜1 + v a = ⎜1 + ⎟ 2.5 = 5 V . (b) m = ⎜ ⎟ = ⎜ ⎟ = 3 ⎟ ⎜ R 3 ⎟ ⎝ 20 ⎠ ⎜ R3 ⎟ V ⎝ 20 ⎠ ⎝ ⎠ ⎝ ⎠ vb − va

=

va − vo

P6.5-15. The circuit shown in Figure 6.5-15 uses a potentiometer to implement a variable resistor having a resistance R that varies over the range

0 ≤ R ≤ 200 kΩ The gain of this circuit is G =

vo vs

. Varying the

resistance R over it’s range causes the value of the gain G to vary over the range vo ≤ G max G min ≤ vs Determine the minimum and maximum values of the gain, G min and G max .

Solution: Let

R eq = 25 + R || 50 Then, recognizing this circuit as a noninverting amplifier, we get R eq 25 + R || 50 G = 1+ = 1+ 25 25 G min corresponds to R = 0:

G min = 1 +

25 + 0 || 50 25 + 0 = 1+ = 2 V/V 25 25

G max corresponds to R = 200 kΩ: G max = 1 +

25 + 200 || 50 25 + 40 = 1+ = 3.6 V/V 25 25

Figure P6.5-15

P6.5-16 The input to the circuit shown in Figure P6.5-16a is the voltage, vs. The output is the voltage vo. The voltage v b is used to adjust the relationship between the input and output. Determine values of R 4 and v b that cause the circuit input and output have the relationship specified by the graph shown in Figure P6.5-21b. Answers: v b = 1.62 V and R 4 = 62.5 kΩ.

(a)

(b) Figure P6.5-16

Solution: Recognize the voltage divider, voltage follower and noninverting amplifier to write R4 ⎞ R4 ⎞ 2 R4 ⎞ R4 ⎞ ⎛ ⎞⎛ ⎛ ⎛ ⎛ 20 × 10 3 vo = ⎜ 3 3 ⎟⎜ − 3 ⎟ v s + ⎜1 + 3 ⎟ vb = ⎜ − 3 ⎟ v s + ⎜1 + 3 ⎟ vb ⎝ 20 × 10 + 5 × 10 ⎠ ⎝ 30 × 10 ⎠ ⎝ 30 × 10 ⎠ ⎝ 75 × 10 ⎠ ⎝ 30 × 10 ⎠

(Alternately, this equation can be obtained by writing two node equations: one at the noninverting node of the left op amp and the other at the inverting node of the right op amp.) The equation of the straight line is

5 vo = − vs + 5 3

Comparing coefficients gives 5 5 75 ×10 3 − =− ⇒ R4 = × = 62.5 × 10 3 = 62.5 kΩ 3 75 ×10 3 3 2 2 R4

and R4 ⎞ ⎛ ⎛ 62.5 × 10 3 ⎞ 5 = ⎜1 + v = ⎜1 + 3 ⎟ b 3 ⎟ v b = 3.08333 v b 30 10 30 10 × × ⎝ ⎠ ⎝ ⎠

⇒ vb =

5 = 1.62 V 3.08333

Figure P6.5-17 P6.5-17 Figure P6.5-17 shows three similar circuits. The outputs of the circuits are the voltages v1, v2 and v3. Determine the values of these three outputs. Solution: 80 80 ⎛ 80 ⎞ v1 = ⎜ 1 + ⎟ (1.8 ) = 7.56 V , v 2 = (1.8) = 1.3714 V and v 3 = − (1.8) = −5.76 V 80 + 25 25 ⎝ 25 ⎠

Figure P6.5-18 P6.5-18 . The input to the circuit shown in Figure P6.5-18 is the source voltage vs. The output is the voltage across the 25 kΩ resistor, vo. The output is related to the input by the equation vo = (g) vi where g is the gain of the circuit. Determine the value of g. Solution: Using equivalent resistance we can draw the circuit as shown. 20 = (15 + 45 ) || 30 and 80 = 64 + ( 80 || 20 )

Recognizing this circuit as an inverting amplifier, we write g=

vo vs

=−

80 V = −4 20 V

Section 6-6: Operational Amplifier Circuits and Linear Algebraic Equations P 6.6-1

Design a circuit to implement the equation z = 4w +

x − 3y 4

The circuit should have one output, corresponding to z, and three inputs, corresponding to w, x, and y. Solution:

P 6.6-2

Design a circuit to implement the equation 0 = 4w + x + 10 – (6y + 2z)

The output of the circuit should correspond to z. Solution:

Section 6-7: Characteristics of the Practical Operational Amplifier 10 kΩ

P 6.7-1 Consider the inverting amplifier shown in Figure P 6.7-1. The operational amplifier is a typical OP-07E (Table 6.7-1). Use the offsets model of the operational amplifier to calculate the output offset voltage. (Recall that the input, vin, is set to zero when calculating the output offset voltage.)

100 kΩ

– + –

vin

+

+ vo –

Answer: 0.45 mV Figure P 6.7-1

Solution:

The node equation at node a is:

vout − vos vos = + ib1 3 100×10 10×103

Solving for vout: ⎛ 100×103 ⎞ vout = ⎜1+ v + (100 × 103 ) ib1 = 11vos + (100 × 103 ) ib1 3 ⎟ os 10 10 × ⎝ ⎠

(

)

= 11 ( 0.03×10−3 )+(100×103 ) 1.2×10−9 = 0.45 mV

+

P 6.7-2 Consider the noninverting amplifier shown in Figure P 6.7-2. The operational amplifier is a typical LF351 (Table 6.7-1). Use the offsets model of the operational amplifier to calculate the output offset voltage. (Recall that the input, vin, is set to zero when calculating the output offset voltage.)



+ 90 kΩ

+ –

vin

vo 10 kΩ –

Figure P 6.7-2 Solution:

The node equation at node a is:

vos v −v + ib1 = 0 os 10000 90000

Solving for vo: ⎛ 90×103 ⎞ vo = ⎜1+ v + ( 90 × 103 ) ib1 = 10 vos + ( 90 × 103 ) i b1 3 ⎟ os ⎝ 10×10 ⎠

= 10(5 ×10−3 )+ ( 90 ×103 ) (.05 × 10−9 ) = 50.0045 × 10−3 − 50 mV

R1

P 6.7-3 Consider the inverting amplifier shown in Figure P 6.7-3. Use the finite gain model of the operational amplifier (Figure 6.7-1c) to calculate the gain of the inverting amplifier. Show that vo Rin ( Ro − AR2 ) = vin ( R1 + Rin )( Ro − R2 ) + R1 Rin (1 + A)

R2

– + –

vin

+

Figure P 6.7-3 Solution:

v1 −vin v1 v1 −v0 ⎫ + + = 0⎪ R1 Rin R2 v0 Rin ( R0 − AR2 ) ⎪ = ⎬ ⇒ v0 + Av1 v0 − v1 vin ( R1 + Rin )( R0 + R2 ) + R1 Rin (1+ A) + =0 ⎪ ⎪⎭ R0 R2

+ vo –

R1

P 6.7-4 Consider the inverting amplifier shown in Figure P 6.7-3. Suppose the operational amplifier is ideal, R1 = 5 kΩ, and R2 = 50 kΩ. The gain of the inverting amplifier will be

R2

– + –

vin

+

vo = −10 vin

Use the results of Problem P 6.7-3P 6.7-3 to find the gain of the inverting amplifier in each of the following cases: (a)

+ vo –

Figure P 6.7-3

The operational amplifier is ideal, but 2 percent resistors are used and R1 = 5.1 kΩ and R2 = 49 kΩ.

(b)

The operational amplifier is represented using the finite gain model with A = 200,000, Ri = 2 MΩ, and Ro = 75 Ω; R1 = 5 kΩ and R2 = 50 kΩ.

(c)

The operational amplifier is represented using the finite gain model with A = 200,000, Ri = 2 MΩ, and Ro = 75 Ω; R1 = 5.1 kΩ and R2 = 49 kΩ.

Solution: a) v0 R 49×103 = − 2 = − = −9.6078 vin R1 5.1×103

(

)

b)

2×106 ) 75−( 200,000 )( 50×103 ) ( v0 = = −9.9957 vin (5×103 + 2×106 )(75+50×103 ) + (5×103 )(2×106 )(1+ 200,000)

c)

v0 2×106 (75− (200,000)(49×103 )) = = −9.6037 vin (5.1×103 +2×106 )(75+49×103 )+(5.1×103 )(2×106 )(1+200,000)

R1

P 6.7-5 The circuit in Figure P 6.7-5 is called a difference amplifier and is used for instrumentation circuits. The output of a measuring element is represented by the common mode signal vcm and the differential signal (vn + vp). Using an ideal operational amplifier, show that R4 (vn + vp ) R1 R4 R3 = R1 R2

vo = −

When

R4

vn + –

– vp + –

+

+

R2 vo

vcm + –

R3 –

Figure P 6.7-5 Solution:

Apply KCL at node b: R3 vb = (vcm − v p ) R2 + R3 Apply KCL at node a: va −v0 va −(vcm + vn ) + = 0 R4 R1 The voltages at the input nodes of an ideal op amp are equal so va = vb . R R +R v0 = − 4 (vcm + vn ) + 4 1 va R1 R1 R v0 = − 4 (vcm + vn ) + R1 ( R4 + R1 ) R3 (vcm − v p ) R1 ( R2 + R3 )

when

R4 R1

R4 +1 ( R4 + R1 ) R3 R3 R R R1 then = = × 3 = 4 R3 R2 R1 ( R2 + R3 ) R1 +1 R2 R2

so v0 = −

R4 R R (vcm + vn ) + 4 (vcm − v p ) = − 4 (vn + v p ) R1 R1 R1

Section 6-10 How Can We Check…? 6 kΩ

4 kΩ

P 6.10-1 Analysis of the circuit in Figure P 6.10-1 shows that io = – 1 mA and vo = 7 V. Is this analysis correct?

io

– + –

Hint: Is KCL satisfied at the output node of the op amp?

+

2V 5V

– +

+ 10 kΩ

Figure P 6.10-1 Solution: Apply KCL at the output node of the op amp io =

v − ( −5 ) vo + o =0 10000 4000

Try the given values: io =−1 mA and vo = 7 V −1×10−3 ≠ 3.7 × 10−3 =

7 − ( −5 ) 7 + 10000 4000

KCL is not satisfied. These cannot be the correct values of io and vo.

vo –

P 6.10-2 Your lab partner measured the output voltage of the circuit shown in Figure P 6.10-2 to be vo = 9.6 V. Is this the correct output voltage for this circuit? Hint: Ask your lab partner to check the polarity of the voltage that he or she measured.

4 kΩ

10 kΩ

+ 2 mA





+

+

vo –

Figure P 6.10-2 Solution:

va = ( 4 ×103 )( 2 ×10−3 ) = 8 V 12 ×103 va = −1.2 ( 8 ) = −9.6 V 10 ×103 So vo = −9.6 V instead of 9.6 V.

vo = −

12 kΩ

4 kΩ 2 kΩ

+

P 6.10-3 Nodal analysis of the circuit shown in Figure P 6.10-3 indicates that vo = – 12 V. Is this analysis correct? Hint: Redraw the circuit to identify an inverting amplifier and a noninverting amplifier.





+

+

2V

vo

6 kΩ

– +

3V

+ –

2 kΩ –

Figure P 6.10-3 Solution: First, redraw the circuit as:

Then using superposition, and recognizing of the inverting and noninverting amplifiers: ⎛ 6 ⎞⎛ 4 ⎞ ⎛ 4⎞ vo = ⎜ − ⎟ ⎜ − ⎟ ( −3) + ⎜ 1 + ⎟ ( 2 ) = −18 + 6 = −12 V ⎝ 2 ⎠⎝ 2 ⎠ ⎝ 2⎠ The given answer is correct.

P 6.10-4 Computer analysis of the circuit in Figure P 6.10-4 indicates that the node voltages are va = – 5 V, vb = 0 V, vc = 2 V, vd = 5 V, ve = 2 V, vf = 2 V, and vg = 11 V. Is this analysis correct? Justify your answer. Hint: Verify that the resistor currents indicated by these node voltages satisfy KCL at nodes b, c, d, and f. a 10 kΩ – +

b

4 kΩ

c

6 kΩ

5 kΩ

40 kΩ

5V

e

10 kΩ

+ –

2V

4 kΩ

d

– f

g

+ 10 kΩ

Figure P 6.10-4 Solution: First notice that ve = v f = vc is required by the ideal op amp. (There is zero current into the input

lead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f, hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal op amp are equal.) The given voltages satisfy all the node equations at nodes b, c and d: node b:

0− (−5) 0 0−2 + + =0 10000 40000 4000

node c:

0− 2 2 −5 = +0 4000 6000

node d:

2 −5 5 5−11 = + 6000 5000 4000

Therefore, the analysis is correct.

a

20 kΩ

b

+ P 6.10-5 Computer analysis of the d – noninverting summing amplifier shown in Figure P 6.10-5 indicates that the node 9 kΩ voltages are va = 2 V, vb = – 0.25 V, vc = – 5 V, c 40 kΩ + vd = – 2.5 V, and ve = – 0.25 V. 2V – e (a) Is this analysis correct? – (b) Does this analysis verify that the circuit + 5V 1 kΩ 40 kΩ is a noninverting summing amplifier? Justify your answers. 1st Hint: Verify that the resistor currents indicated by these node voltages satisfy KCL Figure P 6.10-5 at nodes b and e. 2nd Hint: Compare to Figure 6.5-1e to see that Ra = 10 kΩ and Rb = 1 kΩ. Determine K1, K2, and K4 from the resistance values. Verify that vd = K4(K1va + K2vc).

Solution: The given voltages satisfy the node equations at nodes b and e:

node b:

−.25− 2 −.25 −.25−( −5 ) + + =0 20000 40000 40000

node e:

−2.5−( −0.25 ) −0.25 ≠ +0 9000 1000

Therefore, the analysis is not correct. Notice that

−2.5−( +0.25 ) +0.25 = +0 9000 1000

So it appears that ve = +0.25 V instead of ve = −0.25 V. Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2 = 1/ 4 and K4 = 9. The given node voltages satisfy the equation −2.5 = vd = K

4

( K v + K v ) = 10 ⎛⎜⎝ 12 ( 2 )+ 14 ( −5) ⎞⎟⎠

None-the-less, the analysis is not correct.

1 a

2 c

PSpice Problems SP 6-1 The circuit in Figure SP 6-1 has three inputs: vw, vx, and vy. The circuit has one output, vz. The equation vz = avw + bvx + cvy expresses the output as a function of the inputs. The coefficients a, b, and c are real constants. (a) Use PSpice and the principle of superposition to determine the values of a, b, and c. (b) Suppose vw = 2 V, vx = x, vy = y and we want the output to be vz = z. Express z as a function of x and y. Hint: The output is given by vz = a when vw = 1 V, vx = 0 V, and vy = 0 V. Answer: (a) vz = vw + 4 vx – 5 vy (b) z = 4 x – 5 y + 2 20 kΩ

60 kΩ

– 20 kΩ

+ vx

+

+ –

20 kΩ

+ –

100 kΩ

20 kΩ

– 60 kΩ

vy



20 kΩ

+ vw

+ –

20 kΩ

20 kΩ

Figure SP 6-1 Solution: (a)

vz = a vw + b vx + c v y The following three PSpice simulations show 1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V 4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V -5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V

vz

1 V = vz = a when vw= 1 V, vx = 0 V and vy = 0 V:

4 V = vz = b when vw= 0 V, vx = 1 V and vy = 0 V:

-5 V = vz = c when vw= 0 V, vx = 0 V and vy = 1 V:

Therefore

v z = vw + 4 v x − 5 v y

(b) When vw= 2 V:

vz = 4 vx − 5 v y + 2

25 kΩ

SP 6-2 The input to the circuit in Figure SP 6-2 is vs, and the output is vo.

+ –

(a) Use superposition to express vo as a function of vs.

80 kΩ



vs

+

(b) Use the DC Sweep feature of PSpice to plot vo as a function of vs.

+ vo

2V



+ –

(c) Verify that the results of parts (a) and (b) agree with each other. Figure SP 6-2 Solution:

a) Using superposition and recognizing the inverting and noninverting amplifiers: vo = −

80 ⎛ 80 ⎞ vs + ⎜1 + ⎟ ( −2 ) = −3.2 vs − 8.4 25 ⎝ 25 ⎠

b) Using the DC Sweep feature of PSpice produces the plot show below. Two points have been labeled in anticipation of c). c) Notice that the equation predicts

( −3.2 ) ( −5) − 8.4 = 7.6 and

( −3.2 ) ( 0 ) − 8.4 = −8.4

Both agree with labeled points on the plot.

1

SP 6-3 A circuit with its nodes identified is shown in Figure SP 6-3. Determine v34, v23, v50, and io.

10 kΩ

10 kΩ

30 kΩ 6V

2

3

+–

30 kΩ 10 kΩ

30 kΩ

4

– +

5 io 30 kΩ

+ vo –

Figure SP 6-3 Solution:

VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 V_V2

-3.000E-04 -7.000E-04

v34 = −1.5 − −12 × 10−6 ≅ −1.5 V v23 = 4.5 − ( −1.5 ) = 6 V v50 = 12 − 0 = 12 V io = −7 × 10−4 = −0.7 mA

2 kΩ

SP 6-4 Use PSpice to analyze the VCCS shown in Figure SP 6-4. Consider two cases: (a) The operational amplifier is ideal. (b) The operational amplifier is a typical μA741 represented by the offsets and finite gain model.

10 kΩ – +

2 kΩ

10 kΩ iout

20 mV

+ –

50 kΩ

Figure SP 6-4 Solution: V4 is a short circuit used to measure io.

The input of the VCCS is the voltage of the left-hand voltage source. (The nominal value of the input is 20 mV.) The output of the VCCS is io. A plot of the output of the VCCS versus the input is shown below. The gain of the VCVS is gain =

50 ×10−6 − ( −50 ×10−6 )

1 A = × 10−3 V 100 ×10 − ( −100 ×10 ) 2 −3

−3

Design Problems DP 6-1 Design the operational amplifier circuit in Figure DP 6-1 so that iin iout

1 = ⋅ iin 4

iout Operational Amplifier Circuit

5 kΩ

Figure DP 6-1 Solution: From Figure 6.6-1g, this circuit

is described by vo = R f i in . Since i out =

Notice that i oa = i in +

i in
0 for the circuit of Figure E 7.2-1b when vs(t) is the voltage shown in Figure E 7.2-1a. Hint: Determine iC(t) and iR(t) separately, then use KCL. vs(t)(V) 5 i(t) 4 iR(t)

iC(t)

3 vs(t)

2

+ –



1F

1 1

2

3

4

5

6

7

8

9

t (s)

(a)

(b)

Figure E 7.2-1 Answer:

⎧ 2t − 2 2 < t < 4 ⎪ 4 0

P7.2-20 The input to the circuit shown in Figure P7.2-20 is the voltage: v ( t ) = 3 + 4 e −2 t A for t > 0

The output is the current: i ( t ) = 0.3 − 1.6 e −2 t V for t > 0 Determine the values of the resistance and capacitance Answer: R = 10 Ω and C = 0.25 F. Figure P7.2-20 Solution: Apply KCL at either node to get 3 + 4 e −2 t d +C 3 + 4 e −2 t R dt −2 t 3+ 4e 3 ⎛4 ⎞ = + ( −2 ) 4 C e −2 t = + ⎜ − 8 C ⎟ e −2 t R R ⎝R ⎠

(

0.3 − 1.6 e −2 t =

)

Equating coefficients: 0.3 =

3 R

⇒ R = 10 Ω

and

−1.6 =

4 − 8 C ⇒ C = 0.25 F 10

P7.2-21 Consider the capacitor shown in Figure P7.2-21. The current and voltage are given by ⎧0.5 0 < t < 0.5 ⎧2 t + 8.6 0 ≤ t ≤ 0.5 ⎪ ⎪ i ( t ) = ⎨ 2 0.5 < t < 1.5 and v ( t ) = ⎨ a t + b 0.5 ≤ t ≤ 1.5 ⎪0 ⎪ c t > 1.5 t ≥ 1.5 ⎩ ⎩

where a, b and c are real constants. (The current is given in Amps, the voltage in Volts and the time in seconds.) Determine the values of a, b and c. Answer: a = 8 V/s, b = 5.6 V and c = 17.6 V Solution: At t = 0.5 s

Figure P7.2-21

v ( 0.5 ) = 2 ( 0.5 ) + 8.6 = 9.6 V

For 0.5 ≤ t ≤ 1.5 v (t ) =

1 t 2 dτ + 9.6 = 8τ 0.25 ∫ 0.5

At t = 1.5 s

t 0.5

+ 9.6 = 8 ( t − 0.5 ) + 9.6 = 8 t + 5.6 V

v (1.5 ) = 8 (1.5 ) + 5.6 = 17.6 V

For t ≥ 1.5 v (t ) =

1 t 0 dτ + 17.6 = 17.6 0.25 ∫1.5

Checks: At t = 1.0 s At t = 0.5 s

i (t ) =

1 d 1 d 1 v (t ) = (8 t + 5.6 ) = (8) = 2 A √ 4 dt 4 dt 4 v ( 0.5 ) = 8 ( 0.5 ) + 5.6 = 9.6 V



P7.2-22At time t =0, the voltage across the capacitor shown in Figure P7.2-22 is v(0) = −20 V. Determine the values of the capacitor voltage at times 1 ms, 3 ms and 7 ms.

Figure P7.2-22

Solution:

v (t ) = v (0) +

1 t "area under the curve" "area under the curve" i (τ ) dτ = v ( 0 ) + = −20 + ∫ 0 2.5 × 10−6 C C 1 20 × 10−3 1×10−3 10 v ( 0.001) = −20 + 2 = −20 + = −16 V −6 2.5 × 10 2.5

(

)(

)

(When calculating the value of v(0.001), “area under the curve” indicates the area under the graph of i(t) versus t corresponding to the time interval 0 to 1 ms = 0.001 s.) 1 40 × 10−3 2 ×10−3 + 40 × 10−3 1× 10−3 40 + 40 v ( 0.003) = −20 + 2 = −20 + = 12 V −6 2.5 ×10 2.5

(

)(

) (

)(

)

(When calculating the value of v(0.003), “area under the curve” indicates the area under the graph of i(t) versus t corresponding to the time interval 0 to 3 ms = 0.003 s.) 1 1 40 ×10−3 2 × 10−3 + 40 ×10−3 2 × 10−3 + 40 × 10−3 3 × 10−3 2 v ( 0.007 ) = −20 + 2 = 52 V −6 2.5 ×10

(

)(

) (

)(

)

(

)(

)

(When calculating the value of v(0.007), “area under the curve” indicates the area under the graph of i(t) versus t corresponding to the time interval 0 to 7 ms = 0.007 s.)

Section 7-3: Energy Storage in a Capacitor P 7.3-1 The current, i, through a capacitor is shown in Figure P 7.3-1. When v(0) = 0 and C = 0.5 F, determine and plot v(t), p(t), and w(t) for 0 s < t < 6 s. i(A) 1.0 0.8 0.6 0.4 0.2 0.0

0

2

4

6

8 t (s)

Figure P 7.3-1 Solution: 0 t6 ⎩

Given The capacitor voltage is given by

v (t ) =

t 1 t i (τ ) dτ + v ( 0 ) = 2 ∫ i (τ ) dτ + v ( 0 ) ∫ 0 0.5 0

t

v ( t ) = 2 ∫ 0 dτ + 0 = 0

For t < 2

0

In particular, v ( 2 ) = 0. For 2 < t < 6 v ( t ) = 2 ∫ 2 (τ − 2 ) dτ + 0 = ( 0.2τ 2 − 0.8τ ) = ( 0.2 t 2 − 0.8 t + 0.8 ) V = 0.2 ( t 2 − 4 t + 4 ) V t

t

2

2

In particular, v ( 6 ) = 3.2 V. For 6 < t t

v ( t ) = 2 ∫ 0.8 dτ + 3.2 = 1.6τ 2 + 3.2 = (1.6 t − 6.4 ) V = 1.6 ( t − 4 ) V t

6

Now the power and energy are calculated as 0 ⎧ ⎪ 2 p ( t ) = v ( t ) i ( t ) = ⎨0.04 ( t − 2 ) ⎪ 1.28 ( t − 4 ) ⎩

t 0

d i L ( t ) = 5 e −4000 t V dt

for t > 0

P 8.3-7

Figure P 8.3-7a shows astronaut Dale Gardner using the manned maneuvering unit to dock

with the spinning Westar VI satellite on November 14, 1984. Gardner used a large tool called the apogee capture device (ACD) to stabilize the satellite and capture it for recovery, as shown in Figure P 8.3-7a. The ACD can be modeled by the circuit of Figure P 8.3-7b. Find the inductor current iL for t > 0. Answer: iL(t) = 6e–20t A

Figure P 8.3-7 Solution: At t = 0− (steady-state)

Since the input to this circuit is constant, the inductor will act like a short circuit when the circuit is at steady-state: for t > 0 iL ( t ) = iL ( 0 ) e− ( R L ) t = 6 e−20t A

P 8.3-8 The circuit shown in Figure P 8.3-8 is at steady state before the switch opens at time t = 0. The input to the circuit is the voltage of the voltage source, Vs. This voltage source is a dc voltage source; that is, Vs is a constant. The output of this circuit is the voltage across the capacitor, vo(t). The output voltage is given by

vo(t) = 2 + 8e–0.5t V for t > 0 Determine the values of the input voltage, Vs, the capacitance, C, and the resistance, R.

Figure P 8.3-8

Solution: Before the switch opens, the circuit will be at steady state. Because the only input to this circuit is the constant voltage of the voltage source, all of the element currents and voltages, including the capacitor voltage, will have constant values. Opening the switch disturbs the circuit. Eventually the disturbance dies out and the circuit is again at steady state. All the element currents and voltages will again have constant values, but probably different constant values than they had before the switch opened. Here is the circuit before t = 0, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. The combination of resistor and a short circuit connected is equivalent to a short circuit. Consequently, a short circuit replaces the switch and the resistor R. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the capacitor voltage, vo(t).

Because the circuit is at steady state, the value of the capacitor voltage will be constant. This constant is the value of the capacitor voltage just before the switch opens. In the absence of unbounded currents, the voltage of a capacitor must be continuous. The value of the capacitor voltage immediately after the switch opens is equal to the value immediately before the switch opens. This value is called the initial condition of the capacitor and has been labeled as vo(0). There is no current in the horizontal resistor due to the open circuit. Consequently, vo(0) is equal to the voltage across the vertical resistor, which is equal to the voltage source voltage. Therefore vo ( 0 ) = Vs

The value of vo(0) can also be obtained by setting t = 0 in the equation for vo(t). Doing so gives vo ( 0 ) = 2 + 8 e0 = 10 V

Consequently,

Vs = 10 V

Next, consider the circuit after the switch opens. Eventually (certainly as t →∞) the circuit will again be at steady state. Here is the circuit at t = ∞, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. A capacitor in a steady-state dc circuit acts like an open circuit, so an open circuit replaces the capacitor. The voltage across that open circuit is the steady-state capacitor voltage, vo(∞). There is no current in the horizontal resistor and vo(∞) is equal to the voltage across the vertical resistor. Using voltage division, 10 vo ( ∞ ) = (10 ) R + 10 The value of vo(∞) can also be obtained by setting t = ∞ in the equation for vo(t). Doing so gives vo ( ∞ ) = 2 + 8 e −∞ = 2 V

Consequently, 10 (10 ) ⇒ 2 R + 20 = 100 ⇒ R = 40 Ω R + 10 −t τ Finally, the exponential part of vo(t) is known to be of the form e where τ = R t C and Rt is 2=

the Thevenin resistance of the part of the circuit connected to the capacitor. Here is the circuit that is used to determine Rt. An open circuit has replaced the open switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by a short circuit. R t = 10 +

so

( 40 )(10 ) = 18 40 + 10

Ω

τ = R t C = 18 C

From the equation for vo(t) t −0.5 t = − ⇒ τ =2s

τ

Consequently,

2 = 18 C ⇒ C = 0.111 = 111 mF

P 8.3-9

The circuit shown in Figure P 8.3-9 is

at steady state before the switch closes at time t = 0. The input to the circuit is the voltage of the voltage source, 24 V. The output of this circuit, the voltage across the 3-Ω resistor, is given by vo(t) = 6 – 3e–0.35t V when t > 0 Determine the value of the inductance, L, and of

Figure P 8.3-9

the resistances, R1 and R2. Solution: Before the switch closes, the circuit will be at steady state. Because the only input to this circuit is the constant voltage of the voltage source, all of the element currents and voltages, including the inductor current, will have constant values. Closing the switch disturbs the circuit by shorting out the resistor R1. Eventually the disturbance dies out and the circuit is again at steady state. All the element currents and voltages will again have constant values, but probably different constant values than they had before the switch closed.

The inductor current is equal to the current in the 3 Ω resistor. Consequently, − 0.35 t vo (t ) 6 − 3 e − 0.35 t = = 2− e i (t ) = A when t > 0 3 3

In the absence of unbounded voltages, the current in any inductor is continuous. Consequently, the value of the inductor current immediately before t = 0 is equal to the value immediately after t = 0. Here is the circuit before t = 0, when the switch is open and the circuit is at steady state. The open switch is modeled as an open circuit. An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the inductor. The current in that short circuit is the steady state inductor current, i(0). Apply KVL to the loop to get R1 i ( 0 ) + R 2 i ( 0 ) + 3 i ( 0 ) − 24 = 0 24 R1 + R 2 + 3 The value of i(0) can also be obtained by setting t = 0 in the equation for i(t). Do so gives ⇒ i ( 0) =

i ( 0 ) = 2 − e0 = 1 A

Consequently,

1=

24 ⇒ R1 + R 2 = 21 R1 + R 2 + 3

Next, consider the circuit after the switch closes. Here is the circuit at t = ∞, when the switch is closed and the circuit is at steady state. The closed switch is modeled as a short circuit. The combination of resistor and a short circuit connected is equivalent to a short circuit. Consequently, a short circuit replaces the switch and the resistor R1. An inductor in a steady-state dc circuit acts like a short circuit, so a short circuit replaces the inductor. The current in that short circuit is the steady state inductor current, i(∞). Apply KVL to the loop to get 24 R 2 i ( ∞ ) + 3 i ( ∞ ) − 24 = 0 ⇒ i ( ∞ ) = R2 + 3 The value of i(∞) can also be obtained by setting t = ∞ in the equation for i(t). Doing so gives i ( ∞ ) = 2 − e −∞ = 2 A

Consequently 2=

24 ⇒ R2 = 9 Ω R2 + 3

Then R1 = 12 Ω Finally, the exponential part of i(t) is known to be of the form e

−t τ

where τ =

L and Rt is the Rt

Thevenin resistance of the part of the circuit that is connected to the inductor. Here is shows the circuit that is used to determine Rt. A short circuit has replaced combination of resistor R1 and the closed switch. Independent sources are set to zero when calculating Rt, so the voltage source has been replaced by an short circuit. R t = R 2 + 3 = 9 + 3 = 12 Ω so

τ=

L L = R t 12

From the equation for i(t) −0.35 t = −

t

τ

⇒ τ = 2.857 s

Consequently, 2.857 =

L ⇒ L = 34.28 H 12

P 8.3-10 A security alarm for an office building door is modeled by the circuit of Figure P 8.3-10. The switch represents the door interlock, and v is the alarm indicator voltage. Find v(t) for t > 0 for the circuit of Figure P 8.3-10. The switch has been closed for a long time at t = 0–.

Figure P 8.3-10 Solution: First, use source transformations to obtain the equivalent circuit

for t < 0:

for t > 0:

So iL ( 0 ) = 2 A, I sc

1 L 1 s = 0, Rt = 3 + 9 = 12 Ω, τ = = 2 = 24 Rt 12

and iL ( t ) = 2e−24t

t >0

Finally v ( t ) = 9 iL ( t ) = 18 e−24t

t >0

P 8.3-11 The voltage v(t) in the circuit shown in Figure P 8.3-11 is given by

v(t) = 8 + 4e–2t V for t > 0 Determine the values of R1, R2, and C.

Figure P 8.3-11 Solution: As t → ∞ the circuit reaches steady state and the capacitor acts like an open circuit. Also, from the given equation, v ( t ) → 8 V , as labeled on the drawing to the

right, then 8= After t = 0

4 24 R2 + 4



R2 = 8 Ω

v C ( t ) = 24 − v ( t ) = 16 − 4e −2t

Immediately after t = 0 v C ( 0 + ) = 16 − 4 = 12 V

The capacitor voltage cannot change instantaneously so v ( 0 − ) = 12 V

The circuit is at steady state just before the switch closes so the capacitor acts like an open circuit. Then 12 =

8 24 R1 + 4 + 8



R1 = 4 Ω

After t = 0 the Thevenin resistance seen by the capacitor is 8 Rt = 8 & 4 = Ω 3 1 3 2= ⇒ C= F so 8 16 C 3

Figure P 8.3-12 P 8.3-12 The circuit shown in Figure P 8.3-12 is at steady state when the switch opens at time t = 0. Determine i(t) for t ≥ 0. Solution: Before t = 0, with the switch closed and the circuit at steady state, the inductor acts like a short circuit so we have

Using superposition 9 − 5 ×10−3 = −2 mA 3000 The inductor current is continuous so i ( 0 + ) = i ( 0 − ) = −2 mA . i (0 −) =

After t = 0, the switch is open. Determine the Norton equivalent circuit for the part of the circuit connected to the inductor: 9 = 3 mA 3000 R t = 3000 & 6000 = 2000 Ω i sc =

The time constant is given by τ =

L 5 1 = = 0.0025 so = 400 . R t 2000 τ

The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( −0.002 − 0.003) e−400 t + 0.003 = 3 − 5e−400 t mA for t ≥ 0 (checked: LNAP 6/29/04)

Figure P 8.3-13 P 8.3-13 The circuit shown in Figure P 8.3-13 is at steady state when the switch opens at time t = 0. Determine v(t) for t ≥ 0. Solution: Before t = 0, with the switch closed and the circuit at the steady state, the capacitor acts like an open circuit so we have

Using superposition v (0 −) =

60 & 60 60 & 30 ⎛1⎞ ⎛1⎞ 6+ 36 = ⎜ ⎟ 6 + ⎜ ⎟ 36 = 12 V 30 + ( 60 & 60 ) 60 + ( 60 & 30 ) ⎝2⎠ ⎝4⎠

The capacitor voltage is continuous so v ( 0 + ) = v ( 0 − ) = 12 V . After t = 0 the switch is open. Determine the Thevenin equivalent circuit for the part of the circuit connected to the capacitor:

v oc =

60 6=4V 60 + 30

R t = 30 & 60 = 20 kΩ The time constant is τ = R t C = ( 20 × 103 )( 5 × 10−6 ) = 0.1 s so

1

τ

= 10

1 . s

The capacitor voltage is given by v ( t ) = ( v ( 0 + ) − v oc ) e−t τ + v oc = (12 − 4 ) e−10t + 4 = 4 + 8 e −10t V for t ≥ 0 (checked: LNAP 6/29/04)

Figure P 8.3-14 P 8.3-14 The circuit shown in Figure P 8.3-14 is at steady state when the switch closes at time t = 0. Determine i(t) for t ≥ 0. Solution: Before t = 0, with the switch open and the circuit at steady state, the inductor acts like a short circuit so we have

i (t ) = −

⎤ 18 ⎡ 5 2 ⎥ = 0.29 A ⎢ 4 + 18 ⎣ 5 + 20 + (18 & 4 ) ⎦

After t = 0, we can replace the part of the circuit connected to the inductor by its Norton equivalent circuit. First, performing a couple of source transformations reduces the circuit to

Next, replace the series voltage sources by an equivalent voltage source, replace the series resistors by an equivalent resistor and do a couple of source transformations to get

so The current is given by

2 1 1 = 0.25 ⇒ =5 10 s τ −5t i ( t ) = [ 0.29 − 0.4] e + 0.4 = 0.4 − 0.11e −5t A for t ≥ 0

τ=

P 8.3-15 The circuit in Figure P 8.3-15 is at steady state before the switch closes. Find the inductor current after the switch closes. Hint: i(0) = 0.1 A, Isc = 0.3 A, Rt = 40 Ω Answer: i(t) = 0.3 – 0.2e–2t A t ≥ 0

Solution: At steady-state, immediately before t = 0:

⎞ 12 ⎛ 10 ⎞ ⎛ i ( 0) = ⎜ ⎟ ⎜ ⎟ = 0.1 A ⎝ 10 + 40 ⎠ ⎝ 16 + 40||10 ⎠

After t = 0, the Norton equivalent of the circuit connected to the inductor is found to be

so I sc = 0.3 A, Rt = 40 Ω, τ = Finally:

L 20 1 = = s Rt 40 2

i (t ) = (0.1 − 0.3)e −2t + 0.3 = 0.3 − 0.2e −2t A

P8.3-16. Consider the circuit shown in Figure P8.316. a) Determine the time constant, τ, and the steady state capacitor voltage when the switch is open. b) Determine the time constant, τ, and the steady state capacitor voltage when the switch is closed. Figure P8.3-16. Solution: Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit.

⎛ 120 ⎞ a.) When the switch is open, v oc = ⎜ ⎟ 24 = 16 V and R t = 120 || 60 = 40 Ω . The steady state ⎝ 60 + 120 ⎠ capacitor voltages is voc =16 V. The time constant is τ = (40)(0.02) = 0.8 s. ⎛ 120 ⎞ b.) When the switch is closed, v oc = ⎜ ⎟ 24 = 19.2 V and R t = 120 || 60 || 60 = 24 Ω . The steady ⎝ 30 + 120 ⎠ state capacitor voltages is voc =19.2 V. The time constant is τ = (24)(0.02) = 0.48 s.

LNAPTR, 11/1/07

Figure P 8.3-17 P 8.3-17 The circuit shown in Figure P 8.3-17 is at steady state before the switch closes. The response of the circuit is the voltage v(t). Find v(t) for t > 0. Hint: After the switch closes, the inductor current is i(t) = 0.2(1 – e–1.8t) A Answer: v(t) = 8 + e–1.8t V Solution: Immediately before t = 0, i (0) = 0.

After t = 0, replace the circuit connected to the inductor by its Norton equivalent to calculate the inductor current: I sc = 0.2 A, Rt = 45 Ω, τ =

So i (t ) = 0.2 (1 − e−1.8t ) A Now that we have the inductor current, we can calculate v(t): d i (t ) dt = 8(1 − e −1.8t ) + 5(1.8)e −1.8t

v(t ) = 40 i (t ) + 25

= 8 + e −1.8t V for t > 0

L 25 5 = = Rth 45 9

Figure P 8.3-18 P 8.3-18 The circuit shown in Figure P 8.3-18 is at steady state before the switch closes. The response of the circuit is the voltage v(t). Find v(t) for t > 0. Answer: v(t) = 37.5 – 97.5e–6400t V Solution: At steady-state, immediately before t = 0

so i(0) = 0.5 A. After t > 0: Replace the circuit connected to the inductor by its Norton equivalent to get I sc = 93.75 mA, Rt = 640 Ω,

τ =

L .1 1 = = s Rt 640 6400

i (t ) = 406.25 e−6400t + 93.75 mA Finally:

v (t ) = 400 i (t ) + 0.1

d i (t ) = 400 (.40625e −6400t + .09375) + 0.1 (−6400) (0.40625e −6400t ) dt = 37.5 − 97.5e −6400t V

Figure P 8.3-19 P 8.3-19 ≥ 0.

The circuit shown in Figure P 8.3-19 is at steady state before the switch closes. Find v(t) for t

Solution: Before the switch closes v(t) = 0 so v ( 0 + ) = v ( 0 − ) = 0 V .

For t > 0, we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor, i.e. the part of the circuit to the left of the terminals a – b. Write mesh equations to find voc:

Mesh equations: 12 i1 + 10 i1 − 6 ( i2 − i1 ) = 0 6 ( i2 − i1 ) + 3 i 2 − 18 = 0 28 i1 = 6 i 2 9 i 2 − 6 i1 = 18

36 i1 = 18 ⇒ i1 = i2 = Using KVL, Find Rt:

⎛7⎞ ⎛1⎞ voc = 3 i 2 + 10 i1 = 3 ⎜ ⎟ + 10 ⎜ ⎟ = 12 V ⎝ 3⎠ ⎝ 2⎠

14 ⎛ 1 ⎞ 7 ⎜ ⎟= A 3 ⎝2⎠ 3

1 A 2

Rt =

Then and

12 (10 + 2 ) =6Ω 12 + (10 + 2 )

1 1 ⎛ 1 ⎞ 1 =4 ⎟= s ⇒ s τ ⎝ 24 ⎠ 4 −4t + v oc = ( 0 − 12 ) e + 12 = 12 1 − e−4t

τ = R tC = 6 ⎜ v ( t ) = ( v ( 0 + ) − v oc ) e −t τ

(

)

V for t ≥ 0 (checked: LNAP 7/15/04)

P 8.3-20 The circuit shown in Figure P 8.3-20 is at steady state before the switch closes. Determine i(t) for t ≥ 0.

Figure P 8.3-20 Solution: Before the switch closes the circuit is at steady state so the inductor acts like a short circuit. We have

⎞ 1⎛ 24 i ( t ) = ⎜⎜ ⎟ = 0.8 A 2 ⎝ 5 + ( 20 & 20 ) ⎟⎠ so

i ( 0 + ) = i ( 0 − ) = 0.8 A

After the switch closes, find the Thevenin equivalent circuit for the part of the circuit connected to the inductor. Using voltage division twice ⎛ 20 1 ⎞ v oc = ⎜ − ⎟ 24 = 7.2 V ⎝ 25 2 ⎠

R t = ( 5 & 20 ) + ( 20 & 20 ) = 14 Ω i sc =

v oc Rt

=

7.2 = 0.514 A 14

Then

τ= and

L 3.5 1 = = s R t 14 4



1

τ

=4

1 s

i ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 0.8 − 0.514 ) e −4t + 0.514 = 0.286e−4t + 0.514 A for t ≥ 0 (checked: LNAP 7/15/04)

P8.3-21 The circuit in Figure P8.3-21 at steady state before the switch closes. Determine an equation that represents the capacitor voltage after the switch closes. Figure P8.3-21 Solution: For t < 0, the switch is open and the capacitor acts like an open circuit because the circuit is at steady state. Consequently, the current in the 10 Ω resistor is 0A and so the voltage across this resistor is 0 V. KVL gives v ( t ) = 18 V . Immediately before the switch opens we have v ( 0 − ) = 18 V . The capacitor

voltage does not change instantaneously so v ( 0 + ) = v ( 0 − ) = 18 V . For t > 0, the Thevenin equivalent of the part of the circuit connected to the capacitor is characterized by 40 R t = 10 || 40 = 8 Ω and, using voltage division, voc = (18) = 14.4 V . 10 + 40 A = voc = 14.4 V , B = v ( 0 + ) − voc = 18 − 14.4 = 3.6 V and a =

1

τ

=

1 1 1 = =5 R t C 8 ( 0.025 ) s LNAPTR, 11/3/07

Figure P 8.3-22 P 8.3-22 The circuit shown in Figure P 8.3-22 is at steady state when the switch closes at time t = 0. Determine i(t) for t ≥ 0. Solution: Before t = 0, with the switch open and the circuit at steady state, the inductor acts like a short circuit so we have i (0 +) = i (0 −) = 4 A

After t = 0, we can replace the part of the circuit connected to the inductor by its Norton equivalent circuit. Using superposition, the short circuit current is given by ⎛ ⎞ ⎛ 3+8 ⎞ 8 i sc = ⎜⎜ ⎟⎟ 2 + ⎜⎜ ⎟⎟ 4 = 3.75 A ⎝ 8 + ( 5 + 3) ⎠ ⎝ ( 3 + 8 ) + 5 ⎠

R t = 8 + 3 + 5 = 16 Ω

so

τ=

2 1 1 = 0.125 s ⇒ =8 16 s τ

The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 4 − 3.75 ) e−8t + 3.75 = 3.75 − 0.25 e −8t A for t ≥ 0 (checked: LNAP 7/15/04)

P8.3-23 The circuit in Figure P8.3-23 at steady state before the switch closes. Determine an equation that represents the inductor current after the switch closes. Figure P8.3-23 Solution: For t < 0, the switch is open and the inductor acts like short circuit because the circuit is at steady state. Consequently, the current in inductor just before the switch closes is i ( 0 − ) = 7 Amps . The inductor

current does not change instantaneously so i ( 0 + ) = i ( 0 − ) = 7 Amps . For t > 0, the Thevenin equivalent of the part of the circuit connected to the inductor is characterized by 20 R t = 20 + 60 = 80 Ω and, using current division, i sc = ( 7 ) = 1.75 Amps . 20 + 60 1 R t 80 1 = = 32 A = i sc = 1.75 Amps , B = i ( 0 + ) − i sc = 7 − 1.75 = 5.25 Amps and a = = τ L 2.5 s LNAPTR, 11/3/07

P8.3-24 Consider the circuit shown in Figure 8.3-24a and corresponding plot of the inductor current shown in Figure 8.3-24b. Determine the values of L, R1 and R 2 . Answer: L = 4.8 H, R1 = 200 Ω and R 2 = 300 Ω. Hint: Use the plot to determine values of D, E, F and a such that the inductor current can be represented as ⎧ D for t ≤ 0 i (t ) = ⎨ − at ⎩ E + F e for t ≥ 0

Figure 8.3-24 Solution: From the plot

D = i(t) for t < 0 =120 mA = 0.12 A, E + F = i(0+) = 120 mA = 0.12 A and

E = lim i ( t ) = 200 mA = 0.2 A . t →∞

The point labeled on the plot indicates that i(t) = 160 mA when t = 27.725 ms = 0.027725 s. Consequently 160 = 200 − 80 e − a ( 0.027725) Then ⎧120 mA for t ≤ 0 i (t ) = ⎨ −25 t mA for t ≥ 0 ⎩ 200 − 80 e

⎛ 160 − 200 ⎞ ln ⎜ ⎟ 80 ⎠ = 25 1 ⇒ a= ⎝ −0.027725 s

When t < 0, the circuit is at steady state so the inductor acts like a short circuit. 24 R1 = = 200 Ω 0.12 As t → ∞, the circuit is again at steady state so the inductor acts like a short circuit. R1 || R 2 = 120 = 200 || R 2

24 = 120 Ω 0.2 ⇒ R 2 = 300 Ω

Next, the inductance can be determined using the time constant: 25 = a =

1

τ

=

R1 || R 2 L

=

120 120 ⇒ L= = 4.8 H 25 L

P8.3-25 Consider the circuit shown in Figure P8.3-25a and corresponding plot of the voltage across the 40 Ω resistor shown in Figure P8.3-25b. Determine the values of L and R 2 . Answer:

L = 8 H and R 2 = 10 Ω.

Hint: Use the plot to determine values of D, E, F and a such that the voltage can be represented as

⎧ D for t < 0 v (t ) = ⎨ − at ⎩ E + F e for t > 0

Figure P8.3-25 Solution: From the plot D = v(t) for t < 0 =20 V, E + F = v(0+) = 100 V and E = lim v ( t ) = 20 V . The t →∞

point labeled on the plot indicates that v(t) = 60 V when t = 0.14 s. Consequently

60 = 20 + 80 e − a ( 0.14)

⎛ 60 − 20 ⎞ ln ⎜ ⎟ 1 80 ⎠ ⇒ a= ⎝ =5 −0.14 s

Then ⎧20 V for t ≤ 0 v (t ) = ⎨ −5t ⎩ 20 + 80 e V for t ≥ 0 At t = 0+, i (0 +) =

100 = 2.5 A 40

When t < 0, the circuit is at steady state so the inductor acts like a short circuit. 40 || R 2 =

20 = 8 Ω ⇒ R 2 = 10 Ω 2.5

Next, the inductance can be determined using the time constant: 1 40 40 5=a= = ⇒ L= =8 H 5 τ L

Figure P8.3-26 P8.3-26 Determine vo(t) for t > 0 for the circuit shown in Figure P8.3-26. Solution:

Replace the series inductors with an equivalent inductor and label the current in the inductor:

We will determine the inductor current, i(t) ,first and then use it to determine vo(t) . Determine the initial condition, i(o), by considering the circuit when t < 0 and the circuit is at steady state. Since an inductor in a dc circuit acts like a short circuit , we have

Using current division, we have ⎛ 18 ⎞ i ( 0) = ⎜ ⎟ 2.4 = 1.2 mA ⎝ 18 + 18 ⎠ Next, consider the circuit when t > 0 and the circuit is not at steady state:

To find the Norton equivalent of the part of the circuit connected to the inductor we determine both the Thevenin resistance and the short circuit current:

and

τ=

The time constant is:

L 1.5 1 = = = 0.0556 second R t 27 18

The inductor current is given by i ( t ) = ( i ( 0 ) − i sc ) e −t /τ + i sc = (1.2 − 0.8 ) e −18t − 0.8 mA for t ≥ 0

Using KCL Finally

vo (t ) 9

+ ⎡⎣(1.2 − 0.8 ) e −18t − 0.8⎤⎦ ×10−3 = 2.4 × 10−3 v o ( t ) = 14.4 − 3.6 e −18t mV for t > 0

P8.3-27 The circuit shown in Figure P8.3-27 is at steady state before the switch closes at time t = 0. After the switch closes, the inductor current is given by i ( t ) = 0.6 − 0.2 e −5t A

for t ≥ 0

Determine the values of R1 , R 2 and L. Answers: R1 = 20 Ω , R 2 = 10 Ω and L = 5 H Figure P8.3-27 Solution:

The steady state current before the switch closes is equal to i ( 0 ) = 0.6 − 0.2 e−5( 0 ) = 0.4 A .

The inductor will act like a short circuit when this circuit is at steady state so 0.4 = i ( 0 ) =

12 R1 + R 2

⇒ R1 + R 2 = 30 Ω

After the switch has been open for a long time, the circuit will again be at steady state. The steady state inductor current will be i ( ∞ ) = 0.6 − 0.2 e−5( ∞ ) = 0.6 A

The inductor will act like a short circuit when this circuit is at steady state so 0.6 = i ( ∞ ) =

12 R1

⇒ R1 = 20 Ω

Then R 2 = 10 Ω. After the switch is closed, the Thevenin resistance of the part of the circuit connected to the inductor is R t = R1 . Then 5=

1

τ

=

Rt L

=

R1 L

=

20 ⇒ L=4H L

P8.3-28 After time t = 0, a given circuit is represented by the circuit diagram shown in FigureP8.3-28.

a.) Suppose that the inductor current is i ( t ) = 21.6 + 28.4 e −4 t mA

for t ≥ 0

Determine the values of R1 and R 3 .

Figure P8.3-28

b.) Suppose instead that R1 = 16 Ω , R 3 = 20 Ω and the initial condition is i(0) = 10 mA. Determine the inductor current for t ≥ 0. Solution: The inductor current is given by i ( t ) = i sc + ( i ( 0 ) − i sc ) e− at a. Comparing this to the given equation gives 21.6 = i sc =

4=

Rt 2

(

R1 R1 + 4

( 36 )

for t ≥ 0 where a = ⇒ R1 = 6 Ω and

)

⇒ R t = 8 Ω . Next 8 = R t = R1 + 4 || R 3 = 10 || R 3 ⇒ R 3 = 40 Ω . 10 16 = 5 s . also i sc = ( 36 ) = 28.8 mA . Then τ 2 16 + 4 = 28.8 + (10 − 28.8 ) e−5t = 28.2 − 18.8 e−5t .

b. R t = (16 + 4 ) || 20 = 10 Ω so a =

i ( t ) = i sc + ( i ( 0 ) − i sc ) e − at

1

=

1

τ

=

Rt L

.

P8.3-29 Consider the circuit shown in Figure P8.3-29.

a.) Determine the time constant, τ, and the steady state capacitor voltage, v(∞), when the switch is open. b.) Determine the time constant, τ, and the steady state capacitor voltage, v(∞), when the switch is closed. Answers: a.) τ = 3 s and v(∞) = 24 V; b.) τ = 2.25 s and v(∞) = 2 V; Figure P8.3-29 Solution: a.) When the switch is open we have

After replacing series and parallel resistors by equivalent resistors, the part of the circuit connected to the capacitor is a Thevenin equivalent circuit with R t = 33.33 Ω . The time constant is

τ = R t C = 33.33 ( 0.090 ) = 3 s . Since the input is constant, the capacitor acts like an open circuit when the circuit is at steady state. Consequently, there is zero current in the 33.33 Ω resistor and KVL gives v(∞) = 24 V. b.) When the switch is closed we have

This circuit can be redrawn as

Now we find the Thevenin equivalent of the part of the circuit connected to the capacitor:

So R t = 25 Ω and

τ = R t C = 25 ( 0.090 ) = 2.25 s Since the input is constant, the capacitor acts like an open circuit when the circuit is at steady state. Consequently, there is zero current in the 25 Ω resistor and KVL gives v(∞) = 12 V.

Section 8-4: Sequential Switching P 8.4-1 The circuit shown in Figure P 8.4-1 is at steady state before the switch closes at time t = 0. The switch remains closed for 1.5 s and then opens. Determine the capacitor voltage, v(t), for t > 0. Hint: Determine v(t) when the switch is closed. Evaluate v(t) at time t = 1.5 s to get v(1.5). Use v(1.5) as the initial condition to determine v(t) after the switch opens again. for 0 < t < 1.5 s ⎧ 5 + 5e −0.5t V Answer: v(t ) = ⎨ −2.5( t −1.5) V for 1.5 s < t ⎩10 − 2.64e

Figure P 8.4-1 Solution: Replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get:

Before the switch closes at t = 0 the circuit is at steady state so v(0) = 10 V. For 0 < t < 1.5s, voc = 5 V and Rt = 4 Ω so τ = 4 × 0.05 = 0.2 s . Therefore v (t ) = voc + (v (0) − voc ) e −t τ = 5 + 5e−5 t V for 0 < t < 1.5 s At t =1.5 s, v (1.5) = 5 + 5e ( τ = 8 × 0.05 = 0.4 s . Therefore

−0.05 1.5)

= 5 V . For 1.5 s < t, voc = 10 V and Rt = 8 Ω so

v (t ) = voc + (v (1.5) − voc ) e

−( t −1.5 ) τ

= 10 − 5 e

−2.5 ( t −1.5 )

V for 1.5 s < t

Finally ⎧⎪ 5 + 5 e−5 t V for 0 < t < 1.5 s v (t ) = ⎨ −2.5 ( t −1.5) for 1.5 s < t V ⎪⎩10 − 5 e

P 8.4-2 The circuit shown in Figure P 8.4-2 is at steady state before the switch closes at time t = 0. The switch remains closed for 1.5 s and then opens. Determine the inductor current, i(t), for t > 0. for 0 < 1 < 1.5 s ⎧ 2 + e −0.5t A Answer: v(t ) = ⎨ −0..667( t −1.5) A for 1.5 s < t ⎩3 − 0.53e

Figure P 8.4-2 Solution: Replace the part of the circuit connected to the inductor by its Norton equivalent circuit to get:

Before the switch closes at t = 0 the circuit is at steady state so i(0) = 3 A. For 0 < t < 1.5s, isc = 2 12 A and Rt = 6 Ω so τ = = 2 s . Therefore 6 i (t ) = isc + (i (0) − isc ) e−t τ = 2 + e−0.5 t A for 0 < t < 1.5 s 12 −0.5 1.5 At t =1.5 s, i (1.5) = 2 + e ( ) = 2.47 A . For 1.5 s < t, isc = 3 A and Rt = 8 Ω so τ = = 1.5 s . 8 Therefore −( t −1.5 ) τ −0.667 ( t −1.5 ) i (t ) = isc + (i (1.5) − isc ) e = 3 − 0.53 e V for 1.5 s < t Finally ⎧⎪ 2 + e−0.5 t A for 0 < t < 1.5 s i (t ) = ⎨ −0.667 ( t −1.5 ) for 1.5 s < t A ⎪⎩3 − 0.53 e

P 8.4-3 Cardiac pacemakers are used by people to maintain regular heart rhythm when they have a damaged heart. The circuit of a pacemaker can be represented as shown in Figure P 8.4-3. The resistance of the wires, R, can be neglected since R < 1 mΩ. The heart’s load resistance, RL, is 1 kΩ. The first switch is activated at t = t0, and the second switch is activated at t1 = t0 + 10 ms. This cycle is repeated every second. Find v(t) for t0 ≤ t ≤ 1. Note that it is easiest to consider t0 = 0 for this calculation. The cycle repeats by switch 1 returning to position a and switch 2 returning to its open position. Hint: Use q = Cv to determine v(0–) for the 100-μF capacitor.

Figure P 8.4-3 Solution: At t = 0-: Assume that the circuit has reached steady state so that the voltage across the 100 μF capacitor is 3 V. The charge stored by the capacitor is

q ( 0− ) = (100 × 10−6 ) ( 3) = 300 × 10−6 C 0 < t < 10ms: With R negligibly small, the circuit reaches steady state almost immediately (i.e. at t = 0+). The voltage across the parallel capacitors is determined by considering charge conservation: q ( 0+ ) = (100 μ F) v ( 0+ ) + (400 μ F) v ( 0+ ) v (0

+

) = 100 ×10

( )

q ( 0+ ) −6

+ 400 ×10−6

v 0+ = 0.6 V

10 ms < t < l s: Combine 100 μF & 400 μF in parallel to obtain v(t ) = v ( 0+ ) e − (t −.01) RC 3

= 0.6e − (t −.01) (10 ) (5 x10 v(t ) = 0.6 e−2(t −.01) V

−4 )

=

q ( 0− ) 500 × 10−6

=

300 × 10−6 500 × 10−6

P 8.4-4 An electronic flash on a camera uses the circuit shown in Figure P 8.4-4. Harold E. Edgerton invented the electronic flash in 1930. A capacitor builds a steady-state voltage and then discharges it as the shutter switch is pressed. The discharge produces a very brief light discharge. Determine the elapsed time t1 to reduce the capacitor voltage to one-half of its initial voltage. Find the current, i(t), at t = t1. Solution:

Figure P 8.4-4

v ( 0 ) = 5 V , v ( ∞ ) = 0 and τ = 105 ×10−6 = 0.1 s

∴ v ( t ) = 5 e −10 t V for t > 0

2.5 = 5 e −10 t1 i (t1 ) =

v (t1 ) 100 ×10

3

t 1 = 0.0693 s

=

2.5 = 25 μ A 100 × 103

P 8.4-5 The circuit shown in Figure P 8.4-5 is at steady state before the switch opens at t = 0. The switch remains open for 0.5 second and then closes. Determine v(t) for t ≥ 0.

Figure P 8.4-5 Solution: The circuit is at steady state before the switch closes. The capacitor acts like an open circuit. The initial condition is ⎛ 40 ⎞ v (0 +) = v (0 −) = ⎜ ⎟ 24 = 12 V ⎝ 40 + 40 ⎠

After the switch closes, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit.

R t = 6.67 Ω and v oc = 4 V

Recognize that The time constant is

τ = R t C = ( 6.67 )( 0.05 ) = 0.335 s



1

τ

= 2.988  3

1 s

The capacitor voltage is

v ( t ) = ( v ( 0 + ) − v oc ) e−t τ + v oc = (12 − 4 ) e−3t + 4 = 4 + 8e−3t V for 0 ≥ t ≥ 0.5 s When the switch opens again at time t = 0.5 the capacitor voltage is −3 0.5 v ( 0.5 + ) = v ( 0.5 − ) = 4 + 8e ( ) = 5.785 V

After time t = 0.5 s, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit.

Recognize that

R t = 20 Ω and v oc = 12 V

The time constant is

τ = R t C = 20 ( 0.05 ) = 1

1



τ

=1

1 s

The capacitor voltage is − t −0.5 τ −10 t − 0.5 v ( t ) = ( v ( 0.5 + ) − v oc ) e ( ) + v oc = ( 5.785 − 12 ) e ( ) + 12

= 12 − 6.215e

−10( t − 0.5)

V for t ≥ 0.5 s

so ⎧ 12 V ⎪ v (t ) = ⎨ 4 + 8 e −3t V ⎪ −( t − 0.5) V ⎩12 − 6.215e

for t ≥ 0 for 0 ≤ t ≤ 0.5 s for t ≥ 0.5 s

Section 8.5 Stability of First-Order Circuits P 8.5-1 The circuit in Figure P 8.5-1 contains a current-controlled voltage source. What restriction must be placed on the gain, R, of this dependent source in order to guarantee stability? Answer: R < 400 Ω

Figure P 8.5-1 Solution: This circuit will be stable if the Thèvenin equivalent resistance of the circuit connected to the inductor is positive. The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as

100 ⎫ v T (400− R) 100 iT ⎪ = 100+ 400 ⎬ ⇒ Rt = iT 100+ 400 ⎪ vT = 400 i (t ) − R i (t ) ⎭

i (t ) =

The circuit is stable when R < 400 Ω.

P 8.5-2 The circuit in Figure P 8.5-2 contains a current-controlled current source. What restriction must be placed on the gain, B, of this dependent source in order to guarantee stability?

Figure P 8.5-2 Solution: The Thèvenin equivalent resistance of the circuit connected to the inductor is calculated as Ohm’s law: i ( t ) = −

vT ( t ) 6000

KCL: i ( t ) + B i ( t ) + i T ( t ) =

vT ( t ) 3000

⎛ v ( t ) ⎞ vT ( t ) ∴ i T ( t ) = − ( B + 1) ⎜ − T ⎟+ ⎝ 6000 ⎠ 3000 ( B + 3) vT ( t ) = 6000

Rt =

vT ( t ) 6000 = iT ( t ) B + 3

The circuit is stable when B > −3 A/A.

Section 8.6 The Unit Step Source P 8.6-1

The input to the circuit shown in Figure P 8.6-1

is the voltage of the voltage source, vs(t). The output is the voltage across the capacitor, vo(t). Determine the output of this circuit when the input is vs(t) = 8 – 15 u(t) V. Figure P 8.6-1 Solution: The value of the input is one constant, 8 V, before time t = 0 and a different constant, −7 V, after time t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t

for t > 0

where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuit before and after the input changes value. Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = 0, will be equal to the steady state capacitor voltage before the input changes. At time t = 0 the output voltage is The steady-state circuit for t < 0.

vo ( 0 ) = A + B e

− a ( 0)

= A+ B

Consequently, the capacitor voltage is labeled as A + B. Analysis of the circuit gives A+ B = 8 V Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit.

The steady-state circuit for t > 0.

The value of the capacitor voltage at time t = ∞, will be equal to the steady state capacitor voltage after the input changes. At time t = ∞ the output voltage is vo ( ∞ ) = A + B e

− a (∞)

=A

Consequently, the capacitor voltage is labeled as A. Analysis of the circuit gives Therefore

A = -7 V and B = 15 V

The value of the constant a is determined from the time constant, τ, which is in turn calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit connected to the capacitor. 1 =τ = Rt C a

Here is the circuit used to calculate Rt. Rt = 6 Ω

Therefore

a=

1 1 = 2.5 −3 s ( 6 ) ( 66.7 ×10 )

(The time constant is τ = ( 6 ) ( 66.7 × 10−3 ) = 0.4 s .) Putting it all together: 8 V for t ≤ 0 ⎧ vo ( t ) = ⎨ − 2.5 t V for t ≥ 0 ⎩−7 + 15 e

P 8.6-2 The input to the circuit shown in Figure P 8.6-2 is the voltage of the voltage source, vs(t). The output is the voltage across the capacitor, vo(t). Determine the output of this circuit when the input is vs(t) = 3 + 3 u(t) V. Figure P 8.6-2 Solution: The value of the input is one constant, 3 V, before time t = 0 and a different constant, 6 V, after time t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t

for t > 0

where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuit before and after the input changes value. Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = 0, will be equal to the steady state capacitor voltage before the input changes. At time t = 0 the output voltage is The steady-state circuit for t < 0.

vo ( 0 ) = A + B e

− a ( 0)

= A+ B

Consequently, the capacitor voltage is labeled as A + B. Analysis of the circuit gives

A+ B =

6 ( 3) = 2 V 3+ 6

Capacitors act like open circuits when the input is constant and the circuit is at steady state. Consequently, the capacitor is replaced by an open circuit. The value of the capacitor voltage at time t = ∞, will be equal to the steady state capacitor voltage after the input changes. At time t = ∞ the output voltage is The steady-state circuit for t > 0.

vo ( ∞ ) = A + B e

− a (∞)

=A

Consequently, the capacitor voltage is labeled as A. Analysis of the circuit gives A=

6 ( 6) = 4 V 3+ 6

Therefore

B = −2 V

The value of the constant a is determined from the time constant, τ, which is in turn calculated from the values of the capacitance C and of the Thevenin resistance, Rt, of the circuit connected to the capacitor. 1 =τ = Rt C a Here is the circuit used to calculate Rt. Rt =

( 3)( 6 ) = 2 3+ 6

Ω

Therefore a=

1

( 2 )(.5)

=1

1 s

(The time constant is τ = ( 2 )( 0.5 ) = 1 s .) Putting it all together: for t ≤ 0 ⎧ 2 V vo ( t ) = ⎨ − t ⎩4 − 2 e V for t ≥ 0

P 8.6-3 The input to the circuit shown in Figure P 8.6-3 is the voltage of the voltage source, vs(t). The output is the current across the inductor, io(t). Determine the output of this circuit when the input is vs(t) = – 7 + 13 u(t) V. Figure P 8.6-3 Solution: The value of the input is one constant, −7 V, before time t = 0 and a different constant, 6 V, after time t = 0. The response of the first order circuit to the change in the value of the input will be vo ( t ) = A + B e − a t

for t > 0

where the values of the three constants A, B and a are to be determined. The values of A and B are determined from the steady state responses of this circuit before and after the input changes value. Inductors act like short circuits when the input is constant and the circuit is at steady state. Consequently, the inductor is replaced by a short circuit. The value of the inductor current at time t = 0, will be equal to the steady state inductor current before the input changes. At time t = 0 the output current is The steady-state circuit for t < 0.

io ( 0 ) = A + B e

− a ( 0)

= A+ B

Consequently, the inductor current is labeled as A + B. Analysis of the circuit gives A+ B =

−7 = −1.4 A 5

Inductors act like short circuits when the input is constant and the circuit is at steady state. Consequently, the inductor is replaced by a short circuit.

The steady-state circuit for t > 0.

The value of the inductor current at time t = ∞, will be equal to the steady state inductor current after the input changes. At time t = ∞ the output current is io ( ∞ ) = A + B e

− a (∞)

=A

Consequently, the inductor current is labeled as A. Analysis of the circuit gives

A=

6 = 1.2 A 5

Therefore B = −2.6 V The value of the constant a is determined from the time constant, τ, which is in turn calculated from the values of the inductance L and of the Thevenin resistance, Rt, of the circuit connected to the inductor. 1 L =τ = a Rt Here is the circuit used to calculate Rt. Rt =

( 5) ( 4 ) = 2.22

5+ 4 2.22 1 a= = 1.85 Therefore 1.2 s 1.2 (The time constant is τ = = 0.54 s .) 2.22 Putting it all together:

−1.4 A for t ≤ 0 ⎧ io ( t ) = ⎨ − 1.85 t A for t ≥ 0 ⎩1.2 − 2.6 e

Ω

P8.6-4 Determine vo(t) for t > 0 for the circuit shown in Figure P8.6-4.

Figure P8.6-4 Solution: Determine the initial condition, vo(o), by considering the circuit when t < 0 and the circuit is at steady state. Since a capacitor in a dc circuit acts like a open circuit , we have

Recognizing the inverting amplifier, we have ⎛ 45 ⎞ v o ( 0 ) = ⎜ − ⎟ 2.4 = −3.6 V ⎝ 30 ⎠ Next, consider the circuit when t > 0 and the circuit is not at steady state:

To find the Thevenin equivalent of the part of the circuit connected to the capacitor we determine both the open circuit voltage and the short circuit current:

and

Now we calculate the Thevenin resistance: v oc −5.4 Rt = = = 20 kΩ i sc −0.27 × 10−3 and the time constant:

τ = R t C = ( 20 ×103 )( 5 × 10−6 ) = 0.1 second

The capacitor voltage is given by

v o ( t ) = ( v o ( 0 ) − v oc ) e −t /τ + v oc = ( −3.6 − ( −5.4 ) ) e −10 t − 5.4 = 1.8 e −10 t − 5.4 V for t ≥ 0

P 8.6-5 The initial voltage of the capacitor of the circuit shown in Figure P 8.6-5 is zero. Determine the voltage v(t) when the source is a pulse, described by t 2s ⎩

Solution

Figure P 8.6-5

τ = R C = ( 5 × 105 )( 2 × 10−6 ) = 1 s

Assume that the circuit is at steady state at t = 1−. Then v ( t ) = 4 − 4 e − ( t −1) V for 1 ≤ t ≤ 2

so

v ( 2 ) = 4 − 4 e − (2−1) = 2.53 V

and

v ( t ) = 2.53 e − ( t − 2) V for t ≥ 2

Finally

⎧0 ⎪ v(t ) = ⎨4− 4e − (t −1) ⎪2.53e − (t − 2) ⎩

t ≤1 1≤t ≤ 2 t ≥2

P 8.6-6 Studies of an artificial insect are being used to understand the nervous system of animals. A model neuron in the nervous system of the artificial insect is shown in Figure P 8.6-6. A series of pulses, called synapses, is required. The switch generates a pulse by opening at t = 0 and closing at t = 0.5 s. Assume that the circuit is in steady state and that v(0–) = 10 V. Determine the voltage v(t) for 0 < t < 2 s.

Figure P 8.6-6 Solution: The capacitor voltage is v(0−) = 10 V immediately before the switch opens at t = 0.

For 0 < t < 0.5 s the switch is open: v ( 0 ) = 10 V, v ( ∞ ) = 0 V, τ = 3 ×

1 1 = s 6 2

so v ( t ) = 10 e − 2 t V In particular, v ( 0.5 ) = 10 e For t > 0.5 s the switch is closed:

− 2 ( 0.5)

= 3.679 V

v ( 0 ) = 3.679 V, v ( ∞ ) = 10 V, Rt = 6 || 3 = 2 Ω,

1 6

τ = 2× = so

1 s 3

v ( t ) = 10 + ( 3.679 − 10 ) e = 10 − 6.321 e

− 3 ( t − 0.5 )

− 3 ( t − 0.5)

V

V

P8.6-7 Determine the voltage v o ( t ) in the

circuit shown in Figure P8.6-7.

Figure P8.6-7 Solution: This is a first order circuit containing an inductor. First, determine i L ( t ) . Consider the circuit for time t < 0. Step 1: Determine the initial inductor current.

The circuit will be at steady state before the source voltage changes abruptly at time t = 0 . The source voltage will be 2 V, a constant. The inductor will act like a short circuit. i L ( 0) =

2 2 = = 0.25 A 10 || ( 25 + 15 ) 8

Consider the circuit for time t > 0. Step 2. The circuit will not be at steady state immediately after the source voltage changes abruptly at time t = 0 . Determine the Norton equivalent circuit for the part of the circuit connected to the inductor.

Replacing the resistors by an equivalent resistor, we recognize

v oc = −6 V and R t = 8 Ω Consequently i sc =

−6 = −0.75 A 8

t < 0 , at steady state:

Step 3. The time constant of a first order circuit containing an inductor is given by

τ= Consequently

τ=

L Rt

L 4 1 1 = = 0.5 s and a = = 2 Rt 8 s τ

Step 4. The inductor current is given by:

i L ( t ) = i sc + ( i ( 0 ) − i sc ) e − at = −0.75 + ( 0.25 − ( −0.75 ) ) e −2 t = −0.75 + e −2 t for t ≥ 0 Step 5. Express the output voltage as a function of the source voltage and the inductor current.

Using current division: iR =

10 i L = 0.2 i L 10 + ( 25 + 15 )

Then Ohm’s law gives v o = 15 i R = 3 i L Step 6. The output voltage is given by v o ( t ) = −2.25 + 3 e −2 t for t ≥ 0

P 8.6-8

Determine vc(t) for t > 0 for the circuit of Figure P 8.6-8.

Figure P 8.6-8 Solution: For t < 0, the circuit is:

After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get: vc ( t ) = −15 + ( −6 − ( −15 ) ) e = −15 + 9 e − 5 t V

− t / ( 4000×0.00005 )

P 8.6-9 The voltage source voltage in the circuit shown in Figure P 8.6-9 is

vs(t) = 7 – 14u (t) V Determine v(t) for t > 0.

Figure P 8.6-9 Solution:

The input changes abruptly at time t = 0. The voltage v(t) may not be continuous at t = 0, but the capacitor voltage, vC(t) will be continuous. We will find vC(t) first and then use KVL to find v(t).

The circuit will be at steady state before t = 0 so the capacitor will act like an open circuit. v (0 +) = v (0 −) =

5 7 = 4.375 V 5+3

After t = 0, we replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit. 5 5 v oc = ( 7 − 14 ) = ( −7 ) = −4.375 V 8 8 R t = 5 & 3 = 1.875 Ω The time constant is

τ = R t C = 0.8625 s 1

So

v C ( t ) = ⎡⎣ 4.375 − ( −4.375 ) ⎤⎦ e −1.16t

1 s τ + ( −4.375 ) = −4.375 + 8.75e−1.16t V for t ≥ 0 = 1.16

Using KVL

v ( t ) = v s ( t ) − vC ( t ) = −7 − ⎡⎣ −4.375 + 8.75e −1.16t ⎤⎦ = −2.625 − 8.75e−1.16t V for t > 0

P 8.6-10 Determine the voltage v(t) for t ≥ 0 for the circuit shown in Figure P 8.6-10.

Figure P 8.6-10 Solution: For t < 0

Using voltage division twice v (t ) = so

32 30 5− 5 = 0.25 V 32 + 96 120 + 30 v ( 0 − ) = 0.25 V

and

v ( 0 + ) = v ( 0 − ) = 0.25 V

For t > 0, find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. Using voltage division twice v oc =

32 30 20 − 20 = 5 − 4 = 1 V 32 + 96 120 + 30

R t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48 Ω

then

τ = 48 × 0.0125 = 0.6 s

so 1

τ

Now

= 1.67

1 s

v ( t ) = [ 0.25 − 1] e −1.67t + 1 = 1 − 0.75e −1.67t V for t ≥ 0

(checked: LNAP 7/1/04)

P 8.6-11

The voltage source voltage in the circuit shown in Figure P 8.6-11 is vs(t) = 5 + 20u (t)

Determine i(t) for t ≥ 0.

Figure P 8.6-11 Solution: For t > 0 the circuit is at steady state so the inductor acts like a short circuit:

Apply KVL to the supermesh corresponding to the dependent source to get −5000i b + 1000 ( 3i b ) − 5 = 0



i b = 0.2 mA

Apply KVL to get i ( t ) = 3i b = 0.6 mA i ( 0 − ) = 0.6 mA

so

i ( 0 + ) = i ( 0 − ) = 0.6 mA and For t > 0, find the Norton equivalent circuit for the part of the circuit that is connected to the inductor. Apply KCL at the top node of the dependent source to see that i b = 0 A . Then

( )

v oc = 25 − 5000 i b = 25 V

Apply KVL to the supermesh corresponding to the dependent source to get

( )

−5000 i b + 10000 3 i b − 25 = 0 ⇒ i b = 1 mA Apply KCL to get i sc = 3 i b = 3 mA Then

Rt =

v oc i sc

= 8.33 kΩ

τ=

Then So Now

25 = 3 ms 8333 1 1 = 333 τ s

i ( t ) = [ 0.6 − 3] e −333t + 3 = 3 − 2.4e −333t mA for t ≥ 0

(checked: LNAP 7/2/04)

P 8.6-12

The voltage source voltage in the

circuit shown in Figure P 8.6-12 is vs(t) = 12 – 6u (t) V Determine v(t) for t ≥ 0.

Figure P 8.6-12 Solution: For t > 0, the circuit is at steady state so the capacitor acts like an open circuit. We have the following situation.

Notice that v(t) is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: va = −v(t) Apply KCL at node a: ⎛ 12 − v ( t ) ⎞ v ( t ) ⎛ 3 ⎞ −⎜ + ⎜ − v (t ) ⎟ = 0 ⎟+ 8 4 ⎝ 4 ⎠ ⎝ ⎠ −12 + v ( t ) + 2 v ( t ) − 6 v ( t ) = 0 ⇒ v ( t ) = −4 V

So

v ( 0 + ) = v ( 0 − ) = −4 V

For t > 0, we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor, i.e. the part of the circuit to the left of terminals a – b.

Notice that voc is the node voltage at node a. Express the controlling voltage of the dependent source as a function of the node voltage: va = −voc Apply KCL at node a: ⎛ 6 − voc ⎞ voc ⎛ 3 ⎞ −⎜ + ⎜ − voc ⎟ = 0 ⎟+ ⎝ 8 ⎠ 4 ⎝ 4 ⎠ −6 + voc + 2 voc − 6 voc = 0 ⇒ voc = −2 V Find Rt: We’ll find isc and use it to calculate Rt. Notice that the short circuit forces va = 0 Apply KCL at node a: ⎛ 6−0⎞ 0 ⎛ 3 ⎞ −⎜ ⎟ + + ⎜ − 0 ⎟ + i sc = 0 ⎝ 8 ⎠ 4 ⎝ 4 ⎠ i sc = Rt =

6 3 = A 8 4

voc −2 8 = =− Ω i sc 3 4 3

Then ⎛ 8 ⎞⎛ 3 ⎞

1

τ = R tC = ⎜ − ⎟ ⎜ ⎟ = − s 5 ⎝ 3 ⎠ ⎝ 40 ⎠ and



1

τ

= −5

1 s

v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = ( −4 − ( −2 ) ) e5t + ( −2 ) = −2 (1 + e5t ) V for t ≥ 0

Notice that v(t) grows exponentially as t increases. (checked: LNAP 7/8/04)

P 8.6-13

Determine i(t) for t ≥ 0 for the circuit shown in Figure P 8.6-13.

Figure P 8.6-13 Solution: When t < 0 and the circuit is at steady state, the inductor acts like a short circuit.

The mesh equations are

(

)

(

)

2 i x + 4 i x + i (0 −) + 3i x + 6 i x + i (0 −) = 0 1i ( 0 − ) − 3 i x = 0 so

i (0 +) = i (0 −) = 0

For t ≥ 0 , we find the Norton equivalent circuit for the part of the circuit connected to the inductor. First, simplify the circuit using a source transformation:

Identify the open circuit voltage and short circuit current.

Apply KVL to the mesh to get:

(10 + 2 + 3) i x − 15 = 0

⇒ ix = 1 A

Then v oc = 3 i x = 3 V

Express the controlling current of the CCVS in terms of the mesh currents: i x = i1 − i sc The mesh equations are 10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15 and i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 =

4 i sc 3

so ⎛4 ⎞ 15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A ⎝3 ⎠ The Thevenin resistance is Rt = The time constant is given by τ =

3 =3Ω 1

L 5 1 1 = = 1.67 s so = 0.6 . Rt 3 τ s

The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( 0 − 1) e−0.6 t + 1 = 1 − e −0.6t A for t ≥ 0

P 8.6-14

Determine i(t) for t ≥ 0 for the circuit shown in Figure P 8.6-14.

Figure P 8.6-14 Solution: When t < 0 and the circuit is at steady state, the inductor acts like a short circuit. The initial condition is 0 2 i (0 +) = i (0 −) = + = 0.02 A 150 100

For t ≥ 0 , we find the Norton equivalent circuit for the part of the circuit connected to the inductor. First, simplify the circuit using source transformations: i sc = 20 + 40 = 60 mA

R t = 100 ||150 = The time constant is given by τ =

100 × 150 = 60 Ω 100 + 150

L 2 1 1 = = 0.0333 s so = 30 . R t 60 τ s

The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( 20 − 60 ) e −30 t + 60 = 60 − 40 e −30 t mA for t ≥ 0 (checked: LNAP 7/15/04)

P 8.6-15 Determine v(t) for t ≥ 0 for the circuit shown in Figure P 8.6-15.

Figure P 8.6-15 Solution:

When t < 0 and the circuit is at steady state, the capacitor acts like an open circuit. The 0 A current source also acts like an open circuit. The initial condition is v (0 + ) = v (0 −) = 0 V

For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. 30 ⎡ 170 voc = ⎢ ( 20 ) ⎤⎥ 10 − ⎡⎢ ( 20 ) ⎤⎥ 50 ⎣170 + 30 ⎦ ⎣170 + 30 ⎦ =

170(20)(10) − 30(20)(50) 4000 = = 20 V 200 200

Rt = 8 +

( 20 + 120 ) (10 + 50 ) = 50 Ω ( 20 + 120 ) + (10 + 50 )

The time constant is τ = R t C = ( 50 ) (10−3 ) = 0.05 s s so

1

τ

= 20

1 . s

The capacitor voltage is given by

(

v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = ( 0 − 20 ) e −20t + 20 = 20 1 − e−20t

)

V for t ≥ 0

P 8.6-16

Determine v(t) for t ≥ 0 for the circuit shown in Figure P 8.6-16.

Figure P 8.6-16

Solution: When t < 0 and the circuit is at steady state, the capacitor acts like an open circuit. The 0 A current source also acts like an open circuit.

After a couple of source transformations, the initial condition is calculated as v (0 +) = v (0 −) =

18 16 = 10.667 V 9 + 18

For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. Using source transformations, reduce the circuit as follows.

Now recognize R t = 10 Ω and v oc = 4 V . The time constant is τ = R t C = (10 ) ( 20 ×10−3 ) = 0.2 s s so

1

τ

=5

1 . s

The capacitor voltage is given by v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = (10.667 − 4 ) e−5t + 4 = 4 + 6.667 e −5t V for t ≥ 0 (checked: LNAP 7/15/04)

P 8.6-17

Determine i(t) for t ≥ 0 for the circuit shown in Figure P 8.6-17.

Figure P 8.6-17 Solution:

When t < 0 and the circuit is at steady state, the inductor acts like a short circuit. The 0 V voltage source also acts like a short circuit. After a replacing series and parallel resistors by equivalent resistors, the equivalent resistors, current source and short circuit are all connected in parallel. Consequently i (0 +) = i (0 −) = 2 A

For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. Replace series and parallel resistors by an equivalent resistor. 18 & (12 + 24 ) = 12 Ω

Do a source transformation, then replace series voltage sources by an equivalent voltage source.

Do two more source transformations. Now recognize R t = 8 Ω and i sc = 3 A . The time constant is given by

τ=

L 2 1 1 = = 0.25 s so = 4 . τ Rt 8 s

The inductor current is given by i L ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 2 − 3) e −4t + 3 = 3 − e −4t A for t ≥ 0 (checked: LNAP 7/15/04)

P 8.6-18

The voltage source voltage in the circuit shown in Figure P 8.6-18 is vs(t) = 8 + 12u(t) V

Determine v(t) for t ≥ 0.

Figure P 8.6-18 Solution: Simplify the circuit by replacing the parallel capacitors by an equivalent capacitor and by doing a couple of source transformations:

For t < 0 the circuit is at steady state so the capacitor acts like an open circuit. The voltage source voltage is 6.4 V, so v ( 0 + ) = v ( 0 − ) = 6.4 V

For t > 0 we find the Thevenin equivalent circuit of the part of the circuit connected to the capacitor. In this case we recognize the voc = 16 V and Rt = 50 Ω. The time constant is

τ = R t C = ( 50 ) ( 8 × 10−3 ) = 0.4 s



1

τ

= 2.5

1 s

Then −t

v ( t ) = ( v ( 0 + ) − v oc ) e τ + v oc = ( 6.4 − 16 ) e −2.5t + 16 = 16 − 9.6e −2.5t V

for t ≥ 0

(checked: LNAP 7/12/04)

P8.6-19 Determine the current i o ( t ) in the

circuit shown in Figure P8.6-19.

Figure P8.6-19 Solution: This is a first order circuit containing a capacitor. First, determine v C ( t ) . Consider the circuit for time t < 0. Step 1: Determine the initial capacitor voltage.

t < 0 , at steady state:

The circuit will be at steady state before the source voltage changes abruptly at time t = 0 . The source voltage will be 5 V, a constant. The capacitor will act like an open circuit. Apply KVL to the mesh to get:

(10 + 2 + 3) i x − 5 = 0 Then

⇒ ix =

1 A 3

v C (0) = 3 i x = 1 V

Consider the circuit for time t > 0. Step 2. The circuit will not be at steady state immediately after the source voltage changes abruptly at time t = 0 . Determine the Thevenin equivalent circuit for the part of the circuit connected to the capacitor. First, determine the open circuit voltage, v oc :

Apply KVL to the mesh to get:

(10 + 2 + 3) i x − 15 = 0

⇒ ix = 1 A

Then v oc = 3 i x = 3 V Next, determine the short circuit current, i sc :

Express the controlling current of the CCVS in terms of the mesh currents: i x = i1 − i sc

The mesh equations are 10 i1 + 2 ( i1 − i sc ) + 3 ( i1 − i sc ) − 15 = 0 ⇒ 15 i1 − 5 i sc = 15 i sc − 3 ( i1 − i sc ) = 0 ⇒ i1 =

And

4 i sc 3

⎛4 ⎞ 15 ⎜ i sc ⎟ − 5 i sc = 15 ⇒ i sc = 1 A ⎝3 ⎠

so The Thevenin resistance is

Rt =

3 =3Ω 1

Step 3. The time constant of a first order circuit containing an capacitor is given by

τ = Rt C Consequently 1 1 ⎛1⎞ ⎟ = 0.25 s and a = = 4 τ s ⎝ 12 ⎠

τ = Rt C = 3⎜

Step 4. The capacitor voltage is given by:

(

)

v C ( t ) = v oc + v C ( 0 ) − v oc e − at = 3 + (1 − 3) e−4 t = 3 − 2 e−4 t for t ≥ 0 Step 5. Express the output current as a function of the source voltage and the capacitor voltage.

io (t ) = C

d 1 d v C (t ) = vC (t ) dt 12 dt

Step 6. The output current is given by

io (t ) =

1 d 1 2 3 − 2 e −4 t = ( −2 )( −4 ) e −4 t = e −4 t for t ≥ 0 12 dt 12 3

(

)

P 8.6-20

The voltage source voltage in the circuit shown in Figure P 8.6-20 is vs(t) = 25u(t) – 10 V

Determine i(t) for t ≥ 0.

Figure P 8.6-20 Solution: Simplify the circuit by replacing the series inductors by an equivalent inductor. Then, after a couple of source transformations, we have

For t < 0 the circuit is at steady state and so the inductor acts like a short circuit. The voltage source voltage is −6 V so i ( 0 + ) = i ( 0 − ) = −60 mA

For t > 0 we find the Norton equivalent circuit for the part of the circuit connected to the inductor. In this case we recognize voc = 9V and Rt = 100 Ω so isc = 90 mA. The time constant is Then

τ=

L 20 = = 0.2 s R t 100



1

τ

=5

1 s

i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( −60 − 90 ) e −5t + 90 = 90 − 150e−5t mA for t ≥ 0 (checked: LNAP 7/12/04)

P 8.6-21

The voltage source voltage in the circuit shown in Figure P 8.6-21 is vs(t) = 30 – 24u(t) V

Determine i(t) for t ≥ 0.

Figure P 8.6-21 Solution: Simplify the circuit by replacing the parallel inductors by an equivalent inductor. Then, after doing a couple of source transformations, we have

For t < 0 the circuit is at steady state and the inductor acts like a short circuit. The voltage source voltage is 12 V so i ( 0 + ) = i ( 0 − ) = 0.2 A

For t > 0 we find the Norton equivalent circuit for the part of the circuit connected to the inductor. In this case, we recognize voc = 2.4 V and Rt = 60 Ω so isc = 0.04 A. The time constant is

τ= Then

L 4 1 = = s R t 60 15



1

τ

= 15

1 s

i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = ( 0.2 − 0.04 ) e−15t + 0.04 = 40 + 160e −15t mA (checked: LNAP 7/13/04)

P 8.6-22

The voltage source voltage in the circuit shown in Figure P 8.6-22 is vs(t) = 10 + 40u(t) V

Determine v(t) for t ≥ 0.

Figure P 8.6-22 Solution: Simplify the circuit by replacing the series capacitors by an equivalent capacitor. Then, after doing some source transformations, we have

For t < 0 the circuit is at steady state so the capacitor acts like an open circuit. The voltage source voltage is 8 V so v (0 +) = v (0 −) = 8 V

For t > 0 we find the Thevenin equivalent circuit of the part of the circuit connected to the capacitor. In this case we recognize voc = 40 V and Rt = 8 Ω. The time constant is

τ = R t C = ( 8 ) ( 60 ×10−3 ) = 0.48 Then



1

τ

= 2.08

1 s

v ( t ) = ( v ( 0 + ) − v oc ) e −t τ + v oc = ( 8 − 40 ) e −2.08t + 40 = 40 − 32e−2.08t V for t ≥ 0 (checked: LNAP 7/13/04)

P 8.6-23

Determine v(t) for t > 0 for the circuit shown in Figure P 8.6-23.

Figure P 8.6-23

Solution: The resistor voltage, v(t), may not be continuous at time t = 0. The inductor will be continuous. We will find the inductor current first and then find v(t). Label the inductor current as i(t).

For t < 0 the circuit is at steady state and the inductor acts like a short circuit. The initial condition is i (0 +) = i (0 −) = 0 A

For t > 0 use source transformations to simplify the part of the circuit connected to the inductor until is a Norton equivalent circuit. Recognize that R t = 2 Ω and i sc = 1.333 A

The time constant is

τ=

L 3 = Rt 2



1

τ

= 0.667

1 s

Then

(

i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = 1.333 1 − e−0.667t

)

A for t ≥ 0

Returning to the original circuit we see that d 3 i (t ) v (t ) d = i ( t ) + dt = i (t ) + i (t ) dt 2 3 = 1.333 (1 − e −0.667 t ) + ( −0.667 )(1.333) ( −e −0.667 t ) = 1.333 − 0.4439e−0.667 Finally

v ( t ) = 2.667 − 0.889e0.667 t V for t > 0

(checked: LNAP 7/14/04)

P 8.6-24

The input to the circuit shown in Figure P 8.6-24 is the current source current is(t) = 2 + 4u(t) A

The output is the voltage v(t). Determine v(t) for t > 0.

Figure P 8.6-24 Solution: Label the inductor current, i(t). We will find i(t) first, then find v(t).

For t < 0 the circuit is at steady state and the inductor acts like a short circuit. The initial condition is ⎛ 3 ⎞ i (0 +) = i (0 −) = ⎜ ⎟2 =1 A ⎝ 3+3⎠ For t > 0 use source transformations to simplify the part of the circuit connected to the inductor until it is a Norton equivalent circuit. Recognize that R t = 2 Ω and i sc = 3 A The time constant is Then

τ=

L 0.25 = = 0.125 s Rt 2



1

τ

=8

1 s

i ( t ) = ( i ( 0 + ) − i sc ) e−t τ + i sc = (1 − 3) e−8t + 3 = 3 − 2e −8t A for t ≥ 0

Returning to the original circuit d ⎛ ⎞ 0.25 i ( t ) ⎟ ⎜ d d −8t −8t dt v (t ) = 3⎜ i (t ) + ⎟ + 0.25 i ( t ) = 3i ( t ) + 0.5 i ( t ) = 3 ( 3 − 2e ) + 0.5 (16e ) 3 dt dt ⎜⎜ ⎟⎟ ⎝ ⎠ −8t = 9 + 2 e V for t > 0 (checked: LNAP 7/26/04)

P 8.6-25

The input to the circuit shown in Figure P 8.6-25 is the voltage source voltage vs = 6 + 6u (t)

The output is the voltage vo(t). Determine vo(t) for t > 0.

Figure P 8.6-25 Solution: Label the capacitor voltage, v(t). W will find v(t) first then find vo(t).

For t < 0 the circuit is at steady state and the capacitor acts like an open circuit. The initial condition is v (0 +) = v (0 −) = 6 V

For t > 0 we replace series and then parallel resistors by equivalent resistors in order to replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit. We recognize R t = 4 Ω and v oc = 12 V The time constant is

τ = R t C = 4 ( 0.125 ) = 0.5 s



1

τ

=2

1 s

The capacitor voltage is given by −t

v ( t ) = ( v ( 0 + ) − v oc ) e τ + v oc = ( 6 − 12 ) e −2t + 12 = 12 − 6e−2t V for t ≥ 0

Returning to the original circuit and applying KCL we see 0.125 so

v o ( t ) = 0.25

12 − v ( t ) v o ( t ) d v (t ) = + 6 2 dt

v (t ) d = 0.25 (12e −2t ) − 4 + 4 − 2e−2t = e −2t V for t > 0 v (t ) − 4 + 3 dt (checked: LNAP 7/26/04)

P 8.6-26 Determine v(t) for t > 0 for the circuit shown in Figure P 8.6-26.

Figure P 8.6-26

Solution: Label the inductor current as i(t). We will find i(t) first then use it to find v(t).

For t < 0 the circuit is at steady state and the inductor acts like a short circuit. The initial condition is i (0 +) = i (0 −) = 0 A

For t > 0 we replace series and parallel resistors by equivalent resistors. Then the part of the circuit connected to the inductor will be a Thevenin equivalent circuit. We recognize R t = 2 Ω and v oc = 12 V so

i sc =

v oc Rt

=6A

The time constant is

τ=

L 0.5 = = 0.25 Rt 2



1

τ

=4

1 s

The inductor current is given by

(

i ( t ) = ( i ( 0 + ) − i sc ) e −t τ + i sc = ( 0 − 6 ) e−4t + 6 = 6 1 − e−4t

)

A for t ≥ 0

Returning to the original circuit and applying KCL we see

i (t ) +

0.5

d i ( t ) − 12 v t () dt = 3 4

so v ( t ) = 4i ( t ) +

2d ⎛ 2⎞ i ( t ) − 16 = 24 (1 − e −4t ) + ⎜ ⎟ ( 24e −4t ) − 16 = 8 − 8e −4t V for t > 0 3 dt ⎝ 3⎠ (checked: LNAP 7/26/04)

P 8.6-27 When the input to the circuit shown in Figure P 8.6-27 is the voltage source voltage

vs(t) = 3 – u(t) V the output is the voltage vo(t) = 10 + 5 e–50t V for t ≥ 0 Determine the values of R1 and R2.

Figure P 8.6-27 Solution: Apply KCL at the inverting input of the op amp to get

v o (t ) − v (t ) R2

=

R2 ⎞ ⎛ v (t ) ⇒ v o ( t ) = ⎜1 + ⎟ v (t ) 1000 ⎝ 1000 ⎠

We will determine the capacitor voltage first and then use it to determine the output voltage. When t < 0 and the circuit is at steady state, the capacitor acts like an open circuit. Apply KCL at the noninverting input of the op amp to get 3 − v (0 −) = 0 ⇒ v (0 −) = 3 V R1 The initial condition is v (0 +) = v (0 −) = 3 V

For t ≥ 0 , we find the Thevenin equivalent circuit for the part of the circuit connected to the capacitor.

2 − voc = 0 ⇒ voc = 2 V R1

2 = i sc R1

⇒ Rt =

voc = R1 i sc

The time constant is τ = R t C = R t (10−6 ) . From the given equation for v o ( t ) , R t (10−6 ) =

1

τ

= 50

1 , so s

6

1 10 ⇒ R1 = R t = = 20 kΩ 50 50

The capacitor voltage is given by v ( t ) = ( v ( 0 ) − v oc ) e −t τ + v oc = ( 3 − 2 ) e −50t + 2 = 2 + e−50t V for t ≥ 0 So

v o (t ) = 5 v (t ) ⇒ 5 = 1 +

R2 1000

⇒ R 2 = 4 kΩ (checked LNAPTR 7/31/04)

P8.6-28 The time constant of a particular circuit is τ = 0.25 s. In response to a step input, a capacitor voltage changes from −2.5 V to 4.2 V. How long did it take for the capacitor voltage to increase from −2.0 V to +2.0 V?

Solution: The capacitor voltage can be represented by the equation v ( t ) = A + B e − 4t for t ≥ 0. Given that

A + B = v(0) = −2.5 V and A = v(∞) = 4.2 V we determine v ( t ) = 4.2 − 6.7 e − 4 t . ⎛ −2 − 4.2 ⎞ ln ⎜ ⎟ −6.7 ⎠ Let t1 be the time at which v(t1) = −2.0 V. Then t 1 = ⎝ = 0.01939 s . −4 ⎛ 2 − 4.2 ⎞ ln ⎜ ⎟ −6.7 ⎠ Let t2 be the time at which v(t2) = 2.0 V. Then t 2 = ⎝ = 0.27841 s . −4 The transition requires 0.27841 − 0.01939 = 0.25902 s.

Section 8.7 The Response of a First-Order Circuit to a Nonconstant Source P 8.7-1 Find vc(t) for t > 0 for the circuit shown in Figure P 8.7-1 when v1 = 8e–5tu(t) V. Assume the circuit is in steady state at t = 0–. Answer: vc(t) = 4e–9t + 18e–5t V

Figure P 8.7-1 Solution: Assume that the circuit is at steady state before t = 0:

KVL : 12ix + 3(3 ix ) + 38.5 = 0 ⇒ ix = −1.83 A Then vc (0− ) = −12 ix = 22 V = vc (0+ ) After t = 0:

KVL : 12i (t ) − 8e −5t + v ( t ) = 0 x c dv ( t ) 1 dvc ( t ) KCL : −ix ( t )−2ix ( t ) + (1 36) c = 0 ⇒ ix ( t ) = dt 108 dt ⎡ 1 dvc ( t ) ⎤ −5t ∴ 12 ⎢ + v (t ) = 0 ⎥ − 8e c dt 108 ⎣ ⎦

dv (t ) c + 9v (t ) = 72e −5t ⇒ v (t ) = Ae−9t c cn dt Try v (t ) = Be−5t & substitute into the differential equation ⇒ B = 18 cf

∴ v (t ) = Ae −9t + 18 e−5t c v (0) = 22 = A + 18 ⇒ A = 4 c ∴ v (t ) = 4e−9t + 18e−5t V c

P 8.7-2 Find v(t) for t > 0 for the circuit shown in Figure P 8.7-2. Assume steady state at t = 0–. Answer: v(t) = 20e–10t/3 – 12e–2t V

Figure P 8.7-2 Solution: Assume that the circuit is at steady state before t = 0:

iL (0+ ) = iL (0− ) =

12 = 3A 4

After t = 0:

v( t ) −12 v( t ) + iL ( t ) + = 6 e−2t 4 2 di ( t ) also : v ( t ) = (2 / 5) L dt di ( t ) ⎤ 3⎡ iL ( t ) + ⎢(2 / 5) L ⎥ = 3 + 6 e−2t 4⎣ dt ⎦ KCL :

diL ( t ) 10 + iL ( t ) = 10 + 20 e−2t 3 dt ∴ in (t ) = Ae

− (10 / 3)t

, try i f (t ) = B + Ce−2t , substitute into the differential equation,

and then equating like terms ⇒ B =3, C =15 ⇒ i f (t ) =3+15 e−2t ∴iL (t ) =in (t ) + i f (t ) = Ae −(10 / 3)t + 3+15e−2t , iL (0) = 3 = A + 3 + 15 ⇒ A = −15 ∴ iL (t ) = −15e− (10 / 3) t + 3 + 15e−2t Finally, v(t ) =( 2 / 5 )

diL = 20 e− (10 / 3) t −12 e −2t V dt

P 8.7-3 Find vc(t) for t > 0 for the circuit shown in Figure P 8.7-3 when is = [2 cos 2t] u(t) mA.

Figure P 8.7-3 Solution: Assume that the circuit is at steady state before t = 0:

Replace the circuit connected to the capacitor by its Thevenin equivalent (after t=0) to get:

dvc ( t ) ⎞ ⎛ +v t =0 ⇒ KVL: − 10 cos 2t + 15 ⎜ 1 30 dt ⎟ c ( ) ⎝ ⎠

dvc ( t ) + 2vc ( t ) = 20 cos 2t dt

vn (t ) = Ae−2t , Try v f (t ) = B cos 2t + C sin 2t & substitute into the differential equation to get B = C = 5 ⇒ v f (t ) = 5cos 2t + 5sin 2t. ∴ vc (t ) = vn (t ) + v f (t ) = Ae−2t + 5cos 2t + 5sin 2t Now vc (0) = 0 = A + 5 ⇒ A = −5 ⇒ vc (t ) = −5e −2t + 5cos 2t + 5sin 2t V

P 8.7-4 Many have witnessed the use of an electrical megaphone for amplification of speech to a crowd. A model of a microphone and speaker is shown in Figure P 8.7-4a, and the circuit model is shown in Figure P 8.74b. Find v(t) for

vs = 10(sin 100t)u(t) which could represent a person whistling or singing a pure tone.

Figure P 8.7-4 Solution: Assume that the circuit is at steady state before t = 0. There are no sources in the circuit so i(0) = 0 A. After t = 0, we have: di ( t ) KVL : − 10sin100t + i ( t ) + 5 + v (t ) = 0 dt v( t ) Ohm's law : i ( t ) = 8 dv( t ) ∴ +18 v( t ) = 160sin100t dt

∴ vn (t ) = Ae−18t , try v f (t ) = B cos100t + C sin100t , substitute into the differential equation and equate like terms ⇒ B = −1.55 & C = 0.279 ⇒ v f (t ) = −1.55cos100t + 0.279sin100t ∴ v(t ) = vn (t ) + v f (t ) = Ae−18t −1.55 cos100 t + 0.279 sin100 t v(0) = 8 i (0) = 0 ⇒ v (0) = 0 = A−1.55 ⇒ A = 1.55 so v(t ) = 1.55e−18t −1.55cos100t + 0.279 sin100t V

P 8.7-5 A lossy integrator is shown in Figure P 8.7-5. The lossless capacitor of the ideal integrator circuit has been replaced with a model for the lossy capacitor, namely, a lossless capacitor in parallel with a 1-kΩ resistor. If

and

vs = 15e–2tu(t) V vo(0) = 10 V,

find vo(t) for t > 0. Figure P 8.7-5 Solution: Assume that the circuit is at steady state before t = 0.

vo ( t ) = −vc ( t ) vC (0+ ) = vC (0− ) = −10 V After t = 0, we have i (t ) =

vs ( t ) 8 e−5 t = = 0.533 e−5 t mA 15000 15000

The circuit is represented by the differential dv ( t ) vC ( t ) + . Then equation: i ( t ) = C C dt R

( 0.533 ×10 ) e −3

−5 t

= ( 0.25 ×10−6 )

dvc ( t ) + (10−3 ) vc ( t ) ⇒ dt

dvc ( t ) + 4000 vc ( t ) = 4000 e−5t dt

Then vn ( t ) = Ae−4000t . Try v f ( t ) = Be−5t . Substitute into the differential equation to get

(

d B e−5t dt

) + 4000 ( B e ) = 4000 e −5t

−5t

⇒ B=

4000 = −1.00125 ≅ −1 −3995

vC (t ) = v f ( t ) + vn ( t ) = e −5t + Ae−4000t

vC (0) = −10 = 1 + A ⇒ A = −11 ⇒ vC (t ) = 1 e −2t − 11 e−4000t V Finally

vo (t ) = − vC (t ) = 11e−4000t −1e −5t V , t ≥ 0

P 8.7-6 Determine v(t) for the circuit shown in Figure P 8.7-6.

Figure P 8.7-6 Solution: Assume that the circuit is at steady state before t = 0. 2 v (0+ ) = v (0− ) = 30 = 10 V 4+ 2

After t = 0 we have KVL :

⎛ 1 d v( t ) ⎞ 5 d v( t ) + v (t ) + 4 ⎜ −i ⎟ = 30 2 dt 2 dt ⎝ ⎠ ⎛ 1 d v( t ) ⎞ −3t 2 i ( t ) + 4⎜ i ( t ) − ⎟ + 30 = e 2 dt ⎝ ⎠

The circuit is represented by the differential equation d v( t ) 6 6 2 (10 + e −3t ) v (t ) = + 19 19 3 dt Take vn ( t ) = Ae

−( 6 /19 ) t

. Try , v f ( t ) = B + Ce −3t , substitute into the differential equation to get

−3Ce −3t +

6 60 4 −3t ( B + Ce−3t ) = + e 19 19 19

Equate coefficients to get

4 4 ⇒ v f ( t ) = e−3t + Ae− (6 /19) t 51 51 4 v ( t ) = vn ( t ) + v f ( t ) = 10 − e−3t + Ae− (6 /19) t 51 4 4 vc (0+ ) = 10 V, ⇒ 10 = 10 − + A ⇒ A = 51 51 4 − (6 /19) t −3t ∴ vc (t ) = 10 + (e −e ) V 51 B = 10 , C = −

Then Finally

P 8.7-7 Determine v(t) for the circuit shown in Figure P 8.7-7a when vs varies as shown in Figure P 8.7-7b. The initial capacitor voltage is vc(0) = 0.

Figure P 8.7-7 Solution: We are given v(0) = 0. From part b of the figure: ⎧5t 0 ≤ t ≤ 2 s vs ( t ) = ⎨ t > 2s ⎩10 Find the Thevenin equivalent of the part of the circuit that is connected to the capacitor:

The open circuit voltage:

The short circuit current:

(ix=0 because of the short across the right 2 Ω resistor) Replace the part of the circuit connected to the capacitor by its Thevenin equivalent: dv( t ) + v ( t ) − vs ( t ) = 0 dt dv( t ) v ( t ) vs ( t ) + = dt 2 2 vn ( t ) = Ae−0.5 t

2

For 0 < t < 2 s, vs ( t ) = 5 t . Try v f ( t ) = B + C t . Substituting into the differential equation and equating coefficients gives B = −10 and C =5. Therefore v ( t ) = 5t − 10 + A e − t / 2 . Using v(0) = 0, we determine that A =10. Consequently, v ( t ) = 5t + 10(e−t / 2 − 1) . At t = 2 s, v( 2 ) = 10e−1 = 3.68 . Next, for t > 2 s, vs ( t ) = 10 V . Try v f ( t ) = B . Substituting into the differential equation and equating coefficients gives B = 10. Therefore v ( t ) = 10 + Ae determine that A = −6.32. Consequently, v ( t ) = 10 − 6.32 e

− (t −2) / 2

− (t − 2) / 2

.

. Using v ( 2 ) = 3.68 , we

P 8.7-8 The electron beam, which is used to “draw” signals on an oscilloscope, is moved across the face of a cathode-ray tube (CRT) by a force exerted on electrons in the beam. The basic system is shown in Figure P 8.7-8a. The force is created from a time-varying, ramp-type voltage applied across the vertical or the horizontal plates. As an example, consider the simple circuit of Figure P 8.7-8b for horizontal deflection where the capacitance between the plates is C. Derive an expression for the voltage across the capacitance. If v(t) = kt and Rs = 625 kΩ, k = 1000, and C = 2000 pF, compute vc as a function of time. Sketch v(t) and vc(t) on the same graph for time less than 10 ms. Does the voltage across the plates track the input voltage?

Figure P 8.7-8 Solution:

⎡ d v (t ) ⎤ KVL: − kt + Rs ⎢C C ⎥ + vC ( t ) = 0 dt ⎦ ⎣ d vC ( t ) k 1 vC ( t ) = t ⇒ + Rs C Rs C dt

vc ( t ) = vn ( t ) + v f ( t ) , where vc ( t ) = Ae− t / Rs C . Try v f ( t ) = B0 + B1 t & plug into D.E. ⇒ B1 +

1 k t thus B0 = − kRs C , B1 = k . [ B0 + B1t ] = Rs C Rs C

Now we have vc (t ) = Ae− t / Rs C + k (t − Rs C ). Use vc (0) = 0 to get 0 = A − kRs C ⇒ A = kRs C. ∴ vc (t ) = k[t − Rs C (1− e −t / Rs C )]. Plugging in k =1000 , Rs = 625 kΩ & C = 2000 pF get vc (t ) = 1000[t − 1.25 × 10−3 (1 − e −800 t )]

v(t) and vC(t) track well on a millisecond time scale.

Section 8.10 How Can We Check…? P 8.10-1 Figure P 8.10-1 shows the transient response of a first-order circuit. This transient response was obtained using the computer program PSpice. A point on this transient response has been labeled. The label indicates a time and the capacitor voltage at that time. Placing the circuit diagram on the plot suggests that the plot corresponds to the circuit. Verify that the plot does indeed represent the voltage of the capacitor in this circuit.

Figure P 8.10-1 Solution: First look at the circuit. The initial capacitor voltage is vc(0) = 8 V. The steady-state capacitor voltage is vc = 4 V. We expect an exponential transition from 8 volts to 4 volts. That’s consistent with the plot. Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of ( 2000 )( 4000 ) = 4 kΩ so the time constant is the circuit connected to the capacitor is R t = 2000 + 4000 3 2 ⎛4 ⎞ τ = R t C = ⎜ × 103 ⎟ ( 0.5 × 10−6 ) = ms . Thus the capacitor voltage is 3 ⎝3 ⎠

vc (t ) = 4 e− t

0.67

+4 V

where t has units of ms. To check the point labeled on the plot, let t1 = 1.33 ms. Then

vc (t1 ) = So the plot is correct.

⎛ 1.33 ⎞ −⎜ ⎟ 4 e ⎝ .67 ⎠

+ 4 = 4.541 ~ 4.5398 V

P 8.10-2 Figure P 8.10-2 shows the transient response of a first-order circuit. This transient response was obtained using the computer program PSpice. A point on this transient response has been labeled. The label indicates a time and the inductor current at that time. Placing the circuit diagram on the plot suggests that the plot corresponds to the circuit. Verify that the plot does indeed represent the current of the inductor in this circuit.

Figure P 8.10-2 Solution: The initial and steady-state inductor currents shown on the plot agree with the values obtained from the circuit.

Next, let’s check the shape of the exponential transition. The Thevenin resistance of the part of ( 2000 )( 4000 ) = 4 kΩ so the time constant is the circuit connected to the inductor is R t = 2000 + 4000 3 5 15 L ms . Thus inductor current is τ= = = R t 4 ×103 4 3 iL (t ) − 2 e − t 3.75 + 5 mA

where t has units of ms. To check the point labeled on the plot, let t1 = 3.75 ms. Then iL (t1 ) =

⎛ 3.75 ⎞ −⎜ ⎟ −2 e ⎝ 3.75 ⎠

+ 5 = 4.264 mA ≠ 4.7294 mA

so the plot does not correspond to this circuit.

P 8.10-3 Figure P 8.10-3 shows the transient response of a first-order circuit. This transient response was obtained using the computer program PSpice. A point on this transient response has been labeled. The label indicates a time and the inductor current at that time. Placing the circuit diagram on the plot suggests that the plot corresponds to the circuit. Specify that value of the inductance, L, required to cause the current of the inductor in this circuit to be accurately represented by this plot.

Figure P 8.10-3 Solution: Notice that the steady-state inductor current does not depend on the inductance, L. The initial and steady-state inductor currents shown on the plot agree with the values obtained from the circuit.

After t = 0

So I sc = 5 mA and τ =

The inductor current is given by iL (t ) = −2e −1333t has units of Henries. Let t 1 = 3.75 ms, then

L

+ 5 mA , where t has units of seconds and L

4.836 = iL (t1 ) = −2 e−(1333)⋅(0.00375) L + 5 = −2e−5 L + 5 4.836−5 = e −5 L −2

so and

is the required inductance.

L=

L 1333

−5 =2 H ⎛ 4.836−5 ⎞ ln ⎜ ⎟ ⎝ −2 ⎠

P 8.10-4 Figure P 8.10-4 shows the transient response of a first-order circuit. This transient response was obtained using the computer program PSpice. A point on this transient response has been labeled. The label indicates a time and the capacitor voltage at that time. Assume that this circuit has reached steady state before time t = 0. Placing the circuit diagram on the plot suggests that the plot corresponds to the circuit. Specify values of A, B, R1, R2, and C that cause the voltage across the capacitor in this circuit to be accurately represented by this plot.

Figure P 8.10-4 Solution: First consider the circuit. When t < 0 and the circuit is at steady-state:

For t > 0

So

Voc =

R2 R1 R2 RRC ( A + B) , Rt = and τ = 1 2 R1 + R2 R1 + R2 R1 + R2

Next, consider the plot. The initial capacitor voltage is (vc (0)=) –2 and the steady-state capacitor voltage is (Voc =) 4 V, so vC (t ) = − 6e−t τ + 4

At t 1 = 1.333 ms

3.1874 = vC (t1 ) = − 6 e −0.001333 τ + 4

−0.001333 = 0.67 ms ⎛ −4+ 3.1874 ⎞ ln ⎜ ⎟ −6 ⎝ ⎠

τ =

so

Combining the information obtained from the circuit with the information obtained from the plot gives R2 R2 R1 R2C A = −2, ( A + B ) = 4, = 0.67 ms R1 + R2 R1 + R2 R1 + R2 There are many ways that A, B, R , R , and C can be chosen to satisfy these equations. Here is one convenient way. Pick R = 3000 and R = 6000. Then 1

1

2

2

2A = −2 ⇒ A = − 3 3 2( A+ B) = 4 ⇒ B −3 = 6 ⇒ B = 9 3 2 1 μF = C 2000 ⋅ C = ms ⇒ 3 3

Spice Problems SP 8-1 The input to the circuit shown in Figure SP 8.1 is the voltage of the voltage source, vi(t). The output is the voltage across the capacitor, vo(t). The input is the pulse signal specified graphically by the plot. Use PSpice to plot the output, vo(t), as a function of t. Hint: Represent the voltage source using the PSpice part named VPULSE.

Figure SP 8.1 Solution:

SP 8-2 The input to the circuit shown in Figure SP 8.2 is the voltage of the voltage source, vi(t). The output is the current in the inductor, io(t). The input is the pulse signal specified graphically by the plot. Use PSpice to plot the output, io(t), as a function of t. Hint: Represent the voltage source using the PSpice part named VPULSE.

Figure SP 8.2 Solution:

SP 8-3 The circuit shown in Figure SP 8.3 is at steady state before the switch closes at time t = 0. The input to the circuit is the voltage of the voltage source, 12 V. The output of this circuit is the voltage across the capacitor, v(t). Use PSpice to plot the output, v(t), as a function of t. Use the plot to obtain an analytic representation of v(t) for t > 0. Hint: We expect v(t) = A + Be–t/τ for t > 0, where A, B, and τ are constants to be determined.

Figure SP 8.3

Solution:

v(t ) = A + B e −t / τ 7.2 = v(0) = A + B e 0 8.0 = v(∞) = A + B e −∞

for t > 0

⇒ 7.2 = A + B ⎫⎪ ⎬ ⇒ B = −0.8 V ⇒ A = 8.0 V ⎪⎭

7.7728 = v(0.05) = 8 − 0.8 e −0.05 / τ

⎛ 8 − 7.7728 ⎞ = ln ⎜ ⎟ = −1.25878 τ 0.8 ⎝ ⎠ 0.05 ⇒ τ= = 39.72 ms 1.25878

⇒ −

0.05

Therefore v(t ) = 8 − 0.8 e −t / 0.03972 V for t > 0

SP 8-4 The circuit shown in Figure SP 8.4 is at steady state before the switch closes at time t = 0. The input to the circuit is the current of the current source, 4 mA. The output of this circuit is the current in the inductor, i(t). Use PSpice to plot the output, i(t), as a function of t. Use the plot to obtain an analytic representation of i(t) for t > 0. Hint: We expect i(t) = A + Be–t/τ for t > 0, where A, B, and τ are constants to be determined. Solution:

Figure SP 8.4

i (t ) = A + B e−t / τ 0 = i (0) = A + B e 0

⇒ 0 = A+ B

4 × 10−3 = i (∞) = A + B e −∞



A = 4 × 10−3

2.4514 ×10 = v(5 ×10 ) = ( 4 ×10 −3

⇒ −

−6

5 × 10−6

⇒ τ= Therefore

for t > 0

τ

−3

⎫⎪ −3 ⎬ ⇒ B = −4 × 10 A A ⎪⎭

) − ( 4 ×10 ) e ( −3

)

− 5×10−6 / τ

⎛ ( 4 − 2.4514 ) × 10−3 ⎞ = ln ⎜ ⎟ = −0.94894 −3 4 × 10 ⎝ ⎠

5 × 10−6 = 5.269 μ s 0.94894

i (t ) = 4 − 4 e−t / 5.269×10

−6

mA for t > 0

DESIGN PROBLEMS

DP 8-1 Design the circuit in Figure DP 8.1 so that v(t) makes the transition from v(t) = 6 V to v(t) = 10 V in 10 ms after the switch is closed. Assume that the circuit is at steady state before the switch is closed. Also assume that the transition will be complete after 5 time constants.

Figure DP 8.1

Solution: Steady-state response when the switch is open: 6 =

R3 R1 + R 2 + R 3

Steady-state response when the switch is closed: 10 = 10 ms = 5 τ = ( R 1 || R 3 ) C =

R3 6

R3 R1 + R 3

12 ⇒ R1 + R 2 = R 3 .

12

⇒ R1 =

R3 5

.

C

Let C = 1 μF. Then R 3 = 60 kΩ, R 1 = 30 kΩ and R 2 = 30 kΩ.

DP 8-2 Design the circuit in Figure DP 8.2 so that i(t) makes the transition from i(t) = 1 mA to i(t) = 4 mA in 10 ms after the switch is closed. Assume that the circuit is at steady state before the switch is closed. Also assume that the transition will be complete after 5 time constants. Figure DP 8.2 Solution: steady state response when the switch is open: 0.001 =

12 R1+R 2

steady state response when the switch is closed: 0.004 =

12 R1

⇒ R + R = 12 kΩ . 1

2

⇒ R1 = 3 kΩ .

Therefore, R 2 = 9 kΩ. ⎛ L ⎞ L 10 ms = 5 τ = 5 ⎜ ⇒ L = 240 H ⎟= ⎜ R1 + R 2 ⎟ 2400 ⎝ ⎠ DP 8-3 The switch in Figure DP 8.3 closes at time 0, 2Δt, 4Δt, …2kΔt and opens at times Δt, 3Δt, 5Δt….(2k + 1)Δt. When the switch closes, v(t) makes the transition from v(t) = 0 V to v(t) = 5 V. Conversely, when the switch opens, v(t) makes the transition from v(t) = 5 V to v(t) = 0 V. Suppose we require that Δt = 5τ so that one transition is complete before the next one begins. (a) Determine the value of C required so that Δt = 1 μs. (b) How large must Δt be when C = 2 μF? Answer: (a) C = 4 pF; (b) Δt = 0.5s

Figure DP 8.3

Solution: Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use Rt = 50 kΩ. 10−6 (a) Δt = 5 Rt C ⇒ C = = 4 pF 5 50×103

(

(b) Δ t = 5 50×10

3

)( 2×10

−6

( ) ) = 0.5 s

DP 8-4 The switch in Figure DP 8.3 closes at time 0, 2Δt, 4Δt, …2kΔt and opens at times Δt, 3Δt, 5Δt….(2k + 1)Δt. When the switch closes, v(t) makes the transition from v(t) = 0 V to v(t) = 5 V. Conversely, when the switch opens, v(t) makes the transition from v(t) = 5 V to v(t) = 0 V. Suppose we require that one transition be 95 percent complete before the next one begins. (a) Determine the value of C required so that Δt = 1 μs. (b) How large must Δt be when C = 2 μF? Hint: Show that Δt = – τ ln(1 – k) is required for the transition to be 100 k percent complete. Answer: (a) C = 6.67 pF; (b) Δt = 0.3 s Solution: Rt = 50 kΩ when the switch is open and Rt = 49 kΩ ≈ 50 kΩ when the switch is closed so use Rt = 50 kΩ. Δt −Δt τ When the switch is open: 5 e = (1 − k ) 5 ⇒ ln (1− k ) = − ⇒ Δ t = −τ ln (1− k )

When the switch is open: 5 − 5 e (a) C =

−Δt τ

10−6 = 6.67 pF − ln (1−.95 ) ( 50×103 )

τ

= k 5 ⇒ Δ t = −τ ln (1− k )

(b) Δ t = − ln(1 − .95) ( 50×103 )( 2×10−6 ) = 0.3 s

DP 8-5 A laser trigger circuit is shown in Figure DP 8.5. In order to trigger the laser, we require 60 mA < |i| < 180 mA for 0 < t < 200 μs. Determine a suitable value for R1 and R2.

Figure DP 8.5 Solution:

i (0) =

R1 20 × 40 R1 R1 + 40 40 + 40 + R1

For t > 0: i (t ) = i (0) e

−t

τ

where τ =

L 10−2 = R t 40+ R 2

At t < 200μ s we need i ( t ) > 60 mA and i ( t ) 0 for the circuit of Figure E 9.9-1. Assume there is no initial stored energy.

3 10

+ 1Ω

v1

+ v2 –

1 12 F



Answer: v2(t) = – 15e–2t + 6e– 4t – e–6t + 10 V

H

5 6

F

10u(t) A

Figure E 9.9-1 Solution:

no initial stored energy ⇒ v1 (0+ ) = v2 (0+ ) = i (0+ ) = 0 t = 0+

KVL : − 0 + 3

di(0+ ) di(0+ ) +0=0 ⇒ =0 10 dt dt

0V KCL at A : + i1 (0+ )+0 = 0 1Ω

dv1 (0+ ) =0 ⇒ dt

KCL at B : − 0 + i 2 (0+ ) − 10 = 0 ⇒ i 2 (0+ ) = 5 6

dv 2 (0+ ) dv 2 (0+ ) = 12 V s = 10 ⇒ dt dt

t>0

v1 1 + v1 ' + i = 0 1 12 KCL at B : − i + (5 / 6)v '2 =10

KCL at A :

(1)

KVL : − v1 + ( 3 /10 ) i '+ v2 = 0

( 2) ( 3)

Eliminating i from (1) & (3) yields 1 v '1 + (5 / 6)v2 '− 10 = 0 12 3 ⎛5 ⎞ −v1 + ⎜ v2" ⎟ + v2 = 0 10 ⎝ 6 ⎠

v1 +

(4)

( 5)

From (5)

1 v1 = v2 + v2" ⇒ v1' = v2" + (1/ 4)v2" 4 Now substituting into (4) yields 1 1⎛ 1 ⎞ 5 v2 ' + v2" + ⎜ v2 ' + v2"' ⎟ + v2 ' = 10 4 12 ⎝ 4 ⎠ 6 v2"' + 12v2" + 44v2 ' + 48v2 = 480

Natural Response: Forced Response: Complete Response:

v2 n : s 3 + 12 s 2 + 44s + 48 = 0

⇒ s = − 2, −4, −6

∴ v2 n = A1e −2t + A2 e −4t + A3e−6t

v2 f : try v2 f = B and plug into Diff. Eq. ⇒ B = 10 v2 (t ) = A1e −2t + A2e −4t + A3e −6t + 10

dv2 (0+ ) d 2 v2 (0+ ) Recall v2 (0 ) = 0, = 12 V/s, then from (5) = 4[v1 (0+ ) − v2 (0+ )] = 0 . 2 dt dt + v2 (0 ) = 0 = A1 + A2 + A3 + 10 (6) +

dv2 (0+ ) = 12 = − 2 A1 − 4 A2 − 6 A3 (7) dt d 2 v2 (0+ ) = 0 = 4 A1 + 16 A2 + 36 A3 (8) dt 2 Solving (6)-(80 simultaneously gives A1 = −15, A2 = 6, A3 = −1 Then v2 (t ) = − 15e −2t + 6e −4t − e−6t + 10 V



Exercise 9.10-1 A parallel RLC circuit has L = 0.1 H and C = 100 mF. Determine the roots of the characteristic equation and plot them on the s-plane when (a) R = 0.4 Ω and (b) R = 1.0 Ω.

× –20

× –5

0

Answer: (a) s = – 5, – 20 (Figure E 9.10-1) Figure E 9.10-1

Solution:

s2 +

1 1 s+ = 0 and L = 0.1, C = 0.1 ⇒ RC LC

a) R = 0.4 Ω ⇒ s 2 + 25s + 100 = 0 s = − 5, − 20 b) R = 1Ω

⇒ s 2 + 10s + 100 = 0 s = − 5 ± j5 3

s2 +

10 s + 100 = 0 R

σ

Section 9-2: Differential Equations for Circuits with Two Energy Storage Elements 2Ω

P 9.2-1 Find the differential equation for the circuit shown in Figure P 9.2-1 using the direct method.

1 mH

+ –

vs

100 Ω

Figure P 9.2-1 Solution:

KCL: i L =

v dv +C R2 dt

KVL: Vs = R 1i L + L

L v + C dv OP + L dv + LC d v + v = R M N R dt Q R dt dt L R + 1OPv + LMR C + L OP dv + [LC] d v = M dt N R Q N R Q dt 2

vs

1

2

2

2

2

vs

1

1

2

2

2

R 1 = 2Ω, R 2 = 100Ω, L = 1mH, C = 10μF 2 dv −8 d v . v +.00003 + 1 × 10 v s = 102 dt dt 2 dv d 2 v 8 8 1 × 10 v s = 102 . × 10 v + 3000 + 2 dt dt

di L +v dt

10 μ F

P 9.2-2 Find the differential equation for the circuit shown in Figure P 9.2-2 using the operator method. Answer: d2 d i (t ) + 11, 000 iL (t ) + 1.1 × 108 iL (t ) = 108 is (t ) 2 L dt dt

10 Ω is

10 μ F

100 Ω 1 mH iL

Figure P 9.2-2 Solution:

v + i L + Csv R1 KVL: v = R 2 i L + Lsi L

KCL: i s =

Solving Cramer's rule for i L : iL =

is R 2 Ls + + R 2 Cs + LCs2 +1 R1 R1

LM1 + R OPi + LM L + R COPsi N R Q NR Q 2

L

1

2

L

+ LC s2 i L = i s

1

R 1 = 100Ω, R 2 = 10Ω, L = 1mH, C = 10μF 1.1i L +.00011si L + 1 × 10 −8 s2 i L = i s 11 . × 108 i L + 11000si L + s2 i L = 1 × 108 i s

P 9.2-3 Find the differential equation for iL(t) for t > 0 for the circuit of Figure P 9.2-3.

iL

is

R1

L

t=0

+ –

vs C

+ vc –

Figure P 9.2-3 Solution: t>0

KCL: i L + C

dv c v s + v c + = 0 dt R2

KVL: R 1i s + R 1i L + L

di L − vc − vs = 0 dt

Solving for iL : − R1 d 2iL ⎡ R1 1 ⎤ diL ⎡ R1 1 ⎤ R1 dis 1 dvs + + + + = − + i i s ⎢ ⎥ ⎢ ⎥L dt 2 ⎣ L R2C ⎦ dt ⎣ LR2C LC ⎦ LCR2 L dt L dt

R2

P 9.2-4 The input to the circuit shown in Figure P 9.2-4 is the voltage of the voltage source, Vs. The output is the inductor current i(t). Represent the circuit by a secondorder differential equation that shows how the output of this circuit is related to the input, for t > 0.

t=0

+ –

Hint: Use the direct method.

R1

i(t)

+

L

Vs v(t)

C



Figure P 9.2-4 Solution: After the switch opens, apply KCL and KVL to get

d ⎛ ⎞ R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs dt ⎝ ⎠ Apply KVL to get

v (t ) = L Substituting v ( t ) into the first equation gives

d i (t ) + R2 i (t ) dt

⎛ d⎛ d d ⎞⎞ R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs dt ⎝ dt dt ⎠⎠ ⎝ then R1 C L

d2 dt

2

(

i ( t ) + R1 C R 2 + L

) dtd i ( t ) + ( R1 + R 2 ) i ( t ) = Vs

Dividing by R1 C L :

⎛ R1 C R 2 + L ⎞ d ⎛ R1 + R 2 ⎞ Vs + + = i t i t i t ⎜ ⎟ ⎜ ⎟ ( ) ( ) ( ) ⎜ R1 C L ⎟ dt ⎜ R1 C L ⎟ R1 C L dt 2 ⎝ ⎠ ⎝ ⎠ d2

R3

R2

P 9.2-5 The input to the circuit shown in Figure P 9.2-5 is the voltage of the voltage source, vs. The output is the capacitor voltage v(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0. Hint: Use the direct method. i(t)

t=0

R1 L

+ –

vs

+ R2

C

v(t) –

Figure P 9.2-5 Solution: After the switch closes, use KCL to get i (t ) =

v (t ) d + C v (t ) R2 dt

Use KVL to get

v s = R1 i ( t ) + L

d i (t ) + v (t ) dt

Substitute to get vs =

R1 R2

= CL

v ( t ) + R1C

d L d d2 v (t ) + v ( t ) + CL 2 v ( t ) + v ( t ) dt R 2 dt dt

⎛ R1 + R 2 d2 L ⎞d v t + R1C + v (t ) + v (t ) ⎟ 2 ( ) ⎜ ⎜ ⎟ dt dt R R CL 2 2 ⎝ ⎠

Finally,

vs CL

=

⎛ R1 R1 + R 2 d2 1 ⎞d + + v t v (t ) + v (t ) ( ) ⎜ ⎟ 2 ⎜ L R 2C ⎟ dt dt R 2CL ⎝ ⎠

t=0

P 9.2-6 The input to the circuit shown in Figure P 9.2-6 is the voltage of the voltage source, vs. The output is the inductor current i(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0.

R1

R2 + –

+ C

vs L

i(t)

Hint: Use the direct method. Figure P 9.2-6 Solution: After the switch closes use KVL to get

R2 i (t ) + L

d i (t ) = v (t ) dt

Use KCL and KVL to get

d ⎛ ⎞ v s = R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) dt ⎝ ⎠ Substitute to get

d d2 d i ( t ) + R1CL 2 i ( t ) + R 2i ( t ) + L i ( t ) dt dt dt 2 d d = R1CL 2 i ( t ) + ( R1 R 2C + L ) i ( t ) + ( R1 + R 2 ) i ( t ) dt dt

v s = R1i ( t ) + R1CR 2

Finally

vs R1CL

=

⎛ R2 R1 + R 2 d2 1 ⎞d + i t + i (t ) + i (t ) ⎟ 2 ( ) ⎜ ⎜ L R1C ⎟ dt dt R CL 1 ⎝ ⎠

v(t) –

P 9.2-7 The input to the circuit shown in Figure P 9.2-7 is the voltage of the voltage source, vs. The output is the inductor current i2(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0. Hint: Use the operator method. R2 t=0 L1

R1 + –

i1(t)

L2

i2(t)

vs R3

Figure P 9.2-7 Solution: After the switch opens, KVL gives

L1

d d i1 ( t ) = R 2 i 2 ( t ) + L 2 i 2 ( t ) dt dt

L1

d i1 ( t ) + R1 ( i1 ( t ) + i 2 ( t ) ) = 0 dt

KVL and KCL give

Use the operator method to get

L1s i1 = R 2 i 2 + L 2 s i 2 L1s i1 + R1 ( i1 + i 2 ) = 0

L1s 2i1 + R1s i1 + R1s i 2 = 0 s ( R 2i 2 + L 2 s i 2 ) +

R1 L1

(R i

2 2

+ L 2 s i 2 ) + R1s i 2 = 0

⎛ ⎞ L2 R1 R 2 L 2 s 2 i 2 + ⎜ R 2 + R1 + R1 ⎟ s i 2 + i2 = 0 ⎜ ⎟ L L 1 1 ⎝ ⎠ ⎛ R 2 R1 R1 ⎞ R1 R 2 s 2i 2 + ⎜ + + ⎟ s i2 + i2 = 0 ⎜ L 2 L 2 L1 ⎟ L1 L 2 ⎝ ⎠ so ⎛ R 2 R1 R1 ⎞ d R1 R 2 d2 i t + + + ⎟ i 2 (t ) + i 2 (t ) = 0 2 2( ) ⎜ ⎜ L 2 L 2 L1 ⎟ dt dt L L 1 2 ⎝ ⎠

t=0

P 9.2-8 The input to the circuit shown in Figure P 9.2-8 is the voltage of the voltage source, vs. The output is the capacitor voltage v2(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0.

R2

R1

+

+ + –

Hint: Use the operator method.

C1

vs

v1(t)

C2

– R3

Figure P 9.2-8 Solution: After the switch closes, KVL and KCL give

d ⎛ d ⎞ v1 ( t ) + R 3 ⎜ C 1 v1 ( t ) + C 2 v 2 ( t ) ⎟ = v s dt ⎝ dt ⎠ KVL gives v 1 ( t ) = R 2 C2 Using the operator method

d v 2 (t ) + v 2 (t ) dt

v1 + R 3 ( C 1sv1 + C 2 sv 2 ) = v s

v1 = R 2C 2 sv 2 + v 2 so

(1 + R C s ) v 2

Then

2

v1 = (1 + R 2C2 s ) v 2

2

+ R 3C 1s (1 + R 2C 2 s ) v 2 + R 3C 2 sv 2 = v s

R 2 R 3C 1C 2 s 2 v 2 + ( R 2C 2 + R 3C 1 + R 3C 2 ) sv 2 + v 2 = v s s 2v 2 +

R 2C 2 + R 3C 1 + R 3C 2 R 2 R 3C 1C 2

sv 2 +

vs 1 v2 = R 3 R 2C 1C 2 R 2 R 3C 1C 2

⎛ 1 vs 1 1 ⎞ 1 + + s 2v 2 + ⎜ sv 2 + v2 = ⎟ ⎜ R 3C 1 R 2 C 2 R 2 C 1 ⎟ R 2 R 3C 1C 2 R 2 R 3C1C 2 ⎝ ⎠ so

⎛ 1 1 1 ⎞d 1 d2 = 2 v 2 (t ) + ⎜ + + v (t ) ⎟⎟ v ( t ) + ⎜ R 2 R 3C 1C 2 dt R 2 R 3C 1C 2 ⎝ R 3C1 R 2C 2 R 2C 1 ⎠ dt vs

v2(t) –

t=0

P 9.2-9 The input to the circuit shown in Figure P 9.2-9 is the voltage of the voltage source, vs. The output is the capacitor voltage v(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0.

L + –

vs

R1 R2

Hint: Use the direct method.

C

Figure P 9.2-9 Solution: After the switch closes i (t ) = C

d v (t ) dt

KCL and KVL give ⎛ 1 ⎛ d d ⎞⎞ vs = R2 ⎜ i (t ) + ⎜ L i (t ) + v (t ) ⎟ ⎟ + L i (t ) + v (t ) ⎜ R1 ⎝ dt dt ⎠ ⎟⎠ ⎝

Substituting gives

⎛ R2 ⎞ ⎛ R2 ⎞ d2 d v s = ⎜1 + LC v ( t ) + R 2C v ( t ) + ⎜ 1 + ⎟ ⎟⎟ v ( t ) 2 ⎜ ⎟ ⎜ R dt dt R 1 ⎠ 1 ⎠ ⎝ ⎝ ⎛ R2 ⎞ ⎛ R2 ⎞ d2 d = ⎜1 + LC 2 v ( t ) + R 2C v ( t ) + ⎜ 1 + ⎟ ⎟ v (t ) ⎜ ⎜ R1 ⎟⎠ dt dt R1 ⎟⎠ ⎝ ⎝ Finally

R1v s

LC ( R1 + R 2 )

=

R1 R 2 1 d2 d v (t ) + v (t ) + v (t ) dt LC L ( R1 + R 2 ) dt

i(t) + v(t) –

P 9.2-10 The input to the circuit shown in Figure P 9.2-10 is the voltage of the voltage source, vs. The output is the capacitor voltage v(t). Represent the circuit by a second-order differential equation that shows how the output of this circuit is related to the input, for t > 0. Hint: Find a Thévenin equivalent circuit. ia

i(t)

t=0

L

R1 + –

vs

bia

+

R2

C

v(t) –

Figure P 9.2-10 Solution: Find the Thevenin equivalent circuit for the part of the circuit to the left of the inductor. v s − v oc ⎫ ⎪ R1 ⎪ v s R 2 (1 + b ) ⎬ ⇒ v oc = v oc ⎪ R1 + R 2 (1 + b ) ia + bia = R 2 ⎪⎭ ia =

i sc = i a (1 + b ) =

vs R1

(1 + b )

v s R 2 (1 + b )

Rt =

v oc i sc

=

R1 + R 2 (1 + b ) R1 R 2 = vs R1 + R 2 (1 + b ) (1 + b ) R1

Rt i (t ) + L

d i (t ) + v ( t ) − v oc = 0 dt

i (t ) = C d v (t ) d 2 v (t ) Rt C + LC + v ( t ) = v oc dt d t2 Finally,



d v (t ) dt

d 2 v (t ) Rt d v (t ) v (t ) 1 + + v t = ( ) d t2 L dt LC LC

R1 R 2 d 2 v (t ) d v (t ) v (t ) 1 + + v (t ) = 2 dt LC LC L ( R1 + R 2 (1 + b ) ) d t

P 9.2-11 The input to the circuit shown in Figure P 9.2-11 is the voltage of the voltage source, vs(t). The output is the voltage v2(t). Derive the second-order differential equation that shows how the output of this circuit is related to the input.



R1

R2

+

+ + –

vs(t)

v1(t)

C1



Hint: Use the direct method. Figure P 9.2-11 Solution: KCL gives

v s ( t ) − v1 ( t )

= C1

d v1 ( t ) dt



v s ( t ) = R1C 1

d v1 ( t ) + v1 ( t ) dt

= C2

d v 2 (t ) dt



v1 ( t ) = R 2C 2

d v 2 (t ) + v 2 (t ) dt

R1 and

v1 ( t ) − v 2 ( t ) R2

Substituting gives v s ( t ) = R1C 1

d ⎡ d d ⎤ R 2C 2 v 2 ( t ) + v 2 ( t ) ⎥ + R 2 C 2 v 2 ( t ) + v 2 ( t ) ⎢ dt ⎣ dt dt ⎦

so ⎛ 1 1 d2 1 ⎞ 1 vs (t ) = 2 v 2 (t ) + ⎜ + v 2 (t ) ⎟⎟ v 2 ( t ) + ⎜ R1 R 2C 1C 2 dt R1 R 2C 1C 2 ⎝ R1C 1 R 2C 2 ⎠

+ C2

v2(t) –

C2

P 9.2-12 The input to the circuit shown in Figure P 9.2-12 is the voltage of the voltage source, vs(t). The output is the voltage vo(t). Derive the second-order differential equation that shows how the output of this circuit is related to the input.

– v2(t) + R1

R2

C1 –

Hint: Use the operator method.

+ –

+ v1(t) –

+

vs(t)

+ vo(t) –

Figure P 9.2-12 Solution:

d v1 ( t ) + v1 ( t ) dt v 2 (t ) d d C 1 v1 ( t ) + C 2 v 2 ( t ) + =0 dt dt R2 v s ( t ) = R1C 1

KVL gives KCL gives

v o (t ) = v 2 (t )

KVL gives Using the operator method

v s = R1C 1sv1 + v1 C 1sv1 + C 2 sv 2 +

v2 R2

=0

Solving ⎛ C2 ⎞ 1 v1 = − ⎜ v2 + v2 ⎟ ⎜ C1 R 2C 1s ⎟⎠ ⎝ ⎛ C2 1 ⎞ sv s = ( sR1C 1 + 1) ⎜ s+ ⎟ vo ⎜ C1 R 2C 1 ⎟⎠ ⎝ ⎛ 1 1 1 ⎞ 1 sv s = s 2 v o + ⎜ vo + ⎟⎟ sv o + ⎜ R1C 2 R C R C R R C C 2 2 ⎠ 1 2 1 2 ⎝ 1 1 The corresponding differential equation is ⎛ 1 1 d d2 1 ⎞d 1 vs (t ) = 2 v o (t ) + ⎜ + v o (t ) ⎟⎟ v o ( t ) + ⎜ R1C 2 dt dt R1 R 2C 1C 2 ⎝ R1C 1 R 2C 2 ⎠ dt

t=0

P 9.2-13 The input to the circuit shown in Figure P 9.2-13 is the voltage of the voltage source, vs(t). The output is the voltage vo(t). Derive the second-order differential equation that shows how the output of this circuit is related to the input. Hint: Use the direct method.

– v(t) + C

R1

i(t)

– + –

+

vs(t)

+

L R2

vo(t) –

Figure P 9.2-13 Solution: After the switch opens, KCL gives

vs (t ) R1

+C

d v (t ) = 0 dt

KVL gives v (t ) − v o (t ) = L and Ohm’s law gives

d i (t ) dt

v o (t ) = R2 i (t )

so d 1 v (t ) = − vs (t ) dt R1 C and d d d2 v (t ) − v o (t ) = L 2 i (t ) dt dt dt Then −

1 d d2 d v s (t ) = v (t ) = L 2 i (t ) + R2 i (t ) R1C dt dt dt

or R2 d 1 d2 − vs (t ) = 2 i (t ) + i (t ) R1CL dt L dt

R2

P 9.2-14 The input to the circuit shown in Figure P 9.2-14 is the voltage of the voltage source, vs(t). The output is the voltage v2(t). Derive the second-order differential equation that shows how the output of this circuit is related to the input.

+ v1(t) – C1

R1 – + –

Hint: Use the direct method.

R3 +

+

vs(t)

C2

v2(t) –

Figure P 9.2-14 Solution: KCL gives

vs (t ) R1

and

=

v1 ( t ) R2

v 2 ( t ) + v1 ( t ) R3

+ C1

+ C2

d v1 ( t ) dt

d v 2 (t ) = 0 dt

so v1 ( t ) + R 2C 1

R2 d v1 ( t ) = vs (t ) dt R1

and d ⎛ ⎞ v 1 ( t ) = − ⎜ v 2 ( t ) + R 3C 2 v 2 ( t ) ⎟ dt ⎝ ⎠ Substituting gives R2 ⎡ d d ⎡ d ⎤⎤ ⎢ v 2 ( t ) + R 3C 2 dt v 2 ( t ) + R 2C 1 dt ⎢v 2 ( t ) + R 3C 2 dt v 2 ( t ) ⎥ ⎥ = − R v s ( t ) ⎣ ⎦⎦ ⎣ 1 or ⎛ 1 d2 1 ⎞d 1 1 v t + + v 2 (t ) + v 2 (t ) = − v s (t ) ⎟ 2 2( ) ⎜ ⎜ R 2C 1 R 3C 2 ⎟ dt dt R R C C R R C C 2 3 1 2 1 3 1 2 ⎝ ⎠

P 9.2-15 Find the second-order differential equation for i2 for the circuit of Figure P 9.2-15 using the operator method. Recall that the operator for the integral is 1/s. Answer:

3



vs

+ –

i1

d 2 vs d 2i2 di2 + + i = 4 2 2 dt 2 dt dt 2

Apply KVL to the left mesh : i1 + s(i1 −i2 ) = vs

(1)

dt

⎛1⎞ Apply KVL to the right mesh : 2i2 + 2 ⎜ ⎟ i2 ⎝s⎠ ⎛1⎞ ⇒ i1 = 2⎜ ⎟i2 + ⎝s⎠

+ s(i2 − i1 ) = 0 ⎛1⎞ 2⎜ 2 ⎟i2 + i2 ⎝s ⎠

(2)

Plugging (2) into (1) yields 3s 2i + 4si + 2i = s 2v 2 2 2 s

or

1H

i2

Figure P 9.2-15

Solution:

where s = d



d 2v d 2i di s 2 2 + 4 + 2i = 3 2 2 2 dt dt dt

1 2

F

Section 9-3: Solution of the Second Order Differential Equation - The Natural Response P 9.3-1 Find the characteristic equation and its roots for the circuit of Figure P 9.2-2.

10 Ω is

10 μ F

100 Ω 1 mH iL

Figure P 9.2-2 Solution: From Problem P 9.2-2 the characteristic equation is

−11000 ± (11000)2 − 4(1.1×108 ) 1.1× 108 + 11000s + s 2 = 0 ⇒ s , s = = −5500 ± j8930 1 2 2

100 mH

P 9.3-2 Find the characteristic equation and its roots for the circuit of Figure P 9.3-2. 2

4

Answer: s + 400s + 3 × 10 = 0 roots: s = – 300, – 100

iL is

40 Ω

+ vc –

1 3 mF

Figure P 9.3-2 Solution:

KVL: 40 (is − iL ) = 100 ×10−3

diL + vc dt

⎛1 ⎞ dv i L = i c = ⎜ × 10−3 ⎟ c ⎝3 ⎠ dt di 40 di 100 d 2iL 40 × 10−3 s − ×10−3 L − × 10−6 dt 3 dt 3 3 dt 2 di d 2iL di + 400 L +30000i L = 400 s 2 dt dt dt 2 s + 400s + 30000 = 0 ⇒ ( s + 100)( s + 300) = 0 ⇒ s1 = −100,

iL =

s2 = −300

P 9.3-3 Find the characteristic equation and its roots for the circuit shown in Figure P 9.3-3.





vs

+ vc –

+ –

1 mH iL

Figure P 9.3-3 Solution:

v − vs dv + i L + 10 × 10−6 = 0 1 dt di KVL: v = 2i L +10−3 L dt

KCL:

2 diL diL −6 −6 −3 d iL 0 = 2iL + 10 − vs + iL + 10 × 10 ⋅ 2 + 10 × 10 × 10 dt dt dt 2 di d i vs = 3iL + .00102 L + 1× 10−8 2L dt dt 2 d iL di + 102000 L + 3 × 10−8 iL = 1× 108 vs dt dt 2 s + 102000s + 3 ×108 = 0, ∴ s1 = 3031, s2 = − 98969 −3

10 μ F

P 9.3-4 German automaker Volkswagen, in its bid to make more efficient cars, has come + 10u(t) V – 1 2H up with an auto whose engine saves energy by 5 mF shutting itself off at stoplights. The “stop– start” system springs from a campaign to 10 Ω + – 7u(t) V develop cars in all its world markets that use less fuel and pollute less than vehicles now on the road. The stop–start transmission control Figure P 9.3-4 has a mechanism that senses when the car does not need fuel: coasting downhill and idling at an intersection. The engine shuts off, but a small starter flywheel keeps turning so that power can be quickly restored when the driver touches the accelerator. A model of the stop–start circuit is shown in Figure P 9.3-4. Determine the characteristic equation and the natural frequencies for the circuit. Answer: s2 + 20s + 400 = 0 s = – 10 ± j17.3

Solution: Assume zero initial conditions

1 di1 1 di2 − = 10 − 7 2 dt 2 dt 1 di1 1 di2 + + 200 ∫ i2 dt = 7 loop 2 : − 2 dt 2 dt ⎡⎛ ⎤ 1 ⎞ 1 − s ⎢⎜ 10 + 2 s ⎟ ⎥ 2 ⎝ ⎠ ⎥ determinant : ⎢ ⎢ 1 1 200 ⎛ ⎞⎥ ⎜ s+ ⎟ ⎢ −2s s ⎠ ⎥⎦ ⎝2 ⎣ s 2 + 20s + 400 = 0, ∴ s = − 10 ± j 17.3

loop 1 : 10i1 +

Section 9.4: Natural Response of the Unforced Parallel RLC Circuit P 9.4-1 Determine v(t) for the circuit of Figure P 9.4-1 when L = 1 H and vs = 0 for t ≥ 0. The initial conditions are v(0) = 6 V and dv/dt(0) = – 3000 V/s.

L

vs(t)

+ –

+

80 Ω

v(t)



Answer: v(t) = – 2e–100t + 8e–400t V Figure P 9.4-1 Solution: v ( 0 ) = 6,

dv ( 0 ) = −3000 dt

Using operators, the node equation is: Csv +

v + R

So the characteristic equation is: s 2 + So v ( t ) = Ae −100t + Be−400t

( v −vs ) sL

L ⎛ ⎞ = 0 or ⎜ LCs 2 + s + 1⎟ v = v s ⎝ R ⎠

1 1 s+ = 0 RC LC

⇒ s1,2 = − 250 ± 2502 − 40, 000 = − 100, − 400

v ( 0) = 6 = A + B dv ( 0 ) ⎫ A = −2 = − 3000 = − 100 A − 400 B ⎬ dt ⎭ B = 8 ∴ v ( t ) = − 2e −100t + 8e−400t

t>0

25μ F

P 9.4-2 An RLC circuit is shown in Figure P 9.4-2, where v(0) = 2 V. The switch has been open for a long time before closing at t = 0. Determine and plot v(t).

t=0

+ 1 3F

v(t) –

3 4

Ω

Figure P 9.4-2 Solution: v ( 0 ) = 2, i ( 0 ) = 0 Characteristic equation s 2 + v ( t ) = Ae − t + Be−3t

1 1 s+ = 0 ⇒ s 2 + 4s + 3 = 0 ⇒ s = − 1, − 3 RC LC

Use eq. 9.5 − 12 ⇒ s1 A + s2 B = − −1A − 3B = − also have v ( 0 ) = 2 = A + B

2 −0 = − 8 1 4

From (1) & ( 2 ) get A = −1, B = 3 ∴ v ( t ) = −e − t + 3 e −3t V

v ( 0) i ( 0) − RC C

(1) ( 2)

1H

2H

P 9.4-3 Determine i1(t) and i2(t) for the circuit of Figure P 9.4-3 when i1(0) = i2(0) = 11 A.

1 Ω

i2

i1 3H

2 Ω

Figure P 9.4-3 Solution:

di1 di −3 2 = 0 dt dt di di KVL : − 3 1 + 3 2 + 2i2 = 0 dt dt KVL : i1 + 5

(1) ( 2)

in operator form

(1 + 5s ) i1 + ( −3s ) i2 = 0⎫⎪ ⎬ ( −3s ) i1 + ( 3s + 2 ) i2 = 0 ⎪⎭ Thus i1 ( t ) = Ae i 2 ( t ) = Ce

-t

−t

6

6

thus Δ =

(1 + 5s )( 3s + 2 ) − 9s 2 = 6s 2 + 13s + 2 = 0

+ Be −2t

+ De-2t

Now i1 ( 0 ) = 11 = A + B; i2 ( 0 ) = 11 = C + D from (1) & ( 2 ) get di1 ( 0 ) di2 ( 0 ) A C 33 143 = − = − − 2B ; = − = − − 20 dt 2 6 dt 6 6 which yields A = 3, B = 8, C = − 1, 0 = 12 i1 (t ) = 3e − t /6 + 8e−2t A

&

i2 (t ) = − e− t /6 + 12e−2t A

⇒ s = − 1 −2 6,

P 9.4-4 The circuit shown in Figure P 9.4-4 contains a switch that is sometimes open and sometimes closed. Determine the damping factor, α, the resonant frequency, ω0, and the damped resonant frequency, ωd, of the circuit when (a) the switch is open and (b) the switch is closed.

10 Ω

40 Ω

i(t) 2H

+ –

+ 50 Ω

20 V

v(t)

5 mF



Figure P 9.4-4 Solution: Represent this circuit by a differential equation. (R1 = 50 Ω when the switch is open and R1 = 10 Ω when the switch is closed.)

Use KCL to get i (t ) =

v (t ) d + C v (t ) R2 dt

Use KVL to get

v s = R1 i ( t ) + L

d i (t ) + v (t ) dt

Substitute to get vs =

R1 R2

v ( t ) + R1C

d L d d2 v (t ) + v ( t ) + CL 2 v ( t ) + v ( t ) dt R 2 dt dt

⎛ R1 + R 2 d2 L ⎞d = CL 2 v ( t ) + ⎜ R1C + v (t ) + v (t ) ⎟ ⎜ dt R 2 ⎟⎠ dt R2 ⎝

Finally, Compare to to get

vs CL

=

⎛ R1 R1 + R 2 d2 1 ⎞d v t + + v (t ) + v (t ) ⎟ 2 ( ) ⎜ ⎜ L R 2C ⎟ dt dt R CL 2 ⎝ ⎠

d2

d v ( t ) + ω 02 v ( t ) = f (t ) dt dt R1 R1 + R 2 1 + 2α = and ω 0 2 = L R 2C R 2CL v t + 2α 2 ( )

(a) When the switch is open α = 14.5 , ω 0 = 14.14 rad/s and ω d = j3.2 (the circuit is overdamped). (b) When the switch is closed α = 4.5 , ω 0 = 10.954 rad/s and ω d = 9.987 (the circuit is underdamped).

P 9.4-5 The circuit shown in Figure P 9.4-5 is used to detect smokers in airplanes who surreptitiously light up before they can take a single puff. The sensor activates the switch, and the change in the voltage v(t) activates a light at the flight attendant’s station. Determine the natural response v(t). Answer: v(t) = – 1.16e–2.7t + 1.16e–37.3t V Sensor

t=0

Light bulb 0.4 H 1Ω

1A

+ 1 40

F

v(t) –

Figure P 9.4-5

Solution: 1 1 s + = 0 RC LC s 2 + 40 s + 100 = 0 s2 +

s = − 2.7 , − 37.3

The initial conditions are v(0) = 0 , i (0) 1 A . vn = A1e −2.7 t + A2 e−37.3t , v(0) = 0 = A1 + A2 KCL at t = 0+ yields :

(1)

v(0+ ) 1 dv(0+ ) + i (0+ ) + =0 1 40 dt

dv(0+ ) = − 40v(0+ ) − 40i (0+ ) = − 40(1) = − 2.7 A1 − 37.3 A2 dt from (1) and (2) ⇒ A1 = −1.16 , A2 = 1.16



So v(t ) = vn (t ) = − 1.16e −2.7 t + 1.16e −37.3t

(2)

Section 9.5: Natural Response of the Critically Damped Unforced Parallel RLC Circuit 25 mH

P 9.5-1 Find vc(t) for t > 0 for the circuit shown in Figure P 9.5-1. Answer: vc(t) = (3 + 6000t)e

–2000t

30u(–t) mA

100 Ω

V

vc

Figure P 9.5-1 Solution:

t >0

di c dv + v c = 0, ic = 10−5 c dt dt 2 d vc dv ∴ + 4000 c + 4 × 106 vc = 0 2 dt dt 2 6 s + 4000 s + 4 × 10 = 0 ⇒ s = −2000, − 2000 ∴ vc ( t ) = A1e −2000t + A2te −2000t KVL a : 100ic + .025

t = 0− (Steady − State) iL = ic ( 0



)=0

= ic ( 0

+

vc ( 0− ) = 3 V = vc ( 0+ )

so vc ( 0+ ) = 3 = A1 dvc ( 0+ ) dt vc ( t ) =

= 0 = −2000 A1 + A2 ⇒ A2 = 6000

( 3 + 6000t ) e−2000t

V

)



dvc ( 0+ ) dt

= 0

+ –

10 mF

P 9.5-2 Find vc(t) for t > 0 for the circuit of Figure P 9.5-2. Assume steady-state 20 V conditions exist at t = 0–.

t=0

10 Ω

+ –

1 Ω

1H

+ 1 4

Answer: vc(t) = – 8te–2t V

F

vc



Figure P 9.5-2 Solution:

( )

dvc t =0 KCL at vc : ∫−∞ vc dt + vc + 1 4 dt d 2 vc dv ⇒ + 4 c + 4vc = 0 dt dt

t>0

s 2 + 4 s + 4 = 0, s = −2, − 2

t = 0− (Steady-State)

⇒ vc ( t ) = A1 e −2t + A2 t e −2t

vc ( 0− ) = 0 = vc ( 0+ ) & iL ( 0− ) =

20 V = 2 A = iL ( 0+ ) 10 Ω

Since vc ( 0+ ) = 0 then ic ( 0+ ) = −iL ( 0+ ) = −2 A ∴ So vc ( 0

) dv ( 0 ) +

= 0 = A1

+

= − 8 = A2 dt ∴ vc ( t ) = − 8te −2t V c

dvc ( 0+ ) dt

ic ( 0+ ) = = −8 V S 1 4

10 mH P 9.5-3 Police often use stun guns to incapacitate potentially dangerous felons. The + t=0 hand-held device provides a series of high+ 4 v R = 106 Ω 10 V – voltage, low-current pulses. The power of the C – pulses is far below lethal levels, but it is enough to cause muscles to contract and put Figure P 9.5-3 the person out of action. The device provides a pulse of up to 50,000 V, and a current of 1 mA flows through an arc. A model of the circuit for one period is shown in Figure P 9.5-3. Find v(t) for 0 < t < 1 ms. The resistor R represents the spark gap. Select C so that the response is critically damped.

Solution:

Assume steady − state at t = 0− ∴ vc ( 0− ) = 10 4 V & iL ( 0− ) = 0 t>0

Also ∴ 0.01C

diL + 106 iL = 0 (1) dt ⎡ d 2iL di ⎤ dvc : iL = −C = −C ⎢.01 2 + 106 L ⎥ dt dt dt ⎦ ⎣

KVL a : − vc + .01

( 2)

d 2iL di + 106 C L + iL = 0 2 dt dt

Characteristic eq. ⇒ 0.01C s 2 + 106 s + 1 = 0 ⇒ s =

−106 C ±

(10 C ) 6

2 (.01C )

for critically damped: 1012 C2 − .04C = 0 ⇒ C = 0.04 pF ∴ s = −5 × 107 , −5 × 107 So iL ( t ) = A1e −5×10 t + A2te−5×10 7

7t

diL + 0 ) = 100 ⎡⎣vc ( 0+ ) − 106 iL ( 0+ ) ⎤⎦ = 106 A ( s dt di ( 0 ) 7 So i L ( 0 ) = 0 = A1 and L = 106 = A2 ∴ iL ( t ) = 106 te−5×10 t A dt Now from (1) ⇒

Now v ( t ) = 106 iL ( t ) = 1012 te−5×10 t V 7

2

− 4 (.01C )

P 9.5-4 Reconsider Problem P 9.4-1 when L = 640 mH and the other parameters and conditions remain the same. vs(t) – 250t Answer: v(t) = (6 – 1500t)e V

L

+ –

+

80 Ω

v(t)



Figure P 9.4-1 Solution:

s2 +

1 1 1 1 s+ = 0 with = 500 and = 62.5 × 103 yields s = −250, −250 RC LC RC LC v ( t ) = Ae −250t + Bte −250t v ( 0) = 6 = A dv ( 0 ) = −3000 = − 250 A + B ⇒ B = − 1500 dt ∴ v ( t ) = 6e −250t − 1500te −250t

25μ F

P 9.5-5 An automobile ignition uses an electromagnetic trigger. The RLC trigger circuit shown in Figure P 9.5-5 has a step input of 6 V, and v(0) = 2 V and i(0) = 0. The resistance R must be selected from 2 Ω < R < 7 Ω so that the current i(t) exceeds 0.6 A for greater than 0.5 s in order to activate the trigger. A critically damped response i(t) is required to avoid oscillations in the trigger current. Select R and determine and plot i(t).

1H Trigger

di t + Ri + 2 + 4∫0 idt = 6 

dt v (t )

Figure P 9.5-5

(1)

taking the derivative with respect to t :

d 2i di + R + 4i = 0 2 dt dt

Characteristic equataion: s 2 + Rs + 4 = 0 Let R = 4 for critical damping ⇒

( s + 2)

2

=0

So i ( t ) = Ate −2t + Be −2t i ( 0) = 0 ⇒ B = 0 di ( 0 ) = 4 − R (i ( 0)) = 4 − R ( 0) = 4 = A dt = 4te −2t A

from (1) ∴ i (t )

F

+ – v(t)

6 u(t) V +–

Solution:

KVL :

1 4

i

R

Section 9-6: Natural Response of an Underdamped Unforced Parallel RLC Circuit P 9.6-1 A communication system from a space station uses short pulses to control a robot operating in space. The transmitter circuit is modeled in Figure P 9.6-1. Find the output voltage vc(t) for t > 0. Assume steadystate conditions at t = 0–.

0.8 H

vc

250 Ω

+ –

5 × 10-6 F

250 Ω

t=0

Answer: vc(t) = e– 400t [3 cos 300t + 4 sin 300t] V

+ –

6V

Figure P 9.6-1 Solution: t>0

KCL at vc : also :

Solving for i

L

vc

250

+ iL + 5 ×10−6

vc = 0.8

dvc = 0 dt

diL dt

in (1) & plugging into ( 2 )

d 2 vc dv +800 c + 2.5×105 vc = 0 ⇒ s 2 +800s+ 250,000 = 0, s = −400± j 300 2 dt dt ∴ v (t ) = e c

t = 0−

−400t ⎡ ⎤ ⎣ A1 cos300t + A 2 sin 300t ⎦

(Steady − State)

−6 V −6 = A = iL ( 0+ ) 500 500 Ω − + 6 = 3 V = vc ( 0+ ) vc ( 0 ) = 250 −6 500

iL ( 0− ) =

(

Now from (1) : So vc ( 0

+

)

dvc ( 0+ )

)

dvc ( 0+ )

dt = 3 = A1

= − 2 × 105 iL ( 0+ ) − 800vc ( 0+ ) = 0

= 0 = − 400 A1 + 300 A2 ⇒ A2 = 4 dt ∴ vc ( t ) = e −400t [3cos 300t + 4sin 300t ] V

( 2)

(1)

P 9.6-2 The switch of the circuit shown in Figure P 9.6-2 is opened at t = 0. Determine and plot v(t) when C = 1/4 F. Assume steady state at t = 0–. Answer: v(t) = – 4e–2t sin 2t V 3Ω

6V

t=0

+ –



+ 1 2

H

C



v(t)

Figure P 9.6-2 Solution: t = 0−

i (0) = 2 A v (0) = 0 t = 0− KCL at node a: t

v dv 1 + C + ∫ vdt + i ( 0 ) = 0 (1) dt L 0 1 in operator form have v + Csv + with s 2 + 4s + 8 = 0

1 v + i ( 0 ) = 0 or Ls

1 ⎞ ⎛ 2 1 ⎜s + s+ ⎟v = 0 C LC ⎠ ⎝

⇒ s = −2 ± j 2

v ( t ) = e −2t [ B1 cos 2t + B2 sin 2t ] v ( 0 ) = 0 = B1 dv ( 0 )

1 ⎡ −i ( 0 ) − v ( 0 ) ⎤⎦ = −4 [ 2] = −8 = 2 B2 or B2 = −4 dt C ⎣ So v ( t ) = −4e −2t sin 2 t V From (1) ,

=

P 9.6-3 A 240-W power supply circuit is shown in Figure P 9.6-3a. This circuit employs a large inductor and a large capacitor. The model of the circuit is shown in Figure P 9.6-3b. Find iL(t) for t > 0 for the circuit of Figure P 9.6-3b. Assume steady-state conditions exist at t = 0–. Answer: iL(t) = e–2t(– 4 cos t + 2 sin t) A

4H

iL

t=0 14



F

8Ω 4Ω

7A

(b)

(a)

Figure P 9.6-3 Solution: t>0

1 dvc vc + + iL = 0 4 dt 2 4diL KVL : vc = + 8 iL ( 2) dt KCL at vc :

d 2iL di + 4 L + 5iL = 0 ⇒ s 2 + 4s + 5 = 0 ⇒ s = − 2 ± i 2 dt dt −2 t = e [ A1 cos t + A2 sin t ]

( 2 ) into (1) yields ∴ iL ( t ) t = 0−

( Steady − State )

vc ( 0− ) 2

⎛ 48 ⎞ = 7 ⎜⎜ ⎟⎟ ⎝ 4 8+ 2⎠

⇒ vc ( 0− ) = 8 V = vc ( 0+ ) −8 V = −4 A = iL ( 0+ ) 2Ω diL ( 0+ ) vc ( 0+ ) 8V A + = − 2iL ( 0 ) = − 2 ( −4 ) = 10 dt 4 4 s iL ( 0− ) =

∴ from ( 2 ) So i L ( 0 ∴ iL ( t )

+

) = -4 = A

1

and

diL ( 0+ )

dt −2 t = e [ −4 cos t + 2 sin t

= 10 = −2 A1 + A2 ⇒ A2 = 2

]A

(1)

600

P 9.6-4 The natural response of a parallel RLC circuit is measured and plotted as shown in Figure P 9.6-4. Using this chart, determine an expression for v(t).

500 400 300 200

Hint: Notice that v(t) = 260 mV at t = 5 ms and that v(t) = – 200 mV at t = 7.5 ms. Also, notice that the time between the first and third zerocrossings is 5 ms.

v(t) (mV)

100 0 – 100

Answer: v(t) = 544e– 276t sin 1257t V

– 200 – 300 – 400

0

5

10

15

20 25 Time (ms)

Figure P 9.6-4 Solution: The response is underdamped so

∴ v ( t ) = e − α t [ k1 cos ωt + k2 sin ωt ] + k3 v ( ∞ ) = 0 ⇒ k3 = 0, v ( 0 ) = 0 ⇒ k1 = 0 ∴ v ( t ) = k2 e − α t sin ωt From Fig. P 9.6-4 t ≈ 5ms ↔ v ≈ 260mV (max) t ≈ 7.5ms ↔ v ≈ −200 mV (min) ∴ distance between adjacent maxima is ≈ ω = so 0.26 = k2 e −0.2 = k2 e

−α (.005 )

−α

2π = 1257 rad s T

sin (1257 (.005 ) )

(.0075) sin

(1)

( 1257 (.0075) ) ( 2 )

Dividing (1) by (2) gives

⎛ sin ( 6.29 rad ) ⎞ − 1.3 = eα ( 0.0025) ⎜⎜ ⎟⎟ ⇒ ⎝ sin ( 9.43 rad ) ⎠ From (1) k 2 = 544 so v ( t ) = 544e−267t sin1257t

e0.0025 α = 1.95 ⇒ α = 267

( approx. answer )

30

P 9.6-5 The photovoltaic cells of the proposed space station shown in Figure P 9.6-5a provide the voltage v(t) of the circuit shown in Figure P 9.6-5b. The space station passes behind the shadow of earth (at t = 0) with v(0) = 2 V and i(0) = 1/10 A. Determine and sketch v(t) for t > 0.

Photocells

(a) i 5Ω

+

v

1 10 F

2H



Space station electric motors

The photovoltaic cells connected in parallel

(b)

Figure P 9.6-5 Solution:

v ( 0 ) = 2 V and i ( 0 ) = 1 A 10

Char. eq. ⇒ s 2 +

1 1 s+ = 0 or s 2 + 2 s + 5 = 0 thus the roots are s = −1 ± j 2 RC LC

So have v(t ) = e − t ⎡ B cos 2t + B2 sin 2t ⎤ now v(0+ ) = 2 = B1 ⎣ 1 ⎦

Need So

dv ( 0+ )

dv ( 0 dt

dt +

v ( 0+ ) 1 1 V + + = ic ( 0 ) . KCL yields ic ( 0 ) = − − i ( 0+ ) = − C 5 2 s

) = 10 ⎛ − 1 ⎞ = − B + 2 B ⎜ ⎟ ⎝ 2⎠

1

2

⇒ B2 = − 3

2

3 Finally, we have v ( t ) = 2e − t cos 2t − e− t sin 2t V 2

t>0

Section 9-7: Forced Response of an RLC Circuit P 9.7-1 Determine the forced response for the inductor current if when (a) is = 1 A, (b) is = is u(t) A 0.5t A, and (c) is = 2e–250t A for the circuit of Figure P 9.7-1.

100 65 Ω

10 mH

i

Figure P 9.7-1 Solution:

v dv + iL + C R dt diL KVL : v = L dt d 2iL L diL + iL + LC is = R dt dt 2 KCL : is =

(a)

is = l u (t ) ∴ assume i f = A d 2iL 1 diL 1 iL = is + + 2 dt RC dt LC 1 to get: 0 + 0 + A = 1 ⇒ A = 1× 10−5 = i f −3 .01 1 10 × ( )( )

Let iL = i f = A in

(b)

is = 0.5t u (t ) ∴ assume i f = At + B 0+ A

65 1 + ( At + B ) = 0.5 t (.01)(.001) (100 ) (.001)

⇒ 650 A + 100000 B = 0 and 100000 At = 0.5t A = 5 ×10−6 B = 3.25 × 10−8 i f = 5 × 10−6 t − 3.25 × 10−8 A

(c)

is = 2e −250t Assumming i f = Ae −250t does not work because i f cannot have the same form as is ∴ we choose i f = Bte−250t Be−250t −250 Bte−250t Bte−250t + + = 2 e −250t RC RC LC 150 B = 2 B = 0.0133 if

= 0.0133 te −250t A

1 mF



P 9.7-2 Determine the forced response for the capacitor voltage, vf, for the circuit of Figure P 9.7-2 when

vs u(t) V –+

0.1 H + v –

833.3 μ F

(a) vs = 2 V, (b) vs = 0.2t V, and (c) vs = 1e–30t V. Figure P 9.7-2 Solution:

Represent the circuit by the differential equation: (a)

vs = 2 ∴ assume v f = A Then 0 + 0 + 12000 A = 2 so A = 1

(b)

d 2 v R dv 1 v = vs + + dt L dt LC

6000

= vf

vs = 0.2t ∴ assume v f = At + B 70 A + 12000 At + 12000 B = 0.2t ⇒ 70 A + 12000 B = 0 and 12000 At = 0.2t 1 70 A , B= 60000 12000 t ∴v f = + 350 V 60000 A=

(c)

vs = e −30t



B = 350

∴ assume v f = Ae−30t

900 A − 2100 Ae −30t + 12000 Ae−30t = e−30t vf =

e −30t V 10800

⇒ 10800 Ae−30t = e −30t



A =

1 10800

P 9.7-3

A circuit is described for t > 0 by the equation d 2v dv + 5 + 6v = vs 2 dt dt

Find the forced response vf for t > 0 when (a) vs = 8 V, (b) vs = 3e–4t V, and (c) vs = 2e–2t V. 3 Answer: (a) vf = 8/6 V (b) vf = e– 4t V (c) vf = 2te–2t V 2 Solution:

(a)

(b)

d 2v dv + 5 + 6v = 8 so we try v f = B. 2 dt dt Substituting v f = B into the differential equation gives 6 B = 8 ∴ vf = 8 / 6 V

The differential equation is

d 2v dv + 5 + 6v = 3e −4t so we try v f = Be −4t . 2 dt dt into the differential equation gives

The differential equation is Substituting v f = Be −4t

(−4) 2 B + 5(−4) B + 6 B = 3 ⇒ B = 3 / 2 ∴ vf = 3 / 2e −4t

(c)

d 2v dv + 5 + 6v = 2e −2t so we try vf = Bte−2t 2 dt dt (since − 2 is a natural frequency). Substituting vf = Bte −2t into the differential equation gives The differential equation is

⇒ (4t − 4) B + 5B (1 − 2t ) + 6 Bt = 2 ⇒ B = 2 ∴ vf = 2te −2t

Section 9-8: Complete Response of an RLC Circuit P 9.8-1

Determine i(t) for t > 0 for the circuit shown in Figure P 9.8-1. iL

i 2 kΩ + vc 1 kΩ –

11 mA t=0

6.25 H 1 μF

+ –

4V

Figure P 9.8-1 Solution: First, find the steady state response for t < 0, when the switch is open. Both inputs are constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. After a source transformation at the left of the circuit:

i L ( 0) = and

22 − 4 = 6 mA 3000

v C ( 0) = 4 V

After the switch closes Apply KCL at node a: vC R

+C

d vC + iL = 0 dt

Apply KVL to the right mesh:

L

d d i L + Vs − vC = 0 ⇒ vC = L i L + Vs dt dt

After some algebra: Vs d2 1 d 1 d2 d 16 ⎛ 4 ⎞ i + i + i = − ⇒ i + (103 ) i L + ⎜ × 106 ⎟ i L = − × 103 L L 2 L 2 L dt R C dt LC R LC dt dt 25 ⎝ 25 ⎠ The characteristic equation is ⎛ 4 ⎞ s 2 + (103 ) s + ⎜ ×106 ⎟ = 0 ⇒ s1,2 = −200, − 800 rad/s ⎝ 25 ⎠

After the switch closes the steady-state inductor current is iL(∞) = -4 mA so i L ( t ) = −0.004 + A1 e −200 t + A2 e −800 t 4 ⎛ 4 ⎞d ⎡⎣( −200 ) A1 e −200 t + ( −800 ) A2 e−800 t ⎤⎦ + 4 vC (t ) = ⎜ ⎟ i L (t ) + 4 = 25 ⎝ 25 ⎠ dt = ( −32 ) A1 e −200 t + ( −128 ) A2 e −800 t + 4

Let t = 0 and use the initial conditions:

0.006 = −0.004 + A1 + A2

⇒ 0.01 = A1 + A2

4 = ( −32 ) A1 + ( −128 ) A2 + 4 ⇒

A1 = ( −4 ) A2

So A1 = 8.01 and A2 = 2.00 and i L ( t ) = −0.004 + 8.01 e −200 t + 2.00 e −800 t A

v C ( t ) = ( −104 ) e −200 t + (104 ) e −800 t + 4 V

i (t ) =

vC (t ) 1000

= ( −10 ) e −200 t + (10 ) e−800 t + 0.004 A

Determine i(t) for t > 0 for the circuit shown in Figure P 9.8-2. d2 d Hint: Show that 1 = 2 i (t ) + 5 i (t ) + 5i (t ) for t > 0 dt dt P 9.8-2



2u(t) – 1 V

+ –

4Ω + v(t) –

0.25 F

i(t)

4H

Figure P 9.8-2 Answer:

–3.62t

i(t) = 0.2 + 0.246 e

– 0.646 e–1.38t A

for t > 0.

Solution: First, find the steady state response for t < 0. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit.

i ( 0) =

−1 = 0.2 A 1+ 4

and v (0) =

4 ( −1) = −0.8 V 1+ 4

For t > 0 Apply KCL at node a: v − Vs d +C v+i = 0 R1 dt

Apply KVL to the right mesh: R2 i + L

d d i − v = 0 ⇒ v = R2 i + L iL dt dt

After some algebra: L + R1 R 2C d R1 + R 2 d2 Vs i + i + i= 2 dt R1 L C dt R1 L C R1 L C The forced response will be a constant, if = B so 1 =



d2 d i +5 i +5i =1 2 dt dt

d2 d B + 5 B + 5B ⇒ B = 0.2 A . 2 dt dt

To find the natural response, consider the characteristic equation: 0 = s 2 + 5 s + 5 = ( s + 3.62 )( s + 1.38 )

The natural response is

in = A1 e−3.62 t + A2 e−1.38 t so

i ( t ) = A1 e−3.62 t + A2 e−1.38 t + 0.2 Then d ⎛ ⎞ v ( t ) = ⎜ 4 i ( t ) + 4 i ( t ) ⎟ = −10.48 A1 e−3.62 t − 1.52 A2 e−1.38 t + 0.8 dt ⎝ ⎠ At t=0+

−0.2 = i ( 0 + ) = A1 + A2 + 0.2 −0.8 = v ( 0 + ) = −10.48 A1 − 1.52 A2 + 0.8

so A1 = 0.246 and A2 = -0.646. Finally i ( t ) = 0.2 + 0.246 e−3.62 t − 0.646 e−1.38 t A

P 9.8-3 Answer:

Determine v1(t) for t > 0 for the circuit shown in Figure P 9.8-3. 4

3

v1 (t ) = 10 + e−2.4 × 10 t − 6e−4 × 10 t V for t > 0 1 kΩ

10 V

+ –

1 kΩ

+ v1(t) –

1/6 μ F

+ v2(t) –

t=0 1/16 μ F

Figure P 9.8-3 Solution: First, find the steady state response for t < 0. The input is constant so the capacitors will act like an open circuits at steady state.

v1 ( 0 ) = and

1000 (10 ) = 5 V 1000 + 1000 v2 ( 0 ) = 0 V

For t > 0, Node equations:

v1 − 10 ⎛ 1 v −v ⎞ d + ⎜ × 10−6 ⎟ v1 + 1 2 = 0 1000 ⎝ 6 1000 ⎠ dt ⎛1 ⎞ d ⇒ 2 v1 + ⎜ × 10−3 ⎟ v1 − 10 = v2 ⎝6 ⎠ dt

v1 − v2 ⎛ 1 ⎞ d = ⎜ × 10−6 ⎟ v2 1000 ⎝ 16 ⎠ dt ⎛1 ⎞ d ⇒ v1 − v2 = ⎜ × 10−3 ⎟ v2 ⎝ 16 ⎠ dt After some algebra: d2 d v + ( 2.8 × 104 ) v1 + ( 9.6 × 107 ) v1 = 9.6 × 108 2 1 dt dt The forced response will be a constant, vf = B so d2 d B + ( 2.8 × 104 ) B + ( 9.6 ×107 ) B = 9.6 × 108 ⇒ B = 10 V . 2 dt dt To find the natural response, consider the characteristic equation:

s 2 + ( 2.8 ×104 ) s + ( 9.6 × 107 ) = 0 ⇒ s1,2 = −4 × 103 , −2.4 × 104 The natural response is 3

vn = A1 e −4×10 t + A2 e−2.4×10

4t

so 3

v1 ( t ) = A1 e−4×10 t + A2 e−2.4×10

4t

+ 10

At t = 0 5 = v1 ( 0 ) = A1 e

−4×103 ( 0 )

+ A2 e

−2.4×104 ( 0 )

+ 10 = A1 + A2 + 10

(1)

Next ⎛1 ⎞ d 2 v1 + ⎜ ×10−3 ⎟ v1 − 10 = v2 ⎝6 ⎠ dt



d v1 = 12000v1 + 6000 v2 − 6 × 104 dt

At t = 0 d v1 ( 0 ) = 12000v1 ( 0 ) + 6000 v2 ( 0 ) − 6 × 104 = 12000 ( 5 ) + 6000 ( 0 ) − 6 × 104 = 0 dt so 3 4 d v1 ( t ) = A1 −4 ×103 e−4×10 t + A2 −2.4 × 104 e−2.4×10 t dt

(

)

(

)

At t = 0+ 0=

d −4×103 ( 0 ) −2.4×104 ( 0 ) v1 ( 0 ) = A1 −4 × 103 e + A2 −2.4 × 104 e = A1 −4 × 103 + A2 −2.4 × 104 dt

(

)

(

)

(

so A1 = -6 and A2 = 1. Finally v1 ( t ) = 10 + e −2.4 ×10

4t

3t

− 6 e −4 ×10

V for t > 0

)

(

)



P 9.8-4 Find v(t) for t > 0 for the circuit shown in Figure P 9.8-4 when v(0) = 1 V and iL(0) = 0. +

5 cos t V –

Answer:

v = 25e −3t −

1 ⎡ 429e−4t − 21cos t + 33 sin t ⎤⎦ V 17 ⎣

0.5 H iL



1 12 F

+ –

v

Figure P 9.8-4

Solution: t > 0

di ⎛ ⎞ KCL at top node : ⎜ 0.5 L − 5cos t ⎟ + iL + 1 dv = 0 12 dt dt ⎝ ⎠ di dv KVL at right loop : 0.5 L = 1 +v ( 2) 12 dt dt

d d

d 2iL diL 1 d 2 v + + = −5sin t (3) dt 2 dt 12 dt 2 d 2iL d 2 v dv 1 ⇒ 0.5 2 = + ( 4) 12 dt 2 dt dt

dt

of (1) ⇒ 0.5

dt

of ( 2 )

Solving for

d 2iL di in ( 4 ) and L in ( 2 ) & plugging into ( 3) 2 dt dt

d 2v dv + 7 + 12v = − 30sin t 2 dt dt

⇒ s 2 + 7 s + 12 = 0 ⇒ s = − 3, − 4

so v(t ) = A1e−3t + A2 e−4t + v f

Try v f = B1 cos t + B2 sin t & plug into D.E., equating like terms yields B1 = 21 , B2 = − 33 17 17

t = 0+

5 −1 dv(0+ ) ic ( 0 ) = = 2A ∴ = 2 = 24 V 1 s 11 dt 12 ⎫ So v(0+ ) = 1 = A1 + A2 + 21 A1 = 25 17 ⎪ ⎬ 429 dv(0+ ) ⎪ A2 = − = 24 = − 3 A1 − 4 A2 − 33 17 17 dt ⎭ ∴ v(t ) = 25e −3t − 1 ( 429e −4t − 21cos t + 33sin t ) V 17 +

(1)

Find v(t) for t > 0 for the circuit of Figure P 9.8-5.

P 9.8-5

Answer: v(t) = [ – 16e–t + 16e – 3t + 8]u(t) + [16e–(t–2) – 16e–3(t–2) – 8]u(t – 2) V 1 3

F

+ 2[u(t) – u(t – 2)] A



v

1H



Figure P 9.8-5 Solution: Use superposition − first consider the 2u (t ) source.

KVL at right mesh : vc + siL + 4(iL − 2) = 0 also : iL = (1/ 3) svc ⇒ vc = (3 / s ) iL

(1)

(2)

Plugging (2) into (1) yields ( s 2 + 4s + 3) iL = 0 , roots : s = −1 , −3 So iL (t ) = A1e − t + A2 e−3t −

t = 0 ⇒ circuit is dead ∴ vc (0) = iL (0) = 0 Now from (1)

diL (0+ ) = 8 − 4iL (0+ ) − vC (0+ ) = 8 A/s dt

So iL (0) = 0 = A1 + A2 ⎫ ⎪ ⎬ A1 = 4 , A2 = −4 diL (0) = 8 = − A1 − 3 A2 ⎪ dt ⎭ −t −3t ∴ iL (t ) = 4e − 4e ∴ v1 (t ) = 8 − 4 iL (t ) = 8 − 16e − t + 16e −3t V Now for 2u (t − 2) source, just take above expression and replace t → t − 2 and flip signs ∴ v2 (t ) = −8 + 16e − (t − 2) − 16 e −3(t − 2) V ∴ v(t ) = v1 (t ) + v2 (t ) v(t ) = ⎡⎣8 − 16e − t + 16e −3t ⎤⎦ u (t ) + ⎡⎣ −8 + 16e− (t − 2) − 16 e−3( t − 2) ⎤⎦ u (t − 2) V

P 9.8-6 An experimental space station power supply system is modeled by the circuit shown in Figure P 9.8-6. Find v(t) for t > 0. Assume steady-state conditions at t = 0–. t=0

(10 cos t)u(t) V – +

0.125 F 4Ω

4H

+ v(t) –

+ –



5V

i(t)

Figure P 9.8-6

Solution: First, find the steady state response for t < 0, when the switch is closed. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit.

i ( 0) = − and

5 = −1.25 mA 4

v (0) = 5 V

After the switch closes Apply KCL at node a: v d + 0.125 v = i dt 2 Apply KVL to the right mesh: −10 cos t + v + 4 After some algebra:

d i+4i =0 dt d2 d v + 5 v + 6 v = 20 cos t 2 dt dt

The characteristic equation is s 2 + 5 s + 6 = 0 ⇒ s1,2 = −2, − 3 rad/s Try

vf = A cos t + B sin t

d2 d A cos t + B sin t ) + 5 ( A cos t + B sin t ) + 6 ( A cos t + B sin t ) = 20 cos t 2 ( dt dt ( − A cos t − B sin t ) + 5 ( − A sin t + B cos t ) + 6 ( A cos t + B sin t ) = 20 cos t

( − A + 5 B + 6 A) cos t + ( − B − 5 A + 6 B ) sin t = 20 cos t vf = 2 cos t + 2 sin t

So A =2 and B =2. Then

v ( t ) = 2 cos t + 2 sin t + A1 e −2 t + A2 e−3 t

v (t ) d + 0.125 v ( t ) = i ( t ) ⇒ dt 2

Next

d v (t ) = 8 i (t ) − 4 v (t ) dt

d V ⎛ 5⎞ v ( 0 ) = 8 i ( 0 ) − 4 v ( 0 ) = 8 ⎜ − ⎟ − 4 ( 5 ) = −30 dt s ⎝ 4⎠ Let t = 0 and use the initial conditions: 5 = v ( 0 ) = 2 cos 0 + 2 sin 0 + A1 e −0 + A2 e−0 = 2 + A1 + A2

d v ( t ) = −2 sin t + 2 cos t − 2 A1 e −2 t − 3 A2 e−3 t dt −30 =

d v ( 0 ) = −2 sin 0 + 2 cos 0 − 2 A1 e −0 − 3 A2 e −0 = 2 − 2 A1 − 3 A2 dt

So A1 = -23 and A2 = 26 and v ( t ) = 2 cos t + 2 sin t − 23 e −2 t + 26 e −3 t



P 9.8-7 Find vc(t) for t > 0 in the circuit of Figure P 9.8-7 when (a) C = 1/18 F, (b) C = 1/10 F, and (c) C = 1/20 F.



2u(t) A

Answer: (a) vc(t) = 8e–3t + 24te – 3t – 8 V (b) vc(t) = 10e–t – 2e–5t – 8 V (c) vc(t) = e – 3t(8 cos t + 24 sin t) – 8 V

C

2H a

+ v(t) –

i(t)

Figure P 9.8-7 Solution: First, find the steady state response for t < 0, when the switch is closed. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. i ( 0) = 0 A

and

v ( 0) = 0 V

After the switch closes d v=i dt Apply KVL to the right mesh: d 8 i + v + 2 i + 4 (2 + i) = 0 dt d 12 i + v + 2 i = −8 dt ⎛ 1 ⎞ d2 d 4 After some algebra: v + ( 6) v + ⎜ ⎟v = − 2 dt dt C ⎝2C⎠ The forced response will be a constant, vf = B so Apply KCL at node a:

C

⎛ 1 ⎞ d2 d 4 B + ( 6) B + ⎜ ⎟B = − 2 dt dt C ⎝2C⎠

(a)

⇒ B = −8 V

d2 d v + ( 6 ) v + ( 9 ) v = −72 2 dt dt 2 The characteristic equation is s + 6 s + 9 = 0 ⇒ s1,2 = −3, −3

When C = 1/18 F the differential equation is Then v ( t ) = ( A1 + A2 t ) e−3t − 8 . Using the initial conditions:

0 = v ( 0 ) = ( A1 + A2 ( 0 ) ) e0 − 8 ⇒ 0= So (b)

d v ( 0 ) = −3 ( A1 + A2 ( 0 ) ) e0 + A2 e0 dt

A1 = 8 ⇒

A2 = 24

v ( t ) = ( 8 + 24 t ) e −3t − 8 V for t > 0

d2 d When C = 1/10 F the differential equation is 2 v + ( 6 ) v + ( 5 ) v = −40 dt dt 2 The characteristic equation is s + 6 s + 5 = 0 ⇒ s1,2 = −1, −5 Then v ( t ) = A1 e − t + A2 e−5 t − 8 . Using the initial conditions: 0 = v ( 0 ) = A1 e0 + A2 e0 − 8 ⇒ A1 + A2 = 8 ⎫ ⎪ ⎬ ⇒ A1 = 10 and A2 = −2 d 0 0 0 = v ( 0 ) = − A1 e − 5 A2 e ⇒ − A1 − 5 A2 = 0 ⎪ dt ⎭ So v ( t ) = 10 e − t − 2 e −5 t − 8 V for t > 0

(c)

d2 d v + ( 6 ) v + (10 ) v = −80 2 dt dt 2 The characteristic equation is s + 6 s + 10 = 0 ⇒ s1,2 = −3 ± j

When C = 1/20 F the differential equation is Then v ( t ) = e −3 t ( A1 cos t + A2 sin t ) − 8 .

Using the initial conditions: 0 = v ( 0 ) = e0 ( A1 cos 0 + A2 sin 0 ) − 8 ⇒ 0= So

A1 = 8

d v ( 0 ) = −3 e0 ( A1 cos 0 + A2 sin 0 ) + e0 ( − A1 sin 0 + A2 cos 0 ) ⇒ dt v ( t ) = e −3 t ( 8cos t + 24 sin t ) − 8 V for t > 0

A2 = 24

P 9.8-8 Find vc(t) for t > 0 for the circuit shown in Figure P 9.8-8.





Hint:

d2 d 2 = 2 vc (t ) + 6 vc (t ) + 2vc (t ) for t > 0 dt dt Answer: vc(t) = 0.123e – 5.65t + 0.877e – 0.35t + 1 V for t > 0.

– 1 4 u(t) + 1 2 A

+ 1 4

2H

F

iL(t)

Figure P 9.8-8

Solution:

The circuit will be at steady state for t0: Apply KCL at node b to get: 1 1 d 1 1 d = i L (t ) + v C (t ) ⇒ i L (t ) = − v C (t ) 4 4 dt 4 4 dt Apply KVL to the right-most mesh to get: 4 i L (t ) + 2

d ⎛1 d ⎞ i L (t ) = 8 ⎜ v C (t ) ⎟ + v C (t ) dt ⎝ 4 dt ⎠

Use the substitution method to get d ⎛1 1 d ⎛1 1 d ⎞ ⎞ ⎛1 d ⎞ 4⎜ − vC ( t ) ⎟ + 2 ⎜ − vC ( t ) ⎟ = 8 ⎜ vc ( t ) ⎟ + v ( t ) dt ⎝ 4 4 dt ⎝ 4 4 dt ⎠ ⎠ ⎝ 4 dt ⎠ c d2 d 2= v C (t ) + 6 v C (t ) + 2 v C (t ) or dt dt 2 d d2 The forced response will be a constant, vC= B so 2 = 2 B + 6 B + 2 B ⇒ B = 1 V . dt dt To find the natural response, consider the characteristic equation: 0 = s 2 + 6 s + 2 = ( s + 5.65 )( s + 0.35 )

vC(t) –

The natural response is

vn = A1 e −5.65t + A e −0.35t 2

vC ( t ) = A e

so

1

−5.65 t

+A e

−0.35t

2

+1

Then i L (t ) = At t=0+

1 1 d 1 + vC ( t ) = + 1.41A e −5.65t + 0.0875 A e−0.35t 1 2 4 4 dt 4 2 = vC ( 0 + ) = A + A + 1 1

2

1 1 = i L ( 0 + ) = + 1.41A1 + 0.0875 A 2 2 4 so A1 = 0.123 and A2 = 0.877. Finally vC ( t ) = 0.123 e −5.65 t + 0.877 e −0.35 t + 1 V

P 9.8-9 In Figure P 9.8-9, determine the inductor current i(t) when is = 5u(t) A. Assume that i(0) = 0, vc(0) = 0.

is



Answer: i(t) = 5 + e–2t [ – 5 cos 5t – 2 sin 5t] A

8 29 H

1 8F

i

Figure P 9.8-9 Solution: dv v +i+ =i 2 s dt d 2i ⎛ L ⎞ di LC + i + ⎜ ⎟ = 5 u (t ) dt ⎝ 2 ⎠ dt

KCL : C

1 d 2i ⎛ 4 ⎞ di + i + ⎜ ⎟ = 5 u (t ) 29 dt ⎝ 29 ⎠ d t d 2i di + 4 + i = 145 u ( t ) dt dt

Characteristic eqn: s 2 + 4s + 29 = 0 ⇒

roots : s = − 2 ± j5 ∴ in = e −2t [ A cos 5t + B sin 5t ] and i f = 145 = 5 29 −2t So i (t ) = 5 + e [ A cos 5t + B sin 5t ] Now i (0) = 0 = A + 5 ⇒ A = −5 di (0) = 0 = −2 A + 5 B ⇒ B = − 2 dt

P 9.8-10 Railroads widely use automatic identification of railcars. When a train passes a tracking station, a wheel detector activates a radio-frequency module. The module’s antenna, as shown in Figure P 9.8-10a, transmits and receives a signal that bounces off a transponder on the locomotive. A trackside processor turns the received signal into useful information consisting of the train’s location, speed, and direction of travel. The railroad uses this information to schedule locomotives, trains, crews, and equipment more efficiently. One proposed transponder circuit is shown in Figure P 9.810b with a large transponder coil of L = 5 H. Determine i(t) and v(t). The received signal is

is = 9 + 3e–2tu(t) A.

(a)

Vehicle-mounted transponder tag

Wheel detector input

Antenna

+



(b)

0.5 F

L



v

1.5 Ω

i is

0.5 Ω

Figure P 9.8-10 Solution: t = 0−

t > 0

2 × 9 = 6 A = i (0+ ) 2 +1 1 & v (0− ) = × 9× 1.5 = 4.5 V = v(0+ ) 2 +1 i (0− ) =

dv v + =i dt 1.5 s 5di dv v KVL : v + (0.5 + ) (0.5) = + i dt 1.5 dt KCL at middle node: i + 0.5

(1) (2)

Solving for i in (1) and plugging into (2) yields d d 2 v ⎛ 49 ⎞ dv ⎛ 4 ⎞ ⎛2⎞ is where i = 9 + 3e−2t A 2 + + = + v i ⎜ ⎟ ⎜ ⎟ ⎜ ⎟s s 5 dt 2 ⎝ 30 ⎠ dt ⎝ 5 ⎠ dt ⎝ ⎠ 49 4 So the characteristic equation is s 2 + s + = 0 and its roots are s = − 0.817 ± j 0.365 30 5 −.817 t vn (t ) = e [ A1 cos(0.365 t ) + A2 sin(0.365 t )] Try vf (t ) = B0 + B1e −2t and substitute vf (t ) into the differential equation and equate like terms

to get B0 = 4.5, B1 = −7.04

Then v(t ) = e −.817 t [ A1 cos(0.365t ) + A2 sin(0.365t ) ] + 4.5 − 7.04e−2t Now using the initial conditions gives v(0) = 4.5 = A1 + 4.5 − 7.04 ⇒ A1 = 7.04 4 dv(0) 4 = 2is (0) − 2i (0) − v(0) = 2(9 + 3) − 2(6) − (4.5) = 6 3 dt 3 ∴ 6 = −0.817 A1 + 0.365 A2 + 14.08 ⇒ A2 = −22.82

and

v(t ) dv(t ) − 0.5 so i (t ) = i (t ) − s dt 1.5 −0.817 t i (t ) =e [ 2.37 cos(0.365t ) + 7.14sin(0.365t ) ] + 6 + 0.65e−2t A

i(t)

P 9.8-11 Determine v(t) for t > 0 for the circuit shown in Figure P 9.8-11.

0.1 H + –

Answer:

– va(t) + 4Ω

6u(t) + 10 V

0.625 F 2va(t)

vc(t) = 0.75 e–4t – 6.75 e–36t + 16 V for t > 0

+ v(t) –

Figure P 9.8-11 Solution: First, find the steady state response for t < 0, when the switch is closed. The input is constant so the capacitor will act like an open circuit at steady state, and the inductor will act like a short circuit. va ( 0 ) = −4 i ( 0 ) i ( 0 ) = 2 ( −4 i ( 0 ) ) ⇒ i ( 0 ) = 0 A

and

v ( 0 ) = 10 V

For t > 0 Apply KCL at node 2: va d + K va + C v = 0 R dt KCL at node 1 and Ohm’s Law: va = − R i so

d 1+ K R v= i dt CR

Apply KVL to the outside loop:

L

d i + R i + v − Vs = 0 dt

After some algebra: d2 R d 1+ K R 1+ K R v+ v+ v= Vs 2 dt L dt LC LC



d2 d v + 40 v + 144 v = 2304 2 dt dt

The forced response will be a constant, vf = B so d2 d B + ( 40 ) B + (144 ) B = 2304 ⇒ B = 16 V 2 dt dt

The characteristic equation is s 2 + 40 s + 144 = 0 ⇒ s1,2 = −4, −36 . v ( t ) = A1 e − 4 t + A2 e −36 t + 16 .

Then Using the initial conditions:

⎫ 10 = v ( 0 ) = A1 e0 + A2 e0 + 16 ⇒ A1 + A2 = −6 ⎪ ⎬ ⇒ d 0 0 0 = v ( 0 ) = −4 A1 e − 36 A2 e ⇒ − 4 A1 − 36 A2 = 0 ⎪ dt ⎭

A1 = 0.75 and A2 = −6.75

So v ( t ) = 0.75 e−4 t − 6.75 e−36 t + 16 V for t > 0

P 9.8-12 The circuit shown in Figure P 9.8-12 is at steady state before the switch opens. The inductor current is given to be

i(t) = 240 + 193e–6.25tcos(9.27t – 102°) mA for t ≥ 0. Determine the values of R1, R3, C, and L.

t=0 R1 + –

i(t)

24 V L

+ v(t)

C

20 Ω



Figure P 9.8-12 Solution: Two steady state responses are of interest, before and after the switch opens. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit.

For t > 0, the switch is open. At steady state, inductor 24 current is i ( ∞ ) = . From the given equation, R1 + 20 i ( ∞ ) = lim i ( t ) = 0.24 . Thus, t →∞

0.24 =

24 ⇒ R1 = 80 Ω . R1 + 20

For t < 0, the switch is closed and the circuit is at steady state. 24 = 0.24 + 0.193 cos ( −102° ) = 0.2 80 || R 3 + 20

(

)

Consequently, R 3 = 80 Ω After the switch opens, apply KCL and KVL to get d ⎛ ⎞ R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs dt ⎝ ⎠ Apply KVL to get v (t ) = L

d i (t ) + R2 i (t ) dt

Substituting v ( t ) into the first equation gives

R3

d⎛ d d ⎛ ⎞⎞ R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs dt ⎝ dt dt ⎠⎠ ⎝ then R1 C L

d2 dt

(

i ( t ) + R1 C R 2 + L

2

) dtd i ( t ) + ( R1 + R 2 ) i ( t ) = Vs

Dividing by R1 C L : ⎛ R1 C R 2 + L ⎞ d ⎛ R1 + R 2 ⎞ Vs i t i t + + ⎜ ⎟ ⎜ ⎟ i (t ) = ( ) ( ) 2 ⎜ ⎟ ⎜ ⎟ R1 C L dt ⎝ R1 C L ⎠ dt ⎝ R1 C L ⎠ d2

Compare to d2 dt

to get 2α =

2

i ( t ) + 2α

R1 C R 2 + L R1 C L

d i ( t ) + ω 0 2 i ( t ) = f (t ) dt

, ω 02 =

R1 + R 2 R1 C L

and

f (t ) =

Vs R1 C L

From the given equation, we have α = 6.25 and ω d = 9.27 rad/s . Consequently,

ω 0 = ω d 2 + α 2 = 11.18 rad/s . Next 12.5 =

R1 C R 2 + L R1 C L

=

20 1 + L 80 C

and 125 =

R1 + R 2 R1 C L

=

1.25 1 ⇒ 100 = CL CL

So 12.5 =

20 1 + 1 80 C 100 C

⇒ 0 = 2000 C 2 − 12.5 C + 0.0125 ⇒ C = 1.25, 5 mF

The corresponding values of the inductance are L = 8, 2 H . There are two solutions: R1 = 80 Ω, R 3 = 80 Ω, C = 1.25 mF and L = 8 H

and R1 = 80 Ω, R 3 = 80 Ω, C = 5 mF and L = 2 H

We have used the initial condition i ( 0 ) = 0.2 A but we have not yet used the initial condition v (t ) = L

d i (t ) + R2 i (t ) ⇒ dt

from the given equation,

v (0) R2 i ( 0) 8 4 4 d i ( 0) = − = − = dt L L L L L

i ( t ) = 0.24 + e −6.25 t ( −0.04 cos ( 9.27 t ) + 0.1888sin ( 9.27 t ) ) A for t ≥ 0

d i ( t ) = ( −6.25 ) e −6.25 t ( −0.04 cos ( 9.27 t ) + 0.1888sin ( 9.27 t ) ) dt

+ ( 9.27 ) e −6.25 t ( 0.04sin ( 9.27 t ) + 0.1888cos ( 9.27 t ) ) for t ≥ 0 d i ( 0 ) = ( −6.25 )( −0.04 ) + ( 9.27 )(1.888 ) = 2 dt

Consequently,

2=

d 4 i ( 0) = dt L

⇒ L=2H

and we choose

R1 = 80 Ω, R 3 = 80 Ω, C = 5 mF and L = 2 H

P 9.8-13 The circuit shown in Figure P 9.8-13 is at steady state before the switch opens. Determine the inductor current, i(t), for t > 0.

t=0 8Ω + –

i(t)

18 V + v(t)

0.4 H 25 mF



Figure P 9.8-13 Solution: First, we find the initial conditions;

For t < 0, the switch is closed and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit.

v (0 −) =

12 ×18 = 12 V ( 8 || 24 ) + 12

and

i (0 −) =

24 18 × = 0.75 A 8 + 24 ( 8 || 24 ) + 12

Next, represent the circuit by a differential equation. After the switch opens, apply KCL and KVL to get d ⎛ ⎞ R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs dt ⎝ ⎠

Apply KVL to get v (t ) = L

24 Ω

d i (t ) + R2 i (t ) dt

Substituting v ( t ) into the first equation gives d⎛ d d ⎛ ⎞⎞ R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs dt ⎝ dt dt ⎠⎠ ⎝

12 Ω

then

R1 C L

d2 dt

2

(

i ( t ) + R1 C R 2 + L

) dtd i ( t ) + ( R1 + R 2 ) i ( t ) = Vs

Dividing by R1 C L : ⎛ R1 C R 2 + L ⎞ d ⎛ R1 + R 2 ⎞ Vs i t i t + + ⎜ ⎟ ⎜ ⎟ i (t ) = ( ) ( ) 2 ⎜ R1 C L ⎟ dt ⎜ R1 C L ⎟ R1 C L dt ⎝ ⎠ ⎝ ⎠ d2

Compare to

d2 dt to get

2α =

i t + 2α 2 ( )

R1 C R 2 + L R1 C L

d i ( t ) + ω 02 i ( t ) = f (t ) dt

, ω 02 =

R1 + R 2 R1 C L

and

f (t ) =

Vs R1 C L

With the given element values, we have α = 17.5 and ω 0 2 = 250 . Consequently, the roots of the characteristic equation are s 1 = −α − α 2 − ω 0 2 = −25 and s 2 = −α + α 2 − ω 02 = −10 . The natural response is

i n ( t ) = A1 e−10 t + A 2 e−25 t Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. if =

18 = 0.9 A 8 + 12

So

i ( t ) = i n ( t ) + i f ( t ) = A1 e−10 t + A2 e−25 t + 0.9 It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have 0.75 = i ( 0 ) = A1 + A 2 + 0.9 The other initial condition comes from v (t ) R 2 d i (t ) = − i (t ) ⇒ dt L L then

12 12 d i ( 0) = − × 0.75 = 7.5 0.4 0.4 dt

7.5 =

d i ( 0 ) = −10 A1 − 25 A 2 dt

Solving these equations gives A1 = 0.25 and A2 = −0.4 so

i ( t ) = 0.25 e−10 t − 0.4 e−25 t + 0.9 A for t > 0 (checked using LNAPTR 7/21/04)

*P 9.8-14 The circuit shown in Figure P 9.8-14 is at steady state before the switch closes. Determine the capacitor voltage, v(t), for t > 0. ia

+ –

i(t)

t=0

0.4 H

10 Ω 3ia 20 V

10 Ω

25 mF

+ v(t) –

Figure P 9.8-14 Solution: First, we find the initial conditions;

For t < 0, the switch is open and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit.

v ( 0 − ) = 0 V and i ( 0 − ) = 0 A also

i ( 0) d v ( 0) = =0 0.025 dt

Next, represent the circuit after the switch closes by a differential equation. To do so, we find the Thevenin equivalent circuit for the part of the circuit to the left of the inductor.

v s − v oc ⎫ ⎪ R1 ⎪ v s R 2 (1 + b ) ⎬ ⇒ v oc = v oc ⎪ R1 + R 2 (1 + b ) ia + bia = R 2 ⎪⎭ ia =

i sc = i a (1 + b ) = v s R 2 (1 + b )

Rt =

With the given values, v oc = 16 V and R t = 2 Ω . After the switch closes, apply KVL to get

v oc i sc

=

vs R1

(1 + b )

R1 + R 2 (1 + b ) R1 R 2 = vs R1 + R 2 (1 + b ) (1 + b ) R1

R t i (t ) + L

d i ( t ) + v ( t ) = voc dt

d v (t ) dt Substituting i ( t ) into the first equation gives

Apply KCL to get

i (t ) = C

d2

⎛ 1 ⎞ voc ⎛R⎞ d v (t ) + ⎜ ⎟ v (t ) + ⎜ ⎟ v (t ) = CL dt ⎝ L ⎠ dt ⎝CL⎠ 2

d2

d v ( t ) + ω 02 v ( t ) = f (t ) dt dt Rt v 1 2α = , ω 02 = and f (t ) = oc L CL CL

Compare to

2

to get

v (t ) + 2α

With the given element values, we have α = 2.5 and ω 0 2 = 100 . Consequently, the roots of the characteristic equation are s 1,2 = −α ± α 2 − ω 02 = −2.5 ± j 9.682 and the circuit is underdamped. The damped resonant frequency is ω d = ω 0 2 − α 2 = 9.682 rad/s . The natural response is

(

v n ( t ) = e−2.5 t A1 cos 9.682 t + A 2 sin 9.682 t

)

Next, determine the forced response. The steady state response after the switch closes will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. v f = v oc = 16 V

(

v ( t ) = 16 + e−2.5 t A1 cos 9.682 t + A 2 sin 9.682 t

so

)

It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have 0 = v ( 0 ) = 16 + A1 ⇒ And

0=

d v ( 0 ) = −2.5 A1 + 9.682 A 2 dt



A1 = −16 A2 = −

2.5 × 16 = −4.131 9.682

Finally, v ( t ) = 16 + e −2.5 t ( −16 cos 9.682 t − 4.131sin 9.682 t ) = 16 + 16.525 e −2.5 t cos ( 9.682 t + 165.5° ) V for t ≥ 0

(checked using LNAPTR 7/22/04)

P 9.8-15 The circuit shown in Figure P 9.8-15 is at steady state before the switch closes. Determine the capacitor voltage, v(t), for t > 0. t=0

50 Ω

i(t) 2H

+ –

+ 50 Ω

20 V

v(t)

5 mF



Figure P 9.8-15 Solution: First, we find the initial conditions;

For t < 0, the switch is open and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit.

v ( 0 − ) = 0 V and i ( 0 − ) = 0 A also

i ( 0) v ( 0) d v ( 0) = − =0 0.005 50 × 0.005 dt

Next, represent the circuit after the switch closes by a differential equation. After the switch closes, use KCL to get i (t ) =

v (t ) d + C v (t ) R2 dt

Use KVL to get v s = R1 i ( t ) + L

d i (t ) + v (t ) dt

Substitute to get vs =

R1 R2

= CL

Finally,

v ( t ) + R1C

d L d d2 v (t ) + v ( t ) + CL 2 v ( t ) + v ( t ) dt R 2 dt dt

⎛ R1 + R 2 d2 L ⎞d v t + R1C + v (t ) + v (t ) ⎟ 2 ( ) ⎜ ⎜ ⎟ dt dt R R 2 2 ⎝ ⎠

⎛ R1 R1 + R 2 d2 1 ⎞d = 2 v (t ) + ⎜ + v (t ) + v (t ) ⎟ ⎜ L R 2C ⎟ dt CL dt R 2CL ⎝ ⎠ vs

Compare to

d2 dt

to get

2α =

R1 L

2

+

i ( t ) + 2α

d i ( t ) + ω 02 i ( t ) = f (t ) dt

R1 + R 2 1 , ω 02 = R 2C R 2CL

f (t ) =

and

vs CL

With the given element values, we have α = 14.5 and ω 0 2 = 200 . Consequently, the roots of the characteristic equation are s 1 = −11.3 and s 2 = −17.7 so the circuit is overdamped. The natural response is v n ( t ) = A1 e −11.3 t + A 2 e −17.7 t

Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. vf =

1 v s = 10 V 2

So v n ( t ) = 10 + A1 e −11.3 t + A 2 e−17.7 t

It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have 0 = v ( 0 ) = 10 + A1 + A 2 and 0=

d v ( 0 ) = −11.3 A1 − 17.7 A 2 dt

Solving these equations gives A1 = −27.6 and A 2 = 17.6

Finally, v ( t ) = 10 − 27.6 e−11.3 t + 17.6 e−17.7 t (checked using LNAPTR 7/26/04)

t=0

P 9.8-16 The circuit shown in Figure P 9.8-16 is at steady state before the switch closes. Determine the inductor current, i(t), for t > 0.

16 Ω

9Ω + –

+ 25 mF

20 V 0.4 H

i(t)

Figure P 9.8-16 Solution: First, we find the initial conditions;

For t < 0, the switch is closed and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit.

v ( 0 − ) = 0 V and i ( 0 − ) = 0 A Also 9 i ( 0 ) + 0.4

d i ( 0) = v ( 0) ⇒ dt

d i (0) = 0 dt

Next, represent the circuit by a differential equation. After the switch closes use KVL to get R2 i (t ) + L

d i (t ) = v (t ) dt

Use KCL and KVL to get d ⎛ ⎞ v s = R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) dt ⎝ ⎠ Substitute to get d d2 d v s = R1i ( t ) + R1CR 2 i ( t ) + R1C L 2 i ( t ) + R 2i ( t ) + L i ( t ) dt dt dt 2 d d = R1CL 2 i ( t ) + ( R1 R 2C + L ) i ( t ) + ( R1 + R 2 ) i ( t ) dt dt then ⎛ R2 vs R1 + R 2 d2 1 ⎞d = 2 i (t ) + ⎜ + i (t ) + i (t ) ⎟ ⎜ L R1C ⎟ dt R1CL dt R CL 1 ⎝ ⎠

v(t) –

Compare to

d2 2

dt

to get

2α =

R2 L

+

i ( t ) + 2α

d i ( t ) + ω 02 i ( t ) = f (t ) dt

R1 + R 2 1 , ω 02 = R1C R1 C L

and

f (t ) =

Vs R1 C L

With the given element values, we have α = 12.5 and ω 0 2 = 156.25 . Consequently, the roots of the characteristic equation are s 1,2 = −α ± α 2 − ω 0 2 = −12.5, − 12.5 so the circuit is critically damped. The natural response is

(

)

i n ( t ) = A1 + A 2 t e−12.5 t Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. 20 if = = 0.8 A 16 + 9 So i ( t ) = i n ( t ) + i f ( t ) = A1 + A 2 t e−12.5 t + 0.8

(

)

It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have

And

0=

0 = i ( 0 ) = A1 + 0.8 ⇒

A1 = −0.8

d i ( 0 ) = −12.5 A1 − A 2 dt



A 2 = 10

Thus

i ( t ) = ( −0.8 + 10 t ) e−12.5 t + 0.8 for t > 0 (checked using LNAPTR 7/27/04)

75 Ω

P 9.8-17 The circuit shown in Figure P 9.8-17 is at steady state before the switch opens. Determine the inductor current, i2(t), for t > 0.

t=0 24 Ω + –

4H

i1(t)

20 V

15 Ω

Figure P 9.8-17

Solution: First, we find the initial conditions;

For t < 0, the switch is closed and the circuit is at steady state. At steady state, the inductors act like short circuits.

i1 ( 0 − ) = and

20 = 1.333 A 15

i 2 (0 −) = 0 A

Next, represent the circuit by a differential equation. After the switch opens, KVL gives L1

d d i1 ( t ) = R 2 i 2 ( t ) + L 2 i 2 ( t ) dt dt

KVL and KCL give L1

d i 1 ( t ) + R1 ( i 1 ( t ) + i 2 ( t ) ) = 0 dt

Use the operator method to get

L1s i1 = R 2 i 2 + L 2 s i 2 L1s i1 + R1 ( i1 + i 2 ) = 0

1.6 H

i2(t)

L1s 2i1 + R1s i1 + R1s i 2 = 0 s ( R 2i 2 + L 2 s i 2 ) +

R1 L1

(R i

2 2

+ L 2 s i 2 ) + R1s i 2 = 0

⎛ ⎞ L2 R1 R 2 + R1 ⎟ s i 2 + L 2 s 2 i 2 + ⎜ R 2 + R1 i2 = 0 ⎜ ⎟ L L 1 1 ⎝ ⎠ ⎛ R 2 R 1 R1 ⎞ R1 R 2 + + ⎟ s i2 + s 2i 2 + ⎜ i2 = 0 ⎜ L 2 L 2 L1 ⎟ L1 L 2 ⎝ ⎠

so ⎛ R 2 R1 R1 ⎞ d R1 R 2 d2 i t + + + ⎟ i 2 (t ) + i 2 (t ) = 0 2 2( ) ⎜ ⎜ L 2 L 2 L1 ⎟ dt dt L L 1 2 ⎝ ⎠ Compare to

d2 dt

to get

2α =

R2 L2

2

+

i ( t ) + 2α R1 L2

+

R1 L1

d i ( t ) + ω 02 i ( t ) = f (t ) dt

, ω 02 =

R1R 2 L1L 2

and

f (t ) = 0

With the given element values, we have α = 33.9 and ω 02 = 281.25 . Consequently, the roots of the characteristic equation are s 1,2 = −α ± α 2 − ω 02 = −4.4, − 63.4 so the circuit is overdamped. The natural response is

i n ( t ) = A1 e−4.4 t + A 2 e−63.4 t Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state the inductors act like short circuits. if = 0 A

So

i 2 ( t ) = i n ( t ) + i f ( t ) = A1 e−4.4 t + A2 e−63.4 t It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have

0 = i 2 ( 0 ) = A1 + A 2 L2

d i 2 ( 0 ) + R 2 i 2 ( 0 ) + R1 i 1 ( 0 ) + R 1 i 2 ( 0 ) ⇒ dt

d i 2 ( 0 ) = −20 dt

and −20 =

d i ( 0 ) = −4.4 A1 − 63.4 A 2 dt

Solving these equations gives A1 = −0.339 and A2 = 0.339 so

i 2 ( t ) = −0.339 e−4.4 t + 0.339 e−63.4 t for t ≥ 0 (checked using LNAPTR 7/27/04)

t=0

P 9.8-18 The circuit shown in Figure P 9.8-18 is at steady state before the switch closes. Determine the capacitor voltage, v(t), for t > 0.

2H + –

20 V

50 Ω 5 mF

50 Ω

i(t) + v(t) –

Figure P 9.8-18 Solution: First, we find the initial conditions;

For t < 0, the switch is open and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit.

v ( 0 − ) = 0 V and i ( 0 − ) = 0 A also

i (0) d v ( 0) = =0 0.005 dt

Next, represent the circuit after the switch closes by a differential equation. After the switch closes i (t ) = C

d v (t ) dt

KCL and KVL give ⎛ 1 ⎛ d d ⎞⎞ vs = R2 ⎜ i (t ) + ⎜ L i (t ) + v (t ) ⎟ ⎟ + L i (t ) + v (t ) ⎜ R1 ⎝ dt dt ⎠ ⎟⎠ ⎝

Substituting gives ⎛ R2 ⎞ ⎛ R2 ⎞ ⎛ R2 ⎞ R2 d2 d d2 d vs = LC 2 v ( t ) + R 2C v ( t ) + ⎜ 1 + v t = 1 + LC v ( t ) + R 2C v ( t ) + ⎜1 + ( ) ⎟ ⎜ ⎟ ⎟⎟ v ( t ) 2 ⎜ ⎟ ⎜ ⎟ ⎜ R1 dt dt R R dt dt R 1 1 1 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ So the differential equation is

R1v s

LC ( R1 + R 2 )

=

R1 R 2 d2 d 1 v (t ) + v (t ) + v (t ) dt LC L ( R1 + R 2 ) dt

Compare to

d2 dt

to get

2α =

(

2

i ( t ) + 2α

R1R 2

L R1 + R 2

)

d i ( t ) + ω 02 i ( t ) = f (t ) dt

, ω 02 =

1 and CL

f (t ) =

(

R1v s

LC R1 + R 2

)

With the given element values, we have α = 6.25 and ω 0 2 = 100 . Consequently, the roots of the characteristic equation are s 1,2 = −α ± α 2 − ω 02 = −6.25 ± j 7.806 and the circuit is underdamped. The damped resonant frequency is ω d = ω 0 2 − α 2 = 7.806 rad/s . The natural response is

(

v n ( t ) = e−6.25 t A1 cos 7.806 t + A 2 sin 7.806 t

)

Next, determine the forced response. The steady state response after the switch opens will be used as the forced response. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. vf =

So

50 × 20 = 10 V 50 + 50

(

v ( t ) = 10 + e−6.25 t A1 cos 7.806 t + A 2 sin 7.806 t

)

It remains to evaluate A1 and A2 using the initial conditions. At t = 0 we have 0 = v ( 0 ) = 10 + A1 ⇒ And

0=

d v ( 0 ) = −6.25 A1 + 7.806 A 2 dt



A1 = −10 A2 = −

6.25 × 10 = −8.006 7.806

Finally, v ( t ) = 10 + e −6.25 t ( −10 cos 7.806 t − 8.006sin 7.806 t ) = 10 + 12.81 e −6.25 t cos ( 7.806 t + 141.3° ) V for t ≥ 0

(checked using LNAPTR 7/26/04)

iL(t)

P 9.8-19 Find the differential equation for vc(t) in the circuit of Figure P 9.8-19 using the direct method. Find vc(t) for time t > 0 for each of the following sets of component values:

L

R1

vs(t) = u(t) +–

R2

C

+ vc(t) –

Figure P 9.8-19

(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω (b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω Answer:

(a) vc(t) =

1 2

– e–2t + –2t

1 2

e–4t V; (b) vc(t) =

1 4

–(

1 4

1

+

2

t)e–2t V;

(c) vc(t) = 0.8 – e (0.8 cos 4t + 0.4 sin 4t) V

Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions

vC ( ∞ ) =

R2 R1 + R 2

1

Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives:

vC ( t ) R2 vs ( t ) = L

+C

d vC ( t ) = iL ( t ) dt

d iL ( t ) + R1 iL ( t ) + vC ( t ) dt

Use the substitution method to get vs ( t ) = L

⎞ ⎛ v (t ) ⎞ d ⎛ vC ( t ) d d + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎜⎜ ⎟ ⎜ R2 ⎟ dt ⎝ R 2 dt dt ⎠ ⎝ ⎠

⎛ L ⎞d ⎛ R1 ⎞ d2 = LC 2 vC ( t ) + ⎜ + R1 C ⎟ vC ( t ) + ⎜1 + v t ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠ (a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω Use the steady state response as the forced response: R2 1 v f = vC ( ∞ ) = 1= R1 + R 2 2 The characteristic equation is

R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 s2 + ⎜ + ⎟s+⎜ = s + 6s + 8 = ( s + 2 )( s + 4 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is vn = A1 e −2 t + A2 e−4 t V

The complete response is vc ( t ) = iL ( t ) =

vC ( t ) 1.309

At t = 0+

+

1 + A1 e −2 t + A2 e−4 t V 2

d vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819 dt 0 = vc ( 0 + ) = A1 + A2 + 0.5

0 = iL ( 0 + ) = −1.236 A1 − 3.236 A2 + 0.3819

Solving these equations gives A1 = -1 and A2 = 0.5, so vc ( t ) =

1 −2 t 1 −4 t −e + e V 2 2

(b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω Use the steady state response as the forced response: R2 1 1= v f = vC ( ∞ ) = 4 R1 + R 2 The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 2 s2 + ⎜ + ⎟s+⎜ = s + 4s + 4 = ( s + 2 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is v f = ( A1 + A2 t ) e −2 t V

The complete response is vc ( t ) = iL ( t ) = vC ( t ) + At t = 0+

1 + ( A1 + A2 t ) e −2 t V 4

d 1 vC ( t ) = + dt 4

(( A

2

)

− A1 ) − A2 t e −2 t

0 = vc ( 0 + ) = A1 + 0 = iL ( 0 + ) =

1 4

1 + A2 − A1 4

Solving these equations gives A1 = -0.25 and A2 = -0.5, so 1 ⎛ 1 1 ⎞ −2 t − ⎜ + t ⎟e V 4 ⎝4 2 ⎠ (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω Use the steady state response as the forced response: R2 4 v f = vC ( ∞ ) = 1= R1 + R 2 5 The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 s2 + ⎜ + ⎟s+⎜ = s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is v f = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vc ( t ) =

The complete response is

iL ( t ) = At t = 0+

vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vC ( t ) 4

+

A2 −2 t A1 1d vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 8 dt 2 2 0 = vc ( 0 + ) = 0.8 + A1

0 = iL ( 0 + ) = 0.2 +

A2 2

Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V

P 9.8-20 Find the differential equation for vo(t) in the circuit of Figure P 9.8-20 using the direct method. Find vo(t) for time t > 0 for each of the following sets of component values:

iL(t)

R1

vs(t) = u(t)

+ –

C

L +

+ vc(t) –

R2

vo(t) –

Figure P 9.8-20

(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω (b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω (c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω Answer: 1

– e–2t +

1

(a)

vo(t) =

(b)

vo(t) =

(c)

vo(t) = 0.2 – e–2t(0.2 cos 4t + 0.1 sin 4t) V

2 3 4

–(

3 4

+

2 3 2

e – 4t

V

t)e–2t V

Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions

vC ( ∞ ) =

R2 R1 + R 2

1, iL ( ∞ ) =

1 R1 + R 2

and

vo ( ∞ ) =

R2 R1 + R 2

1

Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives:

d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L

Use the substitution method to get vs ( t ) = R1 C = R1 LC

Using iL ( t ) =

vo ( t ) gives R2

d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt

⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 vs (t ) = LC 2 v o ( t ) + ⎜ + R1 C ⎟ v o ( t ) + ⎜ v t ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ o ( ) R2 dt ⎝ ⎠ ⎝ ⎠ R1

(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω Use the steady state response as the forced response: R2 1 v f = vo ( ∞ ) = 1= R1 + R 2 2 The characteristic equation is R ⎞ ⎛ 1+ 2 ⎟ ⎜ ⎛ 1 R2 ⎞ ⎜ R1 ⎟ 2 s2 + ⎜ + s+ = s + 6 s + 8 = ( s + 2 )( s + 4 ) ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is vn = A1 e −2 t + A2 e −4 t V

The complete response is 1 + A1 e −2 t + A2 e−4 t V 2 A1 −2 t A 2 −4 t vo ( t ) 1 iL ( t ) = e + e V = + 1.309 2.618 1.309 1.309 vo ( t ) =

vC ( t ) = 1.309 iL ( t ) +

1 d 1 iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t 4 dt 2

At t = 0+ 0 = iL ( 0 + ) = 0 = vC ( 0 + ) =

A1 A2 1 + + 2.618 1.309 1.309

1 + 0.6167 A1 + 0.2361 A2 2

Solving these equations gives A1 = -1 and A2 = 0.5, so vo ( t ) =

1 −2 t 1 −4 t −e + e V 2 2

(b) C = 1 F, L = 1 H, R1 = 1 Ω, R2 = 3 Ω Use the steady state response as the forced response: R2 3 v f = vo ( ∞ ) = 1= R1 + R 2 4 The characteristic equation is

R ⎞ ⎛ 1+ 2 ⎟ ⎜ ⎛ 1 R2 ⎞ ⎜ R1 ⎟ 2 2 s2 + ⎜ + s+ = s + 4s + 4 = ( s + 2 ) ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is v f = ( A1 + A2 t ) e −2 t V

The complete response is vo ( t ) = iL ( t ) =

3 + ( A1 + A2 t ) e −2 t V 4

vo ( t )

vC ( t ) = 3 iL ( t ) +

3

=

1 ⎛ A1 A2 ⎞ −2 t t⎟ e V +⎜ + 4 ⎝ 3 3 ⎠

3 ⎛ ⎛ A1 A2 ⎞ A2 ⎞ −2 t d iL ( t ) = + ⎜⎜ ⎜ + t ⎟e ⎟+ 4 ⎝⎝ 3 3 ⎠ 3 ⎟⎠ dt

At t = 0+ 0 = iL ( 0 + ) =

A1

+

1 4

3 3 A1 A2 0 = vC ( 0 + ) = + + 4 3 3 Solving these equations gives A1 = -0.75 and A2 = -1.5, so vo ( t ) =

3 ⎛ 3 3 ⎞ −2 t − ⎜ + t ⎟e V 4 ⎝4 2 ⎠

(c) C = 0.125 F, L = 0.5 H, R1 = 4 Ω, R2 = 1 Ω Use the steady state response as the forced response: R2 1 1= v f = vo ( ∞ ) = 5 R1 + R 2 The characteristic equation is R ⎞ ⎛ 1+ 2 ⎟ ⎜ ⎛ 1 R2 ⎞ ⎜ R1 ⎟ 2 s2 + ⎜ + s+ = s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ so the natural response is v f = e −2 t ( A1 cos 4 t + A2 sin 4 t ) V

The complete response is

vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V iL ( t ) =

vo ( t )

vC ( t ) = iL ( t ) + At t = 0+

1

= 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V

1 d iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e −2 t sin 4 t 2 dt 0 = iL ( 0 + ) = 0.2 + A1 0 = vC ( 0 + ) = 0.2 + 2 A2

Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V

Section 9-9: State Variable Approach to Circuit Analysis P 9.9-1 Find v(t) for t > 0 using the state variable method of Section 9.9 when C = 1/5 F in the circuit of Figure P 9.9-1. Sketch the response for v(t) for 0 < t < 10 s. Answer: v(t) = – 25e–t + e– 5t + 24 V

6Ω + C

4u(t) A 1H

Figure P 9.9-1 Solution:

t = 0− circuit is source free ∴ iL (0) = 0 & v(0) = 0 t>0

⎛1⎞ = 4 KCL at top node: i L + ⎜ ⎟ dv (1) ⎝ 5 ⎠ dt di − 6i L = 0 KVL at right loop: ( v − 1) L dt Solving for i in (1) & plugging into (2) ⇒ 1

d 2v dv + 6 + 5v = 120 2 dt dt

The characteristic equation is: s 2 + 6s + 5 = 0, The roots of the characteristic equation are s = −1, −5 ∴ Tthe natural response is: vn (t ) = A1 e −t + A2 e −5t Try vf = B & plug into D.E. ⇒

B = 24 = vf

dv(0) = 20 − 5iL (0) = 20 V s dt So v(0) = 0 = A1 + A2 + 24 ⎫ A1 = −25, A2 = 1 ⎪ ⎬ ∴ v(t ) = −25 e −t + e−5t + 24 V dv(0) = 20 = − A1 − 5 A2 ⎪ dt ⎭ From (1)



v

P 9.9-2 Repeat Problem P9.9-1when C = 1/10 F. Sketch the response for v(t) for 0 < t < 3 s.

6Ω +

Answer: v(t) = e – 3t(– 24 cos t – 32 sin t) + 24 V

C

4u(t) A



v

1H

Figure P 9.9-1 Solution:

At t = 0− the circuit is source free ∴ iL (0) = 0, & v(0) = 0 At t > 0

( )

i = 4 − 1 dv dt 10 L

KCL at top node :

KVL at right node : v − (1) into (2) yields 2

⇒ s + 6 s + 10 = 0, s = −3 ± j

∴ vn ( t ) = e

−3t

d 2v dt

2

diL

+6

dt

dv +10v = 240 dt

⎡⎣ A1 cos t + A2 sin t ⎤⎦

Try vf = B & plug into D.E. ⇒ vf = B = 24 dv (0) = 40−10 i L (0) = 40 V s dt dv (0) = 40 = −3 A1 + A2 So v (0) = 0 = A1 + 24 ⇒ A1 = −24 & dt From (1)

⇒ ∴ v (t ) = e

−3 t

[ −24cos t

− 32sin t ] + 24 V

− 6i L = 0

A2 = −32

(1) (2)

Determine the current i(t) and the voltage v(t) for the circuit of Figure P 9.9-3.

P 9.9-3

Answer: i(t) = (3.08e–2.57t – 0.08e–97.4t – 6) A

v

0.2 H

–3u(t) A

i

+ –

20 mF

0.5 Ω

3A

Figure P 9.9-3 Solution: i (0) = −3, v(0) = 0 t >0 dv v + +6=0 dt R di KVL: v = L dt KCL: i + C

d 2i dt

2

+ 100

di + 250i = −1500 dt

s = − 2.57, − 97.4 i f (t ) =

−1500 = −6 250

i (t ) = A1e −2.57 t + A2 e−97.4 t −6 ⎫ ⎪ A1 = 3 .081 ⎬ di (0) = 0 = − 2.57 A1 −97.4 A2 ⎪ A2 =−0.081 dt ⎭

i (0) = A1 + A2 − 6 = −3

i (t ) = 3.081 e v (t ) = 0.2

−2.57 t

− 0.081e

di = −1.58e dt

−2.57 t

−97.4 t

−6 A

+ 1.58e−97.4 t

V

P 9.9-4 Clean-air laws are pushing the auto industry toward the development of electric cars. One proposed vehicle using an ac motor is shown in Figure P 9.9-4a. The motor-controller circuit is shown in Figure P 9.9-4b with L = 100 mH and C = 10 mF. Using the state equation approach, determine i(t) and v(t) where i(t) is the motor-control current. The initial conditions are v(0) = 10 V and i(0) = 0.

+

+ –

2vx Transistorized dc to ac inverter

+

Integrated interior permanent magnet Sodium-sulfur battery ac motor and automatic transaxle



2ix

Electric power steering

System controller



C



vx

v

L

ix

i



(a)

(b)

Figure P 9.9-4

Apply KCL to the supernode corresponding to the VCVS to get ix +

vx 2

= 2ix + C

dv dt

⇒ − ix +

Apply KCL to the right node of the CCCS to get i +

vx 2

= 0.01

dv dt

vx

= 2ix . 2 Apply KVL to the mesh consisting of the 2-Ω resistor, inductor and capacitor to get vx + v − L Apply KVL to the outside loop to get

di di = 0 ⇒ v x + v = 0.1 dt dt v + ix + 2vx = 0

(1) (2)

(3) (4)

Combine equations (2) and (4) to get

4 1 2 4 i − v and v x = − i − v 9 9 9 9 Use (5) to eliminate ix and vx from (1) and (3) to get dv 5 1 di 2 5 0.01 = − i − v and 0.1 = − i + v dt 9 9 dt 9 9 Use operators to write 500 100 20 50 sv = − i− v and s i = − i + v 9 9 9 9 The characteristic equation is : s 2 + 13.33s + 333.33 = 0 ⇒ s1 , s2 = −6.67 ± j 17 ix =

(5)

(6)

(7)

The natural response is v ( t ) = e −6.67 t ⎡⎣ A cos (17 t ) + B sin (17 t ) ⎤⎦ and there is no forced because there is no forcing function. The constants A and B are evaluated using the initial conditions: dv(0) v(0) = 10 = A and = −111 = −6.67 A + 17 B ⇒ B = −2.6 dt Then v ( t ) = e −6.67 t ⎡⎣10 cos (17 t ) − 2.6 sin (17 t ) ⎤⎦ V Similarly, the natural response is i ( t ) = e −6.67 t ⎡⎣ A cos (17 t ) + B sin (17 t ) ⎤⎦ and again there is no forced because there is no forcing function. The constants A and B are evaluated using the initial conditions: di (0) i (0) = 0 = A and = 55.6 = −6.67 A + 17 B ⇒ B = −3.27 dt Then i ( t ) = e −6.67 t ⎡⎣3.27 sin (17 t ) ⎤⎦ V

P 9.9-5 Studies of an artificial insect are being used to understand the nervous system of animals. A model neuron in the nervous system of the artificial insect is shown in Figure P 9.9-5. The input signal, vs, is used to generate a series of pulses, called synapses. The switch generates a pulse by opening at t = 0 and closing at t = 0.5 s. Assume that the circuit is at steady state and that v(0 –) = 10 V.

Switch



3Ω vs + –

+ v

1 6F

30 V 1 2H



Determine the voltage v(t) for 0 < t < 2 s. Figure P 9.9-5 Solution First consider t < 0 :

v(0) = 10 V, i (0) = L

10 A 3

Next consider 0 < t < 0.5s

α=

R 3 1 and ω02 = LC = = L 2 12

s =−α ±

α 2 − ω02



s1 = −.028 and s2 = −2.97

The natural response is v(t ) = Ae −0.028t + Be−2.97 t and the forced response is vf = 0. The constants are evaluated using the initial conditions: v(0) = 10 = A + B

⎫ ⎪ A = 16. 89 dv(0) ⎬ = 20 = −0.028 A − 2.97 B ⎪ B = −6.89 dt ⎭ −0.028t −2.97 t so v(t ) = 16. 89 e − 6.89e Similarly i (t ) = − .079e −0.028t + 3.41e−2.97 t At t = 0.5 s, v(0.5) = 15.1 V and i (0.5) = 0.7 A For t > 0.55 s:

v − 30 1 dv + iL + =0 6 6 dt di L KVL: v = 3 i L + 1 2 dt KCL:

Characteristic equation: 0 = s 2 − 7 s − 18 ⇒ s = −1,9

vf = 10 V v(t ) = Ae9t + Be − t + 10

v(0.5) = 15.1 = 90 A + 0.61 B + 10 dv(0.5) = 10.7 = 810 A − 0.61 B dt

t 0 → .5 .5 →2

⎫ ⎪ ⎬ ⎪⎭

A = 17.6 × 10−3 B = 5.77

v(t) 16.89e−0.28τ − 6.89e−2.97 τ V 17.6 × 10-3e9t + 5.77e-t + 10 V

Section 9-10: Roots in the Complex Plane 2 kΩ

P 9.10-1 For the circuit of Figure P 9.10-1, determine the roots of the characteristic equation, and plot the roots on the s-plane.

12 – 6u(t) V

3 kΩ

+ –

2 mH

2 mH

i1

i2

Figure P 9.10-1 Solution:

s 2 + 3.5 × 106 s + 1.5 × 1012 = 0 s = −5 × 105 1

s2 = −3 × 106

P 9.10-2 For the circuit of Figure P 9.6-1, determine the roots of the characteristic equation and plot the roots on the s-plane.

0.8 H

vc

250 Ω t=0

250 Ω + –

6V

Figure P 9.6-1 Solution:

s 2 + 800s + 250000 = 0 s = 400 ± j 300

+ –

5 × 10-6 F

4H

P 9.10-3 For the circuit of Figure P 9.10-3, determine the roots of the characteristic equation and plot the roots on the s-plane.

vs +

1 4



μF

4 kΩ

Figure P 9.10-3 Solution:

KCL: KVL:

Characteristic equation: s 2 + 1×103 s + 1× 106 = 0 s = −500 ± j 866

dv v 1 × 10−6 + dt 4 4000 di vs = 4 + v dt

i=

P 9.10-4 An RLC circuit is shown in Figure P 9.10-4. (a) Obtain the two-node voltage equations using operators. (b) Obtain the characteristic equation for the circuit. (c) Show the location of the roots of the characteristic equation in the s-plane. (d) Determine v(t) for t > 0.

1H a

36u(t) V

+ –

12 Ω

b 6 Ω 1 18 F

+ –

v(t)

Figure P 9.10-4

Solution:

at t = 0 the initial conditions are v(0) = v (0) = 0, b dvb vb − va i (0) = 0 and C + + i =0 (1) dt 6

t=0 Node a:

va (0) − 36 v (0) − vb (0) − i (0) + a = 0 then va (0) + 2va (0) = 36 so va (0) = 12 V 12 6

t ≥ 0 va − vs va − vb 1 + = 0 ∫ (va − vb ) dt + L 12 6 v −v dv 1 Node b : C b + b a + ∫ (va − vb )dt =0 dt 6 L v 1 1⎞ ⎛1 ⎛ 1 1⎞ Using operators ⎜ + + ⎟ va + ⎜ − − ⎟ vb = s 6 s⎠ 12 ⎝ 12 ⎝ 6 s⎠ 1 1 1 1 1 ( − − ) va + ( s + + ) vb = 0 6 s 18 6 s 2 Cramers rule: ( s + 5s + 6)vb = ( s + 6)vs Node a :

Then

vb = 36 + A1 e −2t + A2 e−3t vb (0) = 36 + A1 + A2

need

(2)

dvb (0) = −2 A1 − 3 A2 dt

v ( 0 ) − vb (0) dvb (0) 1 12 = (−2 A1 − 3 A2 ) = a − i (0) = =2 dt 18 6 6 Use (2) and (3) to get

Use 1 above: C

A1 = −72 and A2 = 36 so v ( t ) = vb ( t ) = 36 − 72 e −2t + 36 e −3t , t ≥ 0

(3)

Section 9-11 How Can We Check…? P 9.11-1 Figure P 9.11-1a shows an RLC circuit. The voltage, vs(t), of the voltage source is the square wave shown in Figure P 9.11-1a. Figure P 9.11-1c shows a plot of the inductor current, i(t), which was obtained by simulating this circuit using PSpice. Verify that the plot of i(t) is correct. Answer: The plot is correct.

25 i(t)

100 Ω

vs, V vs

0

4

8

12

+ –

2 μF

12 mH

16

t, ms

(a)

(b)

400 mA (550.562u, 321.886m) (1.6405m, 256.950m)

(3.6854m, 250.035m)

200 mA (1.0787m, 228.510m) I (L1)

0A

–200 mA

0s I (L1)

2.0 ms

4.0 ms Time

6.0 ms

8.0 ms

(c) Figure P 9.11-1a

Solution: This problem is similar to the verification example in this chapter. First, check the steady-state inductor current v 25 i (t ) = s = = 250 mA 100 100

This agrees with the value of 250.035 mA shown on the plot. Next, the plot shows an underdamped response. That requires 12 ⋅10−3 = L < 4 R 2C = 4(100) 2 (2 ⋅10−6 ) = 8 ⋅10−2

This inequality is satisfied, which also agrees with the plot. The damped resonant frequency is given by 2 2 ⎛ ⎞ 1 1 1 ⎛ 1 ⎞ ⎟ = 5.95 ⋅103 ω = −⎜ − ⎜ ⎟ = d 6 − ⎜ ⎟ LC ⎝ 2RC ⎠ 2 ⋅10−6 12 ⋅10−3 ⎝ 2(100) (2 ⋅10 ) ⎠

(

)(

)

The plot indicates a maxima at 550.6μs and a minima at 1078.7μs. The period of the damped oscillation is T = 2 (1078.7 μs − 550.6μs) = 1056.2μs d 2π 2π Finally, check that 5.95 ⋅103 = ω = = = 5.949 ⋅103 d T 1056.2 ⋅10−6 d The value of ωd determined from the plot agrees with the value obtained from the circuit.

The plot is correct

P 9.11-2 Figure P 9.11-2b shows an RLC circuit. The voltage, vs(t), of the voltage source is the square wave shown in Figure P 9.11-2a. Figure P 9.11-2c shows a plot of the inductor current, i(t), which was obtained by simulating this circuit using PSpice. Verify that the plot of i(t) is correct. Answer: The plot is not correct. 15 i(t)

100 Ω

vs, V vs

0

2

4

6

+ –

0.2 μ F

8 mH

8

t, ms

(b)

(a) 300 mA

(426.966u, 172.191m) 200 mA

(1.7753m, 149.952m)

100 mA I (L1) 0A (831.461u, 146.570m) –100 mA 0s

2.0 ms

4.0 ms Time

6.0 ms

8.0 ms

(c)

Figure P 9.11-2 Solution: This problem is similar to the verification example in this chapter. First, check the steady-state inductor current. v 15 i (t ) = s = = 150 mA 100 100 This agrees with the value of 149.952 mA shown on the plot. Next, the plot shows an under damped response. This requires

8 ⋅10−3 = L < 4 R 2C = 4 (100) 2 (0.2 ⋅10−6 ) = 8 ⋅10−3

This inequality is not satisfied. The values in the circuit would produce a critically damped, not underdamped, response. This plot is not correct.

PSpice Problems SP 9-1 The input to the circuit shown in Figure SP 9-1 is the voltage of the voltage source, vi(t). The output is the voltage across the capacitor, vo(t). The input is the pulse signal specified graphically by the plot. Use PSpice to plot the output, vo(t), as a function of t for each of the following cases:

vi(V) 5

0

(a) C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω (b) C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω (c) C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω Plot the output for these three cases on the same axis. Hint: Represent the voltage source using the PSpice part named VPULSE.

10

5 L

+

vi(t) –

15 t (s)

R1

R2

C

+ vo(t) –

Figure SP 9-1

Solution: Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.)

V(C1:2), V(C2:2) and V(C3:2) are the capacitor voltages, listed from top to bottom.

SP 9-2 The input to the circuit shown in Figure SP 9-2 is the voltage of the voltage source, vi(t). The output is the voltage, vo(t), across resistor R2. The input is the pulse signal specified graphically by the plot. Use PSpice to plot the output, vo(t), as a function of t for each of the following cases: (a) (b) (c)

vi(V) 5

0

Plot the output for these three cases on the same axis. Hint: Represent the voltage source using the PSpice part named VPULSE.

R1

+

vi(t) –

15 t (s)

10

5

C = 1 F, L = 0.25 H, R1 = R2 = 1.309 Ω C = 1 F, L = 1 H, R1 = 3 Ω, R2 = 1 Ω C = 0.125 F, L = 0.5 H, R1 = 1 Ω, R2 = 4 Ω

L

C

R2

+ vo(t) –

Figure SP 9-2 Solution: Make three copies of the circuit: one for each set of parameter values. (Cut and paste, but be sure to edit the labels of the parts so, for example, there is only one R1.)

V(R2:2), V(R4:2) and V(R6:2) are the output voltages, listed from top to bottom.

10 Ω

SP 9-3 Determine and plot the capacitor voltage

50 Ω

ig

v(t) for 0 < t < 300 μs for the circuit shown in Figure SP 9-3a. The sources are pulses as shown in Figures SP 9-3b, c.

1 mH +

v



0.1 μ F

+ –

vg

(a) 0.2 A

5V vg

ig 0 0

100 t ( μ s)

200

0 0

100 t ( μ s)

(c)

(b)

Figures SP 9-3 Solution:

50 Ω

200

3 kΩ

SP 9-4 Determine and plot v(t) for the circuit of Figure SP 9-4 when vs(t) = 5u(t) V. Plot v(t) for 0 < t < 0.25 s.

vs(t)

+ –

6 kΩ

+

v(t)



2 kΩ

3 kΩ

2 μF

3 μF

Figure SP 9-4 Solution:

Design Problems DP 9-1 Design the circuit shown in Figure DP 9-1 so that vc(t) =

1 2

+ A1e–2t + A2e–4t V for t > 0

Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is overdamped, and the natural frequencies are 2 and 4 rad/sec. iL(t)

vs(t) = u(t)

R1

L

+ –

R2

C

+ vc(t) –

Figure DP 9-1 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 1 The specifications require that vC ( ∞ ) = so 2 R2 1 = ⇒ R1 = R 2 2 R1 + R 2

Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives:

vC ( t ) R2 vs ( t ) = L

+C

d vC ( t ) = iL ( t ) dt

d iL ( t ) + R1 iL ( t ) + vC ( t ) dt

Use the substitution method to get vs ( t ) = L

⎞ ⎛ v (t ) ⎞ d ⎛ vC ( t ) d d + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎜⎜ ⎟ ⎜ R2 ⎟ dt ⎝ R 2 dt dt ⎠ ⎝ ⎠

= LC

⎛ L ⎞d ⎛ R1 ⎞ d2 v t + v t + R1 C ⎟ vC ( t ) + ⎜1 + 2 C ( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠

The characteristic equation is

R ⎛ 1+ 1 ⎜ ⎛ 1 R1 ⎞ R2 s2 + ⎜ + ⎟s+⎜ ⎜ R 2 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s:

⎞ ⎟ ⎟ = s 2 + 6s + 8 = ( s + 2 )( s + 4 ) ⎟ ⎟⎟ ⎠

R1 1 + = 6 and R2 C L

1+

R1 R2

LC

=8

Using R1 = R 2 = R gives

1 R 1 + = 6 and =4 RC L LC

These equations do not have a unique solution. Try C = 1 F. Then L =

1 H and 4

1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω R 2 4 Pick R = 1.309 Ω. Then vc ( t ) = iL ( t ) = At t = 0+

vC ( t ) 1.309

+

1 + A1 e−2 t + A2 e−4 t V 2

d vC ( t ) = −1.236 A1 e −2 t − 3.236 A2 e−4 t + 0.3819 dt 0 = vc ( 0 + ) = A1 + A2 + 0.5

0 = iL ( 0 + ) = −1.236 A1 − 3.236 A2 + 0.3819

Solving these equations gives A1 = -1 and A2 = 0.5, so vc ( t ) =

1 −2 t 1 −4 t −e + e V 2 2

DP 9-2 Design the circuit shown in Figure DP 9-1 so that

vc(t) =

1 4

+ (A1 + A2t)e–2t V for t > 0

Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is critically damped, and the natural frequencies are both 2 rad/sec. iL(t)

vs(t) = u(t)

L

R1

+ –

R2

C

+ vc(t) –

Figure DP 9-1 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 1 The specifications require that vC ( ∞ ) = so 4 R2 1 = ⇒ 3 R 2 = R1 4 R1 + R 2

Next, represent the circuit by a 2nd order differential equation: vC ( t ) d KCL at the top node of R2 gives: + C vC ( t ) = iL ( t ) R2 dt d KVL around the outside loop gives: vs ( t ) = L iL ( t ) + R1 iL ( t ) + vC ( t ) dt Use the substitution method to get vs ( t ) = L

⎞ ⎛ v (t ) ⎞ d ⎛ vC ( t ) d d + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎜⎜ ⎟ ⎜ R2 ⎟ dt ⎝ R 2 dt dt ⎠ ⎝ ⎠

⎛ L ⎞d ⎛ R1 ⎞ d2 = LC 2 vC ( t ) + ⎜ + R1 C ⎟ vC ( t ) + ⎜1 + v t ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠ The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 2 s2 + ⎜ + ⎟s+⎜ = s + 4s + 4 = ( s + 2 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠

Equating coefficients of like powers of s: 1 R2 C

+

R1 L

1+ = 4 and

R1 R2

LC

=4

Using R 2 = R and R1 = 3R gives

1 3R 1 + = 4 and =1 LC RC L These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω R 3 3 3 Pick R = 1 Ω. Then R1 = 3 Ω and R 2 = 1 Ω . vc ( t ) = iL ( t ) = vC ( t ) +

1 + ( A1 + A2 t ) e −2 t V 4

d 1 vC ( t ) = + dt 4

(( A

2

At t = 0+ 0 = vc ( 0 + ) = A1 + 0 = iL ( 0 + ) =

1 4

1 + A2 − A1 4

Solving these equations gives A1 = -0.25 and A2 = -0.5, so vc ( t ) =

)

− A1 ) − A2 t e −2 t

1 ⎛ 1 1 ⎞ −2 t − ⎜ + t ⎟e V 4 ⎝4 2 ⎠

DP 9-3 Design the circuit shown in Figure DP 9-1 so that vc(t) = 0.8 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0 Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is underdamped and the damped resonant frequency is 4 rad/sec and the damping coefficient is 2. iL(t)

vs(t) = u(t)

R1

L

+ –

R2

C

+ vc(t) –

Figure DP 9-1 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 1 vC ( ∞ ) = R1 + R 2 4 The specifications require that vC ( ∞ ) = so 5 R2 4 = ⇒ 4 R1 = R 2 5 R1 + R 2

Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives:

vC ( t ) R2 vs ( t ) = L

+C

d vC ( t ) = iL ( t ) dt

d iL ( t ) + R1 iL ( t ) + vC ( t ) dt

Use the substitution method to get vs ( t ) = L

⎞ ⎛ v (t ) ⎞ d ⎛ vC ( t ) d d + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎜⎜ ⎟ ⎜ R2 ⎟ dt ⎝ R 2 dt dt ⎠ ⎝ ⎠

⎛ L ⎞d ⎛ R1 ⎞ d2 + R1 C ⎟ vC ( t ) + ⎜1 + v t + v t 2 C ( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠ The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 s2 + ⎜ + ⎟s+⎜ = s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ Equating coefficients of like powers of s: = LC

R1 1 + = 4 and R2 C L

1+

R1 R2

LC

= 20

Using R1 = R and R 2 = 4 R gives R 1 1 + = 4 and = 16 4R C L LC These equations do not have a unique solution. Try C =

1 1 F . Then L = H and 8 2

2 + 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω R Then R1 = 1 Ω and R 2 = 4 Ω . Next

vc ( t ) = 0.8 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V

iL ( t ) = At t = 0+

vC ( t ) 4

+

A2 −2 t A1 1d vC ( t ) = 0.2 + e cos 4 t − e−2 t sin 4 t 8 dt 2 2 0 = vc ( 0 + ) = 0.8 + A1

0 = iL ( 0 + ) = 0.2 +

A2 2

Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.8 − e −2 t ( 0.8cos 4 t + 0.4sin 4 t ) V

DP 9-4 Show that the circuit shown in Figure DP 9-1 cannot be designed so that vc(t) = 0.5 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0 Hint: Show that such a design would require 1/RC + 10RC = 4 where R = R1 = R2. Next, show that 1/RC + 10 RC = 4 would require the value of RC to be complex. iL(t)

vs(t) = u(t)

R1

L

+ –

R2

C

+ vc(t) –

Figure DP 9-1 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 vC ( ∞ ) = 1 R1 + R 2 1 The specifications require that vC ( ∞ ) = so 2 R2 1 = ⇒ R1 = R 2 2 R1 + R 2

Next, represent the circuit by a 2nd order differential equation: KCL at the top node of R2 gives: KVL around the outside loop gives:

vC ( t ) R2 vs ( t ) = L

+C

d vC ( t ) = iL ( t ) dt

d iL ( t ) + R1 iL ( t ) + vC ( t ) dt

Use the substitution method to get ⎞ ⎛ v (t ) ⎞ d ⎛ v (t ) d d vs ( t ) = L ⎜ C + C vC ( t ) ⎟ + R1 ⎜ C + C vC ( t ) ⎟ + vC ( t ) ⎟ ⎜ R2 ⎟ dt ⎜⎝ R 2 dt dt ⎠ ⎝ ⎠ ⎛ L ⎞d ⎛ R1 ⎞ d2 v t + + R1 C ⎟ vC ( t ) + ⎜1 + v t 2 C ( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ C ( ) dt ⎝ ⎠ ⎝ ⎠ The characteristic equation is R ⎞ ⎛ 1+ 1 ⎟ ⎜ ⎛ 1 R1 ⎞ R2 ⎟ 2 s2 + ⎜ + ⎟s+⎜ = s + 4 s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎜ R 2 C L ⎟ ⎜ LC ⎟ ⎝ ⎠ ⎜ ⎟⎟ ⎜ ⎝ ⎠ Equating coefficients of like powers of s: = LC

R1 1 + = 4 and R2 C L

1+

R1 R2

LC

= 20

Using R1 = R 2 = R gives 1 R 1 + = 4 and = 10 RC L LC Substituting L =

1 into the first equation gives 10 C

0.4 ± 0.42 − 4 ( 0.1) 4 1 ( RC ) + = 0 ⇒ RC = 10 10 2 Since RC cannot have a complex value, the specification cannot be satisfied.

( RC )

2



DP 9-5 Design the circuit shown in Figure DP 9-5 so that

vo(t) =

1 2

+ A1 e–2t + A2 e– 4t V for t > 0

Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is overdamped, and the natural frequencies are 2 and 4 rad/sec. iL(t)

R1

L

+ vs(t) = u(t)

+ –

C

+ R2

vc(t) –

vo(t) –

Figure DP 9-5 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions

vC ( ∞ ) =

R2 R1 + R 2

1, iL ( ∞ ) =

The specifications require that vo ( ∞ ) =

1 R1 + R 2

and

vo ( ∞ ) =

R2 R1 + R 2

1

1 so 2 R2

1 = ⇒ R1 = R 2 2 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives:

d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L

Use the substitution method to get vs ( t ) = R1 C = R1 LC

Using vo ( t ) =

iL ( t ) gives R2

d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt

⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 vo ( t ) = LC 2 iL ( t ) + ⎜ + R1 C ⎟ iL ( t ) + ⎜ i t ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ L ( ) R2 dt ⎝ ⎠ ⎝ ⎠ R1

The characteristic equation is

R ⎛ 1+ 2 ⎜ ⎛ 1 R2 ⎞ ⎜ R1 s2 + ⎜ + s+ ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s:

⎞ ⎟ ⎟ = s 2 + 6 s + 8 = ( s + 2 )( s + 4 ) ⎟ ⎟⎟ ⎠

R2 1 + = 6 and R1 C L

1+

R2 R1

LC

=8

Using R1 = R 2 = R gives 1 R 1 + = 6 and =4 RC L LC These equations do not have a unique solution. Try C = 1 F. Then L =

1 H and 4

1 3 1 + 4 R = 6 ⇒ R 2 − R + = 0 ⇒ R = 1.309 Ω or R = 0.191 Ω 2 4 R Pick R = 1.309 Ω. Then 1 + A1 e −2 t + A2 e−4 t V 2 A1 −2 t A 2 −4 t v (t ) 1 iL ( t ) = o e + e V = + 1.309 2.618 1.309 1.309 vo ( t ) =

vC ( t ) = 1.309 iL ( t ) +

1 d 1 iL ( t ) = + 0.6167 A1 e−2 t + 0.2361 A2 e−4 t 4 dt 2

At t = 0+ 0 = iL ( 0 + ) = 0 = vC ( 0 + ) =

A1 A2 1 + + 2.618 1.309 1.309

1 + 0.6167 A1 + 0.2361 A2 2

Solving these equations gives A1 = -1 and A2 = 0.5, so vo ( t ) =

1 −2 t 1 −4 t −e + e V 2 2

DP 9-6 Design the circuit shown in Figure DP 9-5 so that 3

vo(t) =

4

+ (A1 + A2t)e–2t V for t > 0

Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is critically damped, and the natural frequencies are both 2 rad/sec. iL(t)

R1

L

+ + –

vs(t) = u(t)

C

+ R2

vc(t) –

vo(t) –

Figure DP 9-5 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions

vC ( ∞ ) =

R2 R1 + R 2

1, iL ( ∞ ) =

The specifications require that vo ( ∞ ) =

1 R1 + R 2

and

vo ( ∞ ) =

R2 R1 + R 2

1

3 so 4 R2

3 = ⇒ 3R1 = R 2 4 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: KVL around the right-hand mesh gives: KCL at the top node of the capacitor gives:

d iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) R1 dt vC ( t ) = L

Use the substitution method to get vs ( t ) = R1 C = R1 LC

Using vo ( t ) =

iL ( t ) gives R2 vo ( t ) =

d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt

R1 R2

LC

The characteristic equation is

⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 i t + + R1 C ⎟ iL ( t ) + ⎜ i t 2 L( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ L ( ) dt ⎝ ⎠ ⎝ ⎠

R ⎛ 1+ 2 ⎜ ⎛ 1 R2 ⎞ ⎜ R1 s2 + ⎜ + s+ ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s:

⎞ ⎟ ⎟ = s 2 + 4s + 4 = ( s + 2 )2 ⎟ ⎟⎟ ⎠

R2 1 + = 4 and R1 C L

1+

R2 R1

LC

=4

Using R1 = R and R 2 = 3R gives 1 3R 1 + = 4 and =1 RC L LC These equations do not have a unique solution. Try C = 1 F. Then L = 1 H and 1 4 1 1 + 3 R = 4 ⇒ R 2 − R + = 0 ⇒ R = 1 Ω or R = Ω 3 3 3 R Pick R = 1 Ω. Then R1 = 1 Ω and R 2 = 3 Ω . vo ( t ) = iL ( t ) =

3 + ( A1 + A2 t ) e −2 t V 4

vo ( t )

vC ( t ) = 3 iL ( t ) +

3

=

1 ⎛ A1 A2 ⎞ −2 t t⎟ e V +⎜ + 4 ⎝ 3 3 ⎠

d 3 ⎛ ⎛ A1 A2 ⎞ A2 ⎞ −2 t iL ( t ) = + ⎜⎜ ⎜ + t ⎟e ⎟+ dt 4 ⎝⎝ 3 3 ⎠ 3 ⎟⎠

At t = 0+ 0 = iL ( 0 + ) =

A1

+

1 4

3 3 A1 A2 0 = vC ( 0 + ) = + + 4 3 3 Solving these equations gives A1 = -0.75 and A2 = -1.5, so vo ( t ) =

3 ⎛ 3 3 ⎞ −2 t − ⎜ + t ⎟e V 4 ⎝4 2 ⎠

DP 9-7 Design the circuit shown in Figure DP 9-5 so that vc(t) = 0.2 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0 Determine the values of the unspecified constants, A1 and A2. Hint: The circuit is underdamped, the damped resonant frequency is 4 rad/sec, and the damping coefficient is 2. iL(t)

R1

L

+ vs(t) = u(t)

+ –

C

vc(t) –

+ R2

vo(t) –

Figure DP 9-5 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 R2 1 vC ( ∞ ) = and vo ( ∞ ) = 1, iL ( ∞ ) = 1 R1 + R 2 R1 + R 2 R1 + R 2 1 The specifications require that vo ( ∞ ) = so 5 R2 1 = ⇒ R1 = 4 R 2 5 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: d KVL around the right-hand mesh gives: vC ( t ) = L iL ( t ) + R 2 iL ( t ) dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) KCL at the top node of the capacitor gives: R1 dt Use the substitution method to get vs ( t ) = R1 C = R1 LC

Using vo ( t ) =

iL ( t ) gives R2

vo ( t ) =

d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt

R1 R2

LC

The characteristic equation is

⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 i t + + R C i t + i t ( ) ( ) ⎜ ⎟ ⎜ L 1 ⎜ R2 ⎟ dt L ⎜ R 2 ⎟⎟ L ( ) dt 2 ⎝ ⎠ ⎝ ⎠

R ⎛ 1+ 2 ⎜ ⎛ 1 R2 ⎞ ⎜ R1 + s2 + ⎜ s+ ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s:

⎞ ⎟ ⎟ = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎟ ⎟⎟ ⎠

R2 1 + = 4 and R1 C L

1+

R2 R1

LC

= 20

Using R 2 = R and R1 = 4 R gives 1 R 1 + = 4 and = 16 4R C L LC 1 1 These equations do not have a unique solution. Try C = F . Then L = H and 8 2 2 + 2 R = 4 ⇒ R2 − 2R + 2 = 0 ⇒ R = 1 Ω R Then R1 = 4 Ω and R 2 = 1 Ω . Next

vo ( t ) = 0.2 + e −2 t ( A1 cos 4 t + A2 sin 4 t ) V vo ( t )

= 0.2 + e−2 t ( A1 cos 4 t + A2 sin 4 t ) V 1 1 d vC ( t ) = iL ( t ) + iL ( t ) = 0.2 + 2 A2 e−2 t cos 4 t − 2 A1 e −2 t sin 4 t 2 dt iL ( t ) =

At t = 0+

0 = iL ( 0 + ) = 0.2 + A1 0 = vC ( 0 + ) = 0.2 + 2 A2

Solving these equations gives A1 = -0.8 and A2 = -0.4, so vc ( t ) = 0.2 − e −2 t ( 0.2 cos 4 t + 0.1sin 4 t ) V

DP 9-8 Show that the circuit shown in Figure DP 9-5 cannot be designed so that vc(t) = 0.5 + e–2t(A1 cos 4t + A2 sin 4t) V for t > 0 Hint: Show that such a design would require 1/RC + 10RC = 4 where R = R1 = R2. Next, show that 1/RC + 10 RC = 4 would require the value of RC to be complex. iL(t)

R1

L

+ vs(t) = u(t)

+ –

C

vc(t) –

+ R2

vo(t) –

Figure DP 9-5 Solution: When the circuit reaches steady state after t = 0, the capacitor acts like an open circuit and the inductor acts like a short circuit. Under these conditions R2 R2 1 1, iL ( ∞ ) = 1 vC ( ∞ ) = and vo ( ∞ ) = R1 + R 2 R1 + R 2 R1 + R 2 1 The specifications require that vC ( ∞ ) = so 2 R2 1 = ⇒ R1 = R 2 2 R1 + R 2 Next, represent the circuit by a 2nd order differential equation: d vC ( t ) = L iL ( t ) + R 2 iL ( t ) KVL around the right-hand mesh gives: dt vs ( t ) − vC ( t ) d − C vC ( t ) = iL ( t ) KCL at the top node of the capacitor gives: R1 dt Use the substitution method to get vs ( t ) = R1 C = R1 LC

Using vo ( t ) =

iL ( t ) gives R2 vo ( t ) =

d ⎛ d ⎞ ⎛ d ⎞ ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + ⎜ L iL ( t ) + R 2 iL ( t ) ⎟ + R1 iL ( t ) dt ⎝ dt ⎠ ⎝ dt ⎠ d2 d i t + L + R1 R 2 C ) iL ( t ) + ( R1 + R 2 ) iL ( t ) 2 L( ) ( dt dt

R1 R2

LC

The characteristic equation is

⎛ L ⎞d ⎛ R1 + R 2 ⎞ d2 i t + + R1 C ⎟ iL ( t ) + ⎜ i t 2 L( ) ⎜ ⎜ R2 ⎟ dt ⎜ R 2 ⎟⎟ L ( ) dt ⎝ ⎠ ⎝ ⎠

R ⎛ 1+ 2 ⎜ ⎛ 1 R2 ⎞ ⎜ R1 + s2 + ⎜ s+ ⎟ ⎜ R1 C L ⎟ ⎜ LC ⎝ ⎠ ⎜ ⎜ ⎝ Equating coefficients of like powers of s:

⎞ ⎟ ⎟ = s 2 + 4s + 20 = ( s + 2 − j 4 )( s + 2 + j 4 ) ⎟ ⎟⎟ ⎠

R2 1 + = 4 and R1 C L

1+

R2 R1

LC

= 20

Using R1 = R 2 = R gives 1 R 1 + = 4 and = 10 LC RC L Substituting L =

1 into the first equation gives 10 C

( RC )

2

0.4 ± 0.42 − 4 ( 0.1) 4 1 − ( RC ) + = 0 ⇒ RC = 10 10 2

Since RC cannot have a complex value, the specification cannot be satisfied.

L DP 9-9 A fluorescent light uses cathodes (coiled 1 2 tungsten filaments coated with an electron-emitting i t=0 substance) at each end that send current through mercury vapors sealed in the tube. Ultraviolet + 4 Ω radiation is produced as electrons from the cathodes 10 V – 1 3F knock mercury electrons out of their natural orbits. Some of the displaced electrons settle back into orbit, throwing off the excess energy absorbed in Figure DP 9-9 the collision. Almost all of this energy is in the form of ultraviolet radiation. The ultraviolet rays, which are invisible, strike a phosphor coating on the inside of the tube. The rays energize the electrons in the phosphor atoms, and the atoms emit white light. The conversion of one kind of light into another is known as fluorescence. One form of a fluorescent lamp is represented by the RLC circuit shown in Figure DP 99. Select L so that the current i(t) reaches a maximum at approximately t = 0.5 s. Determine the maximum value of i(t). Assume that the switch was in position 1 for a long time before switching to position 2 at t = 0. Hint: Use PSpice to plot the response for several values of L.

Solution:

R 1 s+ =0 L LC Select L so fast response and i achieve maximum at t = 0.5 s Characteristic equation: s 2 +

4 3 s + = 0 Try L = 1 H ⇒ s 2 + 4s + 3 = 0 or s = −3, −1 L L i (t ) = A1 e − t + A2 e −3t s2 +

i (0) = 0 = A1 + A2

⎫ ⎪ ⎬ A1 = 5, A2 = −5 di (0) v(0) 10 = = = − A1 − 3 A2 ⎪ dt L 1 ⎭ −t −3t i (t ) = 5e − 5e at t = 0.5 s i − 1.92

Section 10.2 Sinusoidal Sources    P10.2‐1 Given the sinusoids  v1 ( t ) = 8cos ( 250 t + 15° ) V  and   v 2 ( t ) = 6 cos ( 250 t − 45° ) V ,  determine the time by which  v 2 ( t )  is advanced or delayed with respect to  v1 ( t ) .     Solution: 

T=

The period of both sinusoids is 

2π = 25.1327 ms   250

The difference in the phase angles is     θ 2 − θ1 = −45° − 15° = −60°   td =

The delay time is 

−60 ( 25.1327 ) = −4.188 ms   2π

(The minus sign indicates a delay.)  The voltage  v 2 ( t )  is delayed by 4.188 ms with respect to  v1 ( t ) . 

    P10.2‐2 Given the sinusoids  v1 ( t ) = 8cos (100 t − 54° ) V  and   v 2 ( t ) = 8cos (100 t − 102° ) V ,  determine the time by which  v 2 ( t )  is advanced or delayed with respect to  v1 ( t ) .     Solution:  The period of both sinusoids is 

T=

2π = 62.8319 ms   100

The difference in the phase angles is   θ 2 − θ1 = −102° − ( −54° ) = −48°    

−48° ( 62.8319 ) = −8.3776 ms   360° (The minus sign indicates a delay.)  The voltage  v 2 ( t )  is delayed by 48.3776 ms with respect to  The delay time is  v1 ( t ) . 

   

td =

P10.2‐3 A sinusoidal current is given as    

i ( t ) = 125cos ( 5000 π t − 135° ) mA  

Determine the period, T, and the time, t1, at which the first positive peak occurs.  Answer: T = 0.4 ms and t1 = 0.15 ms.    Solution: The frequency is 5000 π rad/s or 2500 Hertz so the period is   1   T= = 0.0004 = 0.4 ms   2500 ⎛ π ⎞ Converting the angle from degrees to radians, we get  −135° ⎜ ⎟ = −0.75 π radians . The  ⎝ 180° ⎠ −0.75 π = −15 ms . The minus sign  current sinusoid is shifted from  125cos ( 5000 π t ) mA by  5000 π indicates a delay. A positive peak occurs at  t 1 = 0.15 ms . Since 15 ms is less than the period of 

i ( t 1 ) , the positive peak at  t 1 = 0.15 ms is the first positive peak. 

      P10.2‐4 Express the voltage shown in Figure P10.2‐7 in the general form    

v ( t ) = A cos (ω t + θ ) V  

where  A ≥ 0  and  −180° < θ ≤ 180° .   

  Figure P10.2‐4    Solution: The amplitude is A = 45 mv and the period is given by  period is T = 80 ms. The frequency is given by  ω =

 

T = 60 − 20 = 40 ms  so the  2

2π = 78.54 rad/s . Noticing that v(t) is 0  80 ×10−3

at time 0 and is increasing at time 0, we can write    v ( t ) = 45sin ( 78.54 t ) = 45cos ( 78.54 t − 90° ) mV      

  20 V

P 10.2-5 Figure P 10.2-5 shows a sinusoidal voltage, v(t), plotted as a function of time, t. Represent v(t) by a function of the form A cos (ω + θ). v(t)

0V

Answer: v(t) = 18 cos (393t–27°)

–20 V 0s

20 ms t

40 ms

Figure P 10.2-5 Solution:   A = 18 V     T = 18 − 2 = 16 ms     2π 2π ω= = = 393 rad/s   0.016 T   ⎛ 16 ⎞ θ = − cos −1 ⎜ ⎟ = −27°   ⎝ 18 ⎠   v ( t ) = 18 cos ( 393 t − 27° ) V        

20 V

P 10.2-6 Figure P 10.2-6 shows a sinusoidal voltage, v(t), plotted as a function of time, t. Represent v(t) by a function of the form A cos (ωt+θ).

v(t)

0V

–20 V 0s

20 ms

40 ms

60 ms

t

Figure P 10.2-6     Solution:   A = 15 V     T = 43 − 11 = 32 ms     2π 2π = = 196 rad/s   ω= T 0.032   ⎛8⎞ θ = − cos −1 ⎜ ⎟ = −58°   ⎝ 15 ⎠   v ( t ) = 15 cos (196 t − 58° ) V    

   

Section 10.3 Phasors and Sinusoids    P10.3‐1 Express the current     i ( t ) = 2 cos ( 6 t + 120° ) + 2sin ( 6 t − 60° ) mA   In the general form   

i ( t ) = A cos (ω t + θ ) mA  

where  A ≥ 0  and  −180° < θ ≤ 180° .    Solution:  i ( t ) = 2 cos ( 6 t + 120° ) + 2sin ( 6 t − 60° ) mA

= 2 cos ( 6 t + 120° ) + 2 cos ( 6 t − 60° − 90° ) mA

 

Representing the sinusoids using phasors gives:   

I (ω ) = 2∠120° + 4∠ − 150° = ( −1 + j 1.732 ) + ( −3.464 − j 2 ) = −4.464 − j 0.268 = 4.472∠183.4° = 4.472∠ − 176.6° mA

The corresponding sinusoid is: 

i ( t ) = 4.472 cos ( 6 t − 176.6° ) mA  

          P10.3‐2 Express the voltage     v ( t ) = 5 2 cos ( 8 t ) + 2sin ( 8 t + 45° ) V   In the general form   

v ( t ) = A cos (ω t + θ ) V  

where  A ≥ 0  and  −180° < θ ≤ 180° .    Solution:  v ( t ) = 5 2 cos ( 8 t ) + 2sin ( 8 t + 45° ) = 5 2 cos ( 8 t ) + 2 cos ( 8 t + 45° − 90° ) = 5 2 cos ( 8 t ) + 2 cos ( 8 t − 45° ) V Representing the sinusoids using phasors gives:   

V (ω ) = 7.0711 + 10∠ − 45° = 7.0711 + ( 7.0711 − j 7.0711)

The corresponding sinusoid is: 

= 14.1422 − j 7.0711 = 15.811∠ − 26.6° V v ( t ) = 15.811cos ( 8 t − 26.6° ) V  

 

 

 

 

P 10.3-3

Determine the polar form of the quantity (25∠36.9°) ( 80 ∠ − 53.1° ) (4 + j 8) + (6 − j 8)

Answer: 200∠ − 16.2° Solution: (25∠36.9°) ( 80 ∠ − 53.1° ) 25 ⋅ 80∠ ( 36.9° − 53.1° ) 2000∠ − 16.2° = = = 200∠ − 16.2°   (4 + j 8) + (6 − j 8) 10 ( 4 + 6 ) + j (8 − 8)

P 10.3-4

Determine the polar and rectangular form of the expression ⎛ 3 2 ∠ − 45° ⎞ 5 ∠ + 81.87° ⎜⎜ 4 − j 3 + ⎟ 7 − j1 ⎟⎠ ⎝

Answer: 88.162 ∠30.127° = 76.2520 + j 44.2506 Solution:

⎛ ⎛ ⎞ 40 ∠ − 45° ⎞ 40 ∠ − 45° 8 ∠42° ⎜ 8 − j 3 + ⎟ = 8 ∠42° ⎜ 8 − j 3 + ⎟ 7 − j 12 ⎠ 13.892 ∠ − 59.744° ⎠ ⎝ ⎝ = 8 ∠42° ( 8 − j 3 + 2.8793 ∠14.744° ) = 8 ∠42° ( 8 − j 3 + 2.7845 + j 0.7328 ) = 8 ∠42° (10.7845 − j 2.2672 ) = 8 ∠42° (11.02∠ − 11.873° ) = 88.162 ∠30.127° = 76.2520 + j 44.2506

Determine the polar and rectangular form of the expression

P 10.3-5

(60 ∠120°)(−16 + j12 + 20∠15°) 5∠ − 75° Solution: (60 ∠120°)(−16 + j12 + 20∠15°) (60 ∠120°)(−16 + j12 + 19.3185 + j 5.1764) = 5∠ − 75° 5∠ − 75° (60 ∠120°)(3.3185 + j 17.1764) = 5∠ − 75°   (60 ∠120°)(17.494∠79.065°) = 5∠ − 75° 1049.6∠ − 160.93° = = 139.95∠109.07° = 45.714 + j 132.28 5∠ − 75°

    P10.3‐6 The circuit shown in Figure 10.3‐6 is at steady state. The input currents are  i1 ( t ) = 10 cos ( 25 t ) mA  and   i 3 ( t ) = 10 cos ( 25 t + 135° ) mA  

Determine the voltage v2(t). 

  Figure 10.3‐6    Solution:    Using first Ohm’s law and then KCL      v2 ( t ) = 250 i 2 ( t )     and 

i 2 ( t ) = i1 ( t ) − i 3 ( t ) = 10 cos ( 25 t ) − 10 cos ( 25 t + 135° ) mA  

Using phasors  I 2 (ω ) = I 1 (ω ) − I 3 (ω ) = 10 − 10 ∠135 = 10 − ( −7.071 + j 7.071) mA

= 17.071 − j 7.071 = 18.478∠ − 22.5° mA

The corresponding sinusoid is i 2 ( t ) = 18.478cos ( 25 t − 22.5° ) mA   Finally 

v2 ( t ) = 250 i 2 ( t ) = 4.6195cos ( 25 t − 22.5° ) V  

 

P10.3‐7 The circuit shown in Figure 10.3‐7 is at steady state. The inputs to this circuit are the  current source current  i1 ( t ) = 0.12 cos (100 t + 45° ) A    and the voltage source voltage 

v 2 ( t ) = 24 cos (100 t − 60° ) V  

Determine the current i2(t). 

  Figure P10.3‐7    Solution:  Using Ohm’s and Kirchhoff’s laws  v 2 (t ) 24 cos (100 t − 60° ) i 2 ( t ) = i1 ( t ) − = 0.12 cos (100 t + 45° ) −     96 96 = 0.12 cos (100 t + 45° ) − 0.25cos (100 t − 60° ) Using phasors  I 2 (ω ) = 0.12 ∠45° − 0.25 ∠60° = ( 0.0849 + j 0.0849 ) − ( 0.1250 − j 0.2165 ) = −0.0401 + j 0.3014 = 0.3040∠97.6° A

 

The corresponding sinusoid is    i 2 ( t ) = 0.3040 cos (100 t + 97.6° ) A   Checked using LNAPAC 8/16/11        P 10.3-8

Given that

i1 (t)

i2 (t)

i1(t) = 30 cos (4t + 45°) mA and

i2(t) = – 40 cos (4t) mA

Determine v(t) for the circuit shown in

100 Ω 15 H

+

120 Ω

v(t) –

Figure P 10.3-8. Figure P 10.3-8 Solution:  V = j ( 4 )(15 ) ( I 1 + I 2 ) = j 60 ( 0.03∠45° − 0.04∠0° ) = j 60 ( 0.0212 + j 0.0212 − 0.04 ) = −1.273 − j1.127 = 1.7∠ − 138.5° V So                                                        v ( t ) = 1.7 cos ( 4t − 138.5° ) V  

(checked: LNAP 8/7/04)

  P 10.3-9 For the circuit shown in Figure P 10.3-9, find (a) the impedances Z1 and Z2 in polar form, (b) the total combined impedance in polar form, and (c) the steady-state current i(t). Answer: (a) Z1 = 5 ∠53.1°; Z 2 = 8 2 ∠ − 45° (b) Z1 + Z 2 = 11.7 ∠ − 20° (c) i (t ) = (8.55) cos (1250t + 20°) A



100 μF

3.2 mH Z1 i



Z2 +



100 cos (1250t) V

Figure P 10.3-9

Solution: 

(a)  (b)  (c) 

Z1 =3+ j 4 = 5∠53.1° Ω

and

Z 2 =8− j8 = 8 2 ∠− 45° Ω  

  Total impedance = Z1 + Z 2 = 3 + j 4 + 8 − j8 = 11 − j 4 = 11.7∠− 20.0° Ω     100∠0° 100 100 I= = = ∠20.0° ⇒ i (t ) = 8.55 cos (1250 t + 20.0°) A   Z1 +Z 2 11.7 ∠− 20° 11.7

  P 10.3-10 The circuit shown in Figure P 10.3-10 is at steady state. The voltages vs(t) and v2(t) are given by

+ v1(t) –

vs(t) = 7.68 cos (2t + 47°) V and

vs(t)

+ –

C R

v2(t) = 1.59 cos (2t + 125°) V Find the steady-state voltage v1(t).

+ v2(t) –

Figure P 10.3-10

Answer: v1(t) = 7.51 cos (2t + 35°) V   Solution: V1 (ω ) = Vs (ω ) − V2 (ω ) = 7.68∠47° − 1.59∠125°

= ( 5.23 + j 5.62 ) − ( −0.91 + 1.30 ) = ( 5.23 + 0.91) + j ( 5.62 − 1.30 ) = 6.14 + j 4.32 = 7.51∠35° v1 ( t ) = 7.51 cos ( 2 t + 35° ) V  

 

P 10.3-11 The circuit shown in Figure P 10.3-11 is at steady state. The currents i1(t) and i2(t) are given by i1(t) = 744 cos (2t – 118°) mA and

i2(t) = 540.5 cos (2t + 100°) mA

Find the steady-state current i(t). Answer: i(t) = 460 cos (2t + 196°) mA

10 Ω v1(t)

+ –

i1(t)

i2(t)

6H

10 Ω

0.05 F

+ –

10 Ω

v2(t)

i(t)

Figure P 10.3-11 Solution:  I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( −0.094 + j 0.532 ) = ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 ) = −0.443 − j 0.125 = 0.460∠196° i ( t ) = 460 cos (2 t + 196°) mA  

  i(t)

P 10.3-12 Determine i(t) of the RLC circuit shown in Figure P 10.3-12 when

3H 6Ω

vs

vs = 2 cos (4t + 30°) V.

+ –

1 12

Answer: i(t) = 0.185 cos (4t – 26.3°) A

Figure P 10.3-12 Solution: 

Vs = 2 ∠30° V and I =

  2 ∠30° = 0.185 ∠ − 26.3° A 6+ j12+ 3 / j

  i (t ) = 0.185 cos (4 t − 26.3°) A  

 

F

Section 10.4 Impedances P10.4-1 Figure P10.4-1a shows a circuit represented in the time domain. Figure P10.4-1b shows the same circuit represented in the frequency domain, using phasors and impedances. ZR, ZC, ZL1, and ZL2 are the impedances corresponding to the resistor, capacitor, and two inductors in Figure P10.41a. Vs is the phasor corresponding to the voltage of the voltage source. Determine ZR, ZC, ZL1, ZL2, and Vs. + 8Ω

+ Z L1

ZR

2H

Vs

5 sin 5t V vo

+–

1 F 12

4H

Vo

+–

Z L2

ZC –

(a)



(b)

Figure P10.4-1 Hint: 5 sin 5t = 5 cos (5t – 90°) Answer: Z R = 8Ω, Z C =

1 2.4 j 2.4 = = = − j 2.4Ω, Z L1 = j 5(2) = j10Ω, j j× j ⎛ 1⎞ j5 ⎜ ⎟ ⎝ 12 ⎠

ZL2 = j5(4) = j20 Ω, and Vs = 5 ∠–90° V Solution:

ZR = 8 Ω, ZC =

1

=

2.4 j 2.4 = = − j 2.4 Ω, ZL1 = j 5 (2) = j 10 Ω, j j× j

1 12 ZL2 = j 5 (4) = j 20 Ω and VS = 5 ∠-90° V. j5

1

P10.4-2 Figure P10.4-2a shows a circuit represented in the time domain. Figure P10.4-2b shows the

same circuit represented in the frequency domain, using phasors and impedances. ZR, ZC, ZL1, and ZL2 are the impedances corresponding to the resistor, capacitor, and two inductors in Figure P10.4-

2a. Is is the phasor corresponding to the current of the current source. Determine ZR, ZC, ZL1, ZL2, and Is. + 8Ω

+ ZR

2H vo

1 12

Z L1 Vo

4 cos (3t + 15°) A

F 4H

ZC

Is

Z L2





(a)

(b)

Figure P10.4-2 Answer: Z R = 8 Ω, Z C =

1 4 j4 = = = − j 4 Ω, Z L1 = j 3(2) = j 6 Ω, ⎛ 1 ⎞ j j× j j3 ⎜ ⎟ ⎝ 12 ⎠

ZL2 = j3(4) = j12 Ω, and Is = 4 ∠15° A

Solution: ZR = 8 Ω, ZC =

1

=

4 j4 = = − j 4 Ω, ZL1 = j 3 (2) = j 6 Ω, j j× j

1 12 ZL2 = j 3 (4) = j 12 Ω and IS = 4 ∠15° A. j3

2

P10.4-3 Represent the circuit shown in Figure P10.4-3 in the frequency domain using impedances and phasors.

Figure P10.4-3 Solution:

P10.4-4 Represent the circuit shown in Figure P10.4-4 in the frequency domain using impedances and phasors.

Figure P10.4-4 Solution:

3

P10.4-5 Determine the current, i(t), for the circuit shown in Figure P10.4-5.

Figure P10.4-5 Solution: Represent the circuit in the frequency domain using phasors and impedances:

j 96 I − j 312.5 I + 250 I + 7.4∠ − 24° = 0

Using KVL: Solving:

I=

7.4∠ − 24° 7.4∠ − 24° 7.4∠ − 24° = = = 0.0224∠16.9 A j 96 − j 312.5 + 250 − j 216.5 + 250 330.715∠ − 40.9°

In the time domain:

i ( t ) = 22.4 cos ( 200 t + 16.9 ) mA

(Checked using LNAP)

4

P10.4-6 The input to the circuit shown in Figure P10.4-6 is the current i ( t ) = 120 cos ( 4000 t ) mA

Determine the voltage, v(t), across the 40 Ω resistor.

Figure P10.4-6 Solution: Represent the circuit in the frequency domain using phasors and impedances:

Using KCL

0.120∠0° = I 1 + I 2

Using KVL

j 60 I 2 + 40 I 2 − ( − j 12.5 ) I 1 = 0

(

)

j 60 I 2 + 40 I 2 − ( − j 12.5 ) 0.12∠0° − I 2 = 0 j 60 I 2 + 40 I 2 + ( − j 12.5 ) I 2 = ( − j 12.5 )( 0.12∠0° ) I2 =

1.5∠ − 90° 1.5∠ − 90° 1.5∠ − 90° = = = 0.024155∠ − 140° A j 60 + 40 + ( − j 12.5 ) 40 + j 47.5 62.1∠50°

V = 40 I 2 = 0.9662∠ − 140° V In the time domain

v ( t ) = 0.9662 cos ( 4000 t − 140° ) V

(Checked using LNAP)

5

P10.4-7 The input to the circuit shown in Figure P10.4-7 is the current i ( t ) = 82 cos (10000 t ) μA

Determine the voltage, v(t), across the 50 kΩ resistor.

Figure P10.4-7 Solution: Represent the circuit in the frequency domain using phasors and impedances:

Using KCL Using KVL

In the time domain

82 ×10−6 ∠0° = 0 + I 1

( 20 ×103 ) I 1 + ( − j 20 ×103 ) I 1 + V = 0 V = − ( 20 × 103 − j 20 ×103 )( 82 × 10−6 ) = 2.3193∠135°

V

v ( t ) = 2.3193cos (10000 t + 135° ) V

(Checked using LNAP)

6

P 10.4-8 Each of the following pairs of element voltage and element current adheres to the passive convention. Indicate whether the element is capacitive, inductive, or resistive and find the element value. (a) v(t) = 15 cos (400t + 30°); i = 3 sin (400t + 30°) (b) v(t) = 8 sin (900t + 50°); i = 2 sin (900t + 140°) (c) v(t) = 20 cos (250t + 60°); i = 5 sin (250t + 150°) Answer: (a) L = 12.5 mH (b) C = 277.77 μF (c) R = 4 Ω Solution: (a)

v = 15cos (400 t + 30°) V i = 3 sin(400 t+30°) = 3 cos (400 t − 60°) V v leads i by 90° ⇒ element is an inductor v 15 Z L = peak = = 5 = ω L = 400 L ⇒ L = 0.0125 H = 12.5 mH ipeak 3

(b)

i leads v by 90° ⇒ the element is a capacitor

(c)

8 1 1 =4= = ⇒ C = 277.77 μ F ω C 900 C ipeak 2 v = 20 cos (250 t + 60°) V Zc =

vpeak

=

i = 5sin (250 t +150°) =5cos (250 t + 60°) A Since v & i are in phase ⇒ element is a resistor v 20 ∴ R = peak = =4Ω ipeak 5

7

Figure P10.4-9 P10.4-9 This voltage and current for the circuit shown in Figure P10.4-9 are given by v ( t ) = 20 cos (20 t + 15°) V and i ( t ) = 1.49 cos (20 t + 63°) A

Determine the values of the resistance, R, and capacitance, C.

Solution:

1 V 20∠15° 20 =Z= = = ∠ (15° − 63° ) = 13.42∠ − 48° = 8.98 − j9.97 Ω 20 C I 1.49∠63° 1.49 1 Equating real and imaginary parts gives R = 9 Ω and C = = 5 mF . 20 × 9.97 R− j

8

P10.4-10 Figure P10.4-10 shows an ac circuit represented in both the time domain and the frequency domain. Determine the values of A, B, a and b.

Figure P10.4-10 Solution: The impedance between nodes a and b is given by 18 + j (10 )( 2.5 ) = 18 + j 25 = 30.8∠54.2°

To find the impedance between nodes b and c we first find the impedance of the capacitor: −j

then

1 1 =−j = − j 25 0.04 (10 )( 0.004 )

9 ( − j 25 ) − j 225 225∠ − 90° = = = 8.47∠ − 19.8° Ω 9 − j 25 26.57∠ − 70.2° 26.57∠ − 70.2°

The impedance between nodes c and d is given by

( 5 ) ( j (10 )( 0.88) ) j 40 j 40 ⎛ 5 − j 8 ⎞ 1 1 −j = −j = − j 20 5 + j (10 )( 0.8 ) (10 )( 0.005 ) 5 + j 8 0.05 5 + j 8 ⎜⎝ 5 − j 8 ⎟⎠ 320 + j 200 − j 20 25 + 64 = 3.60 + j 2.25 − j 20 = 3.60 − j 17.75 Ω =

So A = 30.8 V, B = 8.47 Ω, , a = 3.57 Ω and b = −17.75 Ω.

9

P 10.4-11 Represent the circuit shown in Figure P 10.4-11 in the frequency domain using impedances and phasors. 6Ω

15 cos 4t V

+ –

2H

5i(t)

0.125 F

i(t)

+ v(t) –

Figure P 10.4-11 Solution:

P 10.4-12 Represent the circuit shown in Figure P 10.4-12 in the frequency domain using impedances and phasors. 12 cos (5t – 30°) V + –

2H 0.25 F

0.05 F 6Ω

+ v(t) –

+ –

i(t)

15 cos (5t + 60°) V

Figure P 10.4-12 Solution:

10

P 10.4-13 Find R and L of the circuit of Figure P 10.4-13 when v(t) = 10 cos (ωt + 40°) V; i(t) = 2 cos (ωt + 15°) mA,

+ v

6

and ω = 2 × 10 rad/s.



Answer: R = 4.532 kω, L = 1.057 mH

R L i

Figure P 10.4-13

Solution: Z=

V 10 ∠40° = = − 5000∠205°Ω = 4532 + j 2113 = R + j ω L −I −2×10−3 ∠−165°

so R = 4532 Ω and L =

2113

ω

=

2113 = 1.057 mH 2×106

11

Section 10.5 Series and Parallel Impedances P10.5-1 Determine the steady state voltage v(t) in the circuit shown in Figure P10.5-1. Answer: v ( t ) = 32 cos ( 250 t − 57.9° ) V

Figure P10.5-1 Solution: Represent the circuits in the frequency domain using phasors and impedances:

Using voltage division: V = −

In the time domain

16 16∠180° ( 40∠ − 15° ) = ( 40∠ − 15° ) = 32∠128.1° V 16 + j 12 20∠36.9° v ( t ) = 32 cos ( 250 t − 57.9° ) V

1

P10.5-2 Determine the voltage v(t) in the circuit shown in Figure P10.5-2. Answer: v ( t ) = 14.57 cos ( 800 t + 111.7° ) V

Figure P10.5-2 Solution: Represent the circuit in the frequency domain:

Replace the series impedances at the right of the circuit by an equivalent impedance Z S = j 112 + ( − j 250 ) + 80 = 80 − j 138 Ω

Replace the parallel impedances at right of the circuit by an equivalent impedance ZP =

(80 − j 138)120 = (80 − j 138 )120

80 − j 138 + 120

200 − j 138

=

(159.51∠ − 59.9 )120

242.99∠ − 34.6 = 78.77∠25.3° Ω Using voltage division

V=−

78.77∠ − 25.3° 78.77∠ − 25.3° 18∠0° = − 18∠0° = 14.57∠111.73° V j 100 + 78.77∠ − 25.3° 97.325∠42.97°

In the time domain

v ( t ) = 14.57 cos ( 800 t + 111.7° ) V

(checked using LNAP)

2

P10.5-3 Determine the voltage v(t) in the circuit shown in Figure P10.5-3.

Answer: v ( t ) = 14.1cos ( 2500 t − 35.2° ) V

Figure P10.5-3

Solution: Represent the circuit in the frequency domain:

Replace the parallel impedances at right of the circuit by an equivalent impedance:

( − j 100 )( j 250 ) = − j166.67 Ω − j 100 + j 250

Using voltage division V=− In the time domain

− j 166.67 166.67∠ − 90° 22∠15° = − 22∠15° = 14.1∠ − 35.2° V 200 − j 166.67 260.3∠ − 39.8° v ( t ) = 14.1cos ( 2500 t − 35.2° ) V

3

P10.5-4 The input to the circuit shown in Figure P10.5-4 is the current i s ( t ) = 48cos ( 25 t ) mA . Determine the current i1(t).

Answer: i1 ( t ) = 144 cos ( 25 t + 180° ) mA

Figure P10.5-4

Solution: Represent the circuits in the frequency domain using phasors and impedances:

Using current division I1 = In the time domain

j 15 15 ( 48∠0° ) = ( 48∠0° ) = 144∠180° mA −5 j 15 − j 20 i1 ( t ) = 144 cos ( 25 t + 180° ) mA

4

P10.5-5 The input to the circuit shown in Figure P10.5-5 is the current i s ( t ) = 48cos ( 25 t ) mA . Determine the current i2(t).

Figure P10.5-5 Solution: Represent the circuits in the frequency domain using phasors and impedances:

Using current division I 2 ( ω) = In the time domain

j15 15∠90° ( 48∠0° ) = ( 48∠0° ) = 28.8∠53.1° mA 20 + j15 25∠36.9° i 2 ( t ) = 28.8cos ( 25 t + 53.1° ) mA

5

P10.5-6 The input to the circuit shown in Figure P10.5-6 is the current i s ( t ) = 48cos ( 25 t ) mA . Determine the current i3(t). Answer: i 3 ( t ) = 16.85cos ( 25 t + 69.4° ) mA

Figure P10.5-6 Solution: Represent the circuits in the frequency domain using phasors and impedances:

Using current division 7.5∠90° j15 || j15 j 7.5 I3 = ( 48∠0° ) = ( 48∠0° ) = ( 48∠0° ) = 16.8539∠69.44° V j 7.5 + 20 21.36∠69.56° ( j15 || j15 ) + 20 In the time domain

i 3 ( t ) = 16.85cos ( 25 t + 69.4° ) mA

6

P10.5-7 Figure P10.5-7 shows a circuit represented in the frequency domain. Determine the voltage phasor V1.

Answer: V1 = 14.59∠ − 13.15° V

Figure P10.5-7

Solution: V1 =

j 80 || − j 50 − j 133.33 133.33∠ − 90° ( 20∠30° ) = ( 20∠30° ) = ( 20∠30° ) 125 + ( j 80 || − j 50 ) 125 − j 133.33 182.8∠ − 46.85° = 14.59∠ − 13.15° V

7

P10.5-8 Figure P10.5-8 shows a circuit represented in the frequency domain. Determine the current phasor I2.

Answer: I 2 = 18.48∠ − 93.7° mA

Solution: I2 =

Figure P10.5-8

− j 50 − j 50 ( 20∠30° ) = ( 20∠30° ) − j 50 + 45 + j 80 45 + j 30 =

50∠ − 90 ( 20∠30° ) = 18.49∠ − 93.7° mA 54.08∠ 33.7

8

P10.5-9 Here’s an ac circuit represented both in the time domain and frequency domain: R1 + –

15 cos 20t V

C

Z1

+ R2

v(t)

+ –

L

15 0° V

+ Z2



V(ω) –

(b)

(a)

Z 1 = 15.3∠ − 24.1° Ω and Z 2 = 14.4∠53.1° Ω

Suppose

Determine the node voltage v(t) and the values of R1, R2, L and C. Solution: Consider Z1: 1 1 R1 − j = 15.3∠ − 24.1° = 14 − j 6.25 ⇒ R1 = 14 Ω and C = = 0.008 F = 8 mF 20 C 20 ( 6.25 ) Next consider Z2: 1 = 14.4∠53.1° ⇒ 1 1 + R 2 j 20 L

1 1 1 1 + = = ∠53.1 = 0.05556 − j 0.04167 R 2 j 20 L 14.4∠53.1° 14.4

Equating coefficients gives R2 =

1 1 = 18 Ω and L = = 1.2 H 0.05556 20 ( 0.04167 )

Next, consider the voltage divider: A∠31.5° =

In the time domain,

(15)(14.4 ) ∠36.9° 14.4∠36.9° (15∠0° ) = 15.3∠ − 24.1° + 14.4∠36.9° (14 − j 6.25)(11.52 + j 8.64 ) =

216∠36.9° 25.52 + j 2.39

=

216∠36.9° = 8.43∠31.5° V 25.63∠5.4°

v ( t ) = 8.43cos ( 20 t + 31.5° ) V .

9

P 10.5-10 Find Z and Y for the circuit of

1 μF

160 μH

36 Ω

Figure P 10.5-10 operating at 10 kHz. Z,Y

Figure P 10.5-10 Solution: ZR

ω = 2π f = 2π (10 ×103 ) = 62832 rad sec 1 1 = R = 36 Ω ⇔ YR = = = 0.0278 S 36 ZR

Z L = jω L = j (62830)(160 ×10−6 ) = j10.053 ≈ j10 Ω ⇔ YL = ZC =

1 = − 0.1 j S ZL

−j −j 1 = = − j15.915 ≈ − j 16 Ω ⇔ YC = = j 0.0625 S −6 ZC ω C (62830)(1×10 ) Yeq = YR + YL + YC = 0.0278 − j0.0375 = 0.0467 ∠ − 53.4° S Z eq =

1 = 21.43∠ 53.4° = 12.75 + j17.22 Ω Yeq

10

300 Ω

P 10.5-11 For the circuit of Figure P 10.5-11, find the value of C required so that Z = 590.7Ω when f = 1 MHz.

C

Answer: C = 0.27 nF

47 μ H

Z

Figure P 10.5-11

Solution: Z L = j ω L = j (6.28 ×106 ) (47 × 10−6 ) = j 295 Ω ⎛ 1 ⎞ ⎜ ⎟( 300 + j 295 ) jω C ⎠ ⎝ Z eq = Z c || ( Z R +Z L ) = = 590.7 Ω 1 +300+ j 295 jω C 300+300 j 590.7 = ⇒ 590.7 − (590.7)(295 ω C ) + j (590.7)(300ω C ) = 300 + j 295 1+300 j ω C −300 ω C

(

Equating imaginary terms ω =2π f = 6.28×106 rad sec

)

(590.7) (300ω C ) = 295 ⇒ C = 0.27 nF

11

P 10.5-12 Determine the impedance Z for the circuit shown in Figure P 10.5-12. 2.5 H

2 mF

2 mF

100 Ω

2 mF

Z 1.5 H

Figure P 10.5-12 Solution: Replace series and parallel capacitors by an equivalent capacitor and series inductors by an equivalent inductor: Then 100 Z = jω 4 + 100 +

4

−j

1

(

jω 5 ×10−3

(

)

1

jω 5 ×10

−3

)

⎛ 200 ⎞ 200 100 ⎜ − j 1+ −j ⎟ ω ⎝ ⎠ = jω 4 + ω × = jω 4 + 2 ⎛ 200 ⎞ 1− j 1+ 100 + ⎜ − j ⎟ ω ω ⎠ ⎝

j j

2

ω 2

ω

2

ω = jω 4 + 100 4 − j 2 ω = 400 + j ⎛ 4 ω − 200 ω ⎞ Z = jω 4 + 100 ω ⎜ ⎟ 4 4 +ω2 4 +ω2 4 +ω2 ⎠ ⎝ 1+ 2 2

ω

12

P 10.5-13 The big toy from the hit movie Speaker

Big is a child’s musical fantasy come true—a sidewalk-sized piano. Like a hopscotch grid,

20 Ω

this once-hot Christmas toy invites anyone v = 12 cos ω t V

+ –

C

i 3 mH

who passes to jump on, move about, and make music. The developer of the “toy” piano Figure P 10.5-13

used a tone synthesizer and stereo speakers as

shown in Figure P 10.5-13 (Gardner, 1988). Determine the current i(t) for a tone at 796 Hz when C = 10 μF. Solution: j (2π ⋅ 796) (3 ⋅10 −3 ) = j15 Ω 12 = 0.48 ∠ − 37° A 20 + j15 i (t ) = 0.48 cos (2π ⋅ 796 t − 37°) A I=

13

P 10.5-14 Determine i(t), v(t), and L for the circuit shown in Figure P 10.5-14. Answer: i(t) = 1.34 cos (2t – 87°) A, v(t) = 7.29 cos (2t – 24°) V, and L = 4 H i(t)

2 cos(3t – 15°) A



L

Figure P 10.5-14

Solution:

Z1 = R = 8 Ω, Z 2 = j 3 L, I = B ∠ − 51.87° and I s = 2 ∠ − 15° A

8 I B ∠−51.87° Z1 = = = = 2 ∠−15° Is Z1 + Z 2 8+ j 3L

8 ∠0° ⎛ 3L ⎞ 82 + (3L) 2 ∠ tan −1 ⎜ ⎟ ⎝ 8 ⎠

Equate the magnitudes and the angles. ⎛ 3L ⎞ angles: + 36.87 = + tan −1 ⎜ ⎟ ⇒ L = 2 H ⎝ 8 ⎠ B 8 magnitudes: = ⇒ B =1.6 2 2 64+ 9 L

14

P 10.5-15 Spinal cord injuries result in paralysis of the lower body and can cause loss of bladder control. Numerous electrical devices have been proposed to replace the normal nerve pathway stimulus for bladder control. Figure P 10.5-15 shows the model of a bladder control system where vs = 20 cos ωt V and ω = 100 rad/s. Find the steady-state voltage across the 10-Ω load resistor. Answer: v(t) = 10 2 cos (100t + 45°) V 1 mF

vs + –

50 Ω

100 μF 10 Ω

Regular nerve pathway load

Figure P 10.5-15 Solution: ⎛ 10 ⎞ V10 = Vs ⎜ ⎟ ⎝ 10− j10 ⎠ 10 ⎛ ⎞ = 20∠0° ⎜ ⎟ ⎝ 10 2∠− 45° ⎠ = 10 2∠45° v10 (t ) = 10 2 cos (100 t + 45°) V

15

P 10.5-16 There are 500 to 1000 deaths each year in

Person's body

the United States from electric shock. If a person makes a

i

good contact with his hands, the circuit can be represented by Figure P 10.5-16, where vs = 160 cos ωt V and ω = 2πf. Find the steady-state current i flowing

Source vs +–

300 Ω 2 μF 100 mH

through the body when (a) f = 60 Hz and (b) f = 400 Hz. Answer: (a)

(b)

i(t) = 0.53 cos (120πt + 5.9°) i(t) = 0.625 cos (800πt + 59.9°) A

Figure P 10.5-16

Solution: (a)

(b)

160 ∠0° 160 ∠0° = (− j1326) (300+ j 37.7) 303 ∠−5.9° − j1326 + 300+ j 37.7 = 0.53 ∠5.9° A i(t ) = 0.53cos (120π t +5.9°) A

I=

I=

160∠0° 160∠0° = (− j199)(300+ j 251) 256∠−59.9° − j199 +300+ j 251 = 0.625∠59.9° A

i(t ) =0.625 cos (800π t +59.9°) A

16

P 10.5-17 Determine the steady-state voltage, v(t), and current, i(t), for each of the circuits shown in Figure P 10.5-17. – v(t)

+

4Ω +

24 V –

40 Ω

10 Ω

i(t)

(a) – v(t)

+

4H 24 cos (4t + 15°) V

+ –

40 Ω

10 mF

i(t)

(b)

Figure P 10.5-17 Solution: (a) v (t ) = − i (t ) =

4 × 24 = −8 V 4 + ( 40 & 10 )

40 24 8 × = = 1.6 A 40 + 10 4 + ( 40 & 10 ) 5

(b) Represent the circuit in the frequency domain using impedances and phasors.

17

V=−

(16∠ − 90° )( 24∠15° ) = 33.66∠ − 65° V j16 × 24∠15° = 40 ( − j 25 ) j16 + ( 40 & − j 25 ) j16 + 40 − j 25 I=

so and

40 × 40 − j 25

24∠15° = 1.78∠57° A 40 ( − j 25 ) j16 + 40 − j 25

v ( t ) = 33.66 cos ( 4t − 65° ) V i ( t ) = 1.78cos ( 4t + 57° ) A

(checked: LNAP 8/1/04)

18

P 10.5-18

Determine the steady-state current, i(t), for the circuit shown in Figure P 10.5-18. 5H

5 mF

20 Ω

2 mF

30 Ω

4H

i(t) +–

5 cos (10t + 30°) V

Figure P 10.5-18 Solution: I=

so

5∠30° 5∠30° 5∠30° + + = 0.100∠ − 23.1° + 0.0923∠98.2° + 0.1667∠ − 60° 30 + j 40 20 − j 50 j 50 − j 20 = 0.186∠ − 29.5° A i ( t ) = 0.186 cos (10t − 29.5° ) A

(checked: LNAP 8/1/04)

19

P10.5-19 Determine the steady state voltage, v(t), for this circuit:

20 Ω

4H

+ v(t)

10 cos(5t + 45°) mA

– 30 Ω

5 mF

2H 4 mF

Solution:

V = 0.01∠45° ⎡⎣( 20 & j 20 ) + ( 30 & ( − j 40 ) ) + ( j10 & ( − j 50 ) ) ⎤⎦ ⎡ 20 ( j 20 ) 30 ( − j 40 ) j10 ( − j 50 ) ⎤ = 0.01∠45° ⎢ + + ⎥ j10 − j 50 ⎦ ⎣ 20 + j 20 30 − j 40 = 0.01∠45° [14.14∠45° + 24∠ − 36.9° + 12.5∠90°] = 0.01∠45° [10 + j10 + 19.2 − j14.4 + j12.5] = 0.303∠60.5° V

so

v ( t ) = 0.303cos ( 5t + 60.5° ) V

(checked: LNAP 8/1/04)

20

P10.5-20 Determine the steady state voltage, v(t), for this circuit: 2.5H 20 Ω 10 mF + –

Solution: Let

and

10 cos(4t + 60°) V

5H

5 mF + v(t) 40 Ω –

⎛ ( 20 − j 25) j10 = 250 − j 200 = 12.81∠75.5° Ω 1 ⎞ Z 1 = ⎜⎜ 20 − j ⎟⎟ & j10 = 4 ( 0.01) ⎠ 20 − j 25 + j10 20 − j15 ⎝ ⎛ ⎞ j 20 ( 40 − j 50 ) 1000 + j800 1 Z 2 = j 20 & ⎜⎜ − j + 40 ⎟⎟ = = = 25.61∠75.5° Ω 40 − j 30 ⎝ 4 ( 0.005 ) ⎠ j 20 + 40 − j 50

Then V=

so

Z2 Z1 + Z 2

× 10∠60° =

25.61∠75.5° × 10∠60° = 6.67∠60° V 12.81∠75.5° + 25.6∠75.5°

v ( t ) = 6.67 cos ( 4t + 60° ) V

(checked: LNAP 8/1/04)

21

P 10.5-21 The input to the circuit shown in Figure P 10.5-21 is the current source current is(t) = 25 cos (10t + 15°) mA The output is the current i1(t). Determine the steady-state response, i1(t). i1(t)

5H

is(t)

5 mF

40 Ω

2H

2 mF

25 Ω

Figure P 10.5-21 Solution: Represent the circuit in the frequency domain using impedances and phasors. Let 40 ( − j 50 ) ⎛ ⎞ 1 = j 50 + = 39.0∠51.3° Ω Z 1 = j 50 + ⎜ 40 & −3 ⎟ 40 − j 50 j10 × 2 × 10 ⎠ ⎝ and Z2 = − j

j 20 ( 25 ) 1 + j 20 & 25 = − j 20 + = 12.5∠ − 38.7 Ω −3 25 + j 20 10 ( 5 ×10 )

Z1 and Z2 are connected in parallel. Current division gives I1 =

so

Z1 Z1 + Z 2

× 0.025∠15° = 0.024∠32.7° A

i1 ( t ) = 0.024 cos (10t + 32.7° ) A

(checked: LNAP 8/1/04)

22

P10.5-22 Determine the steady state voltage, v(t), and current i(t) for each of these circuits: – v(t)

+

4Ω +

24 V –

40 Ω

10 Ω

i(t)

(a) – v(t)

+

4H 24 cos (4t + 15°) V

+ –

40 Ω

10 mF

i(t)

(b)

Solution: i (t ) =

(a)

v (t ) =

80 + 80 0.024 = 19.2 mA 40 + ( 80 + 80 )

80 1 × ( 40 & ( 80 + 80 ) ) 0.024 = ( 32 )( 0.024 ) = 0.384 V 80 + 80 2

(b) Represent the circuit in the frequency domain using impedances and phasors.

I=

80 + j80 × 0.024∠15° = 0.028∠25.5° A − j 25 + ( 80 + j80 )

V=

80 × ⎡ − j 25 & ( 80 + j80 ) ⎤⎦ × 0.024∠15° = 0.494∠ − 109.5° V 80 + j80 ⎣

So

i ( t ) = 28cos (10t + 25.5° ) mA

and

v ( t ) = 0.494 cos (10t − 109.5° ) V

(checked: LNAP 8/1/04)

23

P10.5-23 Determine the steady state current i(t) for this circuit: 25 Ω

2H

2 mF 20 Ω

+ –

40 Ω

i(t)

16 cos(20t + 75°) V 5 mF

2H

Solution: Represent the circuit in the frequency domain using impedances and phasors. Let Z 1 = 25 + j ( 20 ) 2 + Z 2 = 20 +

1 = 25 + j15 = 29.2∠31° Ω j ( 20 )( 0.002 )

1 = 20 − j10 = 22.36∠ − 26.6° Ω j ( 20 )( 0.005 )

Z 3 = 40 + j ( 20 ) 2 = 40 + j 40 = 56.57∠45° Ω

and let Z p = Z 2 & Z 3 = 18.86∠ − 8° = 18.67 − j 2.67 Ω Then I=

so

Z2 16∠75° × = 0.118∠6.1° A Z1 + Z p Z 2 + Z 3 i ( t ) = 0.118cos ( 20t + 6.1° ) A

(checked: LNAP 8/2/04)

24

P 10.5-24 When the switch in the circuit shown in Figure P 10.5-24 is open and the circuit is at steady

R2

R1

state, the capacitor voltage is v(t) = 14.14 cos (100t –45°) V

When the switch is closed and the circuit is at steady

+ –

20 cos (100t) V

state, the capacitor voltage is

0.5 μF

+ v(t) –

Figure P 10.5-24

v(t) = 17.89 cos (100t –26.6°) V

Determine the values of the resistances R1 and R2. Solution: Represent the circuit in the frequency domain using phasors and impedances. The impedance 1 capacitor is = − j 20, 000 . When the switch is closed j (100 ) ( 0.5 ×10−6 ) 17.89∠ − 26.6° = V =

− j 20, 000 × 20∠0° R 2 − j 20, 000

Equating angels gives ⎛ −20, 000 ⎞ −26.6° = −90° − tan −1 ⎜ ⎜ R 2 ⎟⎟ ⎝ ⎠ When the switch is open 14.14∠ − 45° = V =



R2 =

−20, 000 = 10015 Ω tan ( −63.4 )

− j 20, 000 × 20∠0° R1 + R 2 − j 20, 000

Equating angles gives ⎛ −20, 000 ⎞ −45° = −90° − tan −1 ⎜ ⎜ R1 + R 2 ⎟⎟ ⎝ ⎠



R1 + R 2 =

−20, 000 = 20, 000 tan ( −45° )

So R1 = 20, 000 − 10015 = 9985 Ω (checked: LNAP 8/2/04)

25

P10.5-25 Determine the steady state current i(t) for this circuit: 20 cos (5t + 30°) mA

4H

20 Ω

10 mF

15 Ω 2H

8H 40 Ω

8 mF

i(t)

Solution: Represent the circuit in the frequency domain using phasors and impedances. Let Z 1 = ( j 20 & 20 ) +

1 = 10 − j10 = 14.14∠ − 45° Ω j 0.05

⎛ 1 ⎞ Z 2 = j 40 + 40 + ⎜ j10 & ⎟ + 15 = 55 + j 56.67 = 79∠46.3° Ω j 0.04 ⎠ ⎝ Z1 I=− × 20∠30° = 3.535∠129.3° mA Z1 + Z 2

so

i ( t ) = 3.535cos ( 5t + 129.3° ) mA

(checked: LNAP 8/2/04)

26

P10.5-26 Determine the steady state voltage, v(t), and current i(t) for each of these circuits: i(t)

+ –

50 Ω 4i(t)

20 V

40 Ω

+ v(t) –

2H

10 Ω

(a) i(t)

+ –

5 mF

40 Ω

3H

20 cos (10t + 15°) V

4i(t)

+ v(t) –

(b)

Solution: (a) Using KCL and then KVL gives 20 = 80 mA 250 v ( t ) = 40 ( 5i ( t ) ) = 200 ( 0.08 ) = 16 V

20 = 50 i ( t ) + 40 ( 5 i ( t ) ) ⇒ i ( t ) = Then

(b) Represent the circuit in the frequency domain using phasors and impedances.

Where And

Z 1 = 40 + j (10 ) 3 +

1 = 40 + j10 = 41.23∠26.6° Ω j (10 )( 0.005 )

Z 2 = j (10 ) 2 & 10 = 8 + j 4 = 8.944∠26.6° Ω

Using KCL and then KVL gives 20∠15° = Z 1I + 5Z 2 I Then



I = 0.234∠ − 5.6° A

V = Z 2 ( 5I ) = 10.47∠21° A

so

i ( t ) = 0.234 cos (10t − 5.6° ) A

and

v ( t ) = 10.47 cos (10t + 21° ) V

(checked: 8/3/04)

27

P10.5-27 Determine the steady state voltage, v(t), for each of these circuits:

20 Ω

80 Ω

24 V

+

+ –

v(t)

40 Ω



100 Ω

(a) 3H

5 mF 25 Ω

20 Ω

+ + –

v(t)



24 cos (20t + 45°) V 4 mF

4H

20 Ω

2 mF

15 Ω

(b)

Solution: (a) Using voltage division twice v (t ) =

40 100 × 24 − × 24 = −12 V 40 + 80 20 + 100

28

(b) Represent the circuit in the frequency domain using phasors and impedances.

Where

Z 1 = 20 Ω ⎛ ⎞ 1 & 20 ⎟⎟ = 12.2 + j 70.2 = 71.30∠80.2° Ω Z 2 = j ( 20 ) 4 + ⎜⎜ ⎝ j ( 20 )( 0.002 ) ⎠ 1 + 25 = 25 + j 50 = 55.90∠63.4° Ω Z 3 = j ( 20 ) 3 + j ( 20 )( 0.005 )

Z4 =

1 + 15 = 15 − j12.5 = 19.53∠ − 39.8° Ω j ( 20 )( 0.004 )

Using voltage division twice V=

so

Z2 Z1 + Z 2

× 24∠45° −

Z4 Z3 + Z4

× 24∠45° = 24.8∠80° V

v ( t ) = 24.8cos ( 20t + 80° ) V

(Checked using LNAP 10/5/04)

29

P 10.5-28 The input to the circuit shown in Figure P 10.5-28 is the voltage of the voltage source vs(t) = 5 cos (2t + 45°) V

0.1 F

+ –



vs(t)

The output is the inductor voltage, v(t). Determine the steady-state output voltage.

3H

+ v(t) –

Figure P 10.5-28 Solution: Represent the circuit in the frequency domain using phasors and impedances.

4 & j6 =

4 ( j 6 ) 24∠90° = = 3.33∠34° = 2.76 + j1.86 Ω 4 + j 6 7.2∠56°

Using voltage division

V=

3.33∠34° 3.33∠34° 3.33∠34° × 5∠45° = × 5∠45° = × 5∠45° = 3.98∠127° V − j 5 + 2.76 + j1.86 2.76 − j 3.14 4.18∠ − 48°

The corresponding voltage in the time domain is v ( t ) = 3.98cos ( 2t + 127° ) V

30

P10.5-29 Determine the steady state voltage, v(t), for this circuit: + 8Ω

2H 5 sin 5t V v(t)

+–

1 12

F

4H _

Solution: V1 (ω ) = V 2 (ω ) =

j10 5 e − j 90 = 3.9 e − j 51 V 8 + j10 j 20 5 e − j 90 = 5.68 e − j 90 V j 20 − j 2.4

V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90 = 3.58 e j 47 V

31

P10.5-30 Determine the steady state voltage, v(t), for this circuit: + 8Ω

2H v(t)

1 12

4 cos (3t + 15°) A

F 4H –

Solution: V1 (ω ) = V 2 (ω ) =

8 ( j 6 ) j15 4 e = 19.2 e j 68 V 8 + j6 j12 ( − j 4 ) j15 4 e = 24 e − j 75 V j12 − j 4

V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22 V

32

P10.5-31 The input to the circuit in Figure P10.5-31 is the voltage source voltage, v s ( t ) . The output is the voltage v o ( t ) . When the input is v s ( t ) = 8cos ( 40 t ) V , the output is v o ( t ) = 2.5cos ( 40 t + 14° ) V . Determine the values of the resistances R1 and R 2 .

Figure P10.5-31 Solution: Using voltage division in the frequency domain: jω L R 2

L R 2 + jω L R1 Vo (ω ) = = jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L jω

R1 R 2

where R p =

R1 + R 2 Representing the given input and output in the frequency domain: 2.5∠14° Vo (ω ) = = 8∠0° Vi (ω )

In this case the angle of

and the magnitude of

ω

L R1



j ⎜ 90 − tan −1 ω

⎛ L ⎞ 1+ ⎜ω ⎜ R p ⎟⎟ ⎝ ⎠

2

e

⎜ ⎝

L ⎞⎟ R p ⎟⎠

Vo (ω ) L L ( R1 + R 2 ) tan ( 90° − 14° ) = = = 0.1 is specified to be 14° so 40 Rp R1 R 2 Vi (ω )

Vo (ω ) 2.5 is specified to be so 8 Vi (ω )

40

L R1

=

2.5 ⇒ 8

1 + 16 values that satisfies these two equations is L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω .

L = 0.0322 . One set of R1

33

Section 10.6 Mesh and Node Equations    P10.6‐1 The input to the circuit shown in Figure P10.6‐1 is the voltage   v s ( t ) = 48cos ( 2500 t + 45° ) V  

 Write and solve node equations to determine the steady state output voltage  v o ( t ) .  

  Figure P10.6‐1  Solution: Represent the circuit in the frequency domain as 

  The node voltages are  48∠45° = V1 , V2, V3 and Vo. Express the dependent source voltage in  terms of the node voltages:  ⎛ V2 ⎞ V3 − V 2 = 25 I = 25 ⎜ ⎟ ⇒ V3 = 2.25 V 2   ⎝ 20 ⎠ Apply KCL to the supernode corresponding to the CCVS to get    V3 48∠45° − V 2 V 2 V3 − Vo = + + j 50 20 30 − j 40

 

V3 48∠45° V 2 V 2 V3 − Vo = + + + j 50 j 50 20 30 − j 40 ⎛ 1 48∠45° ⎛ 1 1 ⎞ 1 ⎞ 1 =⎜ + ⎟ V2 + ⎜ + ⎟ V3 − V o j 50 30 ⎝ j 50 20 ⎠ ⎝ 30 − j 40 ⎠

 

 

⎛ 1 48∠45° ⎛ 1 1 ⎞ 1 ⎞ 1 =⎜ + ⎟ V2 + ⎜ + ⎟ 2.25V 2 − Vo j 50 30 ⎝ j 50 20 ⎠ ⎝ 30 − j 40 ⎠   48∠45° ⎛ 1 1 2.25 2.25 ⎞ 1 =⎜ + + + ⎟ V 2 − Vo j 50 30 ⎝ j 50 20 30 − j 40 ⎠

Apply KCL at the right node of the 30 Ω resistor to get  V3 − V o 30

=

⎛ 1 1 ⎞ ⎛ 1 ⎞ ⇒ 0 = ⎜ − ⎟ 2.25 V 2 + ⎜ + ⎟ Vo   j 25 ⎝ 30 ⎠ ⎝ 30 j 25 ⎠ Vo

1 2.25 2.25 ⎡ 1 ⎢ j 50 + 20 + 30 + − j 40 ⎢ In matrix form  2.25 ⎢ − ⎢ 30 ⎣ Solving, perhaps using MATLAB,     

1 ⎤ ⎡ 48∠45° ⎤ 30 ⎥ ⎡ V 2 ⎤ ⎢ ⎥⎢ ⎥ = j 50 ⎥   ⎢ ⎥ 1 1 ⎥ ⎣ Vo ⎦ + 0 ⎢ ⎥⎦ ⎣ 30 j 25 ⎥⎦ −

⎡ V 2 ⎤ ⎡10.18∠ − 44.6°⎤ ⎢ ⎥=⎢ ⎥ V  ⎣⎢ Vo ⎦⎥ ⎣ 14.67∠5.6° ⎦

P10.6‐2 Figure P10.6‐2 shows an ac circuit represented in the frequency domain. Determine the  values of the phasor node voltages, Vb and Vc. 

  Figure P10.6‐2  Solution: 

12∠45° − V b

Writing Node equations: 

=

Vb

+

Vb − Vc

20 − j 25 15 − j 30 j 30   V b − Vc 12∠45° − Vc Vc + = 15 − j 30 40 + j 20 j 40

Rearranging: 

 

⎞ ⎛ ⎞ 12∠45° ⎛ 1 1 1 1 =⎜ + + ⎟ Vb − ⎜ ⎟ Vc j 30 ⎝ j 30 20 − j 25 15 − j 30 ⎠ ⎝ 15 − j 30 ⎠ ⎛ ⎞ ⎛ 12∠45° 1 1 1 1 ⎞ = −⎜ + + ⎟ Vb + ⎜ ⎟ Vc 40 + j 20 ⎝ 15 − j 30 ⎠ ⎝ 15 − j 30 40 + j 20 j 40 ⎠

 

In matrix from:  1 1 1 ⎡ 1 ⎤ ⎡ 12∠45° ⎤ + + − ⎢ j 30 20 − j 25 15 − j 30 ⎥ ⎡ V b ⎤ ⎢ j 30 ⎥ 15 − j 30 ⎢ ⎥⎢ ⎥ = ⎢ ⎥    1 1 1 1 ⎥ ⎣⎢ Vc ⎦⎥ ⎢ 12∠45° ⎥ ⎢ − + + ⎢ ⎢ 40 + j 20 ⎥ 15 − j 30 15 − j 30 40 + j 20 j 40 ⎥⎦ ⎣ ⎣ ⎦ Solving using MATLAB:   

V b = 7.69∠ − 19.8° and Vc = 10.18∠7.7° V   Checked using LNAPAC 

P10.6‐3 Figure P10.6‐3 shows an ac circuit represented in the frequency domain. Determine the  value of the phasor node voltage V.  Answer:  V = 71.0346∠ − 39.627° V  

  Figure P10.6‐3  Solution:  Writing a node equation: 

Rearranging  Solving  Finally     

20∠30° − V −20∠30° − V + = 0.25∠15°   j 40 25 − j 50

⎛ 1 ⎞ 1 20∠30° −20∠30° + + − 0.25∠15°   ⎜ ⎟V = 25 − j 50 j 40 ⎝ j 40 25 − j 50 ⎠ ( 0.012042∠ − 48.366° ) V = 0.85537∠87.993°  

V=

0.85537∠87.993° = 71.0346∠ − 39.627° V   0.012042∠ − 48.366°

P10.6‐4 Figure P10.6‐4 shows an ac circuit represented in the frequency domain. Determine the  values of the phasor mesh currents. 

  Figure P10.6‐4  Solution: 

( 40 + j 15) I 1 + ( 25 − j 50 ) ( I 1 − I 3 ) − 48∠75° = 0   ( 65 − j 35 ) I 1 − ( 25 − j 50 ) I 3 = 48∠75°

Mesh 1: 

48∠75° + ( − j 50 ) ( I 2 − I 3 ) + ( 32 + j 16 ) I 2 = 0

Mesh 2: 

( 32 − j 34 ) I 2 + j 50I 3 = −48∠75°

j 40I 3 − ( − j 50 ) ( I 2 − I 3 ) − ( 25 − j 50 ) ( I 1 − I 3 ) = 0

Mesh 3: 

In matrix form: 

 

( −25 + j 50 ) I 1 + j 50 I 2 + ( 25 − j 160 ) I 3 = 0

 

0 −25 + j 50 ⎤ ⎡ I 1 ⎤ ⎡ 48∠75° ⎤ ⎡ 65 − j 35 ⎢ ⎥ ⎢ + j 50 ⎥⎥ ⎢I 2 ⎥ = ⎢⎢ −48∠75° ⎥⎥   0 32 − j 34 ⎢ ⎢⎣ −25 + j 50 ⎥⎦ + j 50 25 − j 60 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ 0

Solving using MATLAB: 

 

I 1 = 0.794∠111°, I 2 = 0.790∠ − 61.7° and I 3 = 0.229∠176° A    

 

P 10.6-5 A commercial airliner has sensing devices to indicate to the cockpit crew that each

R

door and baggage hatch is closed. A device called

vs

a search coil magnetometer, also known as a

b

+ –

proximity sensor, provides a signal indicative of

R a

Door Ls

LR

the proximity of metal or other conducting material to an inductive sense coil. The inductance of the sense coil changes as the metal gets closer to

Figure P 10.6-5

the sense coil. The sense coil inductance is compared to a reference coil inductance with a circuit called a balanced inductance bridge (see Figure P 10.6-5). In the inductance bridge, a signal indicative of proximity is observed between terminals a and b by subtracting the voltage at b, vb, from the voltage at a, va (Lenz, 1990). The bridge circuit is excited by a sinusoidal voltage source vs = sin (800 πt) V. The two resistors, R = 100 ω, are of equal resistance. When the door is open (no metal is present), the sense coil inductance, LS, is equal to the reference coil inductance, LR = 40 mH. In this case, what is the magnitude of the signal Va – Vb? When the airliner door is completely closed, LS = 60 mH. With the door closed, what is the phasor representation of the signal Va – Vb? Solution:   

vs = sin (2π ⋅ 400 t ) V

R = 100 Ω LR = 40 mH ⎧ 40 mH LS = ⎨ ⎩ 60 mH

  door opened door closed

With the door open VA − VB = 0 since the bridge circuit is balanced. With the door closed Z LR = j (800π )(0.04) = j100.5 Ω and Z LS = j (800π )(0.06) = j150.8 Ω. The node equations are: 

 

KCL at node B: KCL at node A :

VB − VC VB j100.5 + = 0 ⇒ VB = VC Z LR R j100.5+100 VA − VC VA + =0 R Z LS

Since VC = Vs =1 V            VB =0.709∠44.86° V and VA = 0.833∠33.55 V  

Therefore 

VA − VB = 0.833∠33.55° − 0.709∠44.86° = (0.694 + j.460) − (0.503 + j 0.500) = 0.191 − j 0.040 = 0.195∠ − 11.83° V  

P 10.6-6 Using a tiny diamond-studded burr operating at 190,000 rpm, cardiologists can remove life-threatening plaque deposits is coronary arteries. The procedure is fast, uncomplicated, and relatively painless (McCarty, 1991). The Rotablator, an angioplasty system, consists of an advancer/catheter, a guide wire, a console, and a power source. The advancer/catheter contains a tiny turbine that drives the flexible shaft that rotates the catheter burr. The model of the operational and control circuit is shown in Figure P 10.6-6. Determine v(t), the voltage that drives the tip, when vs =

2 cos (40t – 135°) V.

Answer: v(t) = 1.414 cos (40t + 135°) V 1 20

vs

+ –

1 80

H

F

1 80

F +

2 Ω

1 80

2i

i

F



v(t)

Figure P 10.6-6   Solution: Represent the circuit in the frequency domain 

V1 − (−1+ j ) V1 V1 − V2 + + =0  −j2 j2 2 V2 − V1 V2 + − IC = 0 − j2 − j2 Also, expressing the controlling signal of the dependent source in terms of the node voltages  ⎡ −1+ j ⎤ −1 + j yields                                 I x = ⇒ IC = 2 Ix = 2 ⎢ ⎥ = −1 − j A   -2 j ⎣ -2 j ⎦ Solving these equations yields  The node equations are:              

V2 =  

−3− j = 2 ∠ − 135° V ⇒ v(t ) = v2 (t ) = 2 cos (40 t − 135°) V 1+ j 2 (checked: LNAP 7/19/04

Figure P 10.6-7, it is known that

1H

v2(t) = 0.7571 cos (2t + 66.7°) V v3(t) = 0.6064 cos (2t–69.8°) V

i1



P 10.6-7 For the circuit of

A cos 2t V

+ –

1F

5H + v2 –

3i1

1 4

F

+ v3 –

Determine i1(t). Figure P 10.6-7 Solution:        V2 = 0.7571∠66.7° V

V3 = 0.6064∠ − 69.8° V

 

 

⎫ I1 = I 2 + I 3 ⎪ ⎪ ⎧ I 3 =0.3032 ∠20.2° A V3 − V2 ⎪ ⎪ I2 = ⎬ yields ⎨ I 2 =0.1267∠−184° A j 10 ⎪ ⎪ I =0.195∠36° A ⎩ 1 ⎪ V3 I3 = ⎪ −j2 ⎭ therefore  i1 (t ) =0.195cos (2 t +36°) A

 

(checked: MATLAB 7/18/04)    

P10.6‐8 The input to the circuit shown in Figure P10.6‐8 is the voltage  vs = 25cos(40t + 45°) V  Determine the mesh currents  i1 and i2 and the voltage vo.   

  Figure P10.6‐8  Solution  Represent the circuit in the frequency domain: 

Apply KVL to mesh 1: 

j 320 I 1 + 400 ( I 1 − I 2 ) − 25∠45° = 0  

Apply KVL to mesh 2: 

50 I 2 + j 240 I 2 + j 120 I 2 − 400 ( I 1 − I 2 ) = 0  

In matrix form: 

−400 ⎤ ⎡ I 1 ⎤ ⎡ 25∠45° ⎤ ⎡ 400 + j 320   ⎢ ⎥= ⎢ −400 450 + j 360 ⎥⎦ ⎣ I 2 ⎦ ⎢⎣ 0 ⎥⎦ ⎣

Solving using MATLAB:  Using Ohm’s Law 

 

⎡ I 1 ⎤ ⎡ 47.5∠ − 24.6°⎤ ⎢I ⎥ = ⎢ ⎥ mA   ⎣ 2 ⎦ ⎣ 33.0∠ − 63.3° ⎦

Vo = 400 ( I 1 − I 2 ) = 12∠18.8° V  

In the time domain    i1 ( t ) = 47.5cos ( 40 t − 24.6° ) mA ,  i 2 ( t ) = 33cos ( 40 t − 63.3° ) mA   and 

v o ( t ) = 12 cos ( 40 t + 18.8° ) V  

   

P10.6‐9 The input to the circuit shown in Figure P10.6‐9 is the voltage  vs(t) = 42cos(800t + 60°) mV  Determine the output voltage vo(t).  Answer: vo(t) = 823.5 cos(800t − 55.6°) mV     

  Figure P10.6‐9  Solution  Represent the circuit in the frequency domain: 

  Apply KCL at the top node of the inductor, node a: 

0.042∠60° − Va

=

Va

+

Va

 

⇒ Va =

4 ( 0.042∠60° )   5− j

500 j 2000 2000 Apply KCL at the inverting input node of the op amp:  Vb Va + = 0 ⇒ V b = −5 V a   2000 10, 000 Apply KCL at the top node of the capacitor, node b:  Vb Vb V b − Vo + + = 0 ⇒ Vo = ( 3 + j 4 ) V b   10, 000 − j 5000 20, 000

Combining these results we get: ( 5∠53.1° )( 20∠ − 180° ) 0.042∠60° = 0.8235∠ − 55.6°   4   V o = ( 3 + j 4 )( −5 ) ( 0.042∠60° ) = ( ) 5− j 5.1∠ − 11.3° In the time domain      v o ( t ) = 832.5cos ( 800 t − 55.6° ) mV  

 

  P 10.6-10

The idea of using an induction coil in a C

lamp isn’t new, but applying it in a commercially available product is. An induction coil in a bulb





induces a high-frequency energy flow in mercury

L

vapor to produce light. The lamp uses about the same amount of energy as a fluorescent bulb but lasts six times longer, with 60 times the life of a conventional incandescent bulb. The circuit model of the bulb and

+

1Ω 2Ω

vs +–

v –

Induction bulb

Figure P 10.6-10

its associated circuit are shown in Figure P 10.6-10. Determine the voltage v(t) across the 2-Ω resistor when C = 40 μF, L = 40 μH, vs = 10 cos (ω0t + 30°), and ω0 = 105 rad/s. Answer: v(t) = 6.45 cos (105t + 44°) V   Solution: Represent the circuit in the frequency domain:  The mesh equations are:    −1 − j 4 ⎤ ⎡ I1 ⎤ ⎡10∠30°⎤ ⎡ (2 + j 4) ⎢ −1 −1 ⎥⎥ ⎢⎢I 2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ (2+1/ j 4) ⎢ ⎢⎣ − j 4 −1 (3+ j 4) ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ 0 ⎥⎦   Using Cramer’s rule yields    2 + j8 I3 = (10∠30° ) = 3.225∠44° A   12+ j 22.5 Then          V = 2 I 3 = 2 ( 3.225∠44° ) = 6.45∠44° V ⇒ v(t ) = 6.45cos (105 t + 44° ) V  

(checked: LNAP 7/19/04)

P 10.6-11 The development of coastal hotels in various parts of the world is a rapidly growing enterprise. The need for environmentally acceptable shark protection is manifest where these developments take place alongside shark-infested waters (Smith, 1991). One concept is to use an electrified line submerged in the water in order to deter the sharks, as shown in Figure P 10.611a. The circuit model of the electric fence is shown in Figure P 10.6-11b, where the shark is represented by an equivalent resistance of 100Ω. Determine the current flowing through the shark’s body, i(t), when vs = 375 cos 400t V.

Electric fence

(a) 100 μ F

25 μF i

vs

+ –

100 Ω

250 mH

Source

Electric fence

Shark

(b) Figure P 10.6-11  

Solution: Represent the circuit in the frequency domain:

Mesh Equations:                                 

j 75 I1 − j 100 I 2 = 375 − j 100 I1 + (100+ j 100) I 2 = 0

Solving for I 2 yields I 2 = 4.5 + j 1.5 = 3 ∠53.1° A



 

i 2 ( t ) 3cos ( 400t + 53.1° ) A

(checked: LNAP 7/19/04)  

P 10.6‐12 Determine the node voltage at nodes a and b  in each of these circuits: 

 

 

Solution (a)    The node equations are                 

24 − v a 40 24 − v b 25

1 1 ⎡1 + ⎢ 40 20 + 15 or                                       ⎢ 1 ⎢ − ⎢⎣ 20

= +

va − vb 20 va − vb 20

+ =

va 15   vb 50

1 ⎤ ⎡ v ⎤ ⎡ 24 ⎤ ⎥ ⎢ a ⎥ ⎢ 40 ⎥ 20 =⎢ ⎥  ⎥ 1 1 1 ⎥ ⎢ ⎥ ⎢ 24 ⎥ ⎢v ⎥ + + 25 20 50 ⎥⎦ ⎣ b ⎦ ⎢⎣ 25 ⎥⎦ −

Solving using MATLAB gives       v a = 8.713 V and v b = 12.69 V    

(b)  Use phasors and impedances to represent the circuit in the frequency domain as 

where 

Z 1 = 25 + j ( 20 ) 4 = 25 + j80 = 83.82∠72.7° Ω ⎛ ⎞ 1 Z 2 = ⎜⎜ 40 & ⎟ + j ( 20 ) 5 = 3.56 + j88.6 = 88.68∠87.7° Ω j ( 20 )( 0.004 ) ⎟⎠ ⎝ Z 3 = 20 Ω Z 4 = 15 + j ( 20 ) 2 = 15 + j 40 = 42.72∠69.4° Z 5 = j ( 20 ) 3 +

1 = j 50 = 50∠90° Ω j ( 20 )( 0.005 )

24∠45° − Va The node equations are            

Z2 24∠45° − V b Z1

+

Va Z4

+

Va − V b

Va − Vb Z3

Z3 =

Vb

 

Z5

⎤ ⎡ 24∠45° ⎤ ⎥ ⎡ Va ⎤ ⎢ Z ⎥ 2 ⎥⎢ ⎥ = ⎢ ⎥ 1 1 1 ⎥ ⎢ ⎥ ⎢ 24∠45° ⎥ + + ⎥ ⎢V ⎥ ⎢ ⎥ Z 1 Z 3 Z 5 ⎥⎦ ⎣ b ⎦ ⎢⎣ Z 1 ⎥⎦

1 1 ⎡ 1 ⎢Z + Z + Z 3 4 ⎢ 2 ⎢ 1 − ⎢ Z3 ⎢⎣



Solving using MATLAB gives                      

So                                                             

=

1 Z3

Va = 7.89∠44.0° V b = 8.45∠45.1°

 

v a ( t ) = 7.89 cos ( 20t + 44° ) V v a ( t ) = 8.45cos ( 20t + 45.1) V

 

(checked: LNAP 8/3/04)    

P 10.6‐13 Determine the voltage v(t): 

 

Solution: Represent the circuit in the frequency domain using impedances and phasors 

The mesh currents are I and  0.05∠ − 30° A .  Apply KVL to the top mesh to get  15∠45° + ( − j 25 ) I + (15 + j 32 )( I − 0.05∠ − 30° ) + 25I = 0

So            I =

−15∠45° + (15 + j 32 )( 0.05∠ − 30° ) = 0.3266∠ − 143.6° = −0.2629 − j 0.1939 A   25 − j 25 + 15 + j 32

Then                               V = ( − j 25 ) I = 8.166∠126.4° = −4.8475 + j 6.5715 V   So                                                      v ( t ) = 8.166 cos ( 8t + 126.4° ) V  

(checked: LNAP 8/3/04)  

P 10.6‐14 Determine the voltage vo(t) when vs(t) = 25 cos (100t‐15°) V. 

 

Solution: Represent the circuit in the frequency domain using impedances and phasors. 

The mesh currents are I and 10I.  Apply KVL to the supermesh corresponding to the dependant  current source to get 

( j500 ) I + ( − j5)(10I ) + 40 (10I ) − 25∠ − 15° = 0 So                                               I =

25∠ − 15° = 0.04152∠ − 63.37° A   400 + j 450

The output voltage is                     V = 40 (10I ) = 16.61∠ − 63.37° V   So                                                      v ( t ) = 16.61cos (100t − 63.37° ) V  

(checked: LNAP 8/3/04)

P 10.6‐15 Determine the mesh current is when i(t) = 0.8394 cos (10t‐138.5°) A. Determine the  values of  L and R. 

 

Solution: Represent the circuit in the frequency domain using phasors and impedances.   Apply KVL to the  center mesh to get  0.8394∠138.5° = I =

8∠210° − 30∠ − 15° R + j10 L

So                                                              R = 35 Ω



R + j10 L = 35 + j 25 = 35 + j (10 ) 2.5

and L = 2.5 H  

(checked: LNAP 8/3/04)

 

P 10.6-16 The circuit shown in Figure P 10.6-16 has two inputs:

50 Ω

5 mF

v1(t) = 50 cos (20t – 75°) V

6H

40 Ω

2H

v2(t) = 35 cos (20t + 110°) V

When the circuit is at steady state, the node voltage is

+ –

v1(t)

R

+ v(t) –

L

+

v2(t) –

v(t) = 21.25 cos (20t – 168.8°) V

Determine the values of R and L.

Figure P 10.6-16

 

Solution: Represent the circuit in the frequency domain using phasors and impedances. Apply KCL at the  top node of R and L to get    ( 50∠ − 75° ) − V + 35∠100° − V = V 40 j 40 R & jω L ⇒

50∠ − 75° 35∠110° ⎛ 1 1 1 1 ⎞ + =⎜ + + −j ⎟V 40∠90° 40 20 L ⎠ ⎝ j 40 40 R

Using the given equation for v(t) we get 

21.25∠ − 168.8° = V =

Then                        

1.587∠161.7° 1 1 0.025 (1 − j ) + − j R 20 L

1 1 1.587∠161.7° −j = − 0.025 (1 − j ) = 0.04 − j 0.01176   R 20 L 21.25∠ − 168.8°

Finally                                R =

1 1 = 25 Ω and L = = 4.25 H   0.04 20 ( 0.01176 )

(checked: LNAP 8/3/04)  

P 10.6‐17 Determine the steady state current i(t): 

  Solution: Represent the circuit in the frequency domain using phasors and impedances.     The node equations are         50∠0° − Va V b − Va Va + =   15 j100 25   50∠0° − V b V b − Va V b = +                − j 20 j100 j 50    

1 1 ⎡1 ⎢15 + j100 + 25 or                           ⎢ 1 ⎢ − ⎢ j100 ⎣

1 ⎤ ⎡ Va ⎤ ⎡ 50∠0° ⎤ ⎥ ⎢ ⎥ j100 ⎥ ⎢ ⎥ = ⎢ 15 ⎥   1 1 1 ⎥ ⎢ ⎥ ⎢ 50∠0° ⎥ ⎢V ⎥ + + j 50 j100 − j 20 ⎥⎦ ⎣ b ⎦ ⎢⎣ − j 20 ⎥⎦ −

⎡ 0.1067 − j 0.010 ⎢ j 0.010 ⎣

j 0.010 ⎤ ⎡ Va ⎤ ⎡3.333⎤ ⎢ ⎥= j 0.020 ⎥⎦ ⎣ V b ⎦ ⎢⎣ j 2.5 ⎥⎦

Solving, e.g. using MATLAB, gives 

Va = 33.05∠ − 12.6° V and V b = 108.9∠1.9° V Then                                                      I =

Va

= 1.322∠ − 12.6° A   25 So                                                     i ( t ) = 1.322 cos ( 25t − 12.6° ) A   (checked: LNAP 8/3/04)

P 10.6‐18 Determine the steady state current i(t): 

 

Solution: Represent the circuit in the frequency domain using phasors and impedances.  Label the node  voltages.      The node equations are               24∠15° − Va Va Va − V b = + 25 j 40 10   24∠15° − V b Va − V b V b + = − j 6.25 10 45    

1 ⎤ ⎡ V ⎤ ⎡ 24∠15° ⎤ ⎥⎢ a⎥ ⎢ ⎥ 10 25 ⎥⎢ ⎥ = ⎢ ⎥  1 1 1⎥ 24∠15° ⎥ ⎢ ⎢V ⎥ + + j 6.25 45 10 ⎥⎦ ⎣ b ⎦ ⎢⎣ 6.25∠ − 90° ⎥⎦

1 1 ⎡1 ⎢ 25 − j 40 + 10 or                         ⎢ 1 ⎢ − ⎢⎣ 10 ⎡ 0.140 − j 0.025 ⎢ −0.10 ⎣



− 0.10

⎤ ⎡ Va ⎤ ⎡ 0.960∠15° ⎤ ⎢ ⎥= 0.1222 + j 0.160 ⎥⎦ ⎣ V b ⎦ ⎢⎣3.840∠105°⎥⎦

Solving gives                    Va = 24.67∠32.6° V and V b = 25.59∠25.2° V   Then                                                 I =

Va − Vb 10

= 0.3347∠134.9° A  

So                                                    i ( t ) = 0.3347 cos (10t + 134.9° ) A  

(checked: LANP 8/4/04)  

P 10.6‐19 Determine the steady state voltage vo(t): 

 

Solution: Represent the circuit in the frequency domain using phasors and impedances.   

20∠0° − V V V − 5V = + j 40 25 − j 20 The node equations are                           5V − Vo Vo = 10 − j10 ⎡1 ⎢ 25 − ⎢ ⎢ ⎢⎣

1 1 j −j 5 40 1 − 2

⎡ 0.04 − j 0.225 ⎢ −0.50 ⎣

⎤ ⎡ V ⎤ ⎡ − j 0.5⎤ ⎥⎢ ⎥ ⎢ ⎥ ⎥⎢ ⎥ = ⎢ ⎥ 1 1 + j ⎥ ⎢⎣ Vo ⎥⎦ ⎢⎣ 0 ⎥⎦ 10 10 ⎥⎦ 0

⎤ ⎡ V ⎤ ⎡ − j 0.5⎤ ⎢ ⎥= 0.10 + j 0.10 ⎥⎦ ⎣ Vo ⎦ ⎢⎣ 0 ⎥⎦ 0

  Solving gives                  V = 2.188∠ − 10.1° V and Vo = 7.736∠ − 55.1° V   So                                                    v o ( t ) = 7.736 cos ( 5t − 55.1° ) V  

(checked: LNAP 8/4/04)  

P 10.6‐20 Determine the steady state current i(t) in each of these circuits:  4 i(t) + – 20 Ω

36 mA

i(t)



(a) 4 i(t) + – 20 Ω

2H i(t)

36 cos (25t) mA

2 mF

15 Ω

(b)

4 mF

 

Solution: (a)  Use KVL to see that the voltage across the 8 Ω resistor is  20i ( t ) − 4i ( t ) = 16i ( t ) .    Apply KCL to the supernode corresponding to the dependent voltage source to get  16i ( t ) 0.036 = i ( t ) + = 3i ( t ) 8 so                                                                         i ( t ) = 12 mA   (b)  Represent the circuit in the frequency domain using phasors and impedances. 

Z 1 = 20 + where                                     

1 = 20 − j 20 Ω j ( 25 )( 0.002 )

⎛ ⎞ 1 Z 2 = j 50 + ⎜⎜ 15 & ⎟⎟ = 43.3∠83.9° Ω j 25 0.004 ( )( ) ⎝ ⎠

 

Use KVL to get                                     V = Z 1I − 4I = ( Z 1 − 4 ) I   Then apply KCL to the supernode corresponding to the dependent source to get 

0.036∠0° = I +

so                                                I =

(Z

1

− 4) I

Z2

Z 2 ( 0.036∠0° ) Z1 + Z 2 − 4

⎛ Z1 + Z 2 − 4 ⎞ =⎜ ⎟⎟ I ⎜ Z 2 ⎝ ⎠

= 50.4∠35.7° mA  

so                                                       i ( t ) = 50.4 cos ( 25t + 35.7° ) mA  

(checked: LNAP 8/4/04)

    10.6‐21 The input to the circuit show in Figure  10.9‐24 is the current    i s ( t ) = 50 cos ( 200 t ) mA     Determine the steady‐state mesh current  i 2 . 

  Figure P10.6‐21 

   

Solution:   Represent the circuit in the  frequency domain using phasors  and impedances:    Apply KVL to the right mesh to  get:   

( j 20 − j 62.5) I 2 + 60 ( I 2 − 0.050∠0° ) = 0 In the time domain 

 

⇒ I2 =

3∠0° = 0.0408∠35.3° A   60 − j 42.5

i 2 ( t ) = 40.8cos ( 200 t + 35.3° ) mA  

MATLAB, 11/20/09   

    10.6‐22 The input to the circuit show in Figure  10.9‐24 is the current  i s ( t ) = 80 cos ( 250 t ) mA  

The steady‐state mesh current in the right  mesh is  i s ( t ) = 66.56 cos ( 250 t + 33.7° ) mA  

  Figure P10.6‐22 

Determine the value of the resistance R. 

Solution: Represent the circuit in the  frequency domain using phasors  and impedances. The mesh  currents are   I 1 = 0.080∠0° A    

and 

I 2 = 0.06656∠33.7° A   Apply KVL to the right to get 

( j 150 − j 200 )( 0.06656∠33.7° ) + R ( 0.06656∠33.7° − 0.080∠0° ) = 0     ( − j 50 )( 0.06656∠33.7° ) + R ( 0.044376∠123.7° ) = 0     ( 50∠90° )( 0.06656∠33.7° ) = 74.9955  75 Ω   R= 0.044376∠123.7°   MATLAB, 11/20/09 

P10.6-23 This circuit shown in Figure P10.6-23 is at steady state. The voltage source voltages are given by v1(t) = 12 cos (2t – 90°) V and v2(t) = 5 cos (2t + 90°) V The currents are given by i1(t) = 744 cos (2t – 118°) mA , i2(t) = 540.5 cos (2t + 100°) mA Determine the values of  R1 ,  R 2 , L and C. 

Figure P10.6-23 Solution: Represent the circuit in the frequency domain using impedances and phasors:

 

I = I 1 + I 2 = 0.744∠ − 118° + 0.5405∠100 = ( −0.349 − j 0.657 ) + ( −0.094 + j 0.532 ) = ( −0.349 − 0.094 ) + j ( −0.657 + 0.532 ) = −0.443 − j 0.125 = 0.460∠ − 164° In the time domain

i ( t ) = 460 cos (2 t − 164°) mA

Replacing series impedances by equivalent impedances gives    Z 1 = R1 + j ω L   and 

Z 2 = R2 − j From KVL 

1   ωC

Z 1 I 1 + 10 I − V1 = 0 ⇒ Z 1 =

V1 − 10 I I1

=

12∠ − 90° − 10 ( 0.460∠ − 164° ) 0.744∠ − 118°

− j 12 − 10 ( −0.443 − j 0.125 ) 0.744∠ − 118° 4.43 − j 10.75 11.63∠ − 67.6° = =   0.744∠ − 118° 0.744∠ − 118° = 15.63∠50.4° =

= 10 + j 12 Ω

and  − Z 2 I 2 + V 2 − 10 I = 0 ⇒ Z 2 =

V 2 − 10 I I2

= =

5∠90° − 10 ( 0.460∠ − 164° ) 0.5405∠100° j 5 − 10 ( −0.443 − j 0.125 )

0.5405∠100° 4.43 + j 6.25 7.66∠54.7°   = = 0.5405∠100° 0.5405∠100° = 14.14∠ − 55.3° = 10 − j 10 Ω

12 =6 H  2 1 ⇒ R 2 = 10 Ω and C = = 0.05 F   2 (10 )

Next                 10 + j 12 = R1 + j ω L = R1 + j 2 L ⇒ R1 = 10 Ω and L = and            10 − j 10 = R 2 − j

1 1 = R2 − j 2C ωC

10.7 Thevenin and Norton Equivalent Circuits    P10.7‐1 Determine the Thevenin equivalent circuit of the circuit shown in Figure P10.7‐1 when  (a) ω = 1000 rad/s, (b) ω = 2000 rad/s and (c) ω = 4000 rad/s. 

  Figure P10.7‐1  Solution: Represent the circuit in the frequency domain as 

  Determine the open circuit voltage and Thevenin impedance: 

 

   

Voc =

ZC 2500 + Z C

(12∠0° )  

 

Zt =

2500 Z C 2500 + Z C

+ ZL 

    a.  Voc = 2.3534∠ − 78.69° V and Z t = 775.22∠82.875° Ω   b.  Voc = 1.194∠ − 84.29° V and Z t = 2252.6∠89.37° Ω   c.  Voc = 0.59925∠ − 87.14° V and Z t = 4875.3∠89.93° Ω     The Thevenin Equivalent Circuit changes whenever the input frequency changes. 

P10.7‐2 Determine the Thevenin equivalent of this circuit when vs(t) = 5 cos (4000t‐30°) V.  1 80

mF 20 mH

vs

+ –

80 Ω

  Solution:  Find  Voc :   

⎛ 80 + j80 ⎞ Voc = ( 5 ∠−30° ) ⎜ ⎟ ⎝ 80 + j80− j 20 ⎠ ⎛ 80 2∠− 45° ⎞ = ( 5 ∠−30° ) ⎜ ⎟ ⎝ 100∠36.90° ⎠

 

= 4 2∠ − 21.9° V   Find  Z t :    Zt =

  The Thevenin equivalent is 

 

( − j 20 )( 80 + j80 ) = 23 ∠ − 81.9° − j 20 + 80 + j80

Ω 

P10.7‐3 Determine the Thevenin equivalent of this circuit  2v

600 Ω +

9 cos 500t –

v

+



+ –

1 150

mF

300 Ω

 

Solution: First, determine  Voc : 

The mesh equations are  600 I1 − j 300 (I1 − I 2 ) = 9 ⇒ (600 − j 300) I1 + j 300 I 2 = 9∠0° −2 V + 300 I 2 − j 300 (I1 − I 2 ) = 0 and V = j 300 (I1 − I 2 ) ⇒

j 3 I1 + (1 − j 3) I 2 = 0

Using Cramer’s rule:                            I 2 = 0.0124∠ − 16° A  

Voc = 300 I 2 = 3.71∠ − 16° V

Then Next, determine  I sc : 

  −2 V − V = 0 ⇒ V = 0 ⇒ I sc =

The Thevenin impedance is 

ZT =   The Thevenin equivalent is 

9∠0° = 0.015∠0° A 600

Voc 3.545∠−16° = = 247∠ − 16° Ω I sc 0.015∠0°

P10.7‐4 Determine the Thevenin equivalent of this circuit when vs(t) = 10 cos (10,000t−53.1°) V.  3i / 2 2Ω 200 μH vs

+ –

a

i 25 μ F b

Solution:   First, determine  Voc :  The node equation is:  Voc Voc − (6 + j8) 3 ⎛ Voc − (6+ j8) ⎞ + − ⎜ ⎟ = 0  − j4 j2 2⎝ j2 ⎠   Voc =3+ j 4 =5∠53.1° V  

Vs = 10∠53° = 6 + j 8 V   Next, determine  I sc :  The node equation is: 

V V V − (6 + j8) 3 ⎡ V − (6 + j8) ⎤ + + − ⎢ ⎥ =0  2 − j4 j2 2⎣ j2 ⎦ V= I sc =

3 + j4   1− j

V 3+ j 4   = 2 2− j 2

The Thevenin impedance is     Z T =

Vs = 10∠53° = 6 + j 8 V ⎛ 2− j 2 ⎞ Voc = 3 + j4 ⎜ ⎟ = 2 − j2 Ω   I sc ⎝ 3+ j 4 ⎠

The Thevenin equivalent is 

(checked: LNAP 7/18/04)

  P10.7‐5 Determine the frequency at which Y is a pure conductance 

20 k Ω

1 nF

Y

 

Solution:

Y = G + YL + YC   1 Y = G when YL + YC = 0 or + jω C = 0   jω L 1 1 1 , fO = ωO = = 2π LC 2π 39.6×10−15 LC = 0.07998×107 Hz =800 kHz (80 on the dial of the radio)

 

39.6 μH

 

 

P 10.7-6 Consider the circuit of Figure P 10.7-6, where we wish to determine the current I. Use a series of source transformations to reduce the part of the circuit connected to the 2-Ω resistor to a Norton equivalent circuit, and then find the current in the 2-Ω resistor by current division. j4 Ω

3 30° A

–j3 Ω



–j2 Ω

2Ω I

Figure P 10.7-6 Solution:  

    Z1 =

(− j 3)(4) = 2.4∠ − 53.1° Ω − j 3+ 4   =1.44 − j1.92 Ω

 

 

  Z 2 = Z1 + j 4 = 1.44 + j 2.08

 

 

 

= 2.53∠55.3° Ω

  Z3 = 3.51∠ − 37.9° Ω

= 2.77 − j 2.16 Ω  

 

⎛ 3.51∠−37.9° ⎞ ( 3.51∠−37.9° ) = 1.9∠ − 92° A I = ( 2.85∠− 78.4° ) ⎜ ⎟ = ( 2.85∠− 78.4° ) ( 5.24∠− 24.4° ) ⎝ 2.77 − j 2.16 + 2 ⎠

(checked: LNAP 7/18/04)

 

P 10.7-7 For the circuit of Figure P 10.7-7,

200 Ω

determine the current I using a series of source 20 45° V

transformations.

+ –

100 Ω 160 μ H

10 μ F

The source has ω = 25 × 103 rad/s.

I

Answer: i(t) = 4 cos (25,000t – 44°) mA

Figure P 10.7-7

 

Solution:  

 

 

 

     

Z2 =

(200)(− j 4) = 4∠ − 88.8° Ω   200− j 4

 

 

   

I=

0.4∠− 44° = 4∠ − 44° mA   −4 j +100+ j 4

  i (t ) = 4 cos (25000 t − 44°) mA    

P10.7‐8 Determine the Thevenin equivalent of the circuit in (a):  a 8Ω

Zt

j10 Ω –j5

a + –

+–

Vt b

j20 Ω

–j2.4 Ω

b

(a)

(b)

 

Solution:   V1 =

j10 5 e − j 90 = 3.9 e− j 51   8 + j10

 

V2 =

j 20 5 e − j 90 = 5.68 e− j 90   j 20 − j 2.4

  Vt = V1 − V 2 = 3.9 e − j 51 − 5.68 e − j 90

= 3.58 e j 47

 

 

 

        8 ( j10 ) − j 2.4 ( j 20 ) Zt = + = 4.9 + j 1.2  8 + j10 − j 2.4 + j 20

   

 

P10.7‐9 Determine the voltage v(t) for this circuit:  + 8Ω

2H 5 sin 5t V v(t)

+–

1 12 F

4 cos (3t + 15°) A

4H –

 

Solution:   V1 (ω ) =

j10 5 e − j 90 = 3.9 e − j 51   8 + j10

 

V 2 (ω ) =

j 20 5 e − j 90 = 5.68 e − j 90   j 20 − j 2.4

  V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90

= 3.58 e j 47   V1 (ω ) =

  V 2 (ω ) =

8 ( j6) 4 e j15 = 19.2 e j 68   8 + j6 j12 ( − j 4 ) 4 e j15 = 24 e − j 75   j12 − j 4

  V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22  

  Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t ‐ 22° )  V.     

 

10.8 Superposition    P10.8‐1 Determine the steady state current i(t) in the circuit shown in Figure P10.8‐1 when the  voltage source voltages are  vs1 (t) = 12 cos (2500 t) V  and   vs2 (t) = 12 cos (4000 t) V 

  Figure 10.8‐1  Solution:   Use superposition in the time domain. These circuits can be used to find the part of io caused by  vs1 and the part of io caused by vs2. 

 

In the frequency domain: 

 

 

 

12∠0° = 0.03578∠26.6° V   300 + j 250 − j 400 12∠0° I2 = = 0.03578∠ − 26.6° V   300 − j 250 + j 400 I1 =

In the time domain  i o1 ( t ) = 35.78cos ( 2500 t + 26.6° ) mA  and   i o2 ( t ) = 35.78cos ( 4000 t − 26.6° ) mA   and 

i o ( t ) = i o1 ( t ) + i o2 ( t ) = 35.78cos ( 2500 t + 26.6° ) + 35.78cos ( 4000 t − 26.6° ) mA V  

P10.8‐2 Determine the steady state voltage v(t) in the circuit shown in Figure P10.8‐2 when the  current source current is (a) 400 rad/s and (b) 200 rad/s. 

  Figure 10.8‐2  Solution:  (a) Represent the circuit in the frequency domain as     Use superposition in the frequency domain  to write  Vo = −

100 j 150 (12∠0° ) + 100 ( 0.1∠0° ) 100 + j 150 100 + j 150

−1200 + j 1500 = = 10.66∠72.35° 100 + j 150

 

In the time domain  v o ( t ) = 10.66cos ( 400 t + 72.35 ) V     (b) Use superposition in the time domain. These circuits can be used to find the part of vo  caused by the current source and the part of vo caused by the voltage source. 

  In the frequency domain: 

   

Vo1 = 100 Vo2 = −

j 75 ( 0.1∠0° ) = 6∠53.1° V   100 + j 75

100 (12∠0° ) = 6.656∠123.7° V   100 + j 150

In the time domain  v o1 ( t ) = 6cos ( 200 t + 53.1° ) V  and   v o2 ( t ) = 6.656cos ( 400 t + 123.7° ) V   and 

v o ( t ) = v o1 ( t ) + v o2 ( t ) = 6cos ( 200 t + 53.1° ) + 6.656cos ( 400 t + 123.7° ) V  

 

P10.8‐3 Determine the steady state current i(t) in the circuit shown in Figure P10.8‐3 when the  voltage source voltage is  vs (t) = 8 + 8 cos (400t − 135° ) V   

  Figure 10.8‐3  Solution:   Use superposition in the time domain:  

  An inductor in a dc circuit acts like a short circuit so: 

i1 ( t ) =

  

8 = 0.533 A   15

Represent the right circuit the frequency domain: 

    In the time domain  and     

  8∠ − 135° = 0.32∠ − 188° A   15 + j 20 i 2 ( t ) = 0.32cos ( 400 t − 188° ) A   I2 =

i ( t ) = i1 ( t ) + i 2 ( t ) = 0.533 + 0.32cos ( 400 t − 188° ) A  

P10.8‐4 Determine the steady state current i(t) in the circuit shown in Figure P10.8‐4 when the  voltage source voltages are  vs1 (t) = 10 cos (800t + 30° ) V   and   vs2 (t) = 15 sin (200t − 30° ) V      

  Figure 10.8‐4  Solution:   Use superposition in the time domain:  

  Represent these circuits the frequency domain: 

 

   

I1 =

10∠30° 15∠ − 120° = 0.1342∠33.4° A   = 0.0447∠ − 33.4° A    and    I 2 = − 100 + j 200 100 + j 50

In the time domain 

 

i ( t ) = i1 ( t ) + i 2 ( t ) = 44.7 cos ( 800 t − 33.4° ) + 134.2cos ( 200 t + 33.4° ) mA  

 

  4i(t)

P 10.8-5

The input to the circuit shown in

+ –

Figure P 10.8-5 is the current source current

20 Ω

is(t) = 36 cos (25t) + 48 cos (50t + 45°) mA Determine the steady-state current, i(t).

2H

i(t)

is(t)

2 mF

15 Ω

4 mF

Figure P 10.8-5 Solution:  Use superposition in the time domain. Let  i s1 ( t ) = 36 cos ( 25 t ) mA and i s 2 ( t ) = 48cos ( 50 t + 45° ) mA

We will find the response to each of these inputs separately.  Let ii(t) denote the response to  isi(t) for i = 1,2.   The sum of the two responses will be i(t), i.e.  i ( t ) = i1 ( t ) + i 2 ( t )

Represent the circuit in the frequency domain as 

Use KVL to get 

Vi = Z i I i − 4I i Apply KCL to the supernode corresponding to the dependent voltage source.  I si = I i +

Vi Z2

=

Z1 + Z 2 − 4 Z2

or  Ii =

Z 2I s i Z1 + Z 2 − 4

Ii

Consider the case i = 1 : is1(t) = 26cos(25t) mA.  Here ω = 25 rad/s and  

I si = 36∠0° mA Z 1 = 20 +

1 = 20 − j 20 Ω j ( 25 )( 0.002 )

⎛ ⎞ 1 Z 2 = j 50 + ⎜⎜15 & ⎟⎟ = 43.3∠83.9° Ω 25 0.004 j ( )( ) ⎝ ⎠ and 

I 1 = 50.4∠35.7° mA so 

i ( t ) = 50.4 cos ( 25t + 35.7° ) mA

Next consider i = 2 : is2 = 48cos(50t + 45°) mA.  Here ω = 50 rad/s and  

I s2 = 48∠45° mA Z 1 = 20 +

1 = 20 − j10 Ω j ( 50 )( 0.002 )

⎛ ⎞ 1 Z 2 = j100 + ⎜⎜15 & ⎟ = 95.5∠89.1° Ω j ( 50 )( 0.004 ) ⎟⎠ ⎝ (Notice that Z1 and Z2 change when ω changes.) 

I 2 = 52.5∠55.7° mA so 

i 2 ( t ) = 52.5cos ( 50t + 55.7° ) mA

Finally, using superposition in the time domain gives  i ( t ) = 50.4 cos ( 25t + 35.7° ) + 52.5cos ( 50t + 55.7° ) mA

(checked: LNAP 8/7/04)  

  20 Ω

20 Ω

4H

5 mF

P 10.8-6 The inputs to the circuit shown in 10 Ω

Figure P 10.8-6 are

i(t)

vs1(t) = 30 cos (20t + 70°) V

and

vs2(t) = 18 cos (10t – 15°) V

+ –

vs1(t)

+

15 Ω

vs2(t) – 2H

Figure P 10.8-6  

Solution: Use superposition in the time domain.  Let i1(t) be the part of i(t) due to vs1(t) and i2(t) be the  part of i(t) due to vs2(t).  To determine i1(t), set vs2(t) = 0.  Represent the resulting circuit in the  frequency domain to get 

where 

Z 1 = 20 + j80 = 82.46∠76° Ω Z 2 = 10 + ( j 40 & 15 ) = 23.15 + j 4.93 = 23.67∠12° Ω Z 3 = 20 +

1 = 20 − j10 = 22.36∠ − 26.6° Ω j ( 20 )( 0.005 )

Next, using Ohm’s law and current division gives 

I1 = so 

Z3 Z 3 ( 30∠70° ) 30∠70° × = = 0.182∠ − 17.6° A Z 1 + ( Z 2 & Z 3 ) Z 2 + Z 3 Z 1Z 2 + Z 2 Z 3 + Z 1Z 3 i ( t ) = 0.182 cos ( 20t − 17.6° ) A

To determine i2(t), set vs1(t) = 0.  Represent the resulting circuit in the frequency domain to get 

where 

Z 4 = 20 + j 40 = 44.72∠63.4° Ω Z 5 = 10 + ( j 20 & 15 ) = 19.6 + j 7.2 = 20.88∠20.2° Ω Z 6 = 20 +

1 = 20 − j 20 = 28.28∠ − 45° Ω j (10 )( 0.005 )

Next, using Ohm’s law and current division gives 

I2 =

Z4 Z 1 (18∠ − 15° ) 18∠ − 15° × = = 0.377∠18° A Z 6 + ( Z 4 & Z 5 ) Z 4 + Z 5 Z 1Z 2 + Z 2 Z 3 + Z 1Z 3

so 

i 2 ( t ) = 0.377 cos (10t + 18° ) A

Using superposition,  i ( t ) = i1 ( t ) + i 2 ( t ) = 0.182 cos ( 20t − 17.6° ) + 0.377 cos (10t + 18° ) A

(checked: LNAP 8/8/04)

 

P 10.8-7 The input to the circuit shown in Figure



25 Ω

P 10.8-7 is the voltage source voltage + –

vs(t) = 5 + 30 cos (100t) V

i(t)

vs(t)

Determine the steady-state current, i(t). 20 μF

50 mH

Figure P 10.8-7 Solution: Use superposition in the time domain.  Let vs1(t) = 5 V and vs2(t) = 30cos(100t) V.    Find the steady state response to vs1(t).  When the input is constant and the circuit is at  steady state, the capacitor acts like an open  circuit and the inductor acts like a short  circuit.  So  5 i1 ( t ) = = 1 A 5   Find the steady state response to vs2(t). 

 

Represent the circuit in the frequency domain  using impedances and phasors.    30∠0° I2 = = 4.243∠ − 45 A 5 + j5 So  i 2 ( t ) = 4.243cos (100t − 45° ) A Using superposition 

i ( t ) = i1 ( t ) + i 2 ( t ) = 1 + 4.243cos (100t − 45° )

P10.8‐8 Determine the voltage v(t) for the circuit  + 8Ω

2H 5 sin 5t V v(t)

+–

1 12 F

4 cos (3t + 15°) A

4H –

 

Solution: V1 (ω ) =

j10 5 e − j 90 = 3.9 e − j 51   8 + j10

 

V 2 (ω ) =

j 20 5 e − j 90 = 5.68 e − j 90   j 20 − j 2.4

  V (ω ) = V1 (ω ) − V 2 (ω ) = 3.9 e − j 51 − 5.68 e − j 90

= 3.58 e j 47   V1 (ω ) =

  V 2 (ω ) =

8 ( j6) 4 e j15 = 19.2 e j 68   8 + j6

j12 ( − j 4 ) 4 e j15 = 24 e − j 75   j12 − j 4

  V (ω ) = V1 (ω ) + V 2 (ω ) = 14.4 e − j 22  

  Using superposition: v(t) = 3.58 cos ( 5t + 47° ) + 14.4 cos ( 3t ‐ 22° )  V.   

 

P10.8‐9 Determine the current i(t) for this circuit when v1(t)=10cos(10t) V  5Ω v1 +

1.5 H

i(t)

10 mF



10 Ω

3A

  Solution:  Use superposition. First, find the response to the voltage source acting alone: 

Z eq =

− j10⋅10 = 5(1 − j ) Ω   10− j10

    Replacing the parallel elements by the equivalent impedance. The write a  mesh equation : 

−10 + 5 I1 + j15 I1 + 5(1 − j ) I1 = 0 ⇒ I1 =

10 = 0.707∠ − 45° A 10+ j10

Therefore:  i1 (t ) = 0.707 cos(10 t − 45° ) A

Next, find the response to the dc current source acting alone:    Current division:     I 2 = −   Using superposition:     i (t ) = 0.707 cos(10 t − 45°) − 2 A  

10 × 3 = −2 A   15

Section 10-9: Phasor Diagrams P 10.9-1

Using a phasor diagram, determine V when

V = V1 – V2 + V3* and V1 = 3 + j3, V2 = 4 + j2, and V3 = –3 – j2. (Units are volts.) Answer: V = 5 ∠143.1° V . Solution:

V = V1 − V2 + V3 = ( 3+ j 3) − ( 4 + j 2 ) + ( −3− j 2 ) = −4 + j 3 *

*

P 10.9-2 Consider the series RLC circuit of Figure P 10.9-2 when R = 10 Ω, L = 1 mH, C = 100 μF, and ω = 103 rad/s. Find I and plot the phasor diagram j ωL

R

10 0° V

+ –

I

1 j ωC

Figure P 10.9-2 Solution:

I=

VR = R I = 7.4∠42° V VL = Z L I = (1∠90°)(0.74∠42°) = 0.74∠132° V VC = Z C I = (10∠−90°)(0.74∠42°) = 7.4∠− 48° V VS = 10∠0° V

10∠0° = 0.74∠42° A 10+ j1− j10

10.10 Op Amps in AC Circuits    P10.10‐1 The input to the circuit shown in Figure P10.10‐1 is the voltage   v s ( t ) = 2.4 cos ( 500 t ) V . 

Determine the output voltage vo(t).  Answer:  v o ( t ) = 6.788cos ( 500 t + 135° ) V  

  Figure P10.10‐1  Solution:  Represent the circuit in the frequency domain as 

  Recognizing this circuit as an inverting amplifier, we can write 

14.14∠ − 45° ⎞ ⎛ 20 || − j 20 ⎞ ⎛ Vo = ⎜ − ⎟ ( 2.4∠0 ) = ⎜ (1∠180° ) ⎟ ( 2.4∠0 ) = 6.788∠135° V   5 5 ⎝ ⎠ ⎝ ⎠ In the time domain  v o ( t ) = 6.788cos ( 500 t + 135° ) V   (Checked using LNAPAC 3/15/12)     

 

P10.10‐2 The input of the circuit shown in Figure P10.10‐2 is the voltage   v s ( t ) = 1.2 cos ( 400 t + 20° ) V . 

Determine the output voltage vo(t). 

  Figure P10.10‐2  Solution:  Represent the circuit in the frequency domain as 

  Recognizing this circuit as a noninverting amplifier, we can write  ⎛ 48 ⎞ V0 = ⎜1 + ⎟ (1.2∠20° ) = (1 + j 4.8 )(1.2∠20° ) = 5.88∠98° V   ⎝ − j 10 ⎠ In the time domain  v o ( t ) = 5.88cos ( 400 t + 98° ) V  

(Checked using LNAPAC 3/15/12)   

 

P10.10‐3 The input of the circuit shown in Figure P10.10‐3 is the voltage   v s ( t ) = 3.2 cos ( 200 t ) V . 

Determine the output voltage vo(t). 

  Figure P10.10‐3  Solution:  Represent the circuit in the frequency domain as 

  Recognizing this circuit as a voltage divider followed by a noninverting amplifier, we can write  ⎞ 20 20 ⎛ 40 ⎞ ⎛ ⎛ ⎞ Vo = ⎜1 + ⎟ ⎜ ⎟ ( 3.2∠0° ) = ⎜ ⎟ ( 9.6∠0° ) = 4.293∠63.4° V   ⎝ 20 ⎠ ⎝ − j 40 + 20 ⎠ ⎝ 44.72∠ − 63.4° ⎠

In the time domain 

v o ( t ) = 4.293cos ( 200 t + 63.4° ) V  

(Checked using LNAPAC 3/15/12)     

 

P10.10‐4 The input of the circuit shown in Figure P10.10‐4 is the voltage   v s ( t ) = 1.2 cos ( 2000 t ) V . 

Determine the output voltage vo(t). 

  Figure P10.10‐4  Solution:  Represent the circuit in the frequency domain as 

  Recognizing this circuit as a voltage divider followed by a voltage follower, we can write  ⎛ − j 2.5 ⎞ ⎛ 2.5∠ − 90° ⎞ Vo = ⎜ ⎟ (1.2∠0° ) = ⎜ ⎟ (1.2∠0° ) = 0.1974∠ − 80.54° V   ⎝ 15.2∠ − 9.46° ⎠ ⎝ 15 − j 2.5 ⎠ v o ( t ) = 0.1974 cos ( 400 t − 80.54° ) V   In the time domain 

(Checked using LNAPAC 3/15/12)   

 

P10.10‐5 The input of the circuit shown in Figure P10.10‐5 is the voltage   v s ( t ) = 1.2 cos ( 2000 t ) V . 

Determine the output voltage vo(t). 

  Figure P10.10‐5  Solution:  Represent the circuit in the frequency domain as 

  Recognizing this circuit as a inverting amplifier, we can write 

In the time domain 

⎛ − j 50 ⎞ Vo = ⎜ − ⎟ (1.2∠0 ) = 12∠90° V   5 ⎠ ⎝ v o ( t ) = 12 cos ( 400 t + 90° ) V   (Checked using LNAPAC 3/15/12) 

 

P 10.10-6

Determine the ratio Vo/Vs for the circuit shown in Figure P 10.10-6. Z2

Z1

+ –

Vs

Z4

Z3





+

+

+

Vo –

Figure P 10.10-6 Solution: Label the nodes: 

The ideal op amps force Va = 0 and Vc = 0.    Apply KCL at node a to get 

Vb =

Z2 Z1 + Z 2

Vs  

  Apply KCL at node c to get  

Vo =

Z4 Z3 + Z4

Vb  

Therefore Z4 Z2 Vo = ×   Vs Z 3 + Z 4 Z 1 + Z 2

   

 

P 10.10‐7  Determine the ratio Vo/Vs for both of the circuits shown in Figure P 10.10‐7.    Z1



Z1

Z3

Z3

+

+

+ + –

Vs

Z4

Z2

+ –

Vo

Vs

Z4

Z2





(a)

                Figure P 10.10-7

Solution: Label a node voltage as Va in each of the  circuits.    In both circuits, we can apply KCL at the node  between Z3 and Z4 to get     Z4 Vo = Va   Z3 + Z4 In (a)  Va = =

(

Z 2 || Z 3 + Z 4

(

)

Z 1 + Z 2 || Z 3 + Z 4

(

)

Vs

Z2 Z3 +Z4

(

)

)

 

(

Z1 Z 2 + Z 3 + Z 4 + Z 2 Z 3 + Z 4

)

Vs

so

Z2 Z4 Va   = Vs Z 1 Z 2 + Z 3 + Z 4 + Z 2 Z 3 + Z 4

(

)

(

In (b)  Va =

Z2 Z1 + Z 2

Vs  

so Z4 Z2 Vo = ×   Vs Z 3 + Z 4 Z 1 + Z 2

     

 

Vo

)

 

(b)

 

  P 10.10‐8  Determine the ratio Vo/Vs for the circuit shown in Figure P 10.10‐8.  Z4

Z3

Z5



Z1

+ + –

Vs

+

Z6

Z2

Vo –

  Figure P 10.10‐8 

Solution: Label the node voltages Va  and Vb as shown:    Apply KCL at the node between Z1 and Z2 to  get   Z2 Va = Vs   Z1 + Z 2   Apply KCL at the node between Z1 and Z2 to  get   Z3 + Z4 Vb = Va Z3   Apply KCL at the node between Z5 and Z6 to  get  Z6 Vo = Vb   Z5 + Z6 so

     

 

Z6 Z3 + Z4 Z2 Vo = × ×   Vs Z 5 + Z 6 Z3 Z1 + Z 2

 

+

P 10.10-9 When the input to the circuit shown in Figure P 10.10-9 is the voltage source voltage

R1

vs(t) = 2 cos (1000t) V

+ –

the output is the voltage



+ R2

vs(t)

vo(t)

+

vo(t) = 5 cos (1000t – 71.6°) V

C=1 mF

Determine the values of the resistances R1 and R2.

v(t)

1 kΩ



Figure P 10.10-9 Solution: The network function of the circuit is  1

1+

R2

1+

R2

R2 ⎞ jω C ⎛ 1000 = 1000 = ⎜1 + = ⎟ 1 V s ⎝ 1000 ⎠ R + 1 + j ω C R1 1 + j 10−3 R1 1 jω C

Vo

Converting the given input and output sinusoids to phasors gives Vo Vs

=

5 ∠71.6° 2

 Consequently  1+

R2

5 ∠71.6° 1000 = 2 1 + j 10−3 R1

Equating angles gives   

(

71.6° = − tan −1 10−3 R1

)

⇒ R1 = tan ( 71.6° ) × 10 3 = 3006 Ω

  Equating magnitudes gives  R2 R2 1+ 1+ 5 1000 1000 = = 2 2 1 + 10−3 R1 1 + 10−3 × 3006

(

     

 

)

(

)

2

⎛5 ⎞ ⇒ R 2 = ⎜ 10 − 1⎟ × 10 3 = 6906 Ω ⎝2 ⎠



P 10.10-10

When the input to the circuit 10 kΩ

source voltage

vs(t)

+ –

C

R

10 kΩ

shown in Figure P 10.10-10 is the voltage

+



v(t)

+



vs(t) = 4 cos (100t) V

+ 10 kΩ



the output is the voltage vo(t) = 8 cos (100t + 135°) V

Figure P 10.10‐10 

Determine the values of C and R.

Solution: Represent the circuit in the frequency domain as   

Apply KCL at the top node of the impedance of the capacitor to get  Vs − V 1 V V = + 4 ⇒ Vs = 1 + j ( 5 × 105 ) C V 4 1 10 10 2 j100C

(

)

Apply KCL at the inverting node of the op amp to get 

V Vo + =0 104 R



Vo = −

R V 104

R 2 ×104 so                                                                     = Vs 1 + j ( 5 ×105 ) C

Vo



Converting the input and output sinusoids to phasors gives  Vo Vs

=

8∠135° = 2∠135° 4∠0°

R R 4 2 ×10 2 ×104 = ∠180° − tan −1 ( 5 × 105 ) C   so                        2∠135° = 5 2 1 + j ( 5 ×10 ) C 1 + ⎡⎣( 5 × 105 ) C ⎤⎦ −

 

vo(t)

(

)

Equating angles gives 

(

135° = 180° − tan −1 ( 5 × 105 ) C

)



C=

tan ( 45° ) = 2 × 10−6 = 2 μ F 5 ×105

Next, equating magnitudes gives 

R R 4 4 2 ×10 2= = 2 ×10 2 1 + ( 5 × 105 )( 2 × 10−6 )        

 



R = 104 = 10 kΩ

  P10.10‐11 The input to the circuit shown in  Figure P10.10‐11 is the voltage source voltage,  v s ( t ) . The output is the voltage  v o ( t ) . The input v s ( t ) = 2.5cos (1000 t ) V  

causes the output to be   v o ( t ) = 8cos (1000 t + 104° ) V . 

Determine the values of the resistances  R1  and 

R 2 .      Answers:  R1 =  1515  Ω  and  R 2 =  20  kΩ. 

Figure P10.10-11

  Solution:   

R2

1 jω C

=

R2 1 + jω CR 2

R2

 

R2

R1 1 + jω CR 2 Vo (ω )   =− =− R1 Vi (ω ) 1 + jω CR 2 R2

R1 Vo (ω ) j (180− tan −1 ω CR 2 ) e =   2 Vi (ω ) 1 + (ω CR 2 )   In this case the angle of  

Vo (ω ) tan (180° − 104° )  is specified to be 104° so  CR 2 = = 0.004  and  Vi (ω ) 1000 R2

the magnitude of 

R1 Vo (ω ) 8 8 = ⇒  is specified to be   so  2.5 Vi (ω ) 1 + 16 2.5

R2 R1

= 13.2 . One set of 

values that satisfies these two equations is  C = 0.2 μ F, R1 = 1515 Ω, R 2 = 20 kΩ .     

 

 

Section 10.11 The Complete Response   

  Figure P10.11‐1    P10.11‐1 The input to the circuit shown in Figure P10.11‐1 is the voltage  v s = 12 cos ( 4000 t ) V .  The output is the capacitor voltage, vo. Determine vo.    Solution:   Before the switch closes the circuit is represented in the frequency domain as  Express the dependent source voltage in  terms of the node voltages:    150 Ι a = Va − Vo   Using Ohm’s law  150

Va 200

so 

= Va − Vo  

Vo =

1 Va   4

Apply KCL to the supernode corresponding to the dependent source to get   

Vs − V a 100

=

Va 200

+

Vo − j 12.5

⇒ Vo =

− j 12.5 ⎛ − j 12.5 − j 12.5 ⎞ Vs − ⎜ + ⎟ Va   100 200 ⎠ ⎝ 100

Vo = − j 0.125 (12∠0° ) + j 0.1875 ( 4 Vo )   − j 0.125 (12∠0° ) = 1.2∠ − 53.1° V   1 − j 0.75 The corresponding sinusoid is  1.2 cos ( 4000 t − 53.1° ) V . The initial capacitor voltage is  Vo (1 − j 0.75 ) = − j 0.125 (12∠0° ) ⇒ Vo =

v o ( 0 ) = 1.2 cos ( −53.1° ) = 0.7205 V . 

  The steady state response after the switch closes is the forced response. The circuit is  represented in the frequency domain as   

Express the dependent source voltage in  terms of the node voltages:    150 Ι a = Va − Vo   Using Ohm’s law  Va 150 = Va − Vo   200 1 Vo = Va   so  4 Apply KCL to the supernode corresponding to the dependent source to get   

Vs − V a 100

=

Va 200

+

Vo − j 12.5

+

⎛ 1 1 ⎞ 1 1 ⎞ ⎛ 1 ⇒ Vo ⎜ + ⎟= + Vs − ⎜ ⎟ Va   25 ⎝ 100 200 ⎠ ⎝ − j 12.5 25 ⎠ 100

Vo

Multiply by 200 to get 

Vo ( 8 + j 16 ) = 2 (12∠0° ) − 3 ( 4 Vo ) ⇒ Vo =

24 = 0.937∠ − 38.7° V   20 + j 16

The corresponding sinusoid is the forced response:    v f ( t ) = 0.937 cos ( 4000 t − 38.7° ) V   The natural response is 

v n ( t ) = k e− t τ V  

To determine the time constant τ we need find to find the Thevenin resistance of the part of  the circuit connected to the capacitor after the switch closes. Here’s the circuit: 

  The terminals separate the capacitor from the part of the circuit connected to the capacitor.   Now (1) remove the capacitor, (2) replace the voltage source by a short circuit to set the input  to zero and (3) connect a current source to the terminals to get      The Thevenin resistance is given by    vt Rt =     it

Express the dependent source voltage in terms of the node voltages to get 

v a − v t = 150 i a = 150

va

⇒ va = 4 vt  

200

Apply KCL to the supernode corresponding to the dependent source to get  it =

va 100

The time constant is  The natural response is 

+

va 200

+

vt 25

=

4vt 100

+

4vt 200

+

vt 25

=

vt 10

⇒ Rt =

vt it

= 10 Ω  

τ = R t C = 10 ( 20 × 10−6 ) = 0.2 × 10−3 = 0.2 ms   v n ( t ) = k e − t τ = k e − 5000 t V  

The complete response is    v o ( t ) = 0.937 cos ( 4000 t − 38.7° ) + k e −5000 t V for t ≥ 0   Using the initial condition we calculate  0.7205 = v o ( 0 ) = 0.937 cos ( −38.7° ) + k ⇒ k = −0.0108   Finally   

v o ( t ) = 0.937 cos ( 4000 t − 38.8° ) − 0.0108 e −5000t V for t ≥ 0  

  Figure P10.11‐2    P10.11‐2 The input to the circuit shown in Figure P10.11‐2 is the voltage  v s = 12 cos ( 4000 t ) V .  The output is the capacitor voltage, vo. Determine vo.  Solution:   Before the switch opens the circuit is represented in the frequency domain as  Express the dependent source voltage in  terms of the node voltages:    150 Ι a = Va − Vo   Using Ohm’s law  Va 150 = Va − Vo   200 1 so  Vo = Va   4 Apply KCL to the supernode corresponding to the dependent source to get   

Vs − V a 100

=

Va 200

+

Vo − j 12.5

+

⎛ 1 1 ⎞ 1 1 ⎞ ⎛ 1 Vs − ⎜ ⇒ Vo ⎜ + ⎟= + ⎟ Va   25 ⎝ 100 200 ⎠ ⎝ − j 12.5 25 ⎠ 100

Vo

Multiply by 200 to get 

Vo ( 8 + j 16 ) = 2 (12∠0° ) − 3 ( 4 Vo ) ⇒ Vo = The corresponding sinusoid is  The initial condition is 

24 = 0.937∠ − 38.7° V   20 + j 16

0.937 cos ( 4000 t − 38.7° ) V  

v o ( 0 ) = 0.937 cos ( −38.7° ) = 0.7313 V  

  The steady state response after the switch closes is the forced response. The circuit is  represented in the frequency domain as 

Express the dependent source voltage in  terms of the node voltages:    150 Ι a = Va − Vo   Using Ohm’s law 

150

Va 200

= Va − Vo  

Vo =

so 

1 Va   4

Apply KCL to the supernode corresponding to the dependent source to get   

Vs − V a 100

=

Va 200

+

Vo − j 12.5

⇒ Vo =

− j 12.5 ⎛ − j 12.5 − j 12.5 ⎞ + Vs − ⎜ ⎟ Va   100 200 ⎠ ⎝ 100

Vo = − j 0.125 (12∠0° ) + j 0.1875 ( 4 Vo )   Vo (1 − j 0.75 ) = − j 0.125 (12∠0° ) ⇒ Vo =

− j 0.125 (12∠0° ) = 1.2∠ − 53.1° V   1 − j 0.75

The corresponding sinusoid is the forced response:  v f ( t ) = 1.2 cos ( 4000 t − 53.1° ) V   The natural response is 

v n ( t ) = k e− t τ V  

To determine the time constant τ we need find to find the Thevenin resistance of the part of  the circuit connected to the capacitor after the switch closes. Here’s the circuit: 

  The terminals separate the capacitor from the part of the circuit connected to the capacitor.   Now (1) remove the capacitor, (2) replace the voltage source by a short circuit to set the input  to zero and (3) connect a current source to the terminals to get 

    The Thevenin resistance is given by    vt   Rt =   it

Express the dependent source voltage in terms of the node voltages to get 

v a − v t = 150 i a = 150

va 200

⇒ va = 4 vt  

Apply KCL to the supernode corresponding to the dependent source to get  it =

va 100

+

va 200

=

4vt 100

+

4vt 200

= 0.06 v t

⇒ Rt =

vt it

= 16.67 Ω  

The time constant is  τ = R t C = 16.67 ( 20 ×10−6 ) = 0.333 × 10−3 = 0.333 ms   The natural response is 

v n ( t ) = k e − t τ = k e − 3000 t V  

The complete response is    v o ( t ) = 1.2 cos ( 4000 t − 53.1° ) + k e −3000 t V for t ≥ 0   Using the initial condition we calculate  0.7313 = v o ( 0 ) = 1.2 cos ( −53.1° ) + k ⇒ k = 0.0108   Finally 

v o ( t ) = 1.2 cos ( 4000 t − 53.1° ) + 0.0108 e −3000 t V for t ≥ 0  

Section 10.12 Using MATLAB to Analyze Electric Circuits    10.12-1 Determine the mesh currents for the circuit shown in Figure P10.12-1 when vs(t) = 12 cos(2500t + 60°) V   and   is(t) = 2 cos(2500t − 15°) mA 

  Figure P10.12‐1  Solution:  Represent the circuit in the frequency domain: 

  Represent the source current in terms of the mesh currents:      I 2 − I 1 = I s = 0.002∠ − 15° A   Apply KVL to the supermesh corresponding to the current source:  

( 2000 − j 2000 ) I 1 + j 1200 ( I 1 − I 3 ) + 1500 ( I 2 − I 3 ) + ( 4000 − j 1000 ) I 2 = 0   Apply KVL to mesh 3:  

( 2400 + j 1875) I 3 + 1500 ( I 3 − I 2 ) + j 1200 ( I 3 − I 1 ) = −Vs = −12∠60°  

−1 1 0 ⎡ ⎤ ⎡ I 1 ⎤ ⎡0.002∠ − 15° ⎤ ⎢ ⎥ ⎢ ⎥ In matrix form: ⎢ 2000 − j 800 5500 − j 1000 −1500 − j 1200 ⎥⎥ ⎢I 2 ⎥ = ⎢⎢ 0 ⎥  ⎢⎣ − j 1200 3900 + j 3075 ⎥⎦ ⎢⎣ I 3 ⎥⎦ ⎢⎣ −12∠60° ⎥⎦ −1500 Solving, using MATLAB, gives 

In the time domain:     

⎡ I 1 ⎤ ⎡1.549∠ − 164° ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ I 2 ⎥ = ⎢ 1.039∠ − 65° ⎥ mA   ⎢ I 3 ⎥ ⎢⎣ 2.904∠ − 148° ⎥⎦ ⎣ ⎦

⎡ i1 ⎤ ⎡1.549 cos ( 2500 t − 164° ) ⎤ ⎢ ⎥ ⎢ ⎥ ⎢i 2 ⎥ = ⎢ 1.039 cos ( 2500 t − 65° ) ⎥ mA   ⎢ i 3 ⎥ ⎢ 2.904 cos ( 2500 t − 148° ) ⎥ ⎦ ⎣ ⎦ ⎣

10.12-2 Determine the node voltages for the circuit shown in Figure P10.12-2 when vs(t) = 12 cos(400t + 45°) V.   

  Figure P10.12‐2  Solution:  Represent the circuit in the frequency domain: 

  Represent the dependent source voltage in terms of the node voltages currents:  V2 V3 − V 2 = 10000 ⇒ V3 = 3.5 V 2   4000 Apply KCL to the supernode corresponding to the dependent voltage source:  

Vs − V 2 j 2000

=

V2 4000

+

V3 j 3200

+

V3 − V 4 4000

 

⎛ 1 ⎛ 1 1 ⎞ 1 ⎞ ⎛ 1 ⎞ =⎜ + + ⎟ V2 + ⎜ ⎟ V3 − ⎜ ⎟ V4   j 2000 ⎝ 4000 j 2000 ⎠ ⎝ 4000 ⎠ ⎝ 4000 j 3200 ⎠ Vs − V 4 V3 − V 4 V4 Apply KCL at node 3:    + = 10, 000 4000 − j 20, 000

Rearranging: 

Vs

Rearranging: 

Vs 10, 000

=−

⎡ ⎢ 3.5 ⎢ ⎢ 1 1 + In matrix form: ⎢ ⎢ 4000 j 2000 ⎢ 0 ⎢ ⎣ Solving, using MATLAB, gives 

In the time domain:   

⎛ 1 ⎞ 1 1 +⎜ + + ⎟ V4   4000 ⎝ 4000 10, 000 − j 20, 000 ⎠ V3

⎡ ⎤ ⎤ ⎢ 0 ⎥ ⎥ −1 0 ⎥ ⎥ ⎡ V2 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ Vs ⎥ 1 1 1 + − ⎥  ⎥ ⎢ V3 ⎥ = ⎢ j 4000 j 3200 4000 2000 ⎥ ⎥ ⎢V ⎥ ⎢ ⎥ ⎣ 4 ⎦ ⎢ Vs ⎥ 1 1 1 1 − + + ⎢ ⎥ ⎥ 4000 4000 10, 000 − j 20, 000 ⎦ ⎣10, 000 ⎦ ⎡ V 2 ⎤ ⎡ 3.236∠34° ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ V3 ⎥ = ⎢11.324∠34° ⎥ V   ⎢ V 4 ⎥ ⎢⎣10.798∠29°⎥⎦ ⎣ ⎦

⎡ v 2 ⎤ ⎡ 3.236 cos ( 400 t + 34° ) ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ v 3 ⎥ = ⎢11.324 cos ( 400 t + 34° ) ⎥ V   ⎢ v 4 ⎥ ⎢10.798cos ( 400 t + 29° ) ⎥ ⎦ ⎣ ⎦ ⎣

10.12-3 Determine the mesh currents for the circuit shown in Figure P10.12-3

  Figure P10.12‐3  Solution:  Represent the source current in terms of the mesh currents:      I 1 − I 2 = 4.2∠30° A   Apply KVL to the supermesh corresponding to the current source:   j 8 ( I 1 − I o ) + 5 ⎡⎣ j 8 ( I 1 − I o ) ⎤⎦ + ( 4 + j 5 ) I 2 + ( 3 − j 8 ) I 1 = 0  

Apply KVL to mesh 3:                        5 I o + 6 ( j 8 ) ( I 1 − I o ) = 0  

In matrix form: 

−1 0 ⎤ ⎡ I 1 ⎤ ⎡ 4.2∠30°⎤ ⎡ 1 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢3 + j 40 4 + j 5 − j 48 ⎥ ⎢ I 2 ⎥ = ⎢ 0 ⎥   ⎢⎣ − j 48 0 5 + j 48⎥⎦ ⎢⎣I o ⎥⎦ ⎢⎣ 0 ⎥⎦

Solving, using MATLAB, gives 

⎡ I 1 ⎤ ⎡ 2.204∠93.1°⎤ ⎢ ⎥ ⎢ ⎥ ⎢ I 2 ⎥ = ⎢ 3.758∠178° ⎥ A   ⎢ I o ⎥ ⎢⎣ 2.192∠99° ⎥⎦ ⎣ ⎦

10.12-4 Determine the node voltages for the circuit shown in Figure P10.12-4.

  Figure P10.12‐4  Solution:  ⎛ 24∠45° − V 2 ⎞ V 2 V 2 − Vo   + 10 ⎜ + ⎟= 2+ j4 2+ j4 j6 ⎝ ⎠ 4 24∠45° − V3 V 2 − Vo Vo + = Apply KCL at node 3:    j6 5− j2 4 11 1 ⎡1 1 ⎤ ⎡ 11 ⎤ − 24∠45° ) ⎥ ( ⎢4 + j 6 + 2+ j 4 ⎥ ⎢ j6 ⎡V ⎤ 2+ j4 ⎥⎢ 2⎥ = ⎢ ⎥  In matrix form:  ⎢ 1 1 1 1 ⎥ ⎣ Vo ⎦ ⎢ 1 ⎢ ⎥ − + + ⎥ ( 24∠45° ) ⎥ ⎢ ⎢ j6 5− j 2 j 6 4⎦ ⎣ ⎣5− j 2 ⎦ V ⎡ 2 ⎤ ⎡ 22.13∠40.8°⎤ Solving, using MATLAB, gives  ⎢V ⎥ = ⎢ ⎥ V  ⎣ o ⎦ ⎣10.07∠30.4° ⎦

Apply KCL at node 2: 

24∠45° − V 2

10.12-5 The input to the circuit shown in Figure P10.12-5 is the voltage source voltage vs(t) = 12 cos(20,000t + 60°) V . and the output is the steady state voltage vo(t). Use MATLAB to plot the input and output  sinusoids.  

  Figure P10.12‐5  Solution:  Represent the circuit in the frequency domain: 

  Write a node equation: 

12∠60° − Vo 200, 000

+

12∠60° − Vo − j 25, 000

+

12∠60° = 0  20, 000

⎛ ⎛ ⎞ 1 1 1 ⎞ 1 1 Rearrange:  ⎜ + + + ⎟12∠60° = ⎜ ⎟ Vo   ⎝ 200, 000 − j 25, 000 20, 000 ⎠ ⎝ 200, 000 − j 25, 000 ⎠   Modify the MATLAB script given in the textbook (and posted on the Student Companion Site for  Introduction to Electric Circuits):  %----------------------------------------------% Describe the input voltage source. %----------------------------------------------w = 20000;

A = 12; theta = (pi/180)*60; Vs = A*exp(j*theta); %-----------------------------------------------% Describe the impedances. %-----------------------------------------------R1=50e3; R2=200e3; ZC=-j*25e3; %-----------------------------------------------% Calculate the phasor corresponding to the % output voltage. %-----------------------------------------------Vo=(1/R2 + 1/ZC + 1/R1)*Vs/(1/R2 + 1/ZC); B = abs(Vo) phi = angle(Vo) %-----------------------------------------------% %-----------------------------------------------T = 2*pi/w; tf = 2*T; N = 100; dt = tf/N; t = 0 : dt : tf; %-----------------------------------------------% Plot the input and output voltages. %-----------------------------------------------for k = 1 : 101 vs(k) = A * cos(w * t(k) + theta); vo(k) = B * cos(w * t(k) + phi); end plot (t, vs, t, vo)

  to get the plot: 

 

10.12-6 The input to the circuit shown in Figure P10.12-6 is the voltage source voltage vs(t) = 3 cos(4000t + 30°) V . and the output is the steady state voltage vo(t). Use MATLAB to plot the input and output  sinusoids.  

  Figure P10.12‐6  Solution:  Represent the circuit in the frequency domain: 

  Recognize this circuit as an inverting amplifier to write:   

20 || − j 12.5 ⎛ 20 || − j 12.5 ⎞ Vo = ⎜ − (12∠60° )   ⎟ Vs = − 5 5 ⎝ ⎠

Modify the MATLAB script given in the textbook (and posted on the Student Companion Site for  Introduction to Electric Circuits):  %----------------------------------------------% Describe the input voltage source. %----------------------------------------------w = 4000; A = 3; theta = (pi/180)*30; Vs = A*exp(j*theta); %-----------------------------------------------% Describe the impedances. %-----------------------------------------------R1=5e3; R2=20e3; ZC=-j*12.5e3;

Zp=R2*ZC/(R2+ZC); %-----------------------------------------------% Calculate the phasor corresponding to the % output voltage. %-----------------------------------------------Vo=(-Zp/R1)*Vs; B = abs(Vo) phi = angle(Vo) %-----------------------------------------------% %-----------------------------------------------T = 2*pi/w; tf = 2*T; N = 100; dt = tf/N; t = 0 : dt : tf; %-----------------------------------------------% Plot the input and output voltages. %-----------------------------------------------for k = 1 : 101 vs(k) = A * cos(w * t(k) + theta); vo(k) = B * cos(w * t(k) + phi); end plot (t, vs, t, vo)

  to get the plot: 

 

 

Section 10.14 How Can We Check…?  P 10.14-1

Computer analysis of the circuit in Figure P 10.14-1 indicates that the values of the

node voltages are V1 = 20 ∠–90° and V2 = 44.7 ∠–63.4°. Are the values correct? Hint: Calculate the current in each circuit element using the values of V1 and V2. Check to see whether KCL is satisfied at each node of the circuit. j10 Ω

V1

10 Ω

2A

V2

10 Ω

3Ix

Ix

Figure P 10.14-1 Solution: Generally, it is more convenient to divide complex numbers in polar form. Sometimes, as in this case, it is more convenient to do the division in rectangular form. Express V1 and V2 as: V1 = − j 20 and V 2 = 20 − j 40

KCL at node 1: 2−

V1 10



V1 − V 2 j 10

= 2−

− j 20 − j 20 − ( 20 − j 40 ) − = 2+ j2−2− j2 = 0 10 j 10

KCL at node 2: V1 − V 2 j 10



⎛ V1 ⎞ − j 20 − ( 20 − j 40 ) 20 − j 40 ⎛ − j 20 ⎞ + 3⎜ ⎟ = − + 3⎜ ⎟ = ( 2 + j 2) − ( 2 − j4) − j 6 = 0 j 10 10 10 ⎝ 10 ⎠ ⎝ 10 ⎠

V2

The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are correct.

P 10.14-2 Computer analysis of the circuit in Figure P 10.14-2 indicates that the mesh currents are i1(t) = 0.39 cos (5t + 39°) A and i2(t) = 0.28 cos (5t + 180°) A.

i1



2H 5 sin 5t V

Is this analysis correct?

+–

Hint: Represent the circuit in the frequency domain using impedances and phasors. Calculate the voltage across each circuit element using the values of I1 and I2. Check to see whether KVL is satisfied for each mesh of the circuit.

1 12

F

i2

4H

Figure P 10.14-2 Solution:

I 1 = 0.390 ∠ 39° and I 2 = 0.284 ∠ 180° Generally, it is more convenient to multiply complex numbers in polar form. Sometimes, as in this case, it is more convenient to do the multiplication in rectangular form. Express I1 and I2 as: I 1 = 0.305 + j 0.244 and I 2 = −0.284 KVL for mesh 1: 8 ( 0.305 + j 0.244 ) + j 10 ( 0.305 + j 0.244 ) − (− j 5) = j 10 ≠ 0

Since KVL is not satisfied for mesh 1, the mesh currents are not correct. Here is a MATLAB file for this problem: Vs = -j*5; Z1 = 8; Z2 = j*10; Z3 = -j*2.4; Z4 = j*20; % Mesh equations in matrix form Z = [ Z1+Z2 0; 0 Z3+Z4 ]; V = [ Vs; -Vs ]; I = Z\V abs(I) angle(I)*180/3.14159 % Verify solution by obtaining the algebraic sum of voltages for % each mesh. KVL requires that both M1 and M2 be zero. M1 = -Vs + Z1*I(1) +Z2*I(1) M2 = Vs + Z3*I(2) + Z4*I(2)

P 10.14-3 Computer analysis of the circuit in Figure P 10.14-3 indicates that the values of the node voltages are v1(t) = 19.2 cos (3t + 68°) V

v1



2H

and 4 cos (3t + 15°) A

v2(t) = 2.4 cos (3t + 105°) V. Is this analysis correct?

1 12

F 4H

Hint: Represent the circuit in the frequency domain using impedances and phasors. Calculate the current in each circuit element using the values of V1 and V2. Check to see whether KCL is satisfied at each node of the circuit.

v2

Figure P 10.14-3

Solution: V1 = 19.2 ∠ 68° and V 2 = 24 ∠ 105° V KCL at node 1 : 19.2 ∠ 68° 19.2 ∠ 68° + − 4∠15 = 0 8 j6 KCL at node 2: 24 ∠105° 24 ∠105° + + 4∠15 = 0 j12 − j4 The currents calculated from V1 and V2 satisfy KCL at both nodes, so it is very likely that the V1 and V2 are correct. Here is a MATLAB file for this problem: Is = 4*exp(j*15*3.14159/180); Z1 = 8; Z2 = j*6; Z3 = -j*4; Z4 = j*12; Y = [ 1/Z1 + 1/Z2 0; 0 1/Z3 + 1/Z4 ]; I = [ Is; -Is ]; V = Y\I abs(V) angle(V)*180/3.14159 % Verify solution by obtaining the algebraic sum of currents for % each node. KCL requires that both M1 and M2 be zero. M1 = -Is + V(1)/Z1 + V(1)/Z2 M2 = Is + V(2)/Z3 + V(2)/Z4

P 10.14-4 A computer program reports that the currents of the circuit of Figure P 10.14-4 are I = 0.2 ∠53.1° A, I1 = 632 ∠–18.4° mA, and I2 = 190 ∠71.6° mA. Verify this result. j500 Ω I 100 0° V

+ –

3000 Ω

–j1000 Ω

I1

I2

Figure P 10.14-4 Solution: First, replace the parallel resistor and capacitor by an equivalent impedance ZP =

(3000)(− j 1000) = 949 ∠ − 72° = 300 − j 900 Ω 3000− j 1000

The current is given by

VS 100 ∠0° = = 0.2∠53° A I= j 500+ Z P j 500+ 300− j 900 Current division yields ⎛ − j 1000 ⎞ I1 = ⎜ ⎟ ( 0.2 ∠53° ) = 63.3 ∠ − 18.5° mA ⎝ 3000 − j 1000 ⎠ ⎛ ⎞ 3000 I2 = ⎜ ⎟ ( 0.2 ∠53° ) = 190∠71.4° mA ⎝ 3000 − j 1000 ⎠

The reported value of I1 is off by an order of magnitude.

P 10.14-5 The circuit shown in Figure P 10.14-5 was built using a 2 percent resistor having a nominal resistance of 500 Ω and a 10 percent capacitor with a nominal capacitance of 5 μF. The steady-state capacitor voltage was measured to be v(t) = 18.3 cos (200t – 24°) V The voltage source represents a “signal generator.” Suppose that the signal generator was adjusted so carefully that errors in the amplitude, frequency, and angle of the voltage source voltage are all negligible. Is the measured response explained by the component tolerances? That is, could the measured v(t) have been produced by this circuit with a resistance R that is within 2 percent of 500 Ω and a capacitance C that is within 5 percent of 5 μF? R + 20 cos (200t) V

+ –

C

v(t) –

Figure P 10.14-5 Solution: Represent the circuit in the frequency domain using phasors and impedances. Use voltage division to get 1 j 200C 18.3∠ − 24° = × 20∠0° 1 R+ j 200C 1 1 0.915∠ − 24° = = ∠ − tan −1 ( 200CR ) So 2 1 + j 200CR 1 + ( 200CR ) Equating angles gives −24° = − tan −1 ( 200CR )



200CR = tan ( 24° ) = 0.4452

The nominal component values cause 200CR = 0.5. So we expect that the actual component values are smaller than the nominal values. Try

C = 5 (1 − 0.10 ) × 10−6 = 4.5 μ F

Then

R=

0.4452 = 494.67 Ω 200 × 4.5 × 10−6

500 − 494.67 = 0.01066 = 1.066% this resistance is within 2% of 500 Ω. We conclude 500 that the measured angle could have been caused by a capacitance that is within 10% of 5 μF and the resistance is within 2% of 500 Ω. Let’s check the amplitude. We require Since

1 1 + ( 0.4452 )

2

= 0.9136  0.915

So the measured amplitude could also have been caused by the given circuit with C = 4.5 μF and R = 494.67 Ω. We conclude that he measured capacitor voltage could indeed have been produced by the given circuit with a resistance that is within 2 % of 500 Ω and a capacitance that is within 10% of 5 μF.

PSpice Problems SP 10-1 The circuit shown in Figure SP 10.1 has two inputs, vs(t) and is(t), and one output, v(t). The inputs are given by vs(t) = 10 sin (6t + 45°) V and

is(t) = 2 sin (6t + 60°) A

1H

vs(t)

+ –



1 12

F

+ v(t) –

is(t)

Figure SP 10.1

Use PSpice to demonstrate superposition. Simulate three versions of the circuit simultaneously. (Draw the circuit in the PSpice workspace. “Cut and paste” to make two copies. Edit the part names in the copies to avoid duplicate names. For example, the resistor will be R1 in the original circuit. Change R1 to R2 and R3 in the two copies.) Use the given vs(t) and is(t) in the first version. Set is(t) = 0 in the second version and vs(t) = 0 in the third version. Plot the capacitor voltage, v(t), for all three versions of the circuit. Show that the capacitor voltage in the first version of the circuit is equal to the sum of the capacitor voltages in the second and third versions. Hint: Use PSpice parts VSIN and ISIN for the voltage and current source. PSpice uses hertz rather than rad/s as the unit for frequency. Remark: Notice that v(t) is sinusoidal and has the same frequency as vs(t) and is(t). Solution:

SP 10-2 The circuit shown in Figure SP 10.1 has two inputs, vs(t) and is(t), and one output, v(t). The inputs are given by vs(t) = 10 sin (6t + 45°) V and

is(t) = 2 sin (18t + 60°) A

1H

vs(t)

+ –



1 12

F

+ v(t) –

is(t)

Figure SP 10.1

Use PSpice to demonstrate superposition. Simulate three versions of the circuit simultaneously. (Draw the circuit in the PSpice workspace. “Cut and paste” to make two copies. Edit the part names in the copies to avoid duplicate names. For example, the resistor will be R1 in the original circuit. Change R1 to R2 and R3 in the two copies.) Use the given vs(t) and is(t) in the first version. Set is(t) = 0 in the second version and vs(t) = 0 in the third version. Plot the capacitor voltage, v(t), for all three versions of the circuit. Show that the capacitor voltage in the first version of the circuit is equal to the sum of the capacitor voltages in the second and third versions. Hint: Use PSpice parts VSIN and ISIN for the voltage and current source. PSpice uses hertz rather than rad/s as the unit for frequency. Remark: Notice that v(t) is not sinusoidal.

Solution:

SP 10-3 The circuit shown in Figure SP 10-1 has two inputs, vs(t) and is(t), and one output, v(t). The inputs are given by vs(t) = 10 sin (6t + 45°) V and

is(t) = 0.8 A

1H

vs(t)

+ –



1 12

F

+ v(t) –

is(t)

Figure SP 10.1

Use PSpice to demonstrate superposition. Simulate three versions of the circuit simultaneously. (Draw the circuit in the PSpice workspace. “Cut and paste” to make two copies. Edit the part names in the copies to avoid duplicate names. For example, the resistor will be R1 in the original circuit. Change R1 to R2 and R3 in the two copies.) Use the given vs(t) and is(t) in the first version. Set is(t) = 0 in the second version and vs(t) = 0 in the third version. Plot the capacitor voltage, v(t), for all three versions of the circuit. Show that the capacitor voltage in the first version of the circuit is equal to the sum of the capacitor voltages in the second and third versions. Hint: Use PSpice part VSIN and IDC for the voltage and current source. PSpice uses hertz rather than rad/s as the unit for frequency. Remark: Notice that v(t) looks sinusoidal, but it’s not sinusoidal because of the dc offset.

Solution:

SP 10-4

The circuit shown in Figure SP 10-

1H



1 has two inputs, vs(t) and is(t), and one output, v(t). When inputs are given by

vs(t)

+ –

1 12

F

vs(t) = Vm sin 6t V and

is(t) = Im A

+ v(t) –

is(t)

Figure SP 10.1

the output will be vo(t) = A sin (6t + θ) + B V Linearity requires that A be proportional to Vm and that B be proportional to Im. Consequently, we can write A = k1Vm and B = k2Im, where k1 and k2 are constants yet to be determined. (a) Use PSpice to determine the value of k1 by simulating the circuit using Vm = 1 V and Im = 0. (b) Use PSpice to determine the value of k2 by simulating the circuit using Vm = 0 V and Im = 1. (c) Knowing k1 and k2, specify the values of Vm and Im that are required to cause vo(t) = 5 sin (6t + θ) + 5 V Simulate the circuit using PSpice to verify the specified values of Vm and Im. Solution: The following simulation shows that k1 = 0.4and k2 = -3 V/A. The required values of Vm and Im are Vm = 12.5 V and Im = -1.667 A.

Design Problems DP 10-1

Design the circuit shown in Figure DP 10-1 to produce the specified output voltage

vo(t) = 8cos(1000t + 104°) V when provided with the input voltage vi(t) = 2.5cos(1000t) V.

Figure DP 10-1 Solution:

R2

1 jω C

=

R2 1 + jω CR 2

R2

R2

1 + jω CR 2 R1 Vo (ω ) =− =− Vi (ω ) R1 1 + jω CR 2 R2

R1 Vo (ω ) j (180 − tan −1 ω CR 2 ) e = 2 Vi (ω ) 1 + (ω CR 2 )

In this case the angle of

the magnitude of

Vo (ω ) tan (180° − 104° ) is specified to be 104° so CR 2 = = 0.004 and 1000 Vi (ω )

Vo (ω ) 8 is specified to be so Vi (ω ) 2.5

R2 R1 1 + 16

=

8 ⇒ 2.5

R2 R1

= 13.2 . One set of

values that satisfies these two equations is C = 0.2 μ F, R1 = 1515 Ω, R 2 = 20 kΩ .

DP 10-2

Design the circuit shown in Figure DP 10-2 to produce the specified output voltage

vo(t) = 2.5cos(1000t − 76°) V when provided with the input voltage vi(t) = 12cos(1000t) V.

Figure DP 10-2 Solution:

R2 Vo (ω ) = Vi (ω ) where K =

1 jω C

=

R2 1 + jω CR 2

R2 1 + jω CR 2 K = R2 1 + jω CR p R1 + 1 + jω CR 2

R1 R1 + R 2

and R p =

R1 R 2 R1 + R 2

Vo (ω ) K − j tan −1 ω CR p = e 2 Vi (ω ) 1 + (ω CR p ) In this case the angle of C Rp = C

R1 R 2 R1 + R 2

=−

Vo (ω ) is specified to be -76° so Vi (ω )

tan ( −76 ) V (ω ) 2.5 is specified to be = 0.004 and the magnitude of o so 1000 Vi (ω ) 12

R2 K 2.5 = ⇒ 0.859 = K = . One set of values that satisfies these two equations is R1 + R 2 1 + 16 12 C = 0.2 μ F, R1 = 23.3 kΩ, R 2 = 142 kΩ .

DP 10-3

Design the circuit shown in Figure DP 10-3 to produce the specified output voltage

vo(t) = 2.5cos(40t + 14°) V when provided with the input voltage vi(t) = 8cos(40t) V.

Figure DP 10-3 Solution: jω L R 2

L R 2 + jω L R1 Vo (ω ) = = jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L where R p =

R1 R 2 R1 + R 2

Vo (ω ) Vi (ω )

In this case the angle of



ω =

L R1

⎛ L ⎞ 1+ ⎜ω ⎜ R p ⎟⎟ ⎝ ⎠

2

e

⎛ L ⎞ j ⎜ 90 − tan −1 ω ⎟ ⎜ ⎟ R p ⎠ ⎝

Vo (ω ) is specified to be 14° so Vi (ω )

V (ω ) L L ( R1 + R 2 ) tan ( 90° − 14° ) 2.5 is specified to be so = = = 0.1 and the magnitude of o 40 Rp R1 R 2 Vi (ω ) 8 40

L R1

=

2.5 ⇒ 8

L = 0.0322 . One set of values that satisfies these two equations is R1

1 + 16 L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω .

DP 10-4 Show that it is not possible to design the circuit shown in Figure DP 10-4 to produce

the specified output voltage vo(t) = 2.5cos(40t − 14°) when provided with the input voltage vi(t) = 8cos(40t) V.

Figure DP 10-4 Solution:

jω L R 2

L R 2 + jω L R1 Vo (ω ) = = jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L where R p =

R1 R 2 R1 + R 2

Vo (ω ) Vi (ω )

In this case the angle of



ω =

L R1

⎛ L ⎞ 1+ ⎜ω ⎜ R p ⎟⎟ ⎝ ⎠

Vo (ω ) is specified to be −14°. This requires Vi (ω )

L L ( R1 + R 2 ) tan ( 90 + 14 ) = = = −0.1 Rp R1 R 2 40 This condition cannot be satisfied with positive element waves.

2

e

⎛ L ⎞ j ⎜ 90 − tan −1 ω ⎟ ⎜ R p ⎟⎠ ⎝

v

DP 10-5 A circuit with an unspecified R, L, and C is shown in Figure DP 10-5. The input source is

is = 10 cos 1000t A,

R is

10 Ω

C L

and the goal is to select the R, L, and C so that the node voltage is v = 80 cos 1000t V.

Figure DP 10-5

Solution: Z1 =10 Ω 1 Z2 = jω C Z3 = R + jω L

1 S 10 Y2 = jω C 1 Y3 = R + jω L Y1 =

v(t ) = 80 cos (1000 t − θ ) V ⇒ V = 80∠ − θ V iS (t ) = 10 cos 1000 t A ⇒ I s =10∠0° A

try θ = 0° . Then ⎡1 ⎤ 1 + jωC ⎥ = 10∠0° ⇒ R + 10 − 10 ω 2 LC + j (ω L + 10 ω RC ) = 1.25 R + j1.25 ω L (80∠−∞ ) ⎢ + ⎣10 R + jω L ⎦ 2 Equate real part: 40 − 40ω LC = R where ω = 1000 rad sec Equate imaginary part: 40 RC = L Solving yields R = 40(1− 4×107 RC 2 )

Now try R = 20 Ω ⇒ 1 − 2(1 − 4 × 107 (20)C 2 ) which yields C = 2.5×10−5 F= 25 μ F so L = 40 RC = 0.02 H=20 mH Now check the angle of the voltage. First Y1 = 1/10 = 0.1 S Y2 = j 0.25 S Y3 = 1/(20+ j 20) = .025− j.025 S then Y = Y1 + Y2 + Y3 = 0.125 , so V =YI s = (0.125∠0°)(10∠0°) = 1.25∠0° V So the angle of the voltage is θ =0° , which satisfies the specifications.

R

DP 10-6 The input to the circuit shown in Figure DP 10-6 is the voltage source voltage

+ + –

vs(t) = 10 cos (1000t) V

vs(t)

C

vo(t) –

The output is the steady-state capacitor voltage vo(t) = A cos (1000t + θ) V

Figure DP 10-6

(a)

Specify values for R and C such that θ = –30°. Determine the resulting value of A.

(b)

Specify values for R and C such that A = 5 V. Determine the resulting values of θ.

(c)

Is it possible to specify values for R and C such that A = 4 and θ = –60°? (If not, justify your answer. If so, specify R and C.)

(d)

Is it possible to specify values of R and C such that A = 7.07 V and θ = –45°? (If not, justify your answer. If so, specify R and C.)

Solution: Represent the circuit in the frequency domain using phasors and impedances. Using voltage division gives 1 10 j1000C ×10∠0° = A∠θ = 1 1 + j103 RC R+ j1000C Equating magnitudes and angles gives 2

A=

10



1 + 106 R 2C 2

⎛ 10 ⎞ ⎜ ⎟ −1 ⎝ A⎠ RC = 1 + j103 RC

and

θ = − tan −1 (103 RC )

(a) Pick C = 1 μF, then R =

θ = −30°





RC =

RC =

tan ( 30° ) 0.577 = . 103 103

0.577 = 577 Ω and A = 8.66 V. 106 ×103 2

(b)

tan ( −θ ) 103

A=5 V



⎛ 10 ⎞ ⎜ ⎟ −1 3 ⎝5⎠ RC = = 3. 3 10 10

Pick C = 1 μF, then R =

3 = 1732 Ω and θ = −60° . 10 × 103 −6

2

(c)

A=4

θ = −60°

⎛ 10 ⎞ ⎜ ⎟ −1 2.29 ⎝ 4⎠ RC = = 3 3 10 10





RC =

tan ( 60° ) 1.73 = 3 103 10

Since RC cannot be both 0.00229 and 0.00173 simultaneously, the specifications cannot be satisfied using this circuit.

2

(d)

A = 7.07

θ = −45°





⎛ 10 ⎞ ⎜ ⎟ −1 ⎝ 7.07 ⎠ RC = = 10−3 3 10 RC =

tan ( 45° ) = 10−3 103

Both specifications can be satisfied by taking R = 1000 Ω and C = 1 μF.

Chapter 11: AC Steady State Power Exercises Exercise 11.3-1 Determine the instantaneous power delivered to an element and sketch p(t) when the element is (a) a resistance R and (b) an inductor L. The voltage across the element is v(t) = Vm cos (ωt + θ) V. V2 V2 Answer: (a) PR = m [1 + cos (2ωt + 2θ)] W (b) PL = m cos (2ωt + 2θ – 90°) W 2R 2ω L Solution: (a) When the element is a resistor, the current has the same phase angle as the voltage:

i (t ) =

v(t ) Vm cos (ω t + θ ) A = R R

The instantaneous power delivered to the resistor is given by 2 2 2 V V Vm V pR (t ) = v(t ) ⋅ i (t ) = Vm cos (ω t + θ ) ⋅ cos (ω t + θ ) = m cos 2 (ω t + θ ) = m + m cos (2ω t + θ ) R R 2R 2R

(b) When the element is an inductor, the current will lag the voltage by 90°. Z L = jω L = ω L∠90° Ω



I=

V V ∠θ V = m = m ∠ (θ −90° ) Z ω L∠90° ω L

The instantaneous power delivered to the inductor is given by 2

V V pL (t ) = i (t ) ⋅ v(t ) = m cos (ω t + θ − 90° ) ⋅Vm cos (ω t + θ ) = m cos ( 2ω t + 2θ −90° ) W ωL 2ω L

Exercise 11.4-1 Find the effective value of the following currents: (a) cos 3t + cos 3t; (b) sin 3t + cos(3t + 60°); (c) 2 cos 3t + 3 cos 5t Answer: (a) 2 (b) 0.366 (c) 2.55

Ex. 11.4-1 (a) I 2 i (t ) = 2 cos 3 t A ⇒ I eff max = = 2A 2 2 (b) i (t ) = cos (3 t − 90° ) + cos (3 t + 60° ) A 1 3 +j = 0.518∠ − 15° A 2 2 0.518 i (t ) = 0.518 cos (3t − 15°) A ⇒ I eff = = 0.366 A 2 I = (1∠−90° ) + (1∠60° ) = − j +

(c)

2

I eff

2

⎛ 2 ⎞ ⎛ 3 ⎞ =⎜ ⎟ +⎜ ⎟ ⎝ 2⎠ ⎝ 2⎠

2

⇒ I eff = 2.55 A

Exercise 11.5-1 Determine the average power delivered to each element of the circuit shown in Figure E 11.5-1. Verify that average power is conserved. Answer: 4.39 + 0 = 4.39 W

Figure E 11.5-1 Solution:

Analysis using Mathcad (ex11_5_1.mcd): A := 12

Enter the parameters of the voltage source: Enter the values of R and L

R := 10

L := 4

The impedance seen by the voltage source is: The mesh current is: I :=

ω := 2

Z := R + j ⋅ω ⋅L

A Z

⎯ I ⋅( I ⋅Z) The complex power delivered by the source is: Sv := Sv = 4.39 + 3.512i 2 ⎯ I ⋅( I ⋅R) The complex power delivered to the resistor is: Sr := Sr = 4.39 2 ⎯ I ⋅( I ⋅j ⋅ω ⋅L) The complex power delivered to the inductor is: Sl := Sl = 3.512i 2 Verify Sv = Sr + Sl :

Sr + Sl = 4.39 + 3.512i

Sv = 4.39 + 3.512i

Exercise 11.5-2 Determine the complex power delivered to each element of the circuit shown in Figure E 11.5-2. Verify that complex power is conserved. Answer: 6.606 + j5.248 – j3.303 + 6.606 = j1.982 VA

Figure E 11.5-2 Solution:

Analysis using Mathcad (ex11_5_2.mcd): Enter the parameters of the voltage source: A := 12 R := 10

Enter the values of R, L an dC

ω := 2

L := 4

C := 0.1

The impedance seen by the voltage source is: Z := R + j ⋅ω ⋅L + The mesh current is:

I :=

A Z

The complex power delivered by the sourceis:

Sv :=

1 j ⋅ω ⋅C

⎯ I ⋅( I ⋅Z) 2

⎯ I ⋅( I ⋅R) The complex power delivered to the resistor is: Sr := 2 ⎯ I ⋅( I ⋅j ⋅ω ⋅L) The complex power delivered to the inductor is: Sl := 2 ⎯⎛ 1 ⎞ I ⋅⎜ I ⋅ ⎟ j ⋅ω ⋅C ⎠ ⎝ The complex power delivered to the capacitor is: Sc := 2 Verify Sv = Sr + Sl + Sc :

Sr + Sl + Sc = 6.606 + 1.982i

Sv = 6.606 + 1.982i Sr = 6.606 Sl = 5.284i

Sc = −3.303i

Sv = 6.606 + 1.982i

Exercise 11.6-1 A circuit has a large motor connected to the ac power lines [ω = (2π)60 = 377 rad/s]. The model of the motor is a resistor of 100 ω in series with an inductor of 5 H. Find the power factor of the motor. Answer: pf = 0.053 lagging

Solution:

( )

(377) (5) ⎤ ⎡ pf = cos (∠ Z) = cos ⎡ tan −1 ω L ⎤ = cos ⎢ tan −1 = 0.053 lagging R ⎥⎦ 100 ⎥⎦ ⎣⎢ ⎣

Exercise 11.6-2 A circuit has a load impedance Z = 50 + j80 Ω, as shown in Figure 11.6-5. Determine the power factor of the uncorrected circuit. Determine the impedance ZC required to obtain a corrected power factor of 1.0. Answer : ZC = –j111.25 Ω

Figure 11.6-5 Solution:

( )

( )

pf = cos (∠ Z) = cos ⎡ tan −1 X ⎤ = cos ⎡ tan −1 80 ⎤ = 0.53 lagging ⎢⎣ ⎢⎣ 50 ⎥⎦ R ⎥⎦

(50)2 + (80)2 XC = = −111.25 Ω ⇒ ZC = − j 111.25 Ω 50 tan (cos −1 1) −80

Exercise 11.6-3 Determine the power factor for the total plant of Example 11.6-1 when the resistive heating load is decreased to 30 kW. The motor load and the supply voltage remain as described in Example 11.6-1. Answer: pf = 0.915

Figure from Example 11.6-1

Solution: PT = 30 + 86 = 116 kW and QT = 51 kVAR S T = PT + j QT = 116 + j 51 = 126.7 ∠23.7° kVA pf plant = cos 23.7° = 0.915

Exercise 11.6-4 A 4-kW, 110-Vrms load, as shown in Figure 11.6-5, has a power factor of 0.82 lagging. Find the value of the parallel capacitor that will correct the power factor to 0.95 lagging when ω = 377 rad/s. Answer: C = 0.324 mF

Figure 11.6-5 Solution: P = V rms I rms cos θ Z=

V rms I rms

⇒ I rms =

P 4000 = = 44.3 A V rms cos θ (110)(.82)

∠ cos −1 (0.82) = 2.48 ∠34.9° = 2.03+ j 1.42 = R + j X

To correct power factor to 0.95 requires (2.03) 2 + (1.42) 2 R2 + X 2 = = − 8.16 Ω (2.03) tan (18.19° ) −1.42 R tan (cos −1 pfc) − X −1 C= =325 μ F ω X1 X1 =

Exercise 11.7-1 Determine the average power absorbed by the resistor in Figure 11.7-2a for these two cases: (a) vA(t) = 12 cos 3t V and vB(t) = 4 cos 3t V; (b) vA(t) = 12 cos 4t V and vB(t) = 4 cos 3t V Answer: (a) 2.66 W (b) 4.99 W

Figure 11.7-2a Solution: (a) I = I1 + I 2 = (1.414∠ − 45° ) + ( 0.4714∠135° ) = 0.9428∠ − 45° A

0.94282 ( 6 ) = 2.66 W 2 (b) 1.22 I1 = 1.2∠53° A ⇒ p1 = ( 6 ) = 4.32 W 2 0.47142 I 2 = 0.4714∠135° A ⇒ p2 = ( 6 ) = 0.666 W 2 ∴ p = p1 + p2 = 4.99 W ⇒

p=

Exercise 11.8-1 For the circuit of Figure 11.8-1, find ZL to obtain the maximum power transferred when the Thévenin equivalent circuit has Vt = 100 ∠0°V and Zt = 10 + j14 Ω. Also determine the maximum power transferred to the load. Answer: ZL = 10–j14 Ω and P = 125 W

Figure 11.8-1

Solution: For maximum power transfer Z L = Z*t = 10 − j14 Ω 100 I= =5 A (10+ j14) + (10− j14) 2

⎛ 5 ⎞ PL = ⎜ ⎟ Re {10− j14} = 125 W ⎝ 2⎠

Exercise 11.8-2 A television receiver uses a cable to connect the antenna to the TV, as shown in Figure E 11.8-2, with vs = 4 cos ωt mV. The TV station is received at 52 MHz. Determine the average power delivered to each TV set if (a) the load impedance is Z = 300 Ω; (b) two identical TV sets are connected in parallel with Z = 300 Ω for each set; (c) two identical sets are connected in parallel and Z is to be selected so that maximum power is delivered at each set. Answer: (a) 9.6 nW (b) 4.9 nW (c) 5 nW

Figure E 11.8-2 Solution: If the station transmits a signal at 52 MHz then ω = 2π f = 104π ×106 rad/sec

so the received signal is

vs (t ) = 4 cos (104π × 106 t ) mV (a) If the receiver has an input impedance of Zin =300 Ω then

Zin 300 1 ⎛ 1 ⎞ ( 2.4 ×10 Vin = Vs = × 4 × 10−3 = 2.4 mV ⇒ P = Vin2 ⎜ ⎟= R + Zin 200+300 2 ⎝ Zin ⎠ 2(300)

)

−3 2

(b) If two receivers are connected in parallel then Z in = 300||300 = 150 Ω and Vin =

total P =

Zin 150 VS = (4 ×10−3 ) = 1.71× 10−3 V R + Zin 200+150

2 Vin ⎛ 1 ⎞ (1.71×10−3 ) 2 = 9.7 nW or 4.85 nW to each set ⎜ ⎟= 2 ⎝ Zin ⎠ 2(150)

(c) In this case, we need Zin = R || R = 200 Ω ⇒ R = 400 Ω , where R is the input impedance of each television receiver. Then Ptotal =

Vin 2 (2×10−3 ) 2 = = 10 nW ⇒ 5 nW to each set 2 Zin 2(200)

= 9.6 nW

Exercise 11.9-1 Determine the voltage vo for the circuit of Figure E 11.9-1. Hint: Write a single mesh equation. The currents in the two coils are equal to each other and equal to the mesh current. Answer: vo = 14 cos 4t V

Figure E 11.9-1 Solution: Coil voltages: V1 = j 24 I 1 + j 16 I 2 = j 40 I

V 2 = j 16 I 1 + j 40 I 2 = j 56 I Mesh equation: 24 = V1 + V 2 = j 40 I + j 56 I = j 96 I 24 1 =−j 4 j 96 ⎛ 1⎞ Vo = V2 = ( j 56 ) ⎜ − j ⎟ = 14 ⎝ 4⎠ vo = 14 cos 4t V

I=

Exercise 11.9-2 Determine the voltage vo for the circuit of Figure E 11.9-2. Hint: This exercise is the same as Exercise 11.9-1, except for the position of the dot on the vertical coil. Answer: vo = 18 cos 4t V

Figure E 11.9-2 Solution: Coil voltages: V1 = j 24 I 1 − j 16 I 2 = j 8 I

V 2 = − j 16 I 1 + j 40 I 2 = j 24 I Mesh equation: 24 = V1 + V 2 = j 8 I + j 24 I = j 32 I 24 3 =−j 4 j 32 ⎛ 3⎞ Vo = V 2 = ( j 24 ) ⎜ − j ⎟ = 18 ⎝ 4⎠ vo = 18 cos 4t V

I=

Exercise 11.9-3 Determine the current io for the circuit of Figure E 11.9-3. Hint: The voltage across the vertical coil is zero because of the short circuit. The voltage across the horizontal coil induces a current in the vertical coil. Consequently, the current in the vertical coil is not zero. Answer: io = 1.909 cos (4t – 90°) A

Solution:

Figure E 11.9-3 0 = V 2 = j 16 I 1 + j 40 I 2 40 I 2 = −2.5 I 2 16 V s = V1 = j 24 I 1 + j 16 I 2 ⇒ I1 = −

= j (24(−2.5) + 16) I 2 = − j 44 I 2 24 6 = j 11 − j 44 I o = I 1 − I 2 = (−2.5 − 1) I 2 I2 =

= −3.5 I 2 ⎛ 6⎞ = −3.5 ⎜ j ⎟ = − j 1.909 ⎝ 11 ⎠ io = 1.909 cos ( 4t - 90° ) A

Exercise 11.9-4 Determine the current io for the circuit of Figure E 11.9-4. Hint: This exercise is the same as Exercise 11.9-3, except for the position of the dot on the vertical coil. Answer: io = 0.818 cos (4t – 90°) A

Figure E 11.9-4 Solution:

0 = V 2 = − j 16 I 1 + j 40 I 2 40 I 2 = 2.5 I 2 16 V s = V1 = j 24 I 1 − j 16 I 2 ⇒ I1 =

= j (24(2.5) − 16) I 2 = j 44 I 2 24 6 =−j j 44 11 I o = I 1 − I 2 = (2.5 − 1) I 2 I2 =

= 1.5 I 2 6⎞ ⎛ = 1.5 ⎜ − j ⎟ = − j 0.818 ⎝ 11 ⎠ io = 0.818 cos ( 4t - 90° ) A

Exercise 11.10-1 Determine the impedance Zab for the circuit of Figure E 11.10-1. All the transformers are ideal. Answer: Zab = 4.063Z

Figure E 11.10-1

Ex. 11.10-1

Z1 =

1 ⎛ Z⎞ Z⎞ 1 Z Z ⎛ = , = = 9 ⎜ Z + ⎟ and Z 3 = 2 (Z + Z 2 ) Z Z + ⎜ 2 2 2 ⎜ 2 ⎟ ⎟ n3 4 n1 n2 ⎝ n3 ⎠ 4⎠ ⎝

then 1⎛ Z ⎞⎞ ⎛ Z ab = Zin = Z + Z3 = Z + ⎜ Z +9⎜ Z+ ⎟ ⎟ = 4.0625 Z 4⎝ 4 ⎠⎠ ⎝

PROBLEMS Section 11.3 Instantaneous Power and Average Power P 11.3-1 An RLC circuit is shown in Figure P 11.3-1. Find the instantaneous power delivered to the inductor when is = 1 cos ωt A and ω = 6283 rad/s.

Figure P 11.3-1 Solution: 1∠0° =

V V V + + ⇒ V = 14.6∠ − 43° V 20 j 63 − j16

V = 0.23∠ − 133° A j 63 p(t ) = i (t )v(t ) = 0.23cos (2π ⋅103 t − 133° ) × 14.6 cos (2π ⋅103 t − 43° ) I=

= 3.36 cos (2π ⋅103 t −133° ) cos (2π ⋅103 t − 43° ) = 1.68 (cos (90° ) + cos (4π ⋅103 t −176° )) =1.68 cos (4π ⋅103 t −176° )

P 11.3-2 Find the average power absorbed by the 0.6-kΩ resistor and the average power supplied by the current source for the circuit of Figure P 11.3-2.

Figure P 11.3-2 Solution: Current division: ⎡ 1800 − j 2400 ⎤ I=4 5⎢ ⎥ ⎣1800− j 2400+ 600 ⎦ =5

5 ∠ − 8.1° mA 2

2

P600Ω

I 600 ⎛5⎞ = = 300(25) ⎜ ⎟ = 1.875 ×104 μW = 18.75 mW 2 ⎝ 2⎠

Psource =

⎛ 5⎞ V I cos θ 1 = (600) ⎜ 5 ⎟ 4 5 cos(−8.1° ) = 2.1× 104 μW = 21 mW 2 2 ⎝ 2⎠

(

)

P 11.3-3 Use nodal analysis to find the average power absorbed by the 20-Ω resistor in the circuit of Figure P 11.3-3. Answer: P = 200 W

Figure P 11.3-3

Solution:

Node equations:

20I X −100 20I X − V + IX + = 0 ⇒ I X (20 − j15) − V = − j 50 10 − j5 V − 20I X V − 3I X + = 0 ⇒ I X (−40 + j 30) + V (2 − j ) = 0 10 − j5

Solving the node equations using Cramer’s rule yields IX =

j 50(2 − j ) 50 5∠63.4° = = 2 5∠10.3° A (40 − j 30) − (20 − j15)(2 − j ) 25∠53.1°

Then 2

PAVE

(

I = X (20) = 10 2 5 2

)

2

= 200 W

P 11.3-4 Nuclear power stations have become very complex to operate, as illustrated by the training simulator for the operating room of the Pilgrim Power Station shown in Figure P 11.34a. One control circuit has the model shown in Figure P 11.3-4b. Find the average power delivered to each element. Answer: Psource current = −12.8 W P8 Ω = 6.4 W PL = 0 W Pvoltage source = 6.4 W

P 11.3-4 Solution: A node equation: (V − 16) V + − (2 2∠45°) = 0 j4 8 ⎛ 2⎞ ⇒ V = ⎜⎜16 ⎟ ∠18.4° V 5 ⎟⎠ ⎝

Then I=

16 − V = 3.2 ∠ − 116.6° A j4

2

PAVE 8Ω

PAVE current source = −

1 V = × 2 8

2

⎛ 2⎞ ⎜16 ⎟ 5⎠ 1 ⎝ = × = 6.4 W absorbed 2 8

1 1⎛ 2⎞ V 2 2 cos θ = − ⎜ 16 ⎟ 2 2 cos ( 26.6° ) = −12.8 W absorbed 2 2⎝ 5⎠

(

)

(

)

PAVE inductor = 0

1 1 PAVE voltage source = − (16) I cosθ = − (16)( 3.2)cos( − 116.6°) = 6.4 W absorbed 2 2

P 11.3-5 Find the average power delivered to each element for the circuit of Figure P 11.3-5.

Figure P 11.3-5 Solution:

A node equation: −20 +

V1 V1 + (3 / 2)V1 + = 0 ⇒ V1 = 50 5 ∠ − 26.6° V 10 15 − j 20

Then I=

V1 + ( 3/2 ) V1 ( 5 / 2 ) V1 = 5 5 ∠26.6° A = 15 − j 20 25∠ − 53.1°

Now the various powers can be calculated: 2

2

PAVE 10Ω PAVE current source = −

1 1 V (20) cosθ = − (50 5)(20) cos (−26.6°) = −1000 W absorbed 2 2

PAVE 15Ω

PAVE voltage source = −

1 V1 1 (50 5) = = = 625 W absorbed 2 10 2 10

(

)

(

)(

2 5 5 I = (15 ) = − 2 2

2

(15 ) = 937.5 W absorbed

)

1 3 1 I V1 cos θ = − 5 5 75 5 cos ( −53.1° ) = −562.5 W absorbed 2 2 2 PAVE capacitor = 0 W

P 11.3-6 A student experimenter in the laboratory encounters all types of electrical equipment. Some pieces of test equipment are battery-operated or operate at low voltage so that any hazard is minimal. Other types of equipment are isolated from electrical ground so that there is no problem if a grounded object makes contact with the circuit. Some types of test equipment, however, are supplied by voltages that can be hazardous or have dangerous voltage outputs. The standard power supply used in the United States for power and lighting in laboratories is the 120, grounded, 60Hz sinusoidal supply. This supply provides power for much of the laboratory equipment, so an understanding of its operation is essential in its safe use (Bernstein, Figure P 11.3-6 1991). Consider the case where the experimenter has one hand on a piece of electrical equipment and the other hand on a ground connection, as shown in the circuit diagram of Figure P 11.3-6a. The hand-to-hand resistance is 200 Ω. Shocks with an energy of 30 J are hazardous to humans. Consider the model shown in Figure P 11.3-6b, which represents the human with R. Determine the energy delivered to the human in 1 s. Solution: Z=

I =

200 ( j 200 ) 200 ∠90° 200 = = ∠45° Ω 200 (1 + j ) 2 ∠45° 2

⎛ 120∠0D j 200 ⎞ = 0.85 ∠ − 45° A, I R = ⎜ ⎟ I = 0.6∠0° A 200 200 + j 200 D ⎝ ⎠ ∠45 2

P = I R = ( 0.6 ) ( 200 ) = 72 W and w = ( 72 )(1) = 72 J 2

2

P 11.3-7 An RLC circuit is shown in Figure P 11.3-7 with a voltage source vs = 7 cos 10t V. (a) Determine the instantaneous power delivered to the circuit by the voltage source. (b) Find the instantaneous power delivered to the inductor. Answer: (a) p = 7.54 + 15.2 cos (20t – 60.3°) W (b) p = 28.3 cos (20t – 30.6°) W

Figure P 11.3-7 Solution: Z = j3 +

4(− j 2) = 0.8 + j1.4 4− j 2 = 1.6 ∠60.3° Ω

∴I =

7∠0° V = = 4.38 ∠ − 60.3° A Z 1.6∠60.3°

i (t ) = 4.38cos (10 t − 60.3° ) A The instantaneous power delivered by the source is given by (7)(4.38) [cos (60.3°) + cos (20 t −60.3°)] 2 = 7.6 + 15.3cos (20 t − 60.3°) W

p(t ) = v(t ) ⋅ i (t ) = (7 cos 10 t )(4.38cos (10 t − 60.3°)) = The inductor voltage is calculated as

VL = I ⋅ Z L = (4.38 ∠ − 60.3°)(j 3) = 13.12 ∠29.69° V

vL (t ) = 13.12 cos (10 t + 29.69°) V The instantaneous power delivered to the inductor is given by pL (t ) = vL (t ) ⋅ i (t ) = ⎡⎣ (13.12 cos (10t + 29.69°)(4.38cos (l0t − 60.3° ) ⎤⎦ 57.47 [ cos (29.69°+ 60.3°)+ cos (20 t + 29.69°−60.3°)] 2 = 28.7 cos (20t − 30.6°) W =

P 11.3-8 (a) Find the average power delivered by the source to the circuit shown in Figure P 11.3-8. (b) Find the power absorbed by resistor R1. Answer: (a) 30 W (b) 20 W

Figure P 11.3-8 Solution: The equivalent impedance of the parallel resistor (1)( j ) = 1 1+ j Ω . Then and inductor is Z = ( ) 1+ j 2 I=

(a) Psource =

I V cos θ = 2

10∠0° 20 20 = = ∠ − 18.4° A 1 j 3 + 10 1+ (1+ j ) 2

⎛ 20 ⎞ ⎟ ⎝ 10 ⎠ cos −18.4° = 30.0 W ( ) 2

(10 ) ⎜ 2

⎛ 20 ⎞ 2 (1) I R1 ⎜⎝ 10 ⎟⎠ (b) PR 1 = = = 20 W 2 2

Section 11.4 Effective Value of a Periodic Waveform P 11.4-1 Find the rms value of the current i for (a) i = 2 – 4 cos 2t A, (b) i = 3 sin π t + π t A, and (c)i = 2 cos 2t + 4 2 cos (2t + 45°) + 12 sin 2t A. Answer: (a) 2 3 (b) 2.35 A (c) 5 2 A Solution:

( Treat i as two sources of different frequencies.)

i = 2 − 4 cos 2t = i1 + i 2

(a)

2A source: I eff = lim

T →∞

1 T



T

o

2

(2) dt = 2 A

and 4 cos 2t source:

Ieff =

4 A 2

The total is calculated as 2

I eff

(b)

2

⎛ 4 ⎞ = ( 2) + ⎜ ⎟ = 12 A ⇒ I rms = I eff = 12 = 2 3 A ⎝ 2⎠ 2

i ( t ) = 3cos (π t − 90° ) + 2 cos π t ⇒ I = ( 3∠ − 90° ) +

(

2 ∠0°

)

= 2 − j 3 = 3.32∠ − 64.8° A I rms = (c)

3.32 = 2.35 A 2

i ( t ) = 2 cos 2t + 4 2 cos ( 2t + 45° ) + 12 cos ( 2 t − 90° )

(

)

I = ( 2∠0° ) + 4 2 ∠45° + (12∠ − 90° ) = ( 2 + 4 ) + ( j 4 − j12 ) =10∠ − 53.1° A I rms =

10 =5 2 A 2

2 cos

P 11.4-2 Determine the rms value for each of the waveforms shown in Figure P 11.4-2. Answer: (a) 4.10 V (b) 4.81 V (c) 4.10 v (V) 6

v (V)

v (V)

6

6

2

2

2 2

5

7

10 t (s)

2

(a)

5

7

10 t (s)

3

5

8

10 t (s)

(c)

(b)

Figure P 11.4-2

Solution: (a)

(b)

(c)

)

1 5

(∫

Vrms =

1 5

( ∫ 2 dt +∫ 6 dt ) = 15 ( ∫ 4 dt +∫ 36 dt ) =

1 116 = 4.81 V (8 + 108) = 5 5

Vrms =

1 5

( ∫ 2 dt +∫ 6 dt ) = 15 ( ∫ 4 dt +∫ 36 dt ) =

1 84 = 4.10 V (12 + 72 ) = 5 5

2 0

2

5

62 dt + ∫ 22 dt = 2

2

0

3

0

5

2

2

2

5

3

2

1 5

(∫

)

Vrms =

2 0

5

36 dt + ∫ 4 dt = 2

2

5

0

2

3

5

0

3

1 84 = 4.10 V ( 72 + 12 ) = 5 5

P 11.4-3 Determine the rms value for each of the waveforms shown in Figure P 11.4-3. Answer: (a) 4.16 V (b) 4.16 V (c) 4.16 Solution: (a)

2

2

Vrms

1 4⎛ 4 2 ⎞ = ⎜ t + ⎟ dt 3 ∫1 ⎝ 3 3 ⎠ =

1

4 4 2 2 t + 1) dt ( ∫ 27 1

4

1

4t 2 + 2

4

1

4

7

10 t (s)

(a)

4 4 2 = ( 4 t + 4 t + 1) dt 27 ∫1 4 ⎛ 4 t3 ⎜ = 27 ⎜ 3 ⎝

v (V) 6

⎞ 4 +t 1 ⎟ ⎟ ⎠

=

4 ( (85.33 − 1.33) + ( 2 )(16 − 1) + 3) 27

=

4 (117 ) = 4.16 V 27

(b)

v (V) 6

2 1

4

7

10 t (s)

(b) v (V) 6

2 2

Vrms =

1 4 ⎛ 4 22 ⎞ ⎜ − t + ⎟ dt 3 ∫1 ⎝ 3 3 ⎠

=

4 4 2 ( −2 t + 11) dt ∫ 1 27

4 4 2 = ( 4 t − 44 t + 121) dt 27 ∫1 4 ⎛ 4 t3 ⎜ = 27 ⎜ 3 ⎝

4

1

44t 2 − 2

4

1

⎞ 4 + 121 t 1 ⎟ ⎟ ⎠

=

4 (84 + ( −22 )15 + (121) 3) 27

=

4 (117 ) = 4.16 V 27

3

6

9

(c) Figure P 11.4-3

t (s)

2

(c)

Vrms

1 3⎛ 4 4 3 4 3 2 2 ⎞ = t + 2 ⎟ dt = 2 t + 3) dt = ( ( 4 t + 12 t + 9 ) dt ⎜ ∫ ∫ 3 0⎝3 27 0 27 ∫ 0 ⎠ 4 ⎛ 4 t3 ⎜ = 27 ⎜ 3 ⎝

3

12t 2 + 2 0

⎞ 3 + 9t 0 ⎟ ⎟ 0 ⎠

3

=

4 ( 36 + 54 + 27 ) 27

=

4 (117 ) = 4.16 V 27

P 11.4-4 Find the rms value for each of the waveforms of Figure P 11.4-4. Answer: Vrms = 1.225 V, Irms = 5 mA v(t) V 2 Sinusoid + Constant

–T/ 2

0

T/ 2

T

t

(a) i(t) mA 10 Sinusoidal

–5

0

5

15 t ( ms)

10

(b) Figure P 11.4-4 Solution: (a)

⎛ 2π v ( t ) = 1 + cos ⎜ ⎝ T

vdc eff

2

t ⎛1 T ⎞ = ⎜ ∫ 0 1dt ⎟ = ⎝T ⎠ T

T

0

⎞ t ⎟ = vdc + vac ⎠

⎛T ⎞ 2 = ⎜ − 0 ⎟ = 1 V and vac eff = T ⎝ ⎠

1 V 2

2

veff = vdc eff + vac eff 2

(b)

2

2

=

⎛ 1 ⎞ 1 +⎜ ⎟ = 1.225 V ⎝ 2⎠ 2

The period of the sinusoid is half the period of i(t). Let T be the period of i(t) and T/2 be the period of the sinusoid. Then

I RMS

4π T ⎧ 0 0 ⎪⎭

Then IA =

24 I A 2 ⇒

cos ( 75 − θ A )

θ A = 75° − 31° = 44°

2 (14.12 ) = 1.37 24 cos ( 31° )

so I A = 1.37∠44° A Also

24∠75° = 137∠16° A 9 + j15 I = I A + I B = 2.662∠30° A IB =

(d)

24∠75° = 1.37∠44° 15 + j 9 24∠75° IB = = 1.37∠16° 9 + j15 I = I A + I B = 2.662∠30° A IA =

Z=

24∠75° = 9.016∠45° Ω 2.662∠30° S = 22.59 + j 22.59 VA

P 11.6-13 Figure P 11.6-13 shows two possible representations of an electrical load. One of these representations is used when the power factor of the load is lagging, and the other is used when the power factor is leading. Consider two cases: (a)

At the frequency ω = 4 rad/s, the load has the power factor pf = 0.8 lagging.

(b)

At the frequency ω = 4 rad/s, the load has the power factor pf = 0.8 leading.

R

R

L

C

Figure P 11.6-13

In each case, choose one of the two representations of the load. Let R = 6 Ω and determine the value of the capacitance, C, or the inductance, L. Solution: ⎛X⎞ Let Z = R + j X = R 2 + X 2 ∠ tan −1 ⎜ ⎟ be the impedance of the load. Further, let V = A∠θ ⎝R⎠ and I = B∠φ be the voltage across and current through the load. Then A∠θ A ⎛X⎞ = ∠ (θ − φ ) R 2 + X 2 ∠ tan −1 ⎜ ⎟ = R + j X = Z = B∠φ B ⎝R⎠ ⎛X⎞ Equating angles gives tan −1 ⎜ ⎟ = θ − φ ⎝R⎠ Also, the complex power delivered to the load is S = ( A∠θ )( B∠φ ) * = So

AB AB cos (θ − φ ) + j sin (θ − φ ) 2 2

pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf )

(Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power factor is leading.) a)

pf = 0.8 lagging ⇒ cos −1 ( 0.8 ) = 36.9°

X = tan ( 36.9 ) = 0.75 ⇒ X = 6 ( 0.75 ) = 4.5 Ω R Choose the inductor to implement the positive reactance. Then

We require

4.5 = ω L = 4 L ⇒ L = 1.125 H b)

pf = 0.8 leading ⇒ cos −1 ( 0.8 ) = −36.9°

X = tan ( −36.9 ) = −0.75 ⇒ X = 6 ( −0.75 ) = −4.5 Ω R Choose the capacitor to implement the negative reactance. Then We require

−4.5 = −

1 1 =− 4C ωC

⇒ C=

8 = 0.889 F 9

P 11.6-14 Figure P 11.6-14 shows two possible representations of an electrical load. One of these representations is used when the power factor of the load L R C R is lagging, and the other is used when the power factor is leading. Consider two cases: (a) At the frequency ω = 4 rad/s, the load has the Figure P 11.6-14 power factor pf = 0.8 lagging. (b) At the frequency ω = 4 rad/s, the load has the power factor pf = 0.8 leading. In each case, choose one of the two representations of the load. Let R = 6 Ω and determine the value of the capacitance, C, or the inductance, L. Solution: 1 1 1 1 1 1 ⎛ R⎞ = −j = + 2 ∠ tan −1 ⎜ − ⎟ be the admittance of the load. Further, Let Y = + 2 R jX R X R X ⎝ X⎠ let V = A∠θ and I = B∠φ be the voltage across and current through the load. Then 1 1 B∠φ B ⎛ R⎞ + 2 ∠ tan −1 ⎜ − ⎟ = Y = = ∠ (φ − θ ) 2 R X A∠θ A ⎝ X⎠ ⎛ R ⎞ tan −1 ⎜ − ⎟ = (φ − θ ) ⎝ ω L⎠ Also, the complex power delivered to the load is

Equating angles gives

S = ( A∠θ )( B∠φ ) * =

AB AB cos (θ − φ ) + j sin (θ − φ ) 2 2

pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf )

So

(Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power factor is leading.) Therefore R ⎛ R⎞ tan −1 ⎜ − ⎟ = (φ − θ ) = − cos −1 ( pf ) ⇒ − = tan ( − cos −1 ( pf ) ) = − tan ( cos −1 ( pf ) ) X ⎝ X⎠ R ⇒ = tan ( cos −1 ( pf ) ) X a)

pf = 0.8 lagging ⇒ cos −1 ( 0.8 ) = 36.9°

R 6 = tan ( 36.9 ) = 0.75 ⇒ X = =8 Ω 0.75 X Choose the inductor to implement the positive reactance. Then

We require

8=ω L = 4L ⇒ L = 2 H

pf = 0.8 leading ⇒ cos −1 ( 0.8 ) = −36.9°

b)

R 6 = tan ( −36.9 ) = −0.75 ⇒ X = = −8 Ω −0.75 X Choose the capacitor to implement the negative reactance. Then

We require

−8 = −

1 1 =− 4C ωC

⇒ C=

1 = 31.25 mF 32

R

P 11.6-15 Figure P 11.6-15 shows two electrical loads. Express the power factor of each load in terms of ω, R, and L.

R

L

L

(a)

(b)

Figure P 11.6-15 Solution: Let V = A∠θ and I = B∠φ be the voltage across and current through the load. The complex power delivered to the load is S = ( A∠θ )( B∠φ ) * = So

AB AB cos (θ − φ ) + j sin (θ − φ ) 2 2

pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf )

(Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power factor is leading.) a.) Let Also Equating angles gives

⎛ω L ⎞ Z = R + j ω L = R 2 + ω 2 L2 ∠ tan −1 ⎜ ⎟ ⎝ R ⎠ A∠θ A Z= = ∠ (θ − φ ) B∠φ B ⎛ω L ⎞ tan −1 ⎜ ⎟ = θ −φ ⎝ R ⎠

⎛ −1 ⎛ ω L ⎞ ⎞ ⎛ω L⎞ −1 tan −1 ⎜ ⎟ = θ − φ = cos ( pf ) ⇒ pf = cos ⎜ tan ⎜ ⎟ ⎟ lagging ⎝ R ⎠ ⎝ R ⎠⎠ ⎝ ⎛ω L ⎞ (The power factor is lagging because tan −1 ⎜ ⎟ is an angle in the first quadrant.) ⎝ R ⎠

so

b.) Let

Y=

1 1 1 1 + = −j = R jω L R ωL

Also

⎛ R ⎞ 1 1 + 2 2 ∠ tan −1 ⎜ − ⎟ 2 R ω L ⎝ ω L⎠

B∠φ B = ∠ (φ − θ ) A∠θ A ⎛ R ⎞ tan −1 ⎜ − ⎟ = (φ − θ ) ⎝ ω L⎠

Y= Equating angles gives

⎛ R ⎞ R −1 = tan ( − cos −1 ( pf ) ) = − tan ( cos −1 ( pf ) ) tan −1 ⎜ − ⎟ = (φ − θ ) = − cos ( pf ) ⇒ − ωL ⎝ ω L⎠ ⎛ ⎛ R ⎞⎞ so pf = cos ⎜ tan −1 ⎜ ⎟ ⎟ lagging ⎝ ω L ⎠⎠ ⎝ ⎛ R ⎞ (The power factor is lagging because tan −1 ⎜ ⎟ is an angle in the first quadrant.) ⎝ω L ⎠

R

P 11.6-16 Figure P 11.6-16 shows two electrical loads. Express the power factor of each load in terms of ω, R, and C.

R

C

C

(b)

(a)

Figure P 11.6-16 Solution: Let V = A∠θ and I = B∠φ be the voltage across and current through the load. The complex power delivered to the load is AB AB S = ( A∠θ )( B∠φ ) * = cos (θ − φ ) + j sin (θ − φ ) 2 2 So pf = cos (θ − φ ) ⇒ θ − φ = cos −1 ( pf ) (Remark: cos −1 ( pf ) is positive when the power factor is lagging and negative when the power factor is leading.) a.) Let Also

Z = R+

1 jω C

= R− j

⎛ 1 1 1 ⎞ = R 2 + 2 2 ∠ tan −1 ⎜ − ⎟ ωC ω C ⎝ ωCR⎠

Z=

A∠θ A = ∠ (θ − φ ) B∠φ B

⎛ 1 ⎞ tan −1 ⎜ − ⎟ =θ −φ ⎝ ωCR⎠

Equating angles gives So

⎛ 1 ⎞ −1 tan −1 ⎜ − ⎟ = θ − φ = cos ( pf ) ⇒ ⎝ ωCR⎠

⎛ ⎛ 1 ⎞⎞ pf = cos ⎜ tan −1 ⎜ − ⎟ ⎟ leading ⎝ ω C R ⎠⎠ ⎝

⎛ −1 ⎞ (The power factor is leading because tan −1 ⎜ ⎟ is an angle in the third quadrant.) ω C R ⎝ ⎠ Y=

b.) Let Also Equating angles gives So

1 1 + jω C = + ω 2 C 2 ∠ tan −1 (ω C R ) 2 R R B∠φ B Y= = ∠ (φ − θ ) A∠θ A tan −1 (ω C R ) = (φ − θ )

tan −1 (ω C R ) = − cos −1 ( pf ) ⇒

pf = cos ( − tan −1 (ω C R ) ) leading

(The power factor is leading because − tan −1 (ω C R ) is an angle in the third quadrant.)

P11.6-17 The source voltage in the circuit shown in Figure P11.6-17 is V s = 24 ∠ 30° V. Consequently I 1 = 3.13 ∠ 25.4° A, I 2 = 1.99 ∠ 52.9° A and V 4 = 8.88 ∠ −10.6° V Determine (a) the average power absorbed by Z 4, (b) the average power absorbed by Z 1, and (c) the complex power delivered by the voltage source. (All phasors are given using peak, not RMS, values.)

Figure P11.6-17 Solution: ⎛ ⎛ −2 ⎞ ⎞ a. The power factor of Z 1 is pf1 = cos ⎜ tan −1 ⎜ ⎟ ⎟ = cos ( −26.6° ) = 0.894 leading . ⎝ 4 ⎠⎠ ⎝ ⎛ ⎛ 8 ⎞⎞ b. The power factor of Z 3 is pf 3 = cos ⎜ tan −1 ⎜ ⎟ ⎟ = cos ( 69.4° ) = 0.352 lagging . ⎝ 3 ⎠⎠ ⎝

c. The power factor of Z 4 is pf 4 = cos ( −10.6° − 52.9° ) = cos ( −63.5° ) = 0.45 leading .

P11.6-18 The current source the circuit shown in Figure P11.6-18 supplies 131.16 – j36.048 VA and the voltage source supplies 64.2275 – 87.8481VA. Determine the values of the impedances Z1 and Z2.

Figure P11.6-18 Solution: The current source this circuit supplies 131.16 – j36.048 VA and the voltage source supplies 64.2275 – 87.8481VA. Determine the values of the impedances Z1 and Z2.

The voltage V1 is calculated from the current source current and complex power: 2S 2 (131.16 − j 36.048 ) = = 136.02∠ − 0.368° V I* ( 2∠15° ) * The current I2 is calculated from the voltage source voltage and complex power:

V1 =

* * ⎛ 2S ⎞ ⎛ 2 ( 64.2275 + j87.8481) ⎞ I2 = ⎜ ⎟ = ⎜ ⎟ = 2.1765∠ − 83.829° A 100∠ − 30° ⎝V⎠ ⎝ ⎠ The impedances are calculated as Z1 =

Z2 =

V1

2∠15° + I 2

=

V1 − 100∠ − 30°

−I 2

136.02∠ − 0.368° = 40 + j 30 Ω 2∠15° + ( 2.1765∠ − 83.829° ) =

136.02∠ − 0.368° − 100∠ − 30° = 20 − j 25 Ω −2.1765∠ − 83.829°

Section 11.7 The Power Superposition Principle 14 A

P 11.7-1 Find the average power absorbed by the 2-Ω resistor in the circuit of Figure P 11.7-1. Answer: P = 413 W

+

110 cos 20t V





12 Ω

1 100 F

Figure P 11.7-1 Solution: Use superposition since we have two different frequency sources. First consider the dc source (ω = 0): ⎛ 12 ⎞ I1 = 14 ⎜ ⎟ = 12 A ⎝ 12 + 2 ⎠ P1 = I12 R = (12) 2 (2) = 288 W Next, consider the ac source (ω = 20 rad/s):

After a source transformation, current division gives − j 60 ⎡ ⎤ ⎢ ⎥ 25 (12 − j 5) I 2 = −9.167 ⎢ ∠116.6° A ⎥ = 5 ⎢ − j 60 + 2+ j 4 ⎥ ⎢⎣ (12 − j 5) ⎥⎦ 2

Then

I (125)(2) P2 = 2 (2) = =125 W 2 2

Now using power superposition P = P1 + P2 = 288 + 125 = 413 W

0.2 H

+

P 11.7-2 Find the average power absorbed by the 8-Ω resistor in the circuit of Figure P 11.7-2. Answer: P = 22 W

5 cos 2000t A 18



40 cos 8000t V mF

1 mH



Figure P 11.7-2 Solution: Use superposition since we have two different frequency sources. First consider ω = 2000 rad/s source: Current division yields ⎡ 8 ⎤ ⎢ −j2 ⎥ 5 I1 = 5 ⎢ ∠63.4° A ⎥= 5 ⎢ 8 +8 ⎥ ⎢⎣ − j 2 ⎥⎦ 2

Then

I 8 P1 = 1 = 20 W 2

Next consider ω = 8000 rad/s source.

Current division yields ⎡ 8 ⎤ ⎢ j7 ⎥ 5 I 2 = − j5 ⎢ ∠ − 171.9° A ⎥= 8 50 ⎢ +8 ⎥ ⎢⎣ j 7 ⎥⎦ 2

Then

I 8 P2 = 2 =2W 2

Now using power superposition

P = P1 + P2 = 22 W

P 11.7-3 For the circuit shown in Figure P 11.7-3, determine the average power absorbed by each resistor, R1 and R2. The voltage source is vs = 10 + 10 cos (5t + 40°) V, and the current source is is = 4 cos (5t – 30°) A. 2H

10 Ω

6i2

R1

R2

5 Ω + –

is 1 10

vs

F

i2

Figure P 11.7-3 P11.7-3 Use superposition since we have two different frequencies. First consider the dc source (ω = 0):

⎛1⎞ i 2 (t ) = 0 and i1 (t ) = 10 ⎜ ⎟ = 1 A ⎝ 10 ⎠

PR1 = i12 R1 = 12 (10) = 10 W PR 2 = 0 W Next consider ω = 5 rad/s sources.

Apply KCL at the top node to get

(10I1 −10∠40°) =0 j10 −10 I1 + (5− j 2) I 2 = 0

−6 I 2 + I1 + I 2 − ( 4 ∠−30° ) + Apply KVL to get Solving these equations gives

I1 = −0.56 ∠ − 64.3° A and I 2 = −1.04∠ − 42.5° A

Then

PR1 =

I12 R1

2

( 0.56 ) =

2

2

(10)

= 1.57 W and PR 2 =

I 22 R 2

2

(1.04 ) = 2

Now using power superposition PR = 10 + 1.57 = 11.57 W and PR 2 = 0 + 2.7 = 2.7 W 1

2

(5)

= 2.7 W

P 11.7-4 For the circuit shown in Figure P 11.7-4, determine the effective value of the resistor voltage vR and the capacitor voltage vC.

+ 4 cos 10t V

+ –

2 Ω

6 sin 5t V

+ –

20 mF

Figure P 11.7-4 Solution: Use superposition since we have two different frequencies. First consider the ω = 10 rad/s source: V1 4∠0° = = 0.28 + j 0.7 A Z 2− j5 V = 2 I1 = 2(0.28 + j.7) = 0.56 + j1.4 = 1.51 ∠68.2° V

I1 = R1

V = − j 5 I1 = 3.77 ∠ − 21.8° V C1

Next consider ω = 5 rad/s source. V2 6∠−90° = = 0.577 − j 0.12 A Z 2 − j10 VR = 2 I 2 = 2(.577 − j 0.12) =1.15− j 0.24

I2 = 2

V

C2

=1.17∠−11.8° V = − j10I 2 =5.9∠258.3° V

Now using superposition vR (t) = 1.51cos (10 t + 68.2°) + 1.17 cos (5 t − 11.8°) V vC (t) = 3.77 cos (10 t − 21.8°) + 5.9 cos (5 t − 258.3°) V Then 2

2

2 Reff

⎛ 1.51 ⎞ ⎛ 1.17 ⎞ =⎜ ⎟ +⎜ ⎟ = 1.82 ⇒ VReff =1.35 V ⎝ 2⎠ ⎝ 2 ⎠

2 Ceff

⎛ 3.77 ⎞ ⎛ 5.9 ⎞ =⎜ ⎟ +⎜ ⎟ = 24.52 ⇒ VCeff = 4.95 V ⎝ 2 ⎠ ⎝ 2⎠

V

2

V

2

vR – + vC –

Section 11.8 The Maximum Power Transfer Theorem P 11.8-1 Determine values of R and L for the circuit shown in Figure P 11.8-1 that cause maximum power transfer to the load. Answer: R = 800 Ω and L = 1.6 H 4000 Ω R

5 cos (1000t + 60°) V

+ –

0.5 μF

L

Source

Load

Figure P 11.8-1 Solution:

Z t = 4000 || − j 2000 = 800 − j1600 Ω Z L = Z*t =800+ j1600 Ω ⎧ R =800 Ω R + j1000 L = 800 + j1600 ⇒ ⎨ ⎩ L =1.6 H

P 11.8-2 Is it possible to choose R and L for the circuit shown in Figure P 11.8-2 so that the average power delivered to the load is 12 mW? Answer: Yes R 2.5 cos 100t mA

0.2 μF

25 kΩ

Source

L

Load

Figure P 11.8-2 Solution:

Z t = 25, 000 || − j 50, 000 = 20, 000 − j10, 000 Ω Z L =Z*t = 20,000+ j10,000 Ω

⎧ R = 20 kΩ ⎪ R + jω L = 20, 000 + j10, 000 ⇒ ⎨100 L =10,000 ⎪ L =100 H ⎩ After selecting these values of R and L, 2

Since Pmax

⎛ 0.14×10−2 ⎞ 3 I = 1.4 mA and Pmax = ⎜ ⎟ ( 20×10 ) = 19.5 mW 2 ⎝ ⎠ > 12 mW , yes, we can deliver 12 mW to the load.

P 11.8-3 The capacitor has been added to the load in the circuit shown in Figure P 11.8-3 in order to maximize the power absorbed by the 4000-Ω resistor. What value of capacitance should be used to accomplish that objective? Answer: 0.1 μF 800 Ω

+ –

0.32 H

5 cos (5000t + 45°) V

Source

4000 Ω

C

Load

Figure P 11.8-3 Solution: ⎛ −j ⎞ R⎜ ω C ⎟⎠ R − jω R 2C Z t = 800 + j1600 Ω and Z L = ⎝ = j 1 + (ω RC ) 2 R− ωC ⎛ −j ⎞ R⎜ ω C ⎟⎠ R − jω R 2C * ⎝ ZL = Zt ⇒ = = 800 − j1600 Ω j 1+ (ω RC ) 2 R− ωC Equating the real parts gives 800 =

R 4000 = 2 1+ (ω RC ) 1+[(5000)(4000)C ]2

⇒ C = 0.1 μ F

P 11.8-4 What is the value of the average power delivered to the 2000-Ω resistor in the circuit shown in Figure P 11.8-4? Can the average power delivered to the 2000-Ω resistor be increased by adjusting the value of the capacitance? Answer: 8 mW. No. 400 Ω

5 cos 1000t V

0.8 H

+ –

2000 Ω

Source

1 μF

Load

Figure P 11.8-4 Solution:

Z t = 400 + j 800 Ω and Z L = 2000 || − j1000 = 400 − j 800 Ω

Since Z L = Z*t the average power delivered to the load is maximum and cannot be increased by adjusting the value of the capacitance. The voltage across the 2000 Ω resistor is

VR = 5

ZL = 2.5 − j 5 = 5.59e − j 63.4 V Zt +ZL 2

⎛ 5.59 ⎞ 1 P=⎜ = 7.8 mW ⎟ ⎝ 2 ⎠ 2000

So

is the average power delivered to the 2000 Ω resistor.

P 11.8-5 What is the value of the resistance R in Figure P 11.8-5 that maximizes the average power delivered to the load? R

5 cos 1000t V

0.8 H

+ –

2000 Ω

Source

1 μF

Load

Figure P 11.8-5 Solution: Notice that Zt,not ZL, is being adjusted .When Zt is fixed, then the average power delivered to the load is maximized by choosing ZL = Zt*. In contrast, when ZL is fixed, then the average power delivered to the load is maximized by minimizing the real part of Zt. In this case, choose R = 0. Since no average power is dissipated by capacitors or inductors, all of the average power provided by source is delivered to the load.

Section 11.9 Coupled Inductors M

a

P 11.9-1 Two magnetically coupled coils are connected as shown in Figure P 11.9-1. Show that an equivalent inductance at terminals a–b is Lab = L1 + L2 – 2M.

L1

L2

b

Figure P 11.9-1 Solution: Vs = I jω L1 − I jω M + I jω L2 − I jω M ⇒ jω ( L1 + L2 − 2 M ) =

Vs I

Therefore Lab = L1 + L2 − 2M

a

P 11.9-2 Two magnetically coupled coils are shown connected in Figure P 11.9-2. Find the equivalent inductance Lab.

M

L1 b

Figure P 11.9-2 Solution: The coil voltages are given by: V1 = I1 jω L1 + I 2 jω M V2 = I 2 jω L2 + I1 jω M From KVL the coil voltages are equal so

I1 jω L1 + I 2 jω M = I 2 jω L2 + I1 jω M

⇒ I1 jω ( L1 − M ) = I 2 jω ( L2 − M ) ⎛ L −M ⇒ I1 = ⎜ 2 ⎝ L1 − M

⎞ ⎟ I2 ⎠

From KCL ⎛ L −M ⎞ ⎛ L + L −2 M ⎞ I s = I1 + I 2 = ⎜ 2 + 1⎟ I 2 = ⎜ 2 1 ⎟ I2 ⎝ L1 − M ⎠ ⎝ L1 − M ⎠

Next

⎛ L1 − M I2 = ⎜ ⎜ L2 + L1 − 2 M ⎝

⎞ ⎛ L −M ⎟⎟ I s and I1 = ⎜ 2 ⎝ L1 − M ⎠

⎞ ⎛ L2 − M ⎟ I2 = ⎜ ⎠ ⎝ L1 − M

⎞ ⎛ L1 − M ⎟ ⎜⎜ ⎠ ⎝ L2 + L1 − 2 M

⎞ ⎟⎟ I s ⎠

Substituting gives ⎛ L2 − M ⎞ ⎛ L1 − M ⎞ V1 = jω L1 I1 + jω M I 2 = jω M ⎜ I + jω M ⎜ I ⎜ L2 + L1 − 2 M ⎟⎟ s ⎜ L2 + L1 − 2 M ⎟⎟ s ⎝ ⎠ ⎝ ⎠ ⎛ M ( L2 − M ) + M ( L1 − M ) ⎞ ⎛ L1 L 2 − M 2 j = jω ⎜ I = ω ⎟⎟ s ⎜⎜ ⎜ L2 + L1 − 2 M ⎝ ⎠ ⎝ L1 + L 2 − 2 M ⎛ L1 L 2 − M 2 ⎞ V1 Z= = jω ⎜ Finally ⎜ L1 + L 2 − 2 M ⎟⎟ Is ⎝ ⎠

⎞ ⎟⎟ I s ⎠

L2

P 11.9-3 The source voltage of the circuit shown in Figure P 11.9-3 is vs = 141.4 cos 100t V. Determine i1(t) and i2(t).



vs + –

M = 0.6 H i1 0.4 H

i2

1.6 H

200 Ω

Figure P 11.9-3 Solution:

Mesh equations:

−141.4∠0° + 2 I1 + j 40 I1 − j 60 I 2 = 0 200 I 2 + j160 I 2 − j 60 I1 = 0 ⇒ I 2 = (0.23∠51°) I1

Solving yields I1 = 4.17∠ − 68° A and I 2 = 0.96∠ − 17°A

Finally

i1 ( t ) = 4.17 cos(100t −68°) A and i 2 ( t ) = 0.96 cos(100t −17°) A (checked: LNAPAC 7/21/04)

P 11.9-4 A circuit with a mutual inductance is shown in Figure P 11.9-4. Find the voltage V2 when ω = 5000. 10 Ω

V1 = 10 0° V

+ –

M = 10 mH

1 mH

100 mH

400 Ω

+ V2 –

Figure P 11.9-4 Solution:

Mesh equations:

(10+ j5 ) I 1 − j50 I 2 = 10 − j 50 I1 + ( 400+ j 500 ) I 2 = 0

Solving the mesh equations using Cramer’s rule: I2 =

Then

(10+ j5)( 0 )−( − j50 )(10 ) 2 (10+ j5) ( 400+ j500 ) − ( − j50 )

= 0.062 ∠29.7° A

V2 = 400 I 2 = 400 ( 0.062 ∠29.7° ) = 24.8 ∠29.7°

1 150

P 11.9-5 Determine v(t) for the circuit of Figure P 11.9-5 when vs = 10 cos 30t V. Answer: v(t) = 23 cos (30t + 9°) V

28 Ω

F +

vs

+ –

v(t)

0.1 H 0.3 H

0.2 H



Figure P 11.9-5 Solution:

Mesh equations:

10 = − j 5I 1 + ⎡⎣ j 9 ( I 1 − I 2 ) + j 3I 2 ⎤⎦

0 = 28I 2 + ⎡⎣ j 6I 2 + j 3 ( I 1 − I 2 ) ⎤⎦ − ⎡⎣ j 9 ( I 1 − I 2 ) + j 3I 2 ⎤⎦

or − j 6 ⎤ ⎡ I 1 ⎤ ⎡10 ⎤ ⎡ j4 ⎢ − j 6 28 + j 9 ⎥ ⎢I ⎥ = ⎢ 0 ⎥ ⎣ ⎦⎣ 2⎦ ⎣ ⎦

Solving the mesh equations, e.g. using MATLAB, yields I1 = 2.62∠ − 72° A and I 2 = 0.53∠0° A

then V = j 9(I1 − I 2 ) + j 3I 2 = j 9I1 − j 6I 2 = 23 ∠10° V

Finally

v (t ) = 23cos (30t + 10°) V (checked: LNAPAC 7/21/04)

P 11.9-6 Find the total energy stored in the circuit shown in Figure P 11.9-6 at t = 0 if the secondary winding is (a) open-circuited, (b) short-circuited, (c) connected to the terminals of a 7-Ω resistor. Answer: (a) 15 J

12 Ω

M = 0.6 H

0.3 H

10 cos 5t A

(b) 0 J (c) 5 J

1.2 H

Figure P 11.9-6

Solution: (a)

I 2 = 0 ⇒ I1 = 10 ∠0° A ⇒ i1 (0) = 10 A 2

w=

(b)

L1 i1 (0) 2

=

(0.3) (10) 2 = 15 J 2

Mesh equations: j 6 I 2 - j 3 I1 = 0 ⇒ I1 = 2 I 2 I1 = 10∠0° A ⇒ I 2 = 5∠0° A

Then w=

1 1 1 1 L1i 2 2 (0) + L 2i12 (0) − M i1 (0) i 2 (0) = (0.3)(10) 2 + (1.2)(5) 2 − (0.6) (10)(5) = 0 2 2 2 2

(c)

(7 + j 6) I 2 − j 3 I1 = 0 I 2 = 3.25 ∠49.4° A

i 2 (t ) = 3.25cos(5t + 49.4°) A

i 2 (0) = 2.12 A Finally 1 1 w = (0.3) (10) 2 + (1.2) (2.12) 2 − (0.6) (10) (2.12) = 5.0 J 2 2

8 mH

P 11.9-7 Find the input impedance, Z, of the circuit of Figure P 11.9-7 when ω = 1000 rad/s. Answer: Z = 8.4 ∠14° Ω

1 6

mF

5 mH

6 mH



Z

Figure P 11.9-7 Solution:

Mesh equations:

− VT + j8 I1 + j 5(I1 − I 2 ) − j 6 I1 + j 6 (I1 − I 2 ) + j 5 I1 = 0 3 I 2 + j 6 (I 2 − I1 ) − j 5 I1 = 0

Solving yields I2 =

j 11 I1 3+ j 6

⎛ j 11 ⎞ VT = I1 ( j 18) + I 2 (− j 11) = ⎜ j 18+ ⎟I 3+ j 6 ⎠ 1 ⎝ Then

Z =

j 11 VT = j 18 + = 8.28 ∠13° = 8.0667 + j 1.8667 Ω I1 3+ j 6 (checked: LNAPAC 7/21/04)

P 11.9-8 A circuit with three mutual inductances is shown in Figure P 11.9-8. When vs = 10 cos 2t V, M1 = 2 H, and M2 = M3 = 1 H, determine the capacitor voltage v(t). M1 5Ω

4H

3H M2

vs



6Ω M3

+ –

+ 1 10

v

F

2H



Figure P 11.9-8 Solution:

The coil voltages are given by V1 = j 6 I1 − j 2 (I1 − I 2 ) − j 4 I 2 = j 4 I1 − j 2 I 2

V2 = j 4 ( I1 − I 2 ) − j 2 I1 + j 2 I 2 = j 2 I1 − j 2 I 2 V3 = j8 I 2 − j 4 I1 + j 2 ( I1 − I 2 ) = − j 2 I1 + j 6 I 2 The mesh equations are 5 I1 + V1 + 6 (I1 − I 2 ) + V2 = 10∠0° − V2 + 6 (I 2 − I1 ) + 2 I 2 + V3 − j 5 I 2 = 0 Combining and solving yields 11 + j 6 10 −6 − j 4 0 60 + j 40 I2 = = = 1.2 ∠0.28° A 11 + j 6 −6 − j 4 50 + j 33 −6 − j 4 8 + j 3 Finally V = − j 5 I 2 = 6.0 ∠ − 89.72° A ⇒ v(t ) = 6 cos(2 t − 89.7°) V

0.25 mF

P 11.9-9 The currents i1(t) and i2(t) in Figure, P 11.9-9 are mesh currents. Represent the circuit in the frequency domain and write the mesh equations.

6H 5H

40 Ω

i1(t)

i2(t) 8H

+ –

15 cos (25t + 30°) V

80 Ω

P 11.9-9 Solution: The equations describing the coupled coils give: V1 = j 200 ( I 1 − I 2 ) + j125 I 2 = j 200I 1 − j 75 I 2

V 2 = j150 I 2 + j125 ( I 1 − I 2 ) = j125 I 1 + j 25 I 2

The mesh equation for the left mesh is − j 160 I 1 + V1 − 15∠30° + 40 I 1 = 0

− j 160 I 1 + j 200I 1 − j 75 I 2 − 15∠30° + 40 I 1 = 0

( 40 + j 40 ) I 1 − j 75 I 2 = 15∠30° The mesh equation for the right mesh is V 2 + 80 I 2 − V1 = 0

j125 I 1 + j 25 I 2 + 80 I 2 − ( j 200I 1 − j 75 I 2 ) = 0

− j 75 I 1 + ( 80 + j100 ) I 2 = 0

P 11.9-10 Determine the mesh currents for the circuit shown in Figure P 11.9-10. 3H 10 Ω + –



+ 4H

v1 12 cos 5t V

6H

+



i1

v2

20 Ω

50 Ω i2

P 11.9-10 Solution: In the frequency domain, the coil voltages are given by V1 = j ω L1 I 1 − j ω M I 2 V2 = j ω L 2 I 2 − j ω M I 1

The mesh equations are R1 I 1 + V1 + R 2 ( I 1 − I 2 ) = Vs

V2 + R3 I 2 − R 2 ( I 1 − I 2 ) = 0

Substituting for the coil voltages gives ⎡ R1 + R 2 + j ω L1 ⎢ ⎢ ⎣ − ( R2 + j ω M )

− ( R 2 + j ω M ) ⎤ ⎡ I 1 ⎤ ⎡ Vs ⎤ ⎥⎢ ⎥ = ⎢ ⎥ R 2 + R 3 + j ω L 2 ⎦⎥ ⎣I 2 ⎦ ⎣ 0 ⎦

Using the given values ⎡ 30 + j 20 − ( 20 + j 15 ) ⎤ ⎡ I 1 ⎤ ⎡12∠0 ⎤ ⎢ ⎥⎢ ⎥ = 70 + j 30 ⎦ ⎣I 2 ⎦ ⎢⎣ 0 ⎥⎦ ⎣ − ( 20 + j 15 ) Solving, for example using MATLAB, gives ⎡ I 1 ⎤ ⎡0.4240∠ − 28.9° ⎤ ⎢ ⎥=⎢ ⎥ ⎣ I 2 ⎦ ⎣ 0.1392∠ − 15.2° ⎦

Back in the time domain, the mesh currents are ⎡ i1 ⎤ ⎡0.4240 cos ( 5t − 28.9° ) ⎤ ⎥ A ⎢i ⎥ = ⎢ − ° t 0.1392 cos 5 15.2 ( ) 2 ⎣ ⎦ ⎣ ⎦ (checked using LNAP, 9/9/04)

P 11.9-11 Determine the coil voltages, v1, v2, v3, and v4, for the circuit shown in Figure P 11.911. v1



2 cos (5t + 45°) A

+

4H 10 Ω

3H

+ 10 Ω

v2

v3

+

6H –

5H

20 Ω

4H



– 5H

v4 +

1.25 cos (5t − 45°) A

2.75 cos (5t) A

Figure P 11.9-11 Solution: In the frequency domain, the coil voltages are given by V1 = − j ω L1 I a + j ω M 1 ( I a − I b ) = j ω ( M 1 − L1 ) I a − j ω M 1 I b

= ( − j 5 )( 2∠45° ) + ( − j 15 )(1.25∠ − 45° ) = 21.25∠ − 106.9° V

V2 = j ω L 2 ( I b − I a ) + j ω M 1 I a

= j ω ( M 1 − L2 ) I a + j ω L2 I b = ( − j 15 )( 2∠45° ) + ( j 30 )(1.25∠ − 45° ) = 48.02∠6.4° V

V3 = j ω L 3 ( I c − I a ) − j ω M 2 I c = − j ω L 2 I a + j ω ( L 3 − M 2 ) I c = ( − j 25 )( 2∠45° ) + ( j 5 )( 2.75∠0° ) = 41.43∠ − 31.4° V

V4 = − j ω L 4 I c + j ω M 2 ( I c − I a ) = − j ω M 2 I a + j ω ( M 2 − L 4 ) I c = ( − j 20 )( 2∠45° ) + ( − j 5 )( 2.75∠0° ) = 50.66∠ − 56.1° V Back in the time domain, the coil voltages are v1 = 21.25 cos ( 5t − 106.9° ) V , v 2 = 48.02 cos ( 5t + 6.3° ) V , v 3 = 41.43 cos ( 5t − 31.4° ) V and v 4 = 50.66 cos ( 5t − 56.1° ) V

(checked using LNAP, 9/9/04)

4H

Solution: (a) In (b) and (c) the coils are coupled by mutual inductance, but not in (a). This circuit can be represented in the frequency domain, using phasors and impedances.

+



5H

vo(t ) –

+

P 11.9-12 Figure P 11.9-12 shows three similar circuits. In each, the input to the circuit is the voltage of the voltage source, vs(t). The output is the voltage across the right-hand coil, vo(t). Determine the steady-state output voltage, vo(t), for each of the three circuits.



vs(t ) = 5.94 cos (3t + 140°) V

(a) 4H +

2H



5H

vo(t )

+

– –

vs(t ) = 5.94 cos (3t + 140°) V

Voltage division gives

Vo =

(b)

j 15 5.94∠140° 5 + j 12 + j 15

4H

= ( 0.5463∠10.5° )( 5.94∠140° ) = 3.245∠150.5° V vo ( t ) = 3.245 cos ( 3 t + 150.5° ) V

(b) The input voltage is a sinusoid and the circuit is at steady state. The output voltage is also a sinusoid and has the same frequency as the input voltage. Consequently, the circuit can be represented in the frequency domain, using phasors and impedances.



5H

vo(t ) –

+

In the time domain, the output voltage is given by

+

2H



vs(t ) = 5.94 cos (3t + 140°) V

(c)

Figure P 11.9-12

This circuit of a single mesh. Notice that the mesh current, I(ω), enters the undotted ends of both coils. Apply KVL to the mesh to get

5 I (ω ) + ( j12 I (ω ) + j 6 I (ω ) ) + ( j 6 I (ω ) + j15 I (ω ) ) − 5.94∠140° = 0 5 I (ω ) + ( j12 + j 6 + j 6 + j15 ) I (ω ) − 5.94∠140° = 0

I (ω ) =

5.94∠140° 5.94∠140° 5.94∠140° = = = 0.151∠57° A 5 + j (12 + 6 + 6 + 15 ) 5 + j 39 39.3∠83

Notice that the voltage, Vo(ω), across the right-hand coil and the mesh current, I(ω), adhere to the passive convention. The voltage across the right-hand coil is given by

Vo (ω ) = j 15 I (ω ) + j 6 I (ω ) = j 21 I (ω ) = j 21 ( 0.151∠57° ) = (21∠90°) ( 0.151∠57° ) = 3.17∠147° V In the time domain, the output voltage is given by vo ( t ) = 3.17 cos ( 3 t + 147° ) V

(c) Circuit (c) is very similar to the circuit (b). There is only one difference: the dot of the lefthand coil is located at the right of the coil in (c) and at the left of the coil in (b). As before, our first step is to represent the circuit in the frequency domain, using phasors and impedances.

This circuit consists of a single mesh. Notice that the mesh current, I(ω), enters the dotted end of the left-hand coil and the undotted end of the right-hand coil. Apply KVL to the mesh to get 5 I (ω ) + ( j12 I (ω ) − j 6 I (ω ) ) + ( − j 6 I (ω ) + j15 I (ω ) ) − 5.94∠140° = 0 5 I (ω ) + ( j12 − j 6 − j 6 + j15 ) I (ω ) − 5.94∠140° = 0

I (ω ) =

5.94∠140° 5.94∠140° 5.94∠140° = = = 0.376∠68.4° A 5 + j (12 − 6 − 6 + 15 ) 5 + j 15 15.8∠71.6

Notice that the voltage, Vo(ω), across the right-hand coil and the mesh current, I(ω), adhere to the passive convention. The voltage across the right-hand coil is given by

Vo (ω ) = j 15 I (ω ) − j 6 I (ω ) = j 9 I (ω ) = j 9 ( 0.376∠68.4° ) = (9∠90°) ( 0.376∠68.4° ) = 3.38∠158.4° V In the time domain, the output voltage is given by vo ( t ) = 3.38 cos ( 3 t + 158.4° ) V

P 11.9-13 Figure P 11.9-13 shows three similar circuits. In each, the input to the circuit is the voltage of the voltage source, vs(t) = 5.7 cos (4t + 158°) V The output in each circuit is the voltage across the right-hand coil, vo(t). Determine the steadystate output voltage, vo(t), for each of the three circuits. +

4Ω + –

vs(t)

5H

4H

2H



vo(t)

+ –

vs(t)

+ 5H

4H



vo(t) –

(a)

(b) 2H

4Ω + –

vs(t)

+ 5H

4H

vo(t) –

(c)

Figure P 11.9-13 Solution: (a) In (b) and (c) the coils are coupled by mutual inductance, but not in (a). This circuit can be represented in the frequency domain, using phasors and impedances.

Voltage division gives j 16 || j 20 5.7∠158° = ( 0.9119∠24° )( 5.7∠158° ) = 5.198∠182° V 4 + ( j 16 || j 20 ) In the time domain, the output voltage is given by

Vo =

vo ( t ) = 5.2 cos ( 4 t + 182° ) V

(b) The input voltage is a sinusoid and the circuit is at steady state. The output voltage is also a sinusoid and has the same frequency as the input voltage. Consequently, the circuit can be represented in the frequency domain, using phasors and impedances.

The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled. Reference directions for these currents and voltages have been selected so that the current and voltage of each coil adhere to the passive convention. Notice that both coil currents, I1(ω) and I2(ω), enter the undotted ends of their respective coils. The device equations for coupled coils are:

and

V1 (ω ) = j 16 I1 (ω ) + j 8 I 2 (ω )

(1)

V2 (ω ) = j 8 I1 (ω ) + j 20 I 2 (ω )

(2)

The coils are connected in parallel, consequently V1 (ω ) = V2 (ω ) . Equating the expressions for

V1(ω) and V2(ω) gives

j 16 I1 (ω ) + j 8 I 2 (ω ) = j 8 I1 (ω ) + j 20 I 2 (ω ) j 8 I1 (ω ) = j 12 I 2 (ω )

I 1 (ω ) =

3 I 2 (ω ) 2

Apply Kirchhoff’s Current Law (KCL) to the top node of the coils to get

I ( ω ) = I 1 ( ω ) + I 2 (ω ) = I 1 (ω ) =

Therefore

3 5 I 2 ( ω ) + I 2 (ω ) = I 2 (ω ) 2 2

3 2 I (ω ) and I 2 (ω ) = I (ω ) 5 5

(3)

Substituting the expressions for I1(ω) and I2(ω) from Equation 3 into Equation 1 gives 16 ( 3) + 8 ( 2 ) ⎛3 ⎞ ⎛2 ⎞ V1 (ω ) = j 16 ⎜ I (ω ) ⎟ + j 8 ⎜ I (ω ) ⎟ = j I (ω ) = j 12.8 I (ω ) 5 ⎝5 ⎠ ⎝5 ⎠

Apply KVL to the mesh consisting of the voltage source, resistor and left-hand coil to get 4 I (ω ) + V1 (ω ) − 5.7∠158° = 0

Using Equation 4 gives Solving for I (ω) gives

4 I (ω ) + j 12.8 I (ω ) − 5.7∠158° = 0

I (ω ) =

5.7∠158° 5.7∠158° = = 0.425∠85° A 4 + j 12.8 13.41∠73°

Now the output voltage can be calculated using Equation 4:

(4)

Vo (ω ) = V1 (ω ) = j 12.8 I (ω ) = j 12.8 ( 0.425∠85° ) = (12.8∠90° )( 0.425∠85° ) = 5.44∠175° V In the time domain, the output voltage is given by vo ( t ) = 5.44 cos ( 4 t + 175° ) V

(c) Circuit (c) is very similar to the circuit (b). There is only one difference: the dot of the righthand coil is located at the bottom of the coil in (b) and at the top of the coil in (c). Here is the circuit from (c) represented in the frequency domain, using impedances and phasors.

The coil currents, I1(ω) and I2(ω), and the coil voltages, V1(ω) and V2(ω), are labeled. Reference directions for these currents and voltages have been selected so that the current and voltage of each coil adhere to the passive convention. Notice that one of coil currents, I1(ω), enters the undotted end of the coil while the other coil current, I2(ω), enters the dotted end of the coil. The device equations for coupled coils are:

and

V1 (ω ) = j 16 I1 (ω ) − j 8 I 2 (ω )

(5)

V2 (ω ) = − j 8 I1 (ω ) + j 20 I 2 (ω )

(6)

The coils are connected in parallel, consequently V1 (ω ) = V2 (ω ) . Equating the expressions for

V1(ω) and V2(ω)gives

j 16 I1 (ω ) − j 8 I 2 (ω ) = − j 8 I1 (ω ) + j 20 I 2 (ω ) j 24 I1 (ω ) = j 28 I 2 (ω )

I 1 (ω ) =

28 7 I 2 (ω ) = I 2 (ω ) 24 6

Apply Kirchhoff’s Current Law (KCL) to the top node of the coils to get 7 13 I 2 (ω ) + I 2 (ω ) = I 2 (ω ) 6 6 7 6 I1 (ω ) = I (ω ) and I 2 (ω ) = I (ω ) Therefore 13 13 Substituting the expressions for I1(ω) and I2(ω) from Equation 7 into Equation 5 gives

I (ω ) = I 1 (ω ) + I 2 (ω ) =

(7)

16 ( 7 ) − 8 ( 6 ) ⎛7 ⎞ ⎛6 ⎞ V1 (ω ) = j 16 ⎜ I (ω ) ⎟ − j 8 ⎜ I (ω ) ⎟ = j I (ω ) = j 4.9 I (ω ) 13 ⎝ 13 ⎠ ⎝ 13 ⎠

Apply KVL to the mesh consisting of the voltage source, resistor and left-hand coil to get 4 I (ω ) + V1 (ω ) − 5.7∠158° = 0

Using Equation 8 gives

4 I (ω ) + j 4.9 I (ω ) − 5.7∠158° = 0

Solving for I (ω) gives

I (ω ) =

5.7∠158° 5.7∠158° = = 0.901∠107° A 4 + j 4.9 6.325∠51°

Now the output voltage can be calculated using Equation 8: Vo (ω ) = V1 (ω ) = j 4.9 I (ω ) = j 4.9 ( 0.901∠107° ) = ( 4.9∠90° )( 0.901∠107° ) = 4.41∠197° V In the time domain, the output voltage is given by vo ( t ) = 4.41 cos ( 4 t + 197° ) V

(8)

P11.9-14 The circuit shown in Figure 11.9-14 is represented in the time domain. Determine coil voltages v1 and v2. Answer: v1 = 104.0 cos (6t + 46.17 °) V and v2 = 100.6 cos (6t + 63.43 °) V.

Figure 11.9-14 Solution: Represent the circuit in the frequency domain as shown. Then V1 = j 48 (1.5∠ − 90° ) + j 30 ( 2.5∠0° ) = j 48 ( − j 1.5 ) + j 30 ( 2.5 ) = 72 + j 75 = 103.97∠46.17° V

and

V 2 = j 36 ( 2.5∠0° ) + j 30 (1.5∠ − 90° ) = j 36 ( 2.5 ) + j 30 ( − j 1.5 ) = 45 + j 90 = 100.62∠63.43° V

P11.9-15 The circuit shown in Figure P11.9-15 is represented in the frequency domain (e.g. –j30 Ω is the impedance due to the mutual inductance of the coupled coils). Suppose V ( ω ) = 70∠0° V. Then I 1 ( ω ) = B∠θ A and I 2 ( ω ) = 0.875∠ − 90° A. Determine the values of B and θ.

Answer: B = 1.75 A and θ = −90 °

Figure 11.9-15 Solution:

Suppose V ( ω ) = 70∠0° V. Then I 1 ( ω ) = B∠θ A and I 2 ( ω ) = 0.875∠ − 90° A. V = − j 25 I 1 + j 50 I 1 + j 30 I 2 = j 25 I 1 + j 30 I 2 = j 25 I 1 + j 30 ( − j 0.875 )

I1 =

V − j 30 ( − j 0.875 ) V − 26.25 70 − 26.25 43.75 = = = = − j 1.75 j 25 j 25 j 25 j 25

P11.9-16 Determine the values of the inductances L1 and L2 in the circuit shown in Figure P11.9-16, given that i ( t ) = 0.319 cos ( 4 t − 82.23° ) A

and

v ( t ) = 0.9285cos ( 4 t − 62.20° ) V .

Figure P11.9-16

Solution: Represent the circuit in the frequency domain as

Apply KVL to the left mesh to get 5∠0° + j 3 ( 0.9285∠ − 62.20° ) ⎛ 0.9285∠ − 62.20° ⎞ 5∠0° = j 4 L1 ( 0.319∠ − 82.23° ) + j 12 ⎜ − ⎟ ⇒ L1 = 4 j 4 ( 0.319∠ − 82.23° ) ⎝ ⎠ = 6.0004 − j 0.0005 ≅6H Apply KVL to the left mesh to get ⎛ 0.9285∠ − 62.20° ⎞ 0.9285∠ − 62.20° = j 4 L 2 ⎜ − ⎟ + j 12 ( 0.319∠ − 82.23° ) 4 ⎝ ⎠ L2 =

( 0.9285∠ − 62.20° ) − j 12 ( 0.319∠ − 82.23° )

⎛ 0.9285∠ − 62.20° ⎞ j 4⎜ − ⎟ 4 ⎝ ⎠ = 3.9998 + j 0.0005 ≅4H

P11.9-17 Determine the complex power supplied by the source in the circuit shown in Figure P11.9-17.

Figure P11.9-17 Solution: Represent the circuit in the frequency domain as

The coil voltages are given by

V1 = j 100 I 1 − j 80 I 2 and V 2 = j 120 I 2 − j 80 I 1 V1 + 40 ( I 1 − I 2 ) + 20 I 1 − 180∠0° = 0

Using KVL

V 2 + 50 I 2 − j 50 I 2 − 40 ( I 1 − I 2 ) = 0

Substituting the coil voltages: ( j 100 I 1 − j 80 I 2 ) + 40 ( I 1 − I 2 ) + 20 I 1 = 180∠0°

( j 120 I

2

− j 80 I 1 ) + 50 I 2 − j 50 I 2 − 40 ( I 1 − I 2 ) = 0

( 60 + j 100 ) I 1 − ( 40 + j 80 ) I 2 = 180∠0° − ( 40 + j 80 ) I 1 + ( 90 + j 70 I 1 ) I 2 = 0

Simplifying

I 1 = 2.731∠ − 26.9° A and I 2 = 2.142∠ − 1.36° A

Solving gives

The complex power delivered by the source is S=

VI * (180∠0° )( 2.731∠ − 26.9° ) * = = 219.19 + j111.20 VA 2 2

P11.9-18 The input to the circuit shown in Figure P11.9-18 is v s ( t ) = 12 cos ( 5 t ) V

The impedance of the load is 20 + j 15 Ω. Determine the complex power (a) supplied by the source, (b) received by the 20 Ω resistor, (c) received by the coupled inductors and (d) received by the load.

. Figure P11.9-18 Solution: Represent the circuit in the frequency domain as

The coil voltages are given by V1 = j 20 I 1 − j 25 I 2 and V 2 = j 40 I 2 − j 25 I 1

Using KVL 20 I 1 + V1 − 12∠45° = 0 and 20 I 2 + j15 I 2 + V 2 = 0 Substituting the coil voltages: 20 I 1 + j 20 I 1 − j 25 I 2 = 12∠0° 20 I 2 + j15 I 2 + j 40 I 2 − j 25 I 1 = 0 Writhing these equations in matrix form − j 25 ⎤ ⎡ I 1 ⎤ ⎡12∠0° ⎤ ⎡ 20 + j 20 ⎢ ⎥= ⎢ − j 25 20 + j 55⎥⎦ ⎣ I 2 ⎦ ⎢⎣ 0 ⎥⎦ ⎣

I 1 = 0.4676∠ − 22.8° A and I 2 = 0.1998∠ − 2.86° A

Solving gives

(a) The complex power delivered by the source is S=

(12∠0° ) I1 * = (12∠0° )( 0.4676∠ − 22.8° ) * = 2.5855 + j1.0893 VA 2

2

(b) The complex power received by the 20 Ω resistor is I S= 1 2

2

( 20 ) =

( 0.4676 ) 2

2

( 20 ) = 2.1865 + j 0 VA

(c) The complex power received by the coupled inductors is S=

V1I1 * V2 I 2 * + = 0 + j 0.79 VA 2 2

(d) The complex power received by the load is I S= 2 2

2

( 20 + j 15)

( 0.1998) = 2

2

( 20 + j 15) = 0.399 + j 0.299 VA (checked using LNAP 2/6/09)

P11.9-19 Figure P11.9-18a shows a source connected to a 160 Ω load. In Figure P11.9-18a an ideal transformer and capacitor have been inserted between the source and the load. (a) Determine the value of the average power delivered to the 160 Ω load in Figure P11.918a. (b) Determine the values of n and C Figure P11.9-18b that maximize the average power in the load. (c) Using the values of n and C from part (b), determine the value of the average power delivered to the 160 Ω load in Figure P11.9-18b.

(a)

(b) Figure P11.9-19 Solution: (a) Represent the circuit in the frequency domain. Using KVL I=

and

120∠0° = 0.7058∠ − 0.674° A 170 + j 2

P=

0.7058 2 (160 ) = 39.85 W 2

(b)

For maximum power transfer to the 160 Ω load

1 1 ⎛ 1 ⎞ 160 ) = 10 ⇒ n = 4 and − ( j 2 ) = ⎜ 2 ⎟ − j 2 ( n 250 C ⎝4 ⎠

⇒ C = 0.125 mF

(c) Represent the circuit in the frequency domain

I1 =

and

120∠0° 2

⎛1⎞ 10 + j 2 + ⎜ ⎟ ( − j 32 + 160 ) ⎝4⎠

=

120∠0° 1 = 6∠0° A and I 2 = I1 = 1.5∠0° A 20 4 P=

1.5 2 (160 ) = 180 W 2

Section 11.10 The Ideal Transformer P 11.10-1

Find V1, V2, I1, and I2 for the circuit of Figure P 11.10-1, when n = 5. I1 2 + 3j 12 0° V

+ –

1:n

I2

+

+

V1

V2



100 – j75

– Ideal

Figure P 11.10-1 Solution:

Z = (2 + j 3) + I1 =

12∠0° 12∠0° = =2A Z 6

⎛ 100− j 75 ⎞ ⎛ 100− j 75 ⎞ V1 = I1 ⎜ ⎟ = (2) ⎜ ⎟ = 10∠ − 36.9° V 2 n ⎝ ⎠ ⎝ 25 ⎠ V2 = nV1 = 5 (10∠−36.9°) = 50 ∠−36.9° V

I2 =

I1 2 = A n 5

(100− j 75) =6Ω 52

P 11.10-2 A circuit with a transformer is shown in Figure P 11.10-2. (a) Determine the turns ratio. (b) Determine the value of Rab. (c) Determine the current supplied by the voltage source. Answer: (a) n = 5 (b) Rab = 400 Ω i

a

5 mA rms 1:n +

10 V rms

+ –

V0 –

Rab b

Ideal

Figure P 11.10-2 Solution: (a) V0 = (5 ×10−3 )(10, 000) = 50 V n= (b)

(c)

V N2 50 = 0 = = 5 N1 V1 10

Rab =

Is =

1 1 R2 = (10 × 103 ) = 400 Ω 2 25 n

10 10 = = 0.025 A = 25 mA Rab 400

10 kΩ

P 11.10-3 Find the voltage Vc in the circuit shown in Figure P 11.10-3. Assume an ideal transformer. The turns ratio is n = 1/3. Answer: Vc = 21.0 ∠ − 105.3° 30 Ω

j20 Ω

1:n

+

5Ω –j8 Ω

80 –50° V –

+ –

Vc

Ideal

Figure P 11.10-3 Solution:

1 Z 2 = 9 Z 2 = 9 (5 − j8) = 45 − j 72 Ω n2 Using voltage division, the voltage across Z 1 is Z1 =

⎛ ⎞ 45− j 72 V1 = ( 80∠ − 50° ) ⎜ ⎟ = 74.4 ∠ − 73.3° V ⎝ 45− j 72+ 30 + j 20 ⎠

Then

V2 = nV1 =

74.4 ∠ − 73.3° = 24.8 ∠ − 73.3° V 3

Using voltage division again yields ⎛ − j8 ⎞ ⎛ 8∠−90° ⎞ Vc = V2 ⎜ = ( 24.8∠−73.3° ) ⎜ ⎟ ⎟ = 21.0∠ − 105.3° V ⎝ 89∠−58° ⎠ ⎝ 5− j 8 ⎠

P 11.10-4 An ideal transformer is connected in the circuit shown in Figure P 11.10-4, where vs = 50 cos 1000t V and n = N2/N1 = 5. Calculate V1 and V2. 2Ω

vs

+ –

1:5 +

+

v1

v2



200 Ω

– Ideal

Figure P 11.10-4 Solution: n = 5, Z1 =

200

( 5)

2

= 8 Ω ⇒ V1 =

8 ( 50∠0° ) = 40∠0° V ⇒ V2 = n V1 = 200∠0° V 8+ 2

P11.10-5 Figure P11.10-5 shows a load connected to a source through an ideal transformer. The input to the circuit is v s ( t ) = 12 cos ( 5 t ) V

Determine (a) The values of the turns ration, n, and load inductance, L, required for maximum power transfer to the load. (b) The complex power delivered to the transformer by the source. (c) The complex power delivered to the load by the transformer.

Figure P11.10-5 Solution: *

(a) For maximum power transfer:

Equating real parts gives n =

1 1 ⎛ ⎞ 288 + j 20 L ) = ⎜ 18 − j ⎟ = 18 + j 5 2 ( 20 × 0.01 ⎠ n ⎝

( )

5 42 288 = 4 . Equating imaginary parts gives L = =4 H. 20 18

(b) Represent the circuit in the frequency domain as

Replace the transformer and load by an equivalent impedance Z equiv =

1 ( 288 + j 80 ) = 18 + j 5 Ω 42

I1 =

and

12∠0° 12∠0° 1 = = ∠0° A (18 − j 5) + (18 + j 5) 36 3

⎛1 ⎞ V1 = (18 + j 5 ) I 1 = (18 + j 5 ) ⎜ ∠0° ⎟ = 6.227∠15.5° V ⎝3 ⎠

The secondary coil current and voltages

and

1 1⎛1 1 ⎞ I 2 = − I 1 = − ⎜ ∠0° ⎟ = − ∠0° = −0.0833∠0° A 4 4⎝3 12 ⎠ 4 V 2 = V1 = 24.91∠15.5° V 1

The complex power delivered to the transformer by the source. 2

I V1I1 * = 1 + j 0.277 VA = 1 (18 + j 5 ) 2 2 (c) The complex power delivered to the load by the transformer is I VI * − 2 2 = 1 + j 0.277 VA = 2 2 2

2

( 288 + j80 )

(checked using LNAP and MATLAB 2/6/09)

a

Ideal

P 11.10-6 Find the Thévenin equivalent at terminals a–b for the circuit of Figure P 11.10-6 v when v = 16 cos 3t V. Answer: Voc = 12 and Zt = 3.75 Ω





+ –

6Ω b

1:2

Figure P 11.10-6 Solution: Z=

1 ( 2 + 6) = 2 Ω 22

⎞ ⎛ 6 ⎞ ⎛⎛ 2 ⎞ Voc = ⎜ ⎟ ( 2) ⎜ ⎜ ⎟16∠0° ⎟ = 12∠0° V ⎝ 6 + 2 ⎠ ⎝⎝ 2 + 2 ⎠ ⎠

Z=

1 1 2 = Ω 2 ( ) 2 2

⎛ ⎞ 1 1 ⎜ 16∠0° ⎟ I sc = −I 2 = I1 = ⎜ ⎟ = 3.25∠0° A 2 2⎜ 2+ 1 ⎟ ⎝ 2 ⎠ Then Zt =

12∠0° = 3.75∠0° Ω 3.2∠0°



P 11.10-7 Find the input impedance Z for the circuit of Figure P 11.10-7. Answer: Z = 6 Ω

2:1 6Ω Z Ideal

Figure P 11.10-7 Solution: 1 V1 2 V − V2 V1 = I3 = 1 2 4 V V I 2 = I3 − 2 = 1 6 6 V 1 I1 = − I 2 = − 1 2 12 V I T = I 3 + I1 = 1 6 V Z= 1 =6 IT V2 =

P 11.10-8 In less developed regions in mountainous areas, small hydroelectric generators are used to serve several residences (Mackay, 1990). Assume each house uses an electric range and an electric refrigerator, as shown in Figure P 11.10-8. The generator is represented as Vs operating at 60 Hz and V2 = 230 ∠0° V. Calculate the power consumed by each home connected to the hydroelectric generator when n = 5. Source and line 1:n

2 + j3

Vs

+

+ –

10 Ω 20 Ω

V2

20 mH –

Ideal

Range

Refrigerator

Figure P 11.10-8 Solution:

ZL =

1 ⎛ 20 (10+ j 7.54) ⎞ 8.1∠23° = = 0.3 + j 0.13 Ω 52 ⎜⎝ 20+10+ j 7.54 ⎟⎠ 25

V V ( 230 ) = 88 kW/home PL = L = 2 = 2 R 2 2 R L 2( 0.3) 2

2

Therefore, 529 kW are required for six homes.

2

P 11.10-9 Three similar circuits are shown in Figure P 11.10-9. In each of these circuits



vs(t)

vs(t) = 5 cos (4t + 45°) V.

+ –

i2(t)

+

+

v1(t)

4H

3H



Determine v2 (t) for each of the three circuits. Answer: v2(t) = 0 V (a) v2(t) = 1.656 cos (4t + 39°) V (b) (c) v2(t) = 2.88 cos (4t + 45°) V

i1(t)

v2(t)

12 Ω



(a) 8Ω

i1(t)

i2(t) 2H

+ vs(t)

+ –

v1(t)

+ 4H

3H



v2(t)

12 Ω



(b) 8Ω

vs(t)

+ –

i1(t)

i2(t) 10:8.66

+

+

v1(t)

v2(t)



– Ideal

(c)

Figure P 11.10-9 Solution: (a)

Coil voltages: V1 = j16 I1 V2 = j12 I 2 Mesh equations: 8 I1 + V1 − 5∠45° = 0 −12 I 2 − V2 = 0

12 Ω

Substitute the coil voltages into the mesh equations and do some algebra: 8 I1 + j16 I1 = 5∠45° ⇒ I1 = 0.28∠ − 18.4° 12 I 2 + j12 I 2 = 0 ⇒ I 2 = 0 V2 = −12 I 2 = 0

(b)

Coil voltages: V1 = j16 I1 + j8 I 2 V2 = j12 I 2 + j8 I1 Mesh equations: 8 I1 + V1 − 5∠45° = 0 −12 I 2 − V2 = 0 Substitute the coil voltages into the mesh equations and do some algebra: 8 I1 + ( j16 I1 + j8 I 2 ) = 5∠45° 12 I 2 + ( j12 I 2 + j8 I1 ) = 0

I1 = −

12 + j12 3 I 2 = ( j − 1) I 2 j8 2

⎡ ⎤ ⎛3⎞ ⎢( 8 + j16 ) ⎜ 2 ⎟ ( j − 1) + j8⎥ I 2 = 5∠45° ⇒ I 2 = 0.138∠ − 141° ⎝ ⎠ ⎣ ⎦ V2 = −12 I 2 = 1.656∠39°

(c )

Coil voltages and currents: 10 V2 8.66 8.66 I1 = − I2 10 V1 =

Mesh equations: 8 I1 + V1 − 5∠45° = 0 −12 I 2 − V2 = 0 Substitute into the second mesh equation and do some algebra: 2

⎛ 10 ⎞ 8.66 ⎛ 10 ⎞ −12 ⎜ − I1 ⎟ = V1 ⇒ V1 = 12 ⎜ ⎟ I1 ⎝ 8.66 ⎠ 10 ⎝ 8.66 ⎠ 2

⎛ 10 ⎞ 8 I1 + 12 ⎜ ⎟ I1 = 5∠45° ⇒ I1 = 0.208∠45° ⎝ 8.66 ⎠ ⎛ 10 ⎞ 12 (10 ) 0.208∠45° = 2.88∠45° V2 = −12 I 2 = −12 ⎜ − I1 ⎟ = 8.66 ⎝ 8.66 ⎠

P 11.10-10

Find V1 and I1 for the circuit of Figure P 11.10-10 when n = 5. I1 1 + j3 10 0° V

+ –

1:n

I2

+ V1

100 – j75

– Ideal

Figure P 11.10-10 Solution:

I1 =

10∠0° 10∠0° = = 2∠0° A 100 − j 75 (1 + j 3) + ( 4 − j 3) 1 j 3 + + ( ) 52 V1 = ( 4 − j 3) 2∠0° = 10∠ − 36.9° V

P 11.10-11 Determine v2 and i2 for the circuit shown in Figure P 11.10-11 when n = 2. Note that i2 does not enter the dotted terminal. Answer: v2 = 0.68 cos (10t + 47.7°) V i2 = 0.34 cos (10t + 42°) A 20 mF

5 cos 10t V

+ –

i1





1:n + v1

– v2



+

20 mH i2

Ideal

Figure P 11.10-11 Solution:

I1 =

5∠0° 5∠0° 5∠0° = = = 0.68∠42° A 2 + j 0.2 5.5 + j 4.95 7.4∠− 42° − + 5 j 5 ( ) 22 1 I 2 = I1 = 0.34∠42° A 2

V2 = ( 2 + j 0.2 ) I 2 = ( 2.01∠5.7° )( 0.34∠42° ) = 0.68∠47.7° V

so v2 (t ) = 0.68cos (10 t + 47.7°) V and i 2 (t ) = 0.34 cos (10 t + 42°) A

P11.10-12 The circuit shown in Figure P11.10-12 is represented in the frequency domain. Given the line current is I Line = 0.5761∠ − 75.88° A , determine PSource , the average power supplied by the source, PLine , the average power delivered to the line and PLoad , the average power delivered to the load. Hint: Use conservation of (average) power to check your answers. Answer: PSource = 42.15 W, PLine = 0.6638 W and PLoad = 41.49 W.

Figure P11.10-12 Solution: ⎛5⎞ ⎜ ⎟120∠0° ⎝1⎠ = 0.5761∠ − 75.88° A I Line = 2 ⎛5⎞ 4 + j 10 + ⎜ ⎟ (10 + j 40 ) ⎝1⎠ ⎛5⎞ I Source = ⎜ ⎟ I Line = 2.8805∠ − 75.88° A ⎝1⎠ ⎛5⎞ I Load = ⎜ ⎟ I Line = 2.8805∠ − 75.88° A ⎝1⎠ (120 )( 2.8805) cos 0 − −75.88° = 42.16 W PSource = )) ( ( 2 0.5761 2 PLine = ( 4 ) = 0.6638 W 2 2.8805 2 PLoad = (10 ) = 41.49 W 2

P11.10-13 The circuit shown in Figure P11.10-13 is represented in the frequency domain. Determine R and X, the real and imaginary parts of the equivalent impedance, Z eq. Answer: R = 180 Ω and X =110 Ω

Figure P11.10-13 Solution:

2

Z eq

⎛6⎞ = − j 250 + ⎜ ⎟ ( 5 + j 10 ) = 180 + j 110 Ω ⎝1⎠

P11.10-14 Figure P11.10-14 shows a load connected to a source through an ideal transformer. Determine the complex power delivered to the transformer by the source. Answer: S = 698.3 + j1745.7 VA

Figure P11.10-14 Solution: Represent the circuit in the frequency domain. Using the element equation of the ideal transformer: 15 V 2 = (120∠15° ) = 450∠15° V 4 Using Ohm’s law: V2 450∠15° = = 8.3563∠ − 53.2° I2 = 20 + j 50 53.852∠68.2° The power supplied by the source is equal to the power received by the load. The power received by the load is calculated as 2

8.35632 S= ( 20 + j 50 ) = ( 20 + j 50 ) = 698.3 + j1745.7 VA 2 2

I2

(checked using LNAP and MATLAB 2/17/12)

Section 11.11 How Can We Check …? P 11.11-1

Computer analysis of the circuit shown in Figure P 11.11-1 indicates that when vs(t) = 12 cos (4t + 30°) V the mesh currents are given by i1(t) = 2.327 cos (4t – 25.22°) A and i2(t) = 1.229 cos (4t – 11.19°) A Check the results of this analysis by checking that the average power supplied by the voltage source is equal to the sum of the average powers received by the other circuit elements. 4Ω

v1(t)

+ –

i1(t)



2H

i2(t)

2H

Figure P 11.11-1 P11.11-1 The average power supplied by the source is Ps =

(12 )( 2.327 ) cos 2

( 30° − ( −25.22° ) ) = 7.96

W

Capacitors and inductors receive zero average power, so the average power supplied by the voltage source should be equal to the sum of the average powers received by the resistors: PR =

2.327 2 1.1292 4 + ( ) ( 2 ) = 10.83 + 1.27 = 12.10 W 2 2

The average power supplied by the voltage source is not equal to the sum of the average powers received by the other circuit elements. The mesh currents cannot be correct. (What went wrong? It appears that the resistances of the two resistors were interchanged when the data was entered for the computer analysis. Notice that PR =

2.327 2 1.1292 ( 2) + ( 4 ) = 5.41 + 2.55 = 7.96 W 2 2

The mesh currents would be correct if the resistances of the two resistors were interchanged. The computer was used to analyze the wrong circuit.)

P 11.11-2

Computer analysis of the circuit shown in Figure P 11.11-2 indicates that when vs(t) = 12 cos (4t + 30°) V the mesh currents are given by i1(t) = 1.647 cos (4t – 17.92°) A and i2(t) = 1.094 cos (4t – 13.15°) A Check the results of this analysis by checking that the complex power supplied by the voltage source is equal to the sum of the complex powers received by the other circuit elements. 4Ω

vs(t)

+ –

i1(t)



2H

i2(t)

4H

Figure P 11.11-2 Solution: The average complex supplied by the source is Ss =

(12∠30° )(1.647∠ − 17.92° ) * = (12∠30° )(1.647∠17.92° ) = 9.88∠47.92° = 6.62 + j 7.33

2 2 The complex power received by the 4 Ω resistor is S4Ω =

( 4 ×1.647∠ − 17.92° )(1.647∠ − 17.92° ) * = 5.43 + j 0 2

VA

The complex power received by the 2 Ω resistor is S2Ω =

( 2 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 1.20 + j 0 2

VA

The current in the 2 H inductor is

(1.647∠ − 17.92° ) − (1.094∠ − 13.15° ) = 0.5640∠ − 27.19° The complex power received by the 2 H inductor is S 2H =

( j 8 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 1.27 2

VA

The complex power received by the 4 H inductor is S 4H =

( j 16 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 9.57 2

VA

W

S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 1.27 ) + ( 0 + j 9.57 ) = 6.63 + j 10.84 ≠ Ss

The complex power supplied by the voltage source is not equal to the sum of the complex powers received by the other circuit elements. The mesh currents cannot be correct. (Suppose the inductances of the inductors were interchanged. Then the complex power received by the 4 H inductor would be S 4H =

( j 16 × 0.5640∠ − 27.19° )( 0.5640∠ − 27.19° ) * = 0 + j 2.54 2

VA

The complex power received by the 2 H inductor would be S 2H =

( j 8 ×1.094∠ − 13.15° )(1.094∠ − 13.15° ) * = 0 + j 4.79 2

VA

S 4 Ω + S 2 Ω + S 2H + S 4H = ( 5.43 + j 0 ) + (1.20 + j 0 ) + ( 0 + j 2.54 ) + ( 0 + j 4.79 ) = 6.63 + j 7.33 ≈ S s

The mesh currents would be correct if the inductances of the two inductors were interchanged. The computer was used to analyze the wrong circuit.)

P 11.11-3

Computer analysis of the circuit shown in Figure P 11.11-3 indicates that when vs(t) = 12 cos (4t + 30°) V the mesh currents are given by i1(t) = 1.001 cos (4t – 47.01°) A and i2(t) = 0.4243 cos (4t – 15.00°) A Check the results of this analysis by checking that the equations describing currents and voltages of coupled coils are satisfied.

vs(t)

+ –

i1(t)

4H

3H

6H

i2(t)

15 Ω

Figure P 11.11-3 Solution: The voltage across the left coil must be equal to the voltage source voltage. Notice that the mesh currents both enter the undotted ends of the coils. In the frequency domain, the voltage across the left coil is

( j 16 )(1.001∠ − 47.01° ) + ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° + 5.092∠75° = (11.715 + j 10.923) + (1.318 + j 4.918 ) = 13.033 + j 15.841 = 20.513∠50.55° The voltage across the left coil isn’t equal to the voltage source voltage so the computer analysis isn’t correct. What happened? A data entry error was made while doing the computer analysis. Both coils were described as having the dotted end at the top. If both coils had the dot at the top, the equation for the voltage across the right coil would be

( j 16 )(1.001∠ − 47.01° ) − ( j12 )( 0.4243∠ − 15° ) = 16.016∠42.99° − 5.092∠75° = (11.715 + j 10.923) − (1.318 + j 4.918 ) = 10.397 + j 6.005 = 12.007∠30.01° This is equal to the voltage source voltage. The computer was used to analyze the wrong circuit.

Computer analysis of the circuit shown in Figure P 11.11-4 indicates that when vs(t) = 12 cos (4t + 30°) V the mesh currents are given by i1(t) = 25.6 cos (4t + 30°) mA and i2(t) = 64 cos (4t + 30°) mA Check the results of this analysis by checking that the equations describing currents and voltages of ideal transformers are satisfied. P 11.11-4

2:5 vs(t)

+ –

i1(t)

i2(t)

75

Figure P 11.11-4 Solution: First check the ratio of the voltages across the coils.

n1 2 12∠30° = 2.5 ≠ = n2 5 ( 75)( 0.064∠30° ) The transformer voltages don’t satisfy the equations describing the ideal transformer. The given mesh currents are not correct. That’ enough but let’s also check the ratio of coil currents. (Notice that the reference direction of the i2(t) is different from the reference direction that we used when discussing transformers.) n1 2 0.064∠30° = 2.5 ≠ = 0.0256∠30° n2 5 The transformer currents don’t satisfy the equations describing the ideal transformer. n1 1 to be 2.5 instead of 0.4 = In both case, we calculated . This suggests that a data entry 2.5 n2 error was made while doing the computer analysis. The numbers of turns for the two coils was interchanged.

PSpice Problems The input to the circuit shown in Figure SP 11-1 is the voltage of the voltage source, vs(t) = 7.5 sin (5t + 15°) V The output is the voltage across the 4-Ω resistor, vo(t). Use PSpice to plot the input and output voltages. Hint: Represent the voltage source using the PSpice part called VSIN. SP 11-1

8Ω 3H vs(t)

+ –

5H

2H

+ vo(t) –

Figure SP 11-1 Solution: The coupling coefficient is k =

3 = 0.94868 . 2×5



8Ω 3H vs(t)

+ –

5H

2H

+ vo(t)





Figure SP 11-1 The input to the circuit shown in Figure SP 11-1 is the voltage of the voltage source, vs(t) = 7.5 sin (5t + 15°) = 7.5 cos (5t – 75°) V The output is the voltage across the 4-Ω resistor, vo(t). Use PSpice to determine the average power delivered to the coupled inductors. Hint: Represent the voltage source using the PSpice part called VAC. Use printers (PSpice parts called IPRINT and VPRINT) to measure the ac current and voltage of each coil. SP 11-2

Solution: Here is the circuit with printers inserted to measure the coil voltages and currents:

Here is the output from the printers, giving the voltage of coil 2 as 2.498∠107.2°, the current of coil 1 as 0.4484∠-94.57°, the current of coil 2 as 0.6245∠-72.77° and the voltage of coil 1 as 4.292∠-58.74°: FREQ 7.958E-01

VM(N00984) 2.498E+00

VP(N00984) 1.072E+02

FREQ IM(V_PRINT1)IP(V_PRINT1) 7.958E-01 4.484E-01 -9.457E+01 FREQ IM(V_PRINT2)IP(V_PRINT2) 7.958E-01 6.245E-01 -7.277E+01 FREQ 7.958E-01

VM(N00959) VP(N00959) 4.292E+00 -5.874E+01

The power received by the coupled inductors is

p=

( 4.292 )( 0.4484 ) cos

2 = 0.78016 − .78000 ≈ 0

( −58.74 − ( −94.57 ) ) +

( 2.498)( 0.6245) cos 2

(107.2 − ( −72.77 ) )

The input to the circuit shown in Figure SP 11-3 is the voltage of the voltage source, vs(t) = 48 cos (4t + 114°) V The output is the voltage across the 9-Ω resistor, vo(t). Use PSpice to determine the average power delivered to the transformer. Hint: Represent the voltage source using the PSpice part called VAC. SP 11-3

i(t)

8Ω 2:3

+ –

+ 9Ω

vs(t)

vo(t) –

Figure SP 11-3 Solution:

The inductance are selected so that

L2 L1

=

N2 N1

=

3 and the impedance of these inductors are 2

much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open circuit while providing a connected circuit.

Here is the output from the printers, giving the voltage of coil 2 as 24.00∠114.1°, the current of coil 1 as 4.000∠114.0°, the current of coil 2 as 2.667∠-65.90° and the voltage of coil 1 as 16.00∠114.1°: FREQ 6.366E-01

VM(N00984) 2.400E+01

VP(N00984) 1.141E+02

FREQ IM(V_PRINT1)IP(V_PRINT1) 6.366E-01 4.000E+00 1.140E+02 FREQ IM(V_PRINT2)IP(V_PRINT2) 6.366E-01 2.667E+00 -6.590E+01 FREQ 6.366E-01

VM(N00959) 1.600E+01

VP(N00959) 1.141E+02

The power received by the transformer is p=

(16 )( 4 ) cos

(114 − 114 ) +

2 = 32 − 32.004 ≈ 0

( 24 )( 2.667 ) cos 2

(114 − ( −66 ) )

SP 11-4 Determine the value of the input impedance, Zt, of the circuit shown in Figure SP 11-4 at the frequency ω = 4 rad/s. Hint: Connect a current source across the terminals of the circuit. Measure the voltage across the current source. The value of impedance will be equal to the ratio of the voltage to the current. 8Ω

2Ω 5:2 4H

Zt

Figure SP 11-4 Solution:

The inductance are selected so that

L2 L1

=

N2 N1

=

2 and the impedance of these inductors are 5

much larger that other impedance in the circuit. The 1 GΩ resistor simulates an open circuit while providing a connected circuit. FREQ 6.366E-01

VM(N00921) 1.011E+02

VP(N00921) 7.844E+01

VR(N00921) 2.025E+01

VI(N00921) 9.903E+01

The printer output gives the voltage across the current source as

20.25 + j 99.03 = 101.1∠78.44° V The input impedance is

Zt =

20.25 + j 99.03 = 20.25 + j 99.03 Ω = 101.1∠78.44° Ω 1

⎛ 52 ⎞ (We expected Z t = 8 + ⎜ 2 ⎟ ( 2 + j ( 4 )( 4 ) ) = 20.5 + j 100 Ω . That’s about 1% error.) ⎝2 ⎠

Design Problems DP 11-1 A 100-kW induction motor, shown in Figure DP 11-1, is receiving 100 kW at 0.8 power factor lagging. Determine the additional apparent power in kVA that is made available by improving the power factor to (a) 0.95 lagging and (b) 1.0. (c) Find the required reactive power in kVAR provided by a set of parallel capacitors for parts (a) and (b). (d) Determine the ratio of kVA released to the kVAR of capacitors required for parts (a) and (b) alone. Set up a table recording the results of this problem for the two values of power factor attained. Distribution line

Capacitor

Induction motor

Figure DP 11-1 Solution:

(a)

P 100 ⎧ = = 125 kVA | S |= P = 100 kW ⎫ ⎪ pf 0.8 ⇒ ⎬ ⎨ pf=0.8 ⎭ ⎪Q =| S |sin ( cos −1 0.8 ) = 125sin ( 36.9° ) = 75 kVAR ⎩ Now pf = 0.95 so P 100 S= = =105.3 kVA pf 0.95 Q = S sin (cos −1 0.95) =105.3sin (18.2°) =32.9 kVAR so an additional 125 − 105.3 = 19.7 kVA is available.

(b)

Now pf = 1 so P 100 = =100 kVA pf 1 Q = S sin (cos −1 1) = 0

S=

and an additional 125−100= 25 kVA is available. (c)

In part (a), the capacitors are required to reduce Q by 75 – 32.9 = 42.1 kVAR. In part (b), the capacitors are required to reduce Q by 75 – 0 = 75 kVAR.

(d) Corrected power factor Additional available apparent power Reduction in reactive power

0.95 19.7 kVA

1.0 25 kVA

42.1 kVAR

75 kVAR

DP 11-2 Two loads are connected in parallel and supplied from a 7.2-kV rms 60-Hz source. The first load is 50-kVA at 0.9 lagging power factor, and the second load is 45 kW at 0.91 lagging power factor. Determine the kVAR rating and capacitance required to correct the overall power factor to 0.97 lagging. Answer: C = 1.01 μF Solution: This example demonstrates that loads can be specified either by kW or kVA. The procedure is as follows: First load: Second load:

Total load:

⎧⎪ P1 = S1 pf =( 50 )( 0.9 ) = 45 W S1 =50 VA ⎫ ⎬ ⇒ ⎨ −1 pf =0.9 ⎭ ⎪⎩Q1 = S1 sin (cos 0.9) =50sin (25.8°) = 21.8 kVAR P 45 ⎧ S2 = 2 = = 49.45 kVA P2 = 45 W ⎫ ⎪ pf 0.91 ⎬ ⇒ ⎨ pf = 0.91 ⎭ ⎪Q = S sin (cos −1 0.91) = 49.45sin (24.5°) = 20.5 kVAR ⎩ 2 2 S L = S1 + S 2 = (45 + 45) + j(21.8+20.5) = 90 + j 42.3 kVA

Specified load: P 90 ⎧ Ss = s = =92.8 kVA Ps =90 W ⎫ ⎪ pf 0.97 ⎬ ⇒ ⎨ pf = 0.97 ⎭ ⎪Q = S sin (cos −1 0.97) =92.8sin (14.1°) = 22.6 kVAR ⎩ s s The compensating capacitive load is Qc = 42.3 − 22.6 = 19.7 kVAR . The required capacitor is calculated as 2

V (7.2 × 103 ) 2 1 Xc = c = = 2626 Ω ⇒ C = = 1.01 μ F 3 Qc 19.7 × 10 377 (2626)

DP 11-3 (a) Determine the load impedance Zab that will absorb maximum power if it is connected to terminals a–b of the circuit shown in Figure DP 11-3. (b) Determine the maximum power absorbed by this load. (c) Determine a model of the load and indicate the element values. 5Ω

a

100 mH

+

10 cos 100 t V

+ –

– +

vab

0.5 vab



b

Figure DP 11-3 Solution: Find the open circuit voltage: −10 + 5I + j10I − 0.5 Voc = 0 and

10 − Voc 5 Voc = 8∠36.9° = 6.4 + j 4.8 V I=

so Find the short circuit current:

I sc =

10∠0° = 2∠0° A 5

The the Thevenin impedance is:

Zt =

Voc = 3.2 + j 2.4 Ω I sc

The short circuit forces the controlling voltage to be zero. Then the controlled voltage is also zero. Consequently the dependent source has been replaced by a short circuit.

(a)

Maximum power transfer requires Z L = Z t * = 3.2 − j 2.4 Ω .

(c)

ZL can be implemented as the series combination of a resistor and a capacitor with 1 R = 3.2 Ω and C = = 4.17 mF . (100 ) (2.4)

(b)

Pmax =

| Voc |2 64 = = 2.5 W 8R 8 ( 3.2 )

DP 11-4 Select the turns ratio n necessary to provide maximum power to the resistor R of the circuit shown in Figure DP 11-4. Assume an ideal transformer. Select n when R = 4 and 8 Ω. 3 Ω

j4 Ω

1:n R

Vs

+ –

j3 Ω Ideal

Figure DP 11-4 Solution:

When n is selected to deliver maximum power to Z3, the value of the maximum power is given as 2 ⎛ R ⎞ Vs ⎜ 2⎟ ⎝n ⎠ 2 P= 2 2 R⎞ ⎛ 3 ⎛ ⎞ 3 + + + 4 ⎜ ⎟ ⎜ ⎟ n2 ⎠ ⎝ n2 ⎝ ⎠ When R = 4 Ω, 2

n 2 R Vs P= 25n 4 + 48n 2 + 25 4 2 2 3 dP 2 ⎡ 2 n(25n + 48n + 25) − n (100n + 96n ) ⎤ = R Vs ⎢ ⎥ dn (25n 4 + 48n 2 + 25) 2 ⎣ ⎦ 5 4 ⇒ − 50n + 50n = 0 ⇒ n = 1 ⇒ n = 1

0=

When R = 8 Ω, a similar calculation gives n = 1.31.

DP 11-5 An amplifier in a short-wave radio operates at 100 kHz. The load Z2 is connected to a source through an ideal transformer, as shown in Figure DP 11-5. The load is a series connection of a 10-Ω resistance and 10-μH inductance. The Zs consists of a 1-Ω resistance and a 1-μH inductance. (a) Select an integer n in order to maximize the energy delivered to the load. Calculate I2 and the energy to the load. (b) Add a capacitance C in series with Z2 in order to improve the energy delivered to the load. Zs Vs

I2

1:n

+ –

Z2

Ideal

Figure DP 11-5 Solution:

Maximum power transfer requires

Equating real parts gives

10 + j 6.28 = Z1 = (1 + j 0.628 ) * n2 10 = 1 ⇒ n = 3.16 n2

Equating imaginary parts requires jX = − j 0.628 ⇒ 3.162

X = −6.28

This reactance can be realized by adding a capacitance C in series with the resistor and inductor that comprise Z2. Then −6.28 = X = −

1 1 + 6.28 ⇒ C = = 0.1267 μ F 5 5 ( 2π ×10 ) C ( 2π ×10 ) (12.56 )

DP 11-6 A new electronic lamp (e-lamp) has been developed that uses a radio-frequency sinusoidal oscillator and a coil to transmit energy to a surrounding cloud of mercury gas as shown in Figure DP 11-6a. The mercury gas emits ultraviolet light that is transmitted to the phosphor coating, which, in turn, emits visible light. A circuit model of the e-lamp is shown in Figure DP 11.6b. The capacitance C and the resistance R are dependent on the lamp’s spacing design and the type of phosphor. Select R and C so that maximum power is delivered to R, which relates to the phosphor coating (Adler, 1992). The circuit operates at ω0 = 107 rad/s. Phosphorus coating 100 Ω

Mercury vapor

1 mH

Coil Coil

V0 sin ω 0 t V

+ –

C

R

(b) (a) Figure DP 11-6 Solution:

Maximum power transfer requires 1 || R = (100 + j107 ×10−6 ) * 7 j10 C R = 100 − j10 1 + j107 R C

R = (100 − j10) (1 + j107 R C ) = 100 + 108 R C + j (109 R C − 10 ) Equating real and imaginary parts yields R = 100 + 108 R C and 109 R C − 10 = 0 then RC = 10−8

⎛ 10−8 ⎞ 10−8 ⇒ R = 100 + 108 R ⎜ = 0.101 nF ⎟ = 99 Ω ⇒ C = 99 ⎝ R ⎠

Chapter 12: Three-Phase Circuits Exercises Exercise 12.2-1 The Y-connected three-phase voltage source has Vc = 120∠ − 240° V rms . Find the line-to-line voltage Vbc. Answer: 207.8 ∠90° V rms Solution: VC = 120∠ − 240° so VA = 120∠0° and VB = 120∠ − 120°

Vbc = 3 (120 ) ∠ − 90°

1

Exercise 12.3-1 Determine complex power delivered to the three-phase load of a four-wire Yto-Y circuit such as the one shown in Figure 12.3-1. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load impedances are ZA = 80 + j50 Ω, ZB = 80 + j80 Ω, and ZC = 100 – j25 Ω. Answer:

SA = 129 + j81 VA, SB = 90 + j90 VA, SC = 136 – j34 VA, and S = 355 + j137 VA Ia A B

b a

A

Ib B –



+

+

Va

ZB

Vb

ZA

n

N

IN n

ZC

+ –

Vc

c

C Ic C

Figure 12.3-1 Solution: Mathcad analysis (12x4_1.mcd): Vp := 120

Describe the three-phase source: j⋅

Va := Vp⋅ e

π 180

⋅0

j⋅

Vb := Va⋅ e

Describe the three-phase load: Calculate the line currents: IaA = 1.079 − 0.674i IaA = 1.272 180 π

⋅ arg( IaA ) = −32.005

IaA :=

π 180

⋅ − 120

Vc := Va⋅ e

ZA := 80 + j⋅ 50 Va

IbB :=

ZA

j⋅

Vb ZB

IbB = 1.061 π

⋅ 120

ZB := 80 + j⋅ 80

IbB = −1.025 − 0.275i

180

π 180

⋅ arg( IbB) = −165

IcC :=

ZC := 100 − j⋅ 25

Vc ZC IcC = −0.809 + 0.837i IcC = 1.164 180 π

⋅ arg( IcC) = 134.036

2

3

Exercise 12.3-2 Determine complex power delivered to the three-phase load of a four-wire Yto-Y circuit such as the one shown in Figure 12.3-1. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load impedances are ZA = ZB = ZC = 40 + j30 Ω. Answer: SA = SB = SC = 230 + j173 VA and S = 691 + j518 VA Ia A B

b a

A

Ib B –



+

+

Va

ZB

Vb

ZA

n

N

IN n

ZC

+ –

Vc

c

C Ic C

Figure 12.3-1 Solution: Mathcad analysis (12x4_2.mcd): Vp := 120

Describe the three-phase source: j⋅

Va := Vp⋅ e

π 180

⋅0

j⋅

Vb := Va⋅ e

Describe the three-phase load: Calculate the line currents: IaA = 1.92 − 1.44i IaA = 2.4 180 π

⋅ arg( IaA ) = −36.87

IaA :=

π 180

⋅ − 120

Vc := Va⋅ e

ZA := 40 + j⋅ 30 Va

IbB :=

ZA

j⋅

Vb ZB

IbB = 2.4 π

180

⋅ 120

ZB := ZA

IbB = −2.207 − 0.943i

180

π

⋅ arg( IbB) = −156.87

IcC :=

ZC := ZA Vc ZC IcC = 0.287 + 2.383i IcC = 2.4 180 π

⋅ arg( IcC) = 83.13

4

5

Exercise 12.3-3 Determine complex power delivered to the three-phase load of a three-wire Yto-Y circuit such as the one shown in Figure 12.3-2. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load impedances are ZA = 80 + j50 Ω, ZB = 80 + j80 Ω, and ZC = 100 – j25 Ω. Intermediate Answer: VnN = 28.89 ∠–150.5 V rms Answer: S = 392 + j142 VA IaA

b

B

a

A

IbB –



+

+

ZB

Va

ZA

Vb n



N

+

VNn

ZC

Vc

+ –

c

C IcC

Figure 12.3-2 Solution: Mathcad analysis (12x4_3.mcd): Vp := 120

Describe the three-phase source: j⋅

Va := Vp⋅ e

π 180

⋅0

j⋅

Vb := Va⋅ e

Describe the three-phase load:

π 180

⋅ − 120

ZA := 80 + j⋅ 50

j⋅

Vc := Va⋅ e

π 180

⋅ 120

ZB := 80 + j⋅ 80

ZC := 100 − j⋅ 25

Calculate the voltage at the neutral of the load with respect to the neutral of the source:

VnN :=

ZA ⋅ ZC⋅ e

4 j⋅ ⋅ π 3

+ ZA ⋅ ZB⋅ e

2 j⋅ ⋅ π 3

+ ZB⋅ ZC

ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC

VnN = −25.137 − 14.236i

VnN = 28.888

⋅ Vp 180 π

⋅ arg( VnN) = −150.475

6

Calculate the line currents:

IaA :=

IaA = 1.385 − 0.687i IaA = 1.546 180 π Check:

⋅ arg( IaA ) = −26.403

Va − VnN ZA

IbB :=

Vb − VnN ZB

IbB = −0.778 − 0.343i IbB = 0.851 180 π

IcC :=

⋅ arg( IbB) = −156.242

Vc − VnN ZC

IcC = −0.606 + 1.03i IcC = 1.195 180 π

⋅ arg( IcC) = 120.475

IaA + IbB + IcC = 0

Calculate the power delivered to the load: ⎯ ⎯ SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 191.168 + 119.48i Total power delivered to the load:

SB = 57.87 + 57.87i

⎯ SC := IcC⋅ IcC⋅ ZC SC = 142.843 − 35.711i

SA + SB + SC = 391.88 + 141.639i

7

Exercise 12.3-4 Determine complex power delivered to the three-phase load of a three-wire Yto-Y circuit such as the one shown in Figure 12.3-2. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. The load impedances are ZA = ZB = ZC = 40 + j30 Ω. Answer: SA = SB = SC = 230 + j173 VA and S = 691 + j518 VA IaA

b

B

a

A

IbB ZB





+

+

Va

ZA

Vb n



N

+

VNn

ZC

Vc

+ –

c

C IcC

Figure 12.3-2 Solution: Mathcad analysis (12x4_4.mcd): Vp := 120

Describe the three-phase source: j⋅

Va := Vp⋅ e

π 180

⋅0

Describe the three-phase load:

j⋅

Vb := Va⋅ e

π 180

⋅ − 120

ZA := 40 + j⋅ 30

j⋅

Vc := Va⋅ e ZB := ZA

π 180

⋅ 120

ZC := ZA

Calculate the voltage at the neutral of the load with respect to the neutral of the source:

VnN :=

ZA ⋅ ZC⋅ e

4 j⋅ ⋅ π 3

+ ZA ⋅ ZB⋅ e

2 j⋅ ⋅ π 3

+ ZB⋅ ZC

ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC

⋅ Vp

8

9

Exercise 12.5-1 Consider the three-phase circuit shown in Figure 12.5-1. The voltages of the Y-connected source are 360 360 360 ∠–30° V rms, Vb = ∠–150° V rms, and Vc = ∠90° V rms Va = 3 3 3 The Δ-connected load is balanced. The impedance of each phase is ZΔ = 180 ∠45° Ω. Determine the phase and line currents when the line-to-line voltage is 360 V rms. Partial Answer: IAB = 2 ∠45° A rms and IaA = 3.46 ∠15° A rms I aA b

a



I bB

+

+

B

Z3





Va

Vb

n

I AB +

– +

VAB

Z1 VBC

Vc

+

A I CA

VCA

I BC –

c



Z2 + C

I cC

Figure 12.5-1 Solution: Balanced delta load:

phase currents:

line currents:

(See Table 12.5-1) Z Δ = 180∠ − 45°

I AB

VAB 360∠0° = = = 2∠45° A D Z Δ 180∠− 45

I BC =

VBC 360∠−120° = = 2∠− 75° A Z Δ 18∠− 45°

I CA =

VCA 360∠120° = = 2∠165° A ° Z Δ 180∠− 45

I A = I AB − I CA = 2∠45° − 2∠165° = 2 3∠15° A I B = 2 3∠−105° A I C = 2 3∠135° A

10

Exercise 12.6-1 Figure 12.6-1a shows a balanced Y-to-Δ three-phase circuit. The phase voltages of the Y-connected source are Va = 110 ∠0° V rms, Vb = 110 ∠–120° V rms, and Vc = 110 ∠120° V rms. The line impedances are each ZL = 10 + j25 Ω. The impedances of the Δconnected load are each ZΔ = 150 + j270 Ω. Determine the phase currents in the Δ-connected load. Answer: IAB = 0.49 ∠–32.5° A rms, IBC = 0.49 ∠–152.5° A rms, ICA = 0.49 ∠87.5° A rms

Figure 12.6-1a Solution: Three-wire Y-to-Delta Circuit with line impedances Mathcad analysis (12x4_4.mcd): Vp := 110

Describe the three-phase source: j⋅

Va := Vp⋅ e

π 180

⋅0

j⋅

Vb := Va⋅ e

π 180

⋅ − 120

Z1 := 150 + j⋅ 270

Describe the delta connected load:

j⋅

Vc := Va⋅ e

π 180

Z2 := Z1

⋅ 120

Z3 := Z1

Convert the delta connected load to the equivalent Y connected load: ZA :=

Z1⋅ Z3

ZB :=

Z1 + Z2 + Z3

ZA = 50 + 90i

Z2⋅ Z3 Z1 + Z2 + Z3

ZB = 50 + 90i

Z1 + Z2 + Z3

ZC = 50 + 90i

ZaA := 10 + j⋅ 25

Describe the three-phase line:

Z1⋅ Z2

ZC :=

ZbB := ZaA

ZcC := ZaA

Calculate the voltage at the neutral of the load with respect to the neutral of the source:

VnN :=

( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e

4 j⋅ ⋅ π 3

+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e

2 j⋅ ⋅ π 3

+ ( ZbB + ZB) ⋅ ( ZcC + ZC)

( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) − 14

VnN = −1.172 × 10

− 14

+ 1.784i× 10

− 14

VnN = 2.135 × 10

180 π

⋅ Vp

⋅ arg( VnN) = 123.304

11

Calculate the line currents:

IaA :=

IaA = 0.392 − 0.752i IaA = 0.848 180 π Check:

⋅ arg( IaA ) = −62.447

Va − VnN

IbB :=

ZA + ZaA

Vb − VnN ZB + ZbB

IbB = −0.847 + 0.036i IbB = 0.848 180 π

IcC :=

⋅ arg( IbB) = 177.553

Vc − VnN ZC + ZcC

IcC = 0.455 + 0.716i IcC = 0.848 180 π

⋅ arg( IcC) = 57.553

IaA + IbB + IcC = 0

Calculate the phase voltages of the Y-connected load: VAN := IaA ⋅ ZA VAN = 87.311 180 π

⋅ arg( VAN) = −1.502

VBN := IbB⋅ ZB VBN = 87.311 180 π

⋅ arg( VBN) = −121.502

VCN := IcC⋅ ZC VCN = 87.311 180 π

⋅ arg( VCN) = 118.498

Calculate the line-to-line voltages at the load: VAB := VAN − VBN VAB = 151.227 180 π

⋅ arg( VAB) = 28.498

VBC:= VBN − VCN VBC = 151.227 180 π

⋅ arg( VBC) = −91.502

VCA := VCN − VAN VCA = 151.227 180 π

⋅ arg( VCA) = 148.498

Calculate the phase currents of the Δ -connected load: IAB :=

VAB Z3

IAB = 0.49 180 π

⋅ arg( IAB) = −32.447

IBC :=

VBC Z1

IBC = 0.49 180 π

⋅ arg( IBC) = −152.447

ICA :=

VCA Z2

ICA = 0.49 180 π

⋅ arg( ICA) = 87.553

12

Exercise 12.7-1 Figure 12.6-1a shows a balanced Y-to-Δ three-phase circuit. The phase voltages of the Y-connected source are Va = 110 ∠0° V rms, Vb = 110 ∠–120° V rms, and Vc = 110 ∠120° V rms. The line impedances are each ZL = 10 + j25 Ω. The impedances of the Δconnected load are each ZΔ = 150 + j 270 Ω. Determine the average power delivered to the Δconnected load. Intermediate Answer: IaA = 0.848 ∠–62.5° A rms and VAN = 87.3 ∠–1.5° V rms Answer: P = 107.9 W

Figure 12.6-1a Solution: Continuing Ex. 12.6-1: Calculate the power delivered to the load: ⎯ ⎯ SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 35.958 + 64.725i Total power delivered to the load:

SB = 35.958 + 64.725i

⎯ SC := IcC⋅ IcC⋅ ZC SC = 35.958 + 64.725i

SA + SB + SC = 107.875 + 194.175i

13

Exercise 12.8-1 The line current to a balanced three-phase load is 24 A rms. The line-to-line voltage is 450 V rms, and the power factor of the load is 0.47 lagging. If two wattmeters are connected as shown in Figure 12.8-2, determine the reading of each meter and the total power to the load. Answer: P1 = –371 W, P2 = 9162 W, and P = 8791 W IB W2

B IA A

+

W1

B

A Z

Z VAC

Z C



C

FIGURE 12.8-2 Solution: P1 = VAB I A cos(θ +30° ) + VCB I C cos(θ − 30° )= P + P2 1

pf = .4 lagging ⇒ θ = 61.97

°

So P T = 450(24) ⎡⎣ cos 91.97° + cos 31.97° ⎤⎦ = 8791 W ∴ P 1 = − 371 W P2 = 9162 W

14

Exercise 12.8-2 The two wattmeters are connected as shown in Figure 12.8-2 with P1 = 60 kW and P2 = 40 kW, respectively. Determine (a) the total power and (b) the power factor. Answers: (a) 100 kW (b) 0.945 leading IB W2

B IA A

W1

+

A

B Z

Z VAC

Z C



C

Figure 12.8-2 Ex. 12.8-2 Consider Fig. 12.8-1 with P1 = 60 kW P2 = 40 kW . (a.) P = P1 + P2 = 100 kW (b.) use equation 12.9-7 to get tan θ = 3

P2 − P1 40−60 = 3 = − .346 ⇒ θ = − 19.11° PL + P2 100

then pf = cos ( − 19.110°) = 0.945 leading

15

Problems Section 12-2: Three Phase Voltages P 12.2-1

A balanced three-phase Y-connected load has one phase voltage: Vc = 277 ∠45° V rms

The phase sequence is abc. Find the line-to-line voltages VAB, VBC, and VCA. Draw a phasor diagram showing the phase and line voltages. Solution: Given VC = 277 ∠45° and an abc phase sequence: VA = 277 ∠ ( 45−120 ) ° = 277 ∠ − 75°

VB = 277 ∠( 45° +120 )° = 277 ∠165° VAB = VA − VB =( 277 ∠− 75° )−( 277 ∠165° ) =( 71.69 − j 267.56 ) −( −267.56+ j 71.69 ) =339.25− j 339.25 = 479.77 ∠− 45°  480 ∠− 45°

Similarly: VBC = 480 ∠ − 165° and VCA = 480 ∠75°

P 12.2-2

A three-phase system has a line-to-line voltage VBA = 12,470 ∠–35° V rms

with a Y load. Find the phase voltages when the phase sequence is abc.

Solution: VAB 3∠30° = 12470 ∠145° V

VAB = VA × 3∠30° ⇒ VA = In our case: So

VAB = −VBA = − (12470 ∠−35° ) VA =

12470 ∠145° = 7200∠115° 3∠30°

Then, for an abc phase sequence: VC = 7200 ∠ (115 + 120 ) ° = 7200 ∠235° = 7200 ∠ − 125° VB = 7200 ∠ (115 − 120 ) ° = 7200 ∠ − 5° V

1

P 12.2-3

A three-phase system has a line-to-line voltage Vab = 1500 ∠30° V rms

with a Y load. Determine the phase voltage.

Solution: Vab = Va × 3∠30° ⇒ Va =

Vab 3∠30°

In our case, the line-to-line voltage is So the phase voltage is

Vab = 1500 ∠30° V 1500 ∠30° Va = = 866∠0° V 3∠30°

2

Section 12.3 P 12.3-1

The Y-to-Y Circuit

Consider a three-wire Y-to-Y circuit. The voltages of the Y-connected source are

Va = (208/ 3 ) ∠0° V rms, Vb = (208/ 3 ) ∠–120° V rms, and Vc = (208/ 3 )∠120° V rms. The Y-connected load is balanced. The impedance of each phase is Z = 12 ∠30° Ω. (a) (b) (c) (d)

Find the phase voltages. Find the line currents and phase currents. Show the line currents and phase currents on a phasor diagram. Determine the power dissipated in the load.

Solution: Balanced, three-wire, Y-Y circuit: where Z A = Z B = Z C = 12∠30 = 10.4 + j 6

MathCAD analysis (12p4_1.mcd): Vp :=

Describe the three-phase source:

208 3

π

j⋅

Va := Vp⋅ e

⋅0

180

j⋅

⋅ − 120

j⋅

180

Vb := Va⋅ e

Describe the balanced three-phase load:

π

Vc := Va⋅ e

ZA := 10.4 + j⋅ 6

π

⋅ 120

180

ZB := ZA

ZC := ZB

Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero:

VnN :=

ZA ⋅ ZC⋅ e

4 j⋅ ⋅ π 3

+ ZA ⋅ ZB⋅ e

IaA :=

Va − VnN ZA

IaA = 8.663 − 4.998i

Check:

− 14

⋅ Vp

IbB :=

VnN = 2.762 × 10 Vb − VnN

IbB = −8.66 − 5.004i

IaA = 10.002 π

+ ZB⋅ ZC

ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC

Calculate the line currents:

180

2 j⋅ ⋅ π 3

IbB = 10.002 180

⋅ arg( IaA ) = −29.982

π − 15

IaA + IbB + IcC = 4.696 × 10

IcC :=

ZB

Vc − VnN ZC −3

IcC = −3.205 × 10

+ 10.002i

IcC = 10.002

⋅ arg( IbB) = −149.982

180 π

⋅ arg( IcC) = 90.018

− 14

− 1.066i× 10

1

Calculate the power delivered to the load: ⎯ ⎯ SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB 3

SA = 1.04 × 10 + 600.222i Total power delivered to the load:

⎯ SC := IcC⋅ IcC⋅ ZC

3

SB = 1.04 × 10 + 600.222i 3

3

SC = 1.04 × 10 + 600.222i 3

SA + SB + SC = 3.121 × 10 + 1.801i× 10

Consequently: (a) The phase voltages are 208 ∠0° = 120∠0° V rms, Vb = 120∠ − 120° V rms and Vc = 120∠120° V rms 3 (b) The currents are equal the line currents (c) Va =

I a = I aA = 10∠ − 30° A rms, I b = I bB = 10∠ − 150° A rms and I c = I cC = 10∠90° A rms

(d) The power delivered to the load is S = 3.121 + j1.801 kVA .

2

P 12.3-2 A balanced three-phase Y-connected supply delivers power through a three-wire plus neutral-wire circuit in a large office building to a three-phase Y-connected load. The circuit operates at 60 Hz. The phase voltages of the Y-connected source are Va = 120 ∠0° V rms, Vb = 120 ∠–120° V rms, and Vc = 120 ∠120° V rms. Each transmission wire, including the neutral wire, has a 2-Ω resistance, and the balanced Y load has a 10-Ω resistance in series with 100 mH. Find the line voltage and the phase current at the load. Solution: Balanced, three-wire, Y-Y circuit: where Va = 120∠0° Vrms, Vb = 120∠ − 120° Vrms and Vc = 120∠120° Vrms

Z A = Z B = Z C = 10 + j ( 2 × π × 60 ) (100 × 10−3 )

and

Z aA

= 10 + j 37.7 Ω = Z bB = Z cC = 2 Ω

Mathcad Analysis (12p4_2.mcd):

3

Consequently, the line-to-line voltages at the source are: Vab = Va × 3∠30° = 120∠0°× 3∠30° = 208∠30° Vrms,

Vbc = 208∠ − 120° Vrms and Vca = 208∠120° Vrms The line-to-line voltages at the load are:

VAB = VA × 3∠30° = 118.3∠3°× 3∠30° = 205∠33° Vrms, Vbc = 205∠ − 117° Vrms and Vca = 205∠123° Vrms and the phase currents are I a = I aA = 10∠ − 72° A rms, I b = I bB = 3∠168° A rms and I c = I cC = 3∠48° A rms

4

P 12.3-3 A Y-connected source and load are shown in Figure P 12.3-3. (a) Determine the rms value of the current iA (t). (b) Determine the average power delivered to the load.

ia(t)

+





+

1H

1H

12 Ω

12 Ω

10 cos 16t V

10 cos (16t – 120°) V Source

Load 12 Ω

– +

10 cos (16t + 120°) V 1H

Figure P 12.3-3

Solution: Balanced, three-wire, Y-Y circuit:

where Va = 10∠0° V = 7.07∠0° V rms, Vb = 7.07∠ − 120° V rms and Vc = 7.07∠120° V rms and

Z A = Z B = Z C = 12 + j (16 )(1) = 12 + j16 Ω

MathCAD analysis (12p4_3.mcd):

5

Vp :=

Describe the three-phase source:

10 2

π

j⋅

Va := Vp⋅ e

⋅0

j⋅

180

π

⋅ − 120

j⋅

180

Vb := Va⋅ e

Vc := Va⋅ e

ZA := 12 + j⋅ 16

Describe the balanced three-phase load:

π

⋅ 120

180

ZB := ZA

ZC := ZB

Check: The voltage at the neutral of the load with respect to the neutral of the source should be zero:

VnN :=

ZA ⋅ ZC⋅ e

4 j⋅ ⋅ π 3

+ ZA ⋅ ZB⋅ e

IaA :=

IaA = 0.212 − 0.283i IaA = 0.354 π

+ ZB⋅ ZC

ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC

Calculate the line currents:

180

2 j⋅ ⋅ π 3

⋅ arg( IaA ) = −53.13

Va − VnN ZA

− 15

⋅ Vp

IbB :=

VnN = 1.675 × 10 Vb − VnN

IbB = −0.351 − 0.042i IbB = 0.354 180 π

⋅ arg( IbB) = −173.13

Calculate the power delivered to the load: ⎯ ⎯ SA := IaA ⋅ IaA ⋅ ZA SB := IbB⋅ IbB⋅ ZB SA = 1.5 + 2i Total power delivered to the load:

SB = 1.5 + 2i

IcC :=

ZB

Vc − VnN ZC

IcC = 0.139 + 0.325i IcC = 0.354 180 π

⋅ arg( IcC) = 66.87

⎯ SC := IcC⋅ IcC⋅ ZC

SC = 1.5 + 2i

SA + SB + SC = 4.5 + 6i

Consequently (a) The rms value of ia(t) is 0.354 A rms. (b) The average power delivered to the load is P = Re {S} = Re {4.5 + j 6} = 4.5 W

6

P 12.3-4 An unbalanced Y-Y circuit is shown in Figure P 12.3-4. Find the average power delivered to the load. Hint:

VNn(ω) = 27.4 ∠–63.6 V

Answer: 436.4 W a

10 Ω

5 mH

A

20 Ω

60 mH

b

10 Ω

5 mH

B

40 Ω

40 mH

C

60 Ω

+



100 cos (377t) –

N

+

n

100 cos (377t +120°) c

10 Ω

5 mH

20 mH

+



100 cos (377t + 240°) Source

Line

Load

Figure P 12.3-4

Solution: Unbalanced, three-wire, Y-Y circuit:

where Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms

Z A = 20 + j ( 377 ) ( 60 ×10−3 ) = 20 + j 22.6 Ω, Z B = 40 + j ( 377 ) ( 40 × 10−3 ) = 40 + j 15.1 Ω Z C = 60 + j ( 377 ) ( 20 × 10−3 ) = 60 + j 7.54 Ω

and

Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω

Mathcad Analysis (12p4_4.mcd):

7

Vp := 100

Describe the three-phase source: π

j⋅

Va := Vp⋅ e

⋅0

j⋅

180

π

⋅ 120

180

Vb := Va⋅ e

π

j⋅

⋅ − 120

180

Vc := Va⋅ e

Enter the frequency of the 3-phase source: ω := 377 Describe the three-phase load:

ZA := 20 + j⋅ ω⋅ 0.06

ZB := 40 + j⋅ ω⋅ 0.04

Describe the three-phase line:

ZaA := 10 + j⋅ ω⋅ 0.005 ZbB := ZaA

ZC := 60 + j⋅ ω⋅ 0.02 ZcC := ZaA

Calculate the voltage at the neutral of the load with respect to the neutral of the source:

VnN :=

( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e

4 j⋅ ⋅ π 3

+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e

2 j⋅ ⋅ π 3

+ ( ZbB + ZB) ⋅ ( ZcC + ZC)

( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC)

VnN = 12.209 − 24.552i Calculate the line currents:

IaA :=

IaA = 2.156 − 0.943i IaA = 2.353 180 π

180

VnN = 27.42 Va − VnN ZA + ZaA

IbB :=

π

IcC :=

ZB + ZbB

180 π

⋅ arg( IbB) = 100.492

SA = 55.382 + 62.637i Total power delivered to the load:

(

ZC + ZcC

IcC = 1.244

Calculate the power delivered to the load: ⎯ ⎯ IbB⋅ IbB IaA ⋅ IaA ⋅ ZA SB := ⋅ ZB SA := 2 2

)

Vc − VnN

IcC = −0.99 − 0.753i

IbB = 2.412

⋅ arg( IaA ) = −23.619

(

⋅ arg( VnN) = −63.561

Vb − VnN

IbB = −0.439 + 2.372i

⋅ Vp

)

SB = 116.402 + 43.884i

180 π

⋅ arg( IcC) = −142.741

SC :=

⎯ (IcC ⋅ IcC) 2

⋅ ZC

SC = 46.425 + 5.834i

SA + SB + SC = 218.209 + 112.355i

The average power delivered to the load is P = Re {S} = Re {218.2 + j112.4} = 218.2 W

8

P 12.3-5 A balanced Y-Y circuit is shown in Figure P 12.3-5. Find the average power delivered to the load. a

10 Ω

5 mH

A

20 Ω

60 mH

b

10 Ω

5 mH

B

20 Ω

60 mH

C

20 Ω

+



100 cos (377t) –

N

+

n

100 cos (377t +120°) c

10 Ω

5 mH

60 mH

+



100 cos (377t + 240°) Source

Line

Load

Figure P 12.3-5

Solution: Balanced, three-wire, Y-Y circuit:

where Va = 100∠0° V = 70.7∠0° V rms, Vb = 70.7∠ − 120° V rms and Vc = 7.07∠120° V rms Z A = Z B = Z C = 20 + j ( 377 ) ( 60 × 10−3 ) = 20 + j 22.6 Ω and

Z aA = Z bB = Z cC = 10 + j ( 377 ) ( 5 × 10−3 ) = 10 + j 1.89 Ω

Mathcad Analysis (12p4_5.mcd):

9

Vp := 100

Describe the three-phase source: π

j⋅

Va := Vp⋅ e

⋅0

j⋅

180

π

⋅ 120

180

Vb := Va⋅ e

π

j⋅

⋅ − 120

180

Vc := Va⋅ e

Enter the frequency of the 3-phase source: ω := 377 Describe the three-phase load:

ZA := 20 + j⋅ ω⋅ 0.06

ZB := ZA

ZC := ZA

Describe the three-phase line:

ZaA := 10 + j⋅ ω⋅ 0.005

ZbB := ZaA

ZcC := ZaA

Calculate the voltage at the neutral of the load with respect to the neutral of the source:

VnN :=

( ZaA + ZA ) ⋅ ( ZcC + ZC) ⋅ e

4 j⋅ ⋅ π 3

+ ( ZaA + ZA ) ⋅ ( ZbB + ZB) ⋅ e

2 j⋅ ⋅ π 3

+ ( ZbB + ZB) ⋅ ( ZcC + ZC)

( ZaA + ZA ) ⋅ ( ZcC + ZC) + ( ZaA + ZA ) ⋅ ( ZbB + ZB) + ( ZbB + ZB) ⋅ ( ZcC + ZC) − 15

VnN = −8.982 × 10

− 14

Calculate the line currents:

IaA :=

IaA = 1.999 − 1.633i IaA = 2.582 180 π

− 14

+ 1.879i× 10

⋅ arg( IaA ) = −39.243

VnN = 2.083 × 10

Va − VnN ZA + ZaA

IbB :=

180 π

⋅ arg( VnN) = 115.55

Vb − VnN

IcC :=

ZB + ZbB

IbB = 0.415 + 2.548i

π

⋅ arg( IbB) = 80.757

Vc − VnN ZC + ZcC

IcC = −2.414 − 0.915i

IbB = 2.582 180

⋅ Vp

IcC = 2.582 180 π

⋅ arg( IcC) = −159.243

Calculate the power delivered to the load: SA :=

⎯ (IaA ⋅ IaA )

⋅ ZA 2 SA = 66.645 + 75.375i

Total power delivered to the load:

SB :=

⎯ (IbB ⋅ IbB)

⋅ ZB 2 SB = 66.645 + 75.375i

SC :=

⎯ (IcC ⋅ IcC)

⋅ ZC 2 SC = 66.645 + 75.375i

SA + SB + SC = 199.934 + 226.125i

The average power delivered to the load is P = Re {S} = Re {200 + j 226} = 200 W

10

P 12.3-6 An unbalanced Y-Y circuit is shown in Figure P 12.3-6 Find the average power delivered to the load. Hint: VNn (ω) = 1.755∠–29.5 V Answer: 436.4 W a

A



1H

b

B



2H

c

C



+



10 cos (4t – 90°) –

N

+

n

10 cos (4t + 150°) 2H

+



10 cos (4t + 30°) Source

Line

Load

Figure P 12.3-6 Solution: Unbalanced, three-wire, Y-Y circuit:

where Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms and Z A = 4 + j ( 4 )(1) = 4 + j 4 Ω, Z B = 2 + j ( 4 )( 2 ) = 2 + j 8 Ω and Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω

Mathcad Analysis (12p4_6.mcd):

11

Vp := 10

Describe the three-phase source: π

j⋅

Va := Vp⋅ e

⋅ − 90

j⋅

180

π

⋅ 150

180

Vb := Vp⋅ e

π

j⋅

Vc := Vp⋅ e

⋅ 30

180

Enter the frequency of the 3-phase source: ω := 4 ZA := 4 + j⋅ ω⋅ 1

Describe the three-phase load:

ZB := 2 + j⋅ ω⋅ 2

ZC := 4 + j⋅ ω⋅ 2

Calculate the voltage at the neutral of the load with respect to the neutral of the source: VnN :=

ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC

VnN = 1.528 − 0.863i Calculate the line currents:

IaA :=

IaA = −1.333 − 0.951i IaA = 1.638 180 π

180

VnN = 1.755 Va − VnN ZA

IbB :=

π Vb − VnN

IbB = 0.39 + 1.371i

180 π

)

⋅ arg( IbB) = 74.116

SA = 5.363 + 5.363i Total power delivered to the load:

(

Vc − VnN ZC

IcC = 0.943 − 0.42i IcC = 1.032

Calculate the power delivered to the load: ⎯ ⎯ IaA ⋅ IaA IbB⋅ IbB SA := ⋅ ZA SB := ⋅ ZB 2 2

(

IcC :=

ZB

IbB = 1.426

⋅ arg( IaA ) = −144.495

⋅ arg( VnN) = −29.466

)

SB = 2.032 + 8.128i

180 π

⋅ arg( IcC) = −24.011

SC :=

⎯ (IcC ⋅ IcC) 2

⋅ ZC

SC = 2.131 + 4.262i

SA + SB + SC = 9.527 + 17.754i

The average power delivered to the load is P = Re {S} = Re {9.527 + j17.754} = 9.527 W

12

P 12.3-7 A balanced Y-Y circuit is shown in Figure P 12.3-7. Find the average power delivered to the load. a

A



2H

b

B



2H

c

C



+



10 cos (4t – 90°) –

N

+

n

10 cos (4t + 150°) 2H

+



10 cos (4t + 30°) Source

Line

Load

Figure P 12.3-7 Solution: Balanced, three-wire, Y-Y circuit:

where Va = 10∠ − 90° V = 7.07∠ − 90° V rms, Vb = 7.07∠150° V rms and Vc = 7.07∠30° V rms and Z A = Z B = Z C = 4 + j ( 4 )( 2 ) = 4 + j 8 Ω

Mathcad Analysis (12p4_7.mcd):

13

Vp := 10

Describe the three-phase source: j⋅

Va := Vp⋅ e

π

⋅ − 90

180

π

j⋅

⋅ 150

j⋅

180

Vb := Vp⋅ e

Vc := Vp⋅ e

π

⋅ 30

180

Enter the frequency of the 3-phase source: ω := 4 Describe the three-phase load:

ZA := 4 + j⋅ ω⋅ 2

ZB := ZA

ZC := ZA

The voltage at the neutral of the load with respect to the neutral of the source should be zero: VnN :=

ZA ⋅ ZC⋅ Vb + ZA ⋅ ZB⋅ Vc + ZB⋅ ZC⋅ Va

Calculate the line currents:

IaA :=

IaA = −1 − 0.5i

π

Va − VnN ZA

IbB :=

Vb − VnN

IbB = 1.118

⋅ arg( IaA ) = −153.435

180 π

)

⋅ arg( IbB) = 86.565

SA = 2.5 + 5i Total power delivered to the load:

(

Vc − VnN ZC

IcC = 0.933 − 0.616i IcC = 1.118

Calculate the power delivered to the load: ⎯ ⎯ IaA ⋅ IaA IbB⋅ IbB SA := ⋅ ZA SB := ⋅ ZB 2 2

(

IcC :=

ZB

IbB = 0.067 + 1.116i

IaA = 1.118 180

− 15

VnN = 1.517 × 10

ZA ⋅ ZC + ZA ⋅ ZB + ZB⋅ ZC

)

SB = 2.5 + 5i

180 π

⋅ arg( IcC) = −33.435

SC :=

⎯ (IcC ⋅ IcC)

2

⋅ ZC

SC = 2.5 + 5i

SA + SB + SC = 7.5 + 15i

The average power delivered to the load is P = Re {S} = Re {7.5 + j15} = 7.5 W

14

Section 12-4: The Δ- Connected Source and Load A balanced three-phase Δ-connected load has one line current:

P 12.4-1

IB = 50 ∠–40° A rms Find the phase currents IBC, IAB, and ICA. Draw the phasor diagram showing the line and phase currents. The source uses the abc phase sequence. Solution: Given I B = 50∠ − 40° A rms and assuming the abc phase sequence we have I A = 50∠80° A rms and I C = 50∠200° A rms From Eqn 12.6-4 I A = I AB × 3∠ − 30° ⇒ I AB = so

IA 3∠ − 30°

50∠80° = 28.9∠110° A rms 3∠−30° = 28.9∠ − 10° A rms and ICA = 28.9∠ − 130° A rms

I AB = I BC

P 12.4-2 A three-phase circuit has two parallel balanced Δ loads, one of 5-Ω resistors and one of 20-Ω resistors. Find the magnitude of the total line current when the line-to-line voltage is 480 V rms. Solution: The two delta loads connected in parallel are equivalent to a single delta load with Z Δ = 5 || 20 = 4 Ω The magnitude of phase current is 480 Ip = = 120 A rms 4 The magnitude of line current is I L = 3 I p = 208 A rms

1

Section 12-5: The Y- to Δ- Circuit P 12.5-1

Consider a three-wire Y-to-Δ circuit. The voltages of the Y-connected source are

Va = (208/ 3 ) ∠–30°V rms, Vb = (208/ 3 ) ∠–150° V rms, and Vc = (208/ 3 ) ∠90°V rms. The Δ-connected load is balanced. The impedance of each phase is Z = 12 ∠30° Ω. Determine the line currents and calculate the power dissipated in the load. Answer: P = 9360 W Solution: We have a delta load with Z = 12∠30° . One phase current is

I AB

⎛ 208 ⎞ ⎛ 208 ⎞ ∠−30° ⎟ −⎜ ∠−150° ⎟ ⎜ V V −V 3 ⎠ ⎝ 3 ⎠ = 208∠0° = 17.31∠ − 30° A rms = AB = A B = ⎝ 12∠30° 12∠30° Z Z

The other phase currents are I BC = 17.31∠ − 150° A rms and I CA = 17.31∠90° A rms One line currents is I A = I AB × 3∠ − 30° = (17.31∠ − 30° ) ×

(

)

3∠ − 30° = 30∠ − 60° A rms

The other line currents are I B = 30∠ − 180° A rms and I C = 30∠60° A rms The power delivered to the load is P = 3(

208 ) (30) cos ( 60° − 30° ) = 9360 W 3

1

P 12.5-2 A balanced Δ-connected load is connected by three wires, each with a 4-Ω resistance, to a Y source with Va = (480/ 3 )∠–30° V rms, Vb = (480/ 3 ) ∠–150° V rms, and Vc = (480/ 3 )∠–90° V rms. Find the line current IA when ZΔ = 39 ∠–40° Ω. Answer: IA = 17 ∠0.9 A Solution: The balanced delta load with Z Δ = 39∠− 40° Ω is equivalent to a balanced Y load with ZY =

ZΔ = 13∠ − 40° = 9.96 − j 8.36 Ω 3

Z T = Z Y + 4 = 13.96 − j 8.36 = 16.3∠ − 30.9 Ω 480 ∠−30° 3 then I A = = 17∠0.9° A rms ° 16.3 ∠−30.9

2

P 12.5-3 The balanced circuit shown in Figure P 12.5-3 has Vab = 380 ∠30°V rms. Determine the phase currents in the load when Z = 3 + j4 Ω. Sketch a phasor diagram. a

A

Z n + –

– +

Vab Z

c

b

– +

Z

B

C

Figure P 12.5-3

Solution: Vab = Va × 3∠30° ⇒ Va =

Vab 3∠30°

In our case, the given line-to-line voltage is

Vab = 380 ∠30° V rms 380 ∠30° So one phase voltage is Va = = 200∠0° V rms 3∠30° So VAB = 380∠30° V rms VA = 220∠0° V rms VBC = 380∠-90° V rms

VB = 220∠−120° V rms

VCA = 380∠150° V rms

VC = 220∠120° V rms

One phase current is

IA =

VA 220∠0° = 44∠ − 53.1° A rms 3+ j4 Z

The other phase currents are

I B = 44∠−173.1° A rms amd I C = 44∠66.9° A rms

3

P 12.5-4 The balanced circuit shown in Figure P 12.5-3 has Vab = 380 ∠0° V rms. Determine the line and phase currents in the load when Z = 9 + j12 Ω.

Solution: Vab = Va × 3∠30° ⇒ Va =

Vab 3∠30°

In our case, the given line-to-line voltage is

Vab = 380 ∠0° V rms Va =

So one phase voltage is So

380 ∠0° = 200∠ − 30° V rms 3∠30°

Vab = 380∠0° V rms

Va = 220∠ − 30° V rms

Vbc = 380∠-120° V rms

Vb = 220∠−150° V rms

Vca = 380∠120° V rms

Vc = 220∠90° V rms

One phase current is

IA =

Va 220∠−30° = = 14.67∠ − 83.1° A rms Z 9+ j12

The other phase currents are

I B = 14.67∠ − 203.1° A rms and I C = 14.67∠36.9° A rms

4

Section 12-6: Balanced Three-Phase Circuits P 12.6-1 The English Channel Tunnel rail link is supplied at 25 kV rms from the United Kingdom and French grid systems. When there is a grid supply failure, each end is capable of supplying the whole tunnel but in a reduced operational mode. The tunnel traction system is a conventional catenary (overhead wire) system similar to the surface main line electric railway system of the United Kingdom and France. What makes the tunnel traction system different and unique is the high density of traction load and the end-fed supply arrangement. The tunnel traction load is considerable. For each half tunnel, the load is 180 MVA (Barnes and Wong, 1991). Assume that each line-to-line voltage of the Y-connected source is 25 kV rms and the three-phase system is connected to the traction motor of an electric locomotive. The motor is a Y-connected load with Z = 150 ∠25° Ω. Find the line currents and the power delivered to the traction motor. Solution: Va =

IA

25 ×103 ∠0° Vrms 3

25 ×103 ∠0° Va 3 = = = 96∠ − 25° A rms 150 ∠25° Z

⎛ 25 ⎞ P = 3 Va I A cos (θ v -θ I ) = 3 ⎜ ×103 ⎟ 96 cos(0 − 25°) = 3.77 MW ⎝ 3 ⎠

P 12.6-2 A three-phase source with a line voltage of 45 kV rms is connected to two balanced loads. The Y-connected load has Z = 10 + j20 Ω, and the Δ load has a branch impedance of 50 Ω. The connecting lines have an impedance of 2 Ω. Determine the power delivered to the loads and the power lost in the wires. What percentage of power is lost in the wires? Solution: Convert the delta load to an equivalent Y connected load: ˆ = 50 Ω ZΔ ⇒ Z Y 3 To get the per-phase equivalent circuit shown to the right: The phase voltage of the source is Z Δ = 50 Ω

Va =

45×103 ∠0° = 26∠0° kV rms 3

The equivalent impedance of the load together with the line is 50 3 + 2 = 12 + j 5 = 13∠22.6° Ω Z eq = 50 10 + j 20 + 3

(10 + j 20 )

The line current is Ι aA

Va 26 × 103 ∠0° = = = 2000∠ − 22.6° A rms Z eq 13∠22.6°

The power delivered to the parallel loads (per phase) is 50 ⎫ ⎧ ⎪ (10 + j 20 ) 3 ⎪ 2 6 PLoads = I aA × Re ⎨ ⎬ = 4 ×10 × 10 = 40 MW 50 ⎪10 + j 20 + ⎪ 3 ⎭ ⎩

The power lost in the line (per phase) is PLine = I aA × Re {Z Line } = 4 × 106 × 2 = 8 MW 2

The percentage of the total power lost in the line is PLine 8 × 100% = × 100% = 16.7% PLoad + PLine 40 +8

P 12.6-3 A balanced three-phase source has a Y-connected source with va = 5 cos (2t + 30°) connected to a three-phase Y load. Each phase of the Y-connected load consists of a 4-Ω resistor and a 4-H inductor. Each connecting line has a resistance of 2 Ω. Determine the total average power delivered to the load. Solution:

Ia =

Va 5∠30° = = 0.5∠ − 23° A ∴ I a = 0.5 A Z T 6 + j8 2

PLoad

I = 3 a Re {Z Load } = 3 × 0.125 × 4 = 1.5 W 2

also (but not required) :

PSource = 3

(5) (0.5) cos(−30 − 23) = 2.25 W 2 2

Pline

I = 3 a Re{Z Line } = 3×0.125× 2 = 0.75 W 2

Section 12.7 Load P 12.7-1

Instantaneous and Average Power in a Balanced Three-Phase

Find the power absorbed by a balanced three-phase Y-connected load when VCB = 208 ∠15° V rms

and

IB = 3 ∠110° A rms

The source uses the abc phase sequence. Answer: P = 620 W Solution: Assuming the abc phase sequence: VCB = 208∠15° V rms ⇒ VBC = 208∠195° V rms ⇒ VAB = 208∠315° V rms Then VA =

VAB 208∠315° 208 = = ∠285° V rms 3∠30° 3∠30° 3

also

I B = 3∠110° A rms ⇒ I A = 3∠230° A rms Finally P = 3 VA I A cos (θ V − θ I ) = 3(

208 ) (3) cos(285° − 230°) = 620 W 3

P 12.7-2 A three-phase motor delivers 20 hp operating from a 480-V rms line voltage. The motor operates at 85 percent efficiency with a power factor equal to 0.8 lagging. Find the magnitude and angle of the line current for phase A. Hint: 1 hp = 745.7 W Solution: Assuming a lagging power factor: cos θ = pf = 0.8 ⇒

θ = 36.9°

The power supplied by the three-phase source is given by

Pin =

Pout

η

=

Pin = 3 I A VA pf

20 ( 745.7 ) = 17.55 kW where 1 hp = 745.7 W 0.85 ⇒

IA =

17.55 ×103 Pin = = 26.4 A rms 3 VA pf ⎛ 480 ⎞ 3⎜ ⎟ ( 0.8 ) ⎝ 3⎠

480 ° I A = 26.4∠ − 36.9° A rms when VA = ∠0 V rms 3 1

P 12.7-3 A three-phase balanced load is fed by a balanced Y-connected source with a line-toline voltage of 220 V rms. It absorbs 1500 W at 0.8 power factor lagging. Calculate the phase impedance if it is (a) Δ-connected and (b) Y-connected. Solution: (a) For a Δ-connected load, Eqn 12.7-5 gives PT 1500 = = 4.92 A rms 3 VP pf 3( 220 )(.8) 3 The phase current in the Δ-connected load is given by PT = 3 VP I L pf

⇒ IL =

I IL 4.92 ⇒ IP = L = = 2.84 A rms 3 3 3 The phase impedance is determined as: IP =

Z=

V VL VL 220 = ∠ (θ V − θ I ) = L ∠ cos −1 pf = ∠ cos −1 0.8 = 77.44∠36.9° Ω IP IP IP 2.84

(b) For a Y-connected load, Eqn 12.7-4 gives PT = 3 VP I L pf ⇒ I L =

PT 1500 = = 4.92 A rms 220 3 VP I L pf 3( )(.8) 3

The phase impedance is determined as: 220 V V V Z = P = P ∠ (θ V − θ I ) = P ∠ cos −1 pf = 3 ∠ cos −1 0.8 = 25.8∠36.9° Ω 4.92 IP IP IP

2

P 12.7-4 A 600-V rms three-phase Y-connected source has two balanced Δ loads connected to the lines. The load impedances are 40 ∠30° Ω and 50 ∠–60° Ω, respectively. Determine the line current and the total average power. Solution:

Parallel Δ loads ZΔ =

Z1Z 2 (40∠30° ) (50∠−60° ) = = 31.2 ∠−8.7° Ω ° ° Ζ1 + Ζ 2 40∠30 + 50∠− 60

VL = VP , Ι P =

VP 600 = = 19.2 A rms, ZΔ 31.2

IL =

3 Ι P = 33.3 A rms

So P = 3 VL I L pf = 3 (600) (33.3) cos ( − 8.7° ) = 34.2 kW P 12.7-5 A three-phase Y-connected source simultaneously supplies power to two separate balanced three-phase loads. The first total load is Δ connected and requires 39 kVA at 0.7 lagging. The second total load is Y connected and requires 15 kW at 0.21 leading. Each line has an impedance 0.038 + j0.072 Ω/phase. Calculate the line-to-line source voltage magnitude required so that the loads are supplied with 208-V rms line-to-line. Solution: We will use In our case:

S = S ∠θ = S cos θ + j S sin θ = S pf + j S sin ( cos −1 pf )

S 1 = 39 (0.7) + j 39 sin ( cos −1 ( 0.7 ) ) = 27.3 + j 27.85 kVA

15 sin ( cos −1 ( 0.21) ) = 15 − j 69.84 kVA S 2 = 15 + 0.21 S S 3φ = S 1 + S 2 = 42.3 − j 42.0 kVA ⇒ S φ = 3φ = 14.1− j 14.0 kVA 3 The line current is *

⎛ S ⎞ (14100+ j 14000) S = Vp I L ⇒ I L = ⎜ ⎟ = = 117.5 + j 116.7 A rms = 167 ∠45° A rms ⎜V ⎟ 208 ⎝ p⎠ 3 208 ∠0° = 120∠0° V rms. The source must The phase voltage at the load is required to be 3 provide this voltage plus the voltage dropped across the line, therefore *

 = 120∠0° + (0.038 + j 0.072)(117.5 + j 116.7) = 115.9 + j 12.9 = 116.6 ∠6.4° V rms V Sφ Finally

 V = 116.6 V rms Sφ 3

P 12.7-6 A building is supplied by a public utility at 4.16 kV rms. The building contains three balanced loads connected to the three-phase lines: (a) (b) (c)

Δ-connected, 500 kVA at 0.85 lagging Y-connected, 75 kVA at 0.0 leading Y connected; each phase with a 150-Ω resistor parallel to a 225-Ω inductive reactance

The utility feeder is five miles long with an impedance per phase of 1.69 + j0.78 Ω/mile. At what voltage must the utility supply its feeder so that the building is operating at 4.16 kV rms? Hint:

41.6 kV is the line-to-line voltage of the balanced Y-connected source.

Solution: The required phase voltage at the load is VP =

4.16 ∠0° = 2.402∠0° kVrms . 3

Let I1 be the line current required by the Δ-connected load. The apparent power per phase 500 kVA required by the Δ-connected load is S1 = = 167 kVA . Then 3

S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 167 ∠ cos−1 ( 0.85) = 167∠31.8° kVA and * 3 ⎛ S1 ⎞ ⎛ (167 ×10 ) ∠31.8° ⎞ ⎜ ⎟ = 69.6∠ − 31.8° = 59 − j36.56 A rms ⇒ I1 = ⎜ ⎟ = 3 ⎜ ⎟ V × ∠ ° 2.402 10 0 ( ) ⎝ P⎠ ⎝ ⎠ *

S1 = VP I1

*

Let I2 be the line current required by the first Y-connected load. The apparent power per phase 75 kVA required by this load is S 2 = = 25 kVA . Then, noticing the leading power factor, 3

S 2 = S 2 ∠θ = S 2 ∠ cos −1 ( pf ) = 25 ∠ cos −1 ( 0 ) = 25∠ − 90° kVA and * 3 ⎛ S 2 ⎞ ⎛ ( 25 ×10 ) ∠ − 90° ⎞ ⎟ = 10.4∠90° = j10.4 A rms ⇒ I2 = ⎜ ⎟ = ⎜ 3 ⎝ VP ⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠ *

S 2 = VP I 2

*

Let I3 be the line current required by the other Y-connected load. Use Ohm’s law to determine I3 to be 2402∠0° 2402∠0° I3 = + = 16 − j 10.7 A rms 150 j 225 The line current is I L = I1 + I 2 + I 3 = 75− j 36.8 A rms

4

4.16 ∠0° = 2.402∠0° kVrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore

The phase voltage at the load is required to be VP =

VSφ = 2402∠0° + (8.45 + j 3.9) (75 − j 36.8) = 3179 ∠−0.3° Vrms Finally VSL = 3 (3179) = 5506 Vrms

5

P 12.7-7 The diagram shown in Figure P 12.7-7 has two three-phase loads that form part of a manufacturing plant. They are connected in parallel and require 4.16 kV rms. Load 1 is 1.5 MVA, 0.75 lag pf, Δ-connected. Load 2 is 2 MW, 0.8 lagging pf, Y-connected. The feeder from the power utility’s substation transformer has an impedance of 0.4 + j0.8 Ω/phase. Determine the following: (a) (b) (c)

The required magnitude of the line voltage at the supply. The real power drawn from the supply. The percentage of the real power drawn from the supply that is consumed by the loads.

Three-phase supply from utility

0.4 Ω

j0.8 Ω

0.4 Ω

j0.8 Ω

0.4 Ω

j0.8 Ω Load 2

Load 1

Figure P 12.7-7 Solution: The required phase voltage at the load is VP =

4.16 ∠0° = 2.402∠0° kVrms . 3

Let I1 be the line current required by the Δ-connected load. The apparent power per phase 1.5 MVA required by the Δ-connected load is S1 = = 0.5 MVA . Then 3

S1 = S1 ∠θ = S1 ∠ cos −1 ( pf ) = 0.5 ∠ cos −1 ( 0.75) = 0.5∠41.4° MVA and * 6 ⎛ S1 ⎞ ⎛ ( 0.5 ×10 ) ∠41.4° ⎞ ⎟ = 2081.6∠ − 41.4° = 1561.4 − j1376.6 A rms ⇒ I1 = ⎜ ⎟ = ⎜ 3 ⎝ VP ⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠ *

S1 = VP I1

*

Let I2 be the line current required by the first Y-connected load. The complex power, per phase, is 0.67 S 2 = 0.67 + sin ( cos −1 ( 0.8 ) ) = 0.67 + j 0.5 MVA 0.8

6

* 6 ⎛ S 2 ⎞ ⎛ ( 0.67 + j 0.5 ) ×106 ⎞ ⎛ ( 0.833 ×10 ) ∠ − 36.9° ⎞ ⎟ =⎜ ⎟ I2 = ⎜ ⎟ = ⎜ 3 3 ⎝ VP ⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠ ⎜⎝ ( 2.402 ×10 ) ∠0° ⎟⎠ = 346.9∠ − 36.9° = 277.4 − j 208.3 A rms The line current is I L = I1 + I 2 = 433.7 − j 345.9 = 554.7∠ − 38.6 A rms *

*

4.16 ∠0° = 2.402∠0° kVrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore

The phase voltage at the load is required to be VP =

VSφ = 2402∠0° + (0.4 + j 0.8) (433.7 − j 345.9) = 2859.6 ∠ − 38.6° Vrms Finally VSL = 3 (2859.6) = 4953 Vrms The power supplied by the source is

PS =

3 (4953) (554.7) cos (4.2° + 38.6° ) = 3.49 MW

The power lost in the line is

PLine = 3 × ( 554.7 2 ) × Re {0.4+ j 0.8} = 0.369 MW The percentage of the power consumed by the loads is 3.49 − 0.369 × 100% = 89.4% 3.49

7

P 12.7-8 The balanced three-phase load of a large commercial building requires 480 kW at a lagging power factor of 0.8. The load is supplied by a connecting line with an impedance of 5 + j25 mΩ for each phase. Each phase of the load has a line-to-line voltage of 600 V rms. Determine the line current and the line voltage at the source. Also, determine the power factor at the source. Use the line-to-neutral voltage as the reference with an angle of 0°.

Solution: The required phase voltage at the load is VP =

600 ∠0° = 346.4∠0° Vrms . 3

Let I be the line current required by the load. The complex power, per phase, is S = 160 + j

160 sin ( cos −1 ( 0.8 ) ) = 160 + j 120 kVA 0.8

The line current is *

*

⎛ S ⎞ ⎛ (160 + j 120 ) ×103 ⎞ I=⎜ ⎟ =⎜ ⎟ = 461.9 − j 346.4 A rms 346.4∠0° ⎝ VP ⎠ ⎝ ⎠ 600 ∠0° = 346.4∠0° Vrms .The source 3 must provide this voltage plus the voltage dropped across the line, therefore The phase voltage at the load is required to be VP =

VSφ = 346.4∠0° + (0.005 + j 0.025) (461.9 − j 346.4) = 357.5 ∠1.6° Vrms Finally VSL = 3 (357.5) = 619.2 Vrms The power factor of the source is pf = cos (θ V − θ I ) = cos (1.6° − ( − 37°)) = 0.78

8

Section 12-8: Two-Wattmeter Power Measurement P 12.8-1 The two-wattmeter method is used to determine the power drawn by a three-phase 440-V rms motor that is a Y-connected balanced load. The motor operates at 20 hp at 74.6 percent efficiency. The magnitude of the line current is 52.5 A rms. The wattmeters are connected in the A and C lines. Find the reading of each wattmeter. The motor has a lagging power factor. Hint: 1 hp = 745.7 W Solution:

W = 14920 W hp P 14920 Pin = out = = 20 kW 0.746 η Pout = 20 hp × 746

Pin = 3 VL I L cos θ

⇒ cos θ =

Pin 20 × 103 = = 0.50 3 VL I L 3 (440) (52.5)

⇒ θ cos -1 ( 0.5 ) = 60° The powers read by the two wattmeters are P1 = VL I L cos (θ + 30° ) = (440) (52.5)cos ( 60° + 30° ) = 0

and

P2 = VL I L cos (θ − 30° ) = (440) (52.5)cos ( 60° − 30° ) = 20 kW

1

P 12.8-2 A three-phase system has a line-to-line voltage of 4000 V rms and a balanced Δ-connected load with Z = 40 + j30 Ω. The phase sequence is abc. Use the two wattmeters connected to lines A and C, with line B as the common line for the voltage measurement. Determine the total power measurement recorded by the wattmeters. Answer: P = 768 kW Solution: VP = VL = 4000 V rms IP =

VP 4000 = = 80 A rms ZΔ 50

Z

Δ

= 40 + j 30 = 50 ∠36.9°

Ι L = 3 I P = 138.6 A rms

pf = cos θ = cos (36.9° ) = 0.80 P1 = VL I L cos (θ + 30° ) = 4000 (138.6) cos 66.9° = 217.5 kW P2 = VL I L cos (θ −30° ) = 4000 (138.6) cos 6.9° = 550.4 kW PT = P1 + P2 = 767.9 kW Check : PT = 3 Ι L VL cos θ =

3 (4000) (138.6) cos 36.9°

= 768 kW which checks

2

P 12.8-3 A three-phase system with a sequence abc and a line-to-line voltage of 200 V rms feeds a Y-connected load with Z = 70.7 ∠45° Ω. Find the line currents. Find the total power by using two wattmeters connected to lines B and C. Answer: P = 400 W Solution: Vp = Vp =

200 = 115.47 Vrms 3

VA =115.47∠0° V rms, VB = 115.47∠−120° V rms and VC = 115.47∠120° V rms

IA =

VA 115.47∠0° = = 1.633∠ − 45° A rms Z 70.7∠45°

I B = 1.633 ∠ − 165° A rms and I C = 1.633 ∠75° A rms PT =

3 VL I L cos θ = 3 (200) (1.633) cos 45° = 400 W

PB = VAC I A cos θ1 = 200 (1.633) cos (45° − 30° ) = 315.47 W PC = VBC I B cos θ 2 = 200 (1.633) cos (45° + 30° ) = 84.53 W

3

P 12.8-4 A three-phase system with a line-to-line voltage of 208 V rms and phase sequence abc is connected to a Y-balanced load with impedance 10 ∠–30° Ω and a balanced Δ load with impedance 15 ∠30° Ω. Find the line currents and the total power using two wattmeters. Solution:

ZY = 10∠ − 30° Ω and Z Δ = 15∠30° Ω Convert Z Δ to Z Yˆ → Z Yˆ = then Zeq =

ZΔ = 5∠30° Ω 3

(10∠−30° ) ( 5∠30° ) =

10∠−30°+5∠30° 208 Vp = Vp = = 120 V rms 3 VA = 120∠0° V rms ⇒ I A =

50∠0° = 3.78∠10.9° Ω 13.228 ∠−10.9°

120∠0° = 31.75 ∠−10.9° 3.78 ∠10.9°

I B = 31.75∠−130.9° I C = 31.75∠109.1° PT = 3VL I L cos θ = 3 ( 208 ) ( 31.75 ) cos (10.9 ) =11.23 kW W1 = VL I L cos (θ −30°) = 6.24 kW W2 = VL I L cos (θ + 30°) = 4.99 kW

P 12.8-5 The two-wattmeter method is used. The wattmeter in line A reads 920 W, and the wattmeter in line C reads 460 W. Find the impedance of the balanced Δ-connected load. The circuit is a three-phase 120-V rms system with an abc sequence. Answer: ZΔ = 27.1 ∠–30° Ω P12.8-5

PT = PA + PC = 920 + 460 = 1380 W

tan θ = 3

( −460 ) = −0.577 ⇒ θ = −30° PC − PA = 3 PC + PA 1380

PT = 3 VL I L cos θ so I L =

PT 1380 = =7.67 A rms 3 VL cos θ 2 ×120×cos( −30 )

IL = 4.43 A rms 3

120 = 27.1 Ω ο r Z Δ = 27.1 ∠−30° 4.43

IP =

∴ ZΔ =

4

P 12.8-6 Using the two-wattmeter method, determine the power reading of each wattmeter and the total power for Problem P 12.5-1 when Z = 0.868 + j4.924 Ω. Place the current coils in the Ato-a and C-to-c lines. Solution:

Z = 0.868 + j 4.924 = 5∠80° VL = 380 V rms, VP = I L = I P and I P =



θ = 80°

380 = 219.4 V rms 3

VP = 43.9 A rms Z

P1 = ( 380 ) ( 43.9 ) cos (θ −30° ) = 10,723 W P2 = ( 380 ) ( 43.9 ) cos (θ + 30° ) = −5706 W PT = P1 + P2 = 5017 W

5

Section 12.9 How Can We Check …? P 12.9-1 A Y-connected source is connected to a Y-connected load (Figure 12.3-1) with Z = 10 + j4 Ω. The line voltage is VL = 416 V rms. A student report states that the line current IA = 38.63 A rms and that the power delivered to the load is 16.1 kW. Verify these results. Solution:

416 = 240 V = VA 3 Z = 10 + j4 = 10.77 ∠21.8° Ω VA =

VA 240 = = 22.28 A rms ≠ 38.63 A rms Z 10.77 38.63 = 22.3 . It appears that the line-to-line voltage was The report is not correct. (Notice that 3 mistakenly used in place of the phase voltage.) IA =

P 12.9-2 A Δ load with Z = 40 + j30 Ω has a three-phase source with VL = 240 V rms (Figure 12.3-2). A computer analysis program states that one phase current is 4.8 ∠–36.9° A. Verify this result. Solution:

VL = VP = 240∠0° Vrms Z = 40 + j 30 = 50 ∠36.9° Ω IP =

VP 240∠0° = = 4.8 ∠−36.9° A rms ° Z 50∠36.9

The result is correct.

1

PSpice Problems SP 12-1 Use PSpice to determine the power delivered to the load in the circuit shown in Figure SP 12-1. a



5 mH

A

20 Ω

60 mH

b



5 mH

B

20 Ω

60 mH

C

20 Ω

+



110 cos (377t)

N

+

n



110 cos (377t +120°) c



5 mH

60 mH

+



110 cos (377t + 240°) Source

Line

Load

Figure SP 12-1 Solution:

FREQ 6.000E+01

IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3) 3.142E+00 -1.644E+02 -3.027E+00 -8.436E-01

FREQ

IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1)

1

6.000E+01

3.142E+00

-4.443E+01

FREQ 6.000E+01

VM(N01496) 2.045E-14

FREQ 6.000E+01

IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2) 3.142E+00 7.557E+01 7.829E-01 3.043E+00

VP(N01496) 2.211E+01

2.244E+00 VR(N01496) 1.895E-14

-2.200E+00 VI(N01496) 7.698E-15

3.1422 20 = 98.7 W 2 3.1422 20 = 98.7 W I B = 3.142∠75.57° A and RB = 20 Ω ⇒ PB = 2 3.1422 20 = 98.7 W I C = 3.142∠ − 164.4° A and RC = 20 Ω ⇒ PC = 2 I A = 3.142∠ − 43.43° A and

RA = 20 Ω ⇒ PA =

P = 3 ( 98.7 ) = 696.1 W

2

SP 12-2 2.

Use PSpice to determine the power delivered to the load in the circuit shown in SP 12a

10 Ω

5 mH

A

20 Ω

60 mH

b

10 Ω

5 mH

B

30 Ω

25 mH

C

60 Ω

+



110 cos (377t) –

N

+

n

110 cos (377t +120°) c

10 Ω

5 mH

20 mH

+



110 cos (377t + 240°) Source

Line

Load

Figure SP 12-2 Solution:

FREQ 6.000E+01

IM(V_PRINT3)IP(V_PRINT3)IR(V_PRINT3)II(V_PRINT3) 1.612E+00 -1.336E+02 -1.111E+00 -1.168E+00

FREQ 6.000E+01

IM(V_PRINT1)IP(V_PRINT1)IR(V_PRINT1)II(V_PRINT1) 2.537E+00 -3.748E+01 2.013E+00 -1.544E+00

3

FREQ 6.000E+01

VM(N01496) VP(N01496) 1.215E+01 -1.439E+01

VR(N01496) VI(N01496) 1.177E+01 -3.018E+00

FREQ 6.000E+01

IM(V_PRINT2)IP(V_PRINT2)IR(V_PRINT2)II(V_PRINT2) 2.858E+00 1.084E+02 -9.023E-01 2.712E+00

2.537 2 20 = 64.4 W 2 2.8582 30 = 122.5 W I B = 2.858∠108.4° A and RB = 30 Ω ⇒ PB = 2 1.6122 60 = 78 W I C = 1.612∠ − 133.6° A and RC = 600 Ω ⇒ PC = 2

I A = 2.537∠ − 37.48° A and

RA = 20 Ω ⇒ PA =

P = 64.4 + 122.5 + 78 = 264.7 V

4

Design Problems DP 12-1 A balanced three-phase Y source has a line voltage of 208 V rms. The total power delivered to the balanced Δ load is 1200 W with a power factor of 0.94 lagging. Determine the required load impedance for each phase of the Δ load. Calculate the resulting line current. The source is a 208-V rms ABC sequence. Solution: P = 400 W per phase, 0.94 = pf = cos θ

θ = cos -1 ( 0.94 ) =20°



208 I L 0.94 ⇒ I L = 3.5 A rms 3 I I Δ = L = 2.04 A rms 3 V 208 Z = L = = 101.8 Ω IΔ 2.04

400 =

Z = 101.8 ∠20° Ω

DP 12-2 A three-phase 240-V rms circuit has a balanced Y-load impedance Z. Two wattmeters are connected with current coils in lines A and C. The wattmeter in line A reads 1440 W, and the wattmeter in line C reads zero. Determine the value of the impedance. Solution:

VL = 240 V rms PA = VL I L cos (30° + θ ) = 1440 W PC = VL I L cos (30° − θ ) = 0 W ⇒

30−θ = 90° or θ = −60°

then 1440 = 240 I L cos (−30° )

I L = 6.93 A rms

IL

= IP =

VP Z





240 V Z = P = 3 = 20 Ω IP 6.93

Finally, Z = 20 ∠ − 60° Ω

1

DP 12-3 A three-phase motor delivers 100 hp and operates at 80 percent efficiency with a 0.75 lagging power factor. Determine the required Δ-connected balanced set of three capacitors that will improve the power factor to 0.90 lagging. The motor operates from 480-V rms lines. Solution: Pin =

Pout

η

100 hp × (746 =

W ) hp

0.8

= 93.2 kW, P =

φ

Pin = 31.07 kW 3

VL = 480 V rms, pfc = 0.9 and pf = 0.75. We need the impedance of the load so that we can use Eqn 11.6-7 to calculate the value of capacitance needed to correct the power factor. 0.75 = pf = cos θ



θ = cos-1 ( 0.75) = 41.4°

480 I p ( 0.75 ) ⇒ I p = 149.5 Arms 3 480 VP 3 = 1.85 Ω Z = = IP 149.5

31070 =

Z = 1.85 ∠41.4° Ω = 1.388 + j1.223 Ω

The capacitance required to correct the power factor is given by ⎡⎣ tan (cos −1 0.75) − tan (cos −1 0.9) ⎤⎦ 1.365 C= = 434 μ F × 1.3652 +1.2042 377

(Checked using LNAPAC 6/12/03)

2

DP 12-4 A three-phase system has balanced conditions so that the per-phase circuit representation can be utilized as shown in Figure DP 12-4. Select the turns ratio of the step-up and step-down transformers so that the system operates with an efficiency greater than 99 percent. The load voltage is specified as 4 kV rms, and the load impedance is 4/3 Ω. 2.5 Ω

1 : n1

+

20 kV –

j40 Ω

IL

n2 : 1

+

+

+

V1

V2

VL







Load

Figure DP 12-4 Solution: VL = 4∠0° kV rms n2 25 VL = 4000∠0° = 100∠0° kVrms n1 1

Try n2 = 25 then V2 =

VL 4×103∠0° = = 3∠0° kA rms 4 ZL 3 3000∠0° = 120∠0° A rms The line current in 2.5 Ω is I = 25 Thus V1 = ( R + j X ) I + V2

IL =

= (2.5 + j 40) (120∠0°) + 100×103 = 100.4 ∠2.7° kV Step need : n1 = Ploss = I

2

100.4 kV = 5.02 ≅ 5 20 kV

R = 120

2

(2.5) = 36 kW, P = (4×103 ) (3× 103 ) = 12 MW

12 − .036 × 100% = 99.7 % of the power supplied by the source 12 is delivered to the load.

∴η =

3

Chapter 13: Frequency Response Exercises Exercise 13.2-1 The input to the circuit shown in Figure E 13.2-

R

1 is the source voltage, vs, and the response is the capacitor voltage, vo. Suppose R = 10 kΩ and C = 1 μF. What are the values

+ vs

+ –

vo

C –

of the gain and phase shift when the input frequency is ω = 100 rad/s?

Figure E 13.2-1

Answer: 0.707 and –45° Solution: H (ω ) = gain =

1 Vo (ω ) = 1 + jω C R Vs (ω ) 1 1 + (ω C R) 2

phase shift = − tan −1 (ω C R ) When R = 104 Ω, ω = 100 rad/s, and C = 10−6 F, then gain =

1 = 0.707 and phase shift = − 45o 2

13Ex-1

Exercise 13.2-2 The input to the circuit shown in Figure E 13.2-2

L

is the source voltage, vs, and the response is the resistor voltage, vo.

+

R = 30 Ω and L = 2 H. Suppose the input frequency is adjusted until

vs +

vo

R



i

the gain is equal to 0.6. What is the value of the frequency?



Figure E 13.2-2

Answer: 20 rad/s Solution: H (ω ) = gain =

Vo (ω ) R = Vs (ω ) R + jω L R R 2 + (ω L) 2 2

0.6 =

30 30 + (2ω ) 2 2

⎛ 30 ⎞ 2 ⎜ ⎟ − 30 ⎝ .6 ⎠ ⇒ ω= = 20 rad s 2

Exercise 13.2-3 The input to the circuit shown in Figure E 13.2-2

L

is the source voltage, vs, and the response is the mesh current, i. R = 30 Ω and L = 2 H. What are the values of the gain and phase shift when the input frequency is ω = 20 rad/s?

+ vs + –

R i

vo –

Figure E 13.2-2

Answer: 0.02 A/V and –53.1° Solution:

H (ω ) = gain =

I (ω ) 1 = R + jω L Vs (ω ) 1 R + (ω L) 2 2

phase shift = − tan −1

ωL R

When R = 30 Ω, L = 2 H, and ω = 20 rad/s, then gain =

1 30 + 40 2

2

= 0.02

A ⎛ 40 ⎞ and phase shift = − tan −1 ⎜ ⎟ = − 53.1° V ⎝ 30 ⎠

13Ex-2

Exercise 13.2-4 The input to the circuit shown in Figure E 13.2-1

R

is the source voltage, vs, and the response is the capacitor voltage,

+

vo. Suppose C = 1 μF. What value of R is required to cause a phase

+ –

vs

vo

C –

shift equal to –45° when the input frequency is ω = 20 rad/s? Answer: R = 50 kΩ

Figure E 13.2-1

Solution:

H (ω ) = gain =

Vo (ω ) 1 = Vs (ω ) 1 + jω C R 1 1 + (ω C R) 2

phase shift = − tan −1 ω C R −1

−6

−45° = − tan (20 ⋅10 ⋅ R )

tan (45° ) ⇒ R= = 50 ⋅103 = 50 kΩ −6 20⋅10

Exercise 13.2-5 The input to the circuit shown in Figure E 13.2-1

R

is the source voltage, vs, and the response is the capacitor voltage, vo. Suppose C = 1 μF. What value of R is required to cause a gain

+ + –

vs

vo

C –

equal to 1.5 when the input frequency is ω = 20 rad/s?

Figure E 13.2-1

Answer: No such value of R exists. The gain of this circuit will

never be greater than 1. Solution: H (ω ) = gain =

Vo (ω ) 1 = Vs (ω ) 1 + jω C R 1 1 + (ω C R) 2

ω , C , and R are all positive, or at least nonnegative, so gain ≤ 1. These specifications cannot be met.

13Ex-3

Exercise 13.3-1 (a) Convert the gain|Vo/Vs| = 2 to decibels. (b) Suppose |Vo/Vs| = –6.02 dB.

What is the value of this gain “not in dB”? Answer: (a) + 6.02 dB (b) 0.5

Solution: (a) dB = 20 log (2) = 6.02 dB −6.02 20 (b) 10 = 0.5

Exercise 13.3-2 In a certain frequency range, the magnitude of the network function can be

approximated as H = 1/ω2. What is the slope of the Bode plot in this range, expressed in decibels per decade? Answer:–40 dB/decade

Solution: ⎛ 1 ⎞ 20 log H = 20 log ⎜ 2 ⎟ = 20 log (ω )-2 = −40 log ω ⎝ω ⎠ ⎛ω ⎞ = − 40 log ω2 + 40 log ω1 = − 40 log⎜ 2 ⎟ ⎝ ω1 ⎠ let ω2 = 10 ω1 to consider 1 decade, then slope = − 40 log10 = − 40 dB decade slope = 20 log H (ω2 ) − 20 log H (ω1 )

13Ex-4

Exercise 13.3-3 Consider the network function

H (ω ) =

jω A B + jωC

Find (a) the corner frequency, (b) the slope of the asymptotic magnitude Bode plot for ω above the corner frequency in decibels per decade, (c) the slope of the magnitude Bode plot below the corner frequency, and (d) the gain for ω above the corner frequency in decibels. Answer: (a) ω0 = B/C (b) zero (c) 20 dB/decade (d) 20 log10 =

A C

Solution: When ω C >> B, H (ω ) − (d ) (b)

jω A A = jω C C

⎛ A⎞ H (ω ) in dB = 20 log10 H (ω ) = 20 log10 ⎜ ⎟ ⎝C ⎠ H (ω ) does not depend on ω so slope = 0

When ω C 0 and p > 0. When the input to

v s ( t ) = 12 cos (120 t + 30° ) V

this circuit is

v o ( t ) = 42.36 cos (120 t − 48.69° ) V .

the output is Determine the values of k and p.

Answers: k = 18 and p = 24 rad/s.

Solution: Vo 42.36∠ − 48.69° ⎛ 120 ⎞ k k = ∠ − tan −1 ⎜ and = = 3.53∠ − 78.69° ⎟ 2 120 12 30 V ∠ ° p ⎝ ⎠ s ⎛ 120 ⎞ 1+ j 1+ ⎜ p ⎟ ⎝ p ⎠

so ⎛ 120 ⎞ 120 − tan −1 ⎜ = tan ( 78.69° ) = 5 ⇒ ⎟ = −78.69° ⇒ p ⎝ p ⎠

p=

120 = 24 rad/s 5

and

k ⎛ 120 ⎞ 1+ ⎜ ⎟ ⎝ p ⎠

2

=

k 1 + ( 5)

2

= 3.53 ⇒ k = 3.53 26 = 18

13.2-31

P13.2-28

The network function of a circuit is H (ω ) =

20 . When the input to this circuit is sinusoidal, the 8+ jω

output is also sinusoidal. Let ω 1 be the frequency at which the output sinusoid is twice as large as the input sinusoid and let ω 2 be the frequency at which output sinusoid is delayed by one tenth period with respect to the input sinusoid. Determine the values of ω 1 and ω 2.

Solution:

The gain is 2 at the frequency ω 1 so 2 = When the frequency is ω 2, the period is

20 8 2 + ω12 2π

ω2

2

⎛ 20 ⎞ and ω1 = ⎜ ⎟ − 8 2 = 6 rad/s . ⎝ 2 ⎠

. Also a delay t o corresponds to a phase shift −ω 2 t o. In this

⎛ 2π ⎞ ⎛ ω2 ⎞ case, t 0 = 0.1⎜ so the phase shift is -0.2 π. Then −0.2 π = − tan −1 ⎜ ⎟ ⎟ so ⎜ ω2 ⎟ ⎝ 8 ⎠ ⎝ ⎠ ω 2 = 8 tan ( 0.2 π ) = 5.8123 rad/s .

13.2-32

P13.2-29 The input to the circuit in Figure P13.3-29 is the voltage source voltage, v s ( t ) . The output is the voltage v o ( t ) . When the input is v s ( t ) = 8cos ( 40 t ) V , the output is v o ( t ) = 2.5cos ( 40 t + 14° ) V . Determine

the values of the resistances R1 and R 2 .

Figure P13.2-29 Solution:

jω L R 2

L R 2 + jω L R1 Vo (ω ) = = jω L R 2 L Vi (ω ) 1 + jω R1 + Rp R 2 + jω L where R p = Vo (ω ) Vi (ω )

In this case the angle of

the magnitude of



R1 R 2 R1 + R 2

ω =

L R1



j ⎜ 90 − tan −1 ω

⎛ L ⎞ 1+ ⎜ω ⎜ R p ⎟⎟ ⎝ ⎠

2

e

⎜ ⎝

L ⎞⎟ R p ⎟⎠

Vo (ω ) L L ( R1 + R 2 ) tan ( 90° − 14° ) = = = 0.1 and is specified to be 14° so 40 Rp R1 R 2 Vi (ω )

Vo (ω ) 2.5 is specified to be so 8 Vi (ω )

40

L R1

=

2.5 ⇒ 8

1 + 16 that satisfies these two equations is L = 1 H, R1 = 31 Ω, R 2 = 14.76 Ω .

L = 0.0322 . One set of values R1

13.2-33

P13.2-30 The input to the circuit shown in Figure P13.2-30 is the voltage source voltage, v s ( t ) . The output is the

voltage v o ( t ) . The input v s ( t ) = 2.5cos (1000 t ) V causes the output to be v o ( t ) = 8cos (1000 t + 104° ) V . Determine the values of the resistances R1 and R 2 .

Answers: R1 = 1515 Ω and R 2 = 20 kΩ.

Figure P13.2-30 Solution:

R2

1 jω C

=

R2

R2 1 + jω CR 2 R2

1 + jω CR 2 R1 Vo (ω ) =− =− Vi (ω ) R1 1 + jω CR 2 R2

R1 Vo (ω ) j (180− tan −1 ω CR 2 ) e = 2 Vi (ω ) 1 + (ω CR 2 )

In this case the angle of

magnitude of

Vo (ω ) tan (180° − 104° ) is specified to be 104° so CR 2 = = 0.004 and the 1000 Vi (ω )

Vo (ω ) 8 so is specified to be 2.5 Vi (ω )

R2 R1 1 + 16

=

8 ⇒ 2.5

R2 R1

= 13.2 . One set of values that

satisfies these two equations is C = 0.2 μ F, R1 = 1515 Ω, R 2 = 20 kΩ .

13.2-34

Section 13-3: Bode Plots

P 13.3-1

Sketch the magnitude Bode plot of H (ω ) =

4(5 + jω ) ω⎞ ⎛ ⎜1 + j ⎟ 50 ⎠ ⎝

Solution:

ω⎞ ⎛ 20 ⎜ 1 + j ⎟ 5⎠ H (ω )= ⎝ ω⎞ ⎛ ⎜1 + j ⎟ 50 ⎠ ⎝

⎧ ⎪ ⎪ 20 ⎪ ⎪ ⎪ ⎪ 20 ⎛ j ω ⎞ ⎪ ⎜ ⎟ ≈⎨ ⎝ 5⎠ ⎪ ⎪ ⎪ ⎛ ω⎞ ⎪ 20 ⎜ j 5 ⎟ ⎠ = 200 ⎪ ⎝ ⎪ ⎛jω⎞ ⎪⎩ ⎜⎝ 50 ⎟⎠

ω p1 ⎪−(C1 R2 ) ( jωC1 R1 )( jωC2 R2 ) jωC2 R1 ⎩

Here’s the Bode plot:

13.3-3

C1

C2

R1

R2

P 13.3-4 The input to the circuit shown in Figure P 13.3-4 is the source voltage, vs(t), and the response is the voltage across R3, vo(t). Determine H(ω) and sketch the Bode diagram.



vs +

+

+



R3

vo –

Figure P 13.3-4 Solution:

R2 R (1+ jω C1 R1 ) R 1 1 1+ jω C2 R2 H (ω ) = − =− 2 so K = − 2 , z = and p = R1 R1 (1+ jω C2 R2 ) R1 C1 R1 C2 R2 1+ jω C1 R1 When z < p

When z > p

13.3-4

P 13.3-5 The input to the circuit shown in Figure P 13.3-5a is the voltage, vi(t), of the independent voltage source. The output is the voltage, vo(t), across the capacitor. Design this circuit to have the Bode plot shown in Figure P 13.3-5b. Hint: First show that the network function of the circuit is ⎛ AL R4 ⎞ jω ⎜ ⎟ R1 ( R3 + R4 ⎠ Vo (ω ) ⎝ = H (ω ) = Vi (ω ) ⎛ C R3 R4 ⎞ L( R1 + R2 ) ⎞ ⎛ ⎟ ⎜1 + jω ⎟ ⎜1 + jω R1 R2 ⎠ ⎝ R3 + R4 ⎠ ⎝

R3

R1

+ –

v1(t)

L

+ v2(t) –

+ –

R2

Av2(t)

R4

+ vo (t) –

C

20 log10 |H(ω)|, (dB)

(a)

20

0

200

20

20k

200k

ω (rad/s, log scale)

(b)

Figure P 13.3-5 Solution: Using voltage division twice gives: V2 (ω ) = Vi (ω )

and

jω L R 2 R 2 + jω L jω L R 2 L jω = = jω L R 2 R1 R 2 + jω L ( R1 + R 2 ) R1 L ( R1 + R 2 ) R1 + + ω j 1 R 2 + jω L R1 R 2

Vo (ω ) = V2 (ω )

R4

A R4

A R4 R3 + R 4 1 + jω C R 4 A= = R4 C R3 R 4 R3 + R 4 + jω C R3 R 4 R3 + 1 + jω R3 + R 4 1 + jω C R 4 13.3-5

Combining these equations gives H (ω ) =

ALR 4 Vo (ω ) jω = Vi (ω ) R1 ( R 3 + R 4 ) ⎛ L ( R1 + R 2 ) ⎞ ⎛ CR 3 R 4 ⎞ ⎜1 + jω ⎟ ⎜ 1 + jω ⎟ ⎜ ⎟ ⎜⎝ R1 R 2 R 3 + R 4 ⎟⎠ ⎝ ⎠

The Bode plot corresponds to the network function: H (ω ) =

k jω k jω = ω ⎞⎛ ω ⎞ ⎛ ω ⎞⎛ ω ⎞ ⎛ ⎜1 + j ⎟ ⎜1 + j ⎟ ⎜⎝1 + j 200 ⎟⎠ ⎜⎝1 + j 20000 ⎟⎠ p1 ⎠ ⎝ p2 ⎠ ⎝

⎧ ⎪ ⎪ ⎪ k jω = k jω ⎪ 1 ⋅1 ⎪ ⎪ k jω H (ω ) ≈ ⎨ = k p1 jω ⎪ ⋅1 p1 ⎪ ⎪ k jω k p1 p2 ⎪ = jω ⎪ jω ⋅ jω ⎪ p p ⎩ 1 2

ω ≤ p1 p1 ≤ ω ≤ p2

ω ≥ p2

This equation indicates that |H(ω)|=k p1 when p1 ≤ ω ≤ p2. The Bode plot indicates that |H(ω)|=20 dB = 10 when p1 ≤ ω ≤ p2. Consequently 10 10 k= = = 0.05 p1 200

Finally,

H (ω ) =

0.05 jω ω ⎞⎛ ω ⎞ ⎛ ⎜1 + j ⎟ ⎜1 + j ⎟ 200 ⎠ ⎝ 20000 ⎠ ⎝

Comparing the equation for H(ω) obtained from the circuit to the equation for H(ω)obtained from the Bode plot gives: R3 + R 4 ALR 4 R1 R 2 , 200 = 0.05 = and 20000 = C R3 R 4 R1 ( R 3 + R 4 ) L ( R1 + R 2 )

Pick L = 1 H, and R1 = R2 , then R1 = R2 = 400 Ω. Let C = 0.1 μF and R3 = R4 , then R3 = R4 = 1000 Ω. Finally, A=40. (Checked using ELab 3/5/01)

13.3-6

P 13.3-6

The input to the circuit shown in Figure P 13.3-6b is the voltage of the voltage source, vi(t).

The output is the voltage vo(t). The network function of this circuit is H(ω) = Vo(ω)/Vi(ω). Determine the values of R2, C1, and C2 that are required to make this circuit have the magnitude Bode plot shown in Figure P 13.3-6a.

20 log10|H(ω)|, dB

Answer: R2 = 400 kΩ, C1 = 25 nF, and C2 = 6.25 nF

32

v1(t)

R1 = 10 kΩ

R2

C1

C2 –

+ –

+

12

10 kΩ 40

400

4k

+ vo(t) –

40 k

ω (rad/s, log scale)

(a)

(b)

Figure P 13.3-6 Solution: From Table 13.3-2:

R2 R1

= k = 32 dB = 40 R 2 = 40 (10 × 103 ) = 400 kΩ

1 1 = p = 400 rad/s ⇒ C 2 = = 6.25 nF C 2 R2 ( 400 ) ( 400 ×103 ) 1 1 = z = 4000 rad/s ⇒ C 1 = = 25 nF C 1 R1 ( 4000 ) (10 ×103 )

13.3-7

The input to the circuit shown in Figure P 13.3-7b is the voltage of the voltage source, vi(t).

P 13.3-7

The output is the voltage vo(t). The network function of this circuit is H(ω) = Vo(ω)/Vi(ω). The magnitude Bode plot is shown in Figure p 13.3-7a. Determine values of the corner frequencies, z and p. Determine value of the low-frequency gain, k. 20 log10|H( ω)|, dB

8Ω +

0

2Ω vi(t)

20 log10(k)

+ –

vo(t) 0.4 H

z

p



ω (rad/s, log scale)

(a)

(b)

Figure P 13.3-7 Solution: H (ω ) =

R 2 + jω L Vo (ω ) = Vi (ω ) R + R 2 + jω L L ⎛ 1 + jω ⎜ ⎛ R2 ⎞ R2 =⎜ ⎟⎜ ⎜ R + R2 ⎟⎜ L ⎝ ⎠ ⎜ 1 + jω R + R2 ⎝

H (ω ) =

( 0.2 ) (1 + j ( 0.2 ) ω ) 1 + j ( 0.04 ) ω

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

⎧ ⎪ k = 0.2 ⎪ 1 ⎪ ⇒ ⎨ z= =5 0.2 ⎪ 1 ⎪ ⎪⎩ p = 0.04 = 25

13.3-8

P 13.3-8

Determine H( jω) from the asymptotic Bode diagram in Figure P 13.3-8. 40

20

dB

0

–20

–40 0.1

1

10

100

1000

10,000

ω (rad/s)

Figure P 13.3-8 Solution • The slope is 40dB/decade for low frequencies, so the numerator will include the factor (jω)2 . • The slope decreases by 40 dB/decade at ω = 0.7rad/sec. So there is a second order pole at ω 0 = 0.7 rad/s. The damping factor of this pole cannot be determined from the asymptotic Bode plot; call it δ 1. The denominator of the network function will contain the factor

ω

⎛ ω ⎞ −⎜ 1 + 2 δ1 j ⎟ 0.7 ⎝ 0.7 ⎠ • • •

2

The slope increases by 20 dB/decade at ω = 10 rad/s, indicating a zero at 10 rad/s. The slope decreases by 20 dB/decade at ω = 100 rad/s, indicating a pole at 100 rad/s. The slope decreases by 40 dB/decade at ω = 600 rad/s, indicating a second order pole at ω 0 = 600rad/s. The damping factor of this pole cannot be determined from an asymptotic Bode plot; call it δ 2. The denominator of the network function will contain the factor

ω

⎛ ω ⎞ −⎜ 1 + 2δ 2 j ⎟ 600 ⎝ 600 ⎠

H (ω ) =

K (1+ j

ω

2

)( jω ) 2

10 2 2 ⎛ ⎞⎛ ω ⎛ω ⎞ ω ⎛ ω ⎞ ⎞⎛ ω ⎞ −⎜ 1+ 2δ 2 j −⎜ ⎜⎜ 1+ 2δ1 j ⎟ ⎟⎜ ⎟ ⎟⎟⎜1+ j ⎟ ⎟⎜ 0.7 ⎝ 0.7 ⎠ ⎠⎝ 600 ⎝ 600 ⎠ ⎠⎝ 100 ⎠ ⎝

To determine K , notice that H (ω ) = 1 dB=0 when 0.7 < ω < 10. That is 1=

K (1) ω 2 2

⎛ ω ⎞ −⎜ ⎟ (1)(1) ⎝ 0.7 ⎠

= K (0.7) 2 ⇒ K = 2

13.3-9

P 13.3-9 A circuit has a voltage ratio H (ω ) =

k (1 + jω / z ) jω

(a)

Find the high- and low-frequency asymptotes of the magnitude Bode plot.

(b)

The high- and low-frequency asymptotes comprise the magnitude Bode plot. Over what ranges of frequencies is the asymptotic magnitude Bode plot of H(ω) within 1 percent of the actual value of H(ω) in decibels?

Solution: (a)

ω⎞ ⎛ K ⎜1+ j ⎟ z⎠ ⎝ H (ω ) = jω H (ω ) =

K

ω

⎛ω ⎞ 1+ ⎜ ⎟ ⎝z⎠

H (ω ) dB = 20 log10

K

ω

2

⎛ω ⎞ 1+⎜ ⎟ ⎝z⎠

2

= 20 log10 K − 20 log10 ω + 20 log10 Let

⎛ω ⎞ 1+⎜ ⎟ ⎝z⎠

2

H L (ω ) dB = 20 log10 K − 20 log10 ω

K z ⎪⎧ H L (ω ) dB Then H (ω ) dB ~_ ⎨ ⎪⎩ H H (ω ) dB and H H (ω ) dB = 20 log10

ω > z

So H L (ω ) dB and H H (ω ) dB are the required low and high-frequency asymptotes.

13.3-10

The Bode plot will be within 1% of |H(ω)| dB both for ω > z. The range when ω > z is characterized by H H (ω ) = .99 H (ω )

(gains not in dB)

or equivalently 20 log10 0.99 = H H (ω ) dB − H (ω ) dB

(gains in dB)

⎛ω ⎞ = 20 log10 K − 20 log10 z − 20 log10 1+ ⎜ ⎟ ω ⎝z⎠ K

2

2

1 ⎛ω ⎞ = − 20 log10 1+ ⎜ ⎟ = 20 log10 2 ω ⎝z⎠ ⎛z⎞ ⎜ ⎟ +1 ⎝ω ⎠ z

Therefore 2

⎛ 1 ⎞ = ⎜ ⎟ −1 ⇒ ω ⎝ .99 ⎠ z

The error is less than 1% when ω
7 z. 7

13.3-11

P 13.3-10

Physicians use tissue electrodes to form the

R1

interface that conducts current to the target tissue of the human body. The electrode in tissue can be modeled by the RC circuit shown in Figure P 13.3-10. The value of each

+ vs

+ –

C

Rt

vo –

element depends on the electrode material and physical Figure P 13.3-10

construction as well as the character of the tissue being probed. Find the Bode diagram for Vo/Vs = H(jω) when R1 = 1 kΩ, C = 1 μF, and the tissue resistance is Rt = 5 kΩ. Solution: H (ω ) =

=

Vo (ω ) = Vs (ω )

Rt R t + R1

1 jω C

= Rt +

⎛ Rt =⎜ R1 + R t + jω C R1 R t ⎜⎝ R1 + R t R t (1+ jω C R1 )

Rt R1 1+ jω C R1 ⎞ 1+ jω C R1 ⎟⎟ ⎠ 1+ jω ⎛ C R1 R t ⎜⎜ ⎝ R1 + R t

⎞ ⎟⎟ ⎠

When R1 = 1 kΩ, C = 1 μ F and R t = 5 kΩ

ω ⎞ ⎛ 1+ j 5⎜ 1000 ⎟ ⇒ H (ω ) = ⎜ ⎟ 6 ⎜ 1+ j ω ⎟ 1200 ⎠ ⎝

⎧5 ω 5 ⎩

The slope of the asymptotic magnitude Bode plot is −20 db/decade for ω < 2 and ω > 5 rad/s and is −40 db/decade for 2 < ω < 5 rad/s. Also, at ω = 1 rad/s ⎧ 7.5 ⎪ 1 = 7.5 at ω = 1 rad/s ⎧ 20 log10 ( 7.5 ) = 17.5 dB at ω = 1 rad/s ⎪ ⎪ 7.5 ⎪ = 3.75 at ω = 2 rad/s ⇒ 20 log10 H = ⎨ 20 log10 ( 3.75 ) = 11.5 dB at ω = 2 rad/s H =⎨ ⎪ 2 ⎪20 log ( 0.6 ) = −4.44 dB at ω = 5 rad/s 10 ⎩ ⎪ 3 0.6 at 5 rad/s = = ω ⎪ 5 ⎩ The asymptotic magnitude Bode plot for H is

13.3-23

13.3-24

P 13.3-21

The network function of a circuit is H (ω ) =

( jω )3 . Sketch the asymptotic magnitude (4 + j 2ω )

Bode plot corresponding to H. Solution:

1 3 3 ω j ( ) ω j ( ) =4 H (ω ) = ( 4 + j 2ω ) 1 + j ω 2

There is a pole at 2 rad/s and three zeros at 0 rad/s. To obtain the asymptotic magnitude Bode plot, use ⎧1 ⎪ 1 + j = ⎨ω p ⎪ ⎩p

ω

Then

⎧1 3 ⎪ 4ω 1 = ω3 ⎪ 4 ⎪⎪ (1) H = H = ⎨1 3 ⎪ 4ω 1 = ω2 ⎪ ⎪ ⎛⎜ ω ⎞⎟ 2 ⎪⎩ ⎝ 2 ⎠

for ω < p for ω > p

for ω < 2

for ω > 2

⎧⎪ 20 log10 ( 0.25 ) + 3 ⎡⎣ 20 log10 (ω ) ⎤⎦ for ω < 2 20 log10 H = ⎨ ⎩⎪20 log10 ( 0.50 ) + 2 ⎡⎣ 20 log10 (ω ) ⎤⎦ for ω > 2

The slope of the asymptotic magnitude Bode plot is 60 db/decade for ω < 2 rad/s and is 40 db/decade for ω > 2 rad/s. Also, 20 log10 H = 20 log10 ( 0.25 ) + 3 ⎡⎣ 20 log10 (1) ⎤⎦ = −12 dB at ω = 1 rad/s 20 log10 H = 20 log10 ( 0.25 ) + 3 ⎡⎣ 20 log10 ( 2 ) ⎤⎦ = 6 dB at ω = 2 rad/s The asymptotic magnitude Bode plot for H is

13.3-25

P 13.3-22

The network function of a circuit is H (ω ) =

2( j 2ω + 5) . Sketch the asymptotic (4 + j 3ω )( jω + 2)

magnitude Bode plot corresponding to H.

Solution: 5 ⎛⎜ ω ⎞⎟ 1+ j 5 ⎟ 4⎜ 2 ( j 2ω + 5) 2⎠ ⎝ H (ω ) = = ⎞ ( 4 + j 3ω )( j ω + 2 ) ⎛ ⎜1 + j ω ⎟ ⎛⎜ 1 + j ω ⎞⎟ 4 ⎟⎝ 2⎠ ⎜ 3⎠ ⎝

There is a zero at 2.5 rad/s and poles at 1.33 and 2 rad/s. To obtain the asymptotic magnitude Bode plot, use ⎧ 1 for ω < p ω ⎪ 1 + j = ⎨ω for ω > p p ⎪ ⎩p Then 5 ⎧ (1) ⎪ 4 for ω < 4 rad/s =5 ⎪ 4 3 1 1 ( )( ) ⎪ ⎪ 5 5 (1) ⎪ ⎪ 4 = 3 for 4 5 10 10 ⎪⎩ 3 2

( ) ( ) ( )

The slope of the asymptotic magnitude Bode plot is −20 db/decade for 4/3 < ω < 2 rad/s and ω > 5/2 rad/s and is −40 db/decade for 2 < ω < 5/2 rad/s. Also,

( 4 ) = 1.9 dB

20 log10 H = 20 log10 5

( 3 ) − 20 log

20 log10 H = 20 log10 5

10

for ω ≤ 4

( 2 ) = −1.6 dB

( 3 ) − 40 log ( 5 2 ) = −5.4 dB

20 log10 H = 20 log10 10

10

3

rad/s

at ω = 2 rad/s at ω = 5

2

rad/s

The asymptotic magnitude Bode plot for H is

13.3-27

P 13.3-23 The network function of a circuit is H (ω ) =

4(20 + jω )(20, 000 + jω ) Sketch the asymptotic (200 + jω )(2000 + jω )

magnitude Bode plot corresponding to H.

Solution:

ω ⎞⎛ ω ⎞ ⎛ 4 ⎜1 + j ⎟ ⎜1 + j 4 ( 20 + j ω ) ( 20, 000 + j ω ) 20 ⎠ ⎝ 20, 000 ⎟⎠ ⎝ H (ω ) = = ω ⎞⎛ ω ⎞ ⎛ ( 200 + j ω ) ( 2000 + j ω ) ⎜1 + j ⎟ ⎜1 + j ⎟ 200 ⎠ ⎝ 2000 ⎠ ⎝ There are zeros at 20 and 20,000 rad/s and poles at 200 and 2000 rad/s. To obtain the asymptotic magnitude Bode plot, use ⎧ 1 for ω < p ω ⎪ 1 + j = ⎨ω for ω > p p ⎪ ⎩p Then 4 (1)(1) ⎧ =4 for ω < 20 rad/s ⎪ 1 1 ( )( ) ⎪ ⎪ ⎛ω ⎞ 4 ⎜ ⎟ (1) ⎪ ω ⎝ 20 ⎠ ⎪ for 20 < ω < 200 rad/s = ⎪ 5 (1)(1) ⎪ ⎪ ⎛ω ⎞ 4 ⎜ ⎟ (1) ⎪ ⎝ 20 ⎠ for 200 < ω < 2000 rad/s = 40 ⎪ ⎛ ω ⎞ ⎪⎪ ⎜ ⎟ (1) H = H =⎨ ⎝ 200 ⎠ ⎪ ⎪ 4 ⎛ ω ⎞ (1) ⎜ ⎟ ⎪ 80000 ⎝ 20 ⎠ for 2000 < ω < 20, 000 rad/s = ⎪ ω ω ω ⎛ ⎞ ⎛ ⎞ ⎪⎜ ⎟⎜ ⎟ ⎪ ⎝ 200 ⎠ ⎝ 2000 ⎠ ⎪ ⎪ 4 ⎛⎜ ω ⎞⎟ ⎛ ω ⎞ ⎜ ⎟ ⎪ ⎝ 20 ⎠ ⎝ 20, 000 ⎠ ⎪ ⎛ ω ⎞ ⎛ ω ⎞ = 4 for ω > 2000 rad/s ⎪ ⎜ ⎟ ⎟⎜ ⎩⎪ ⎝ 200 ⎠ ⎝ 2000 ⎠

⎧ 20 log10 ( 4 ) for ω < 20 ⎪ 20 log10 (ω ) − 20 log10 ( 5 ) for 20 < ω < 200 ⎪⎪ 20 log10 H = ⎨ 20 log10 ( 40 ) for 200 < ω < 2000 ⎪20 log ( 80000 ) − 20 log (ω ) for 2000 < ω < 20, 000 10 10 ⎪ 20 log10 ( 4 ) for ω > 20, 000 ⎪⎩ 13.3-28

The slope of the asymptotic magnitude Bode plot is 20 db/decade for 20 < ω < 200 rad/s and is −20 db/decade for 2000 < ω < 20,000 rad/s and is 0 db/decade for ω < 20 and 200 < ω < 2000 rad/s, and ω > 20,000 rad/s. Also, 20 log10 H = 20 log10 ( 4 ) = 12 dB for ω ≤ 20 and ω ≥ 20, 000 rad/s 20 log10 H = 20 log10 ( 40 ) = 32 dB for 200 ≤ ω ≤ 2000 rad/s

The asymptotic magnitude Bode plot for H is

13.3-29

R3 – +

+ –

vs

R1

+

+

0.5 μF

va

vb





R4

+

0.5 μF

vo

20 log10|H(ω)| (dB)

R2

−20

32

dB decade

0 −40

dB decade



320

8

ω (rad/s logarithmic scale)

(a)

(b)

Figure P 13.3-24 P 13.3-24

The input to the circuit shown in Figure P 13.3-24a is the voltage of the voltage source, vs.

The output of the circuit is the capacitor voltage, vo. The network function of the circuit is H (ω ) =

Vo (ω ) Vs (ω )

Determine the values of the resistances R1, R2, R3, and R4 required to cause the network function of the circuit to correspond to the asymptotic Bode plot shown in Figure P 13.3-24b. Solution: From Figure P13.3-24b, H (ω ) has poles at 8 and 320 rad/s and has a low frequency gain equal to 32 dB

= 40. Consequently, the network function corresponding to the Bode plot is H (ω ) =

±40 ω ⎞⎛ ω ⎞ ⎛ ⎜1 + j ⎟ ⎜1 + j ⎟ 8 ⎠⎝ 320 ⎠ ⎝

Next, we find the network function corresponding to the circuit. Represent the circuit in the frequency domain.

Apply KCL at the top node of the left capacitor, C1, to get 13.3-30

V a − Vs R1

+ j ω C 1 Va = 0 ⇒ Va =

1 Vs 1 + j ω C 1 R1

The op amp, together with resistors R2 and R3, comprise a noninverting amplifier so ⎛ R3 ⎞ Vb = ⎜1 + V ⎜ R 2 ⎟⎟ a ⎝ ⎠ (Alternately, this equation can be obtained by applying KCL at the inverting input node of the op amp.) Apply KCL at the top node of the right capacitor, C2, to get Vo − V b R4

+ j ω C 2 Vo = 0 ⇒ Vo =

1 Vb 1+ j ω C 2 R4

Combining these equations gives H (ω ) =

V o (ω ) V s (ω )

1+ =

R3 R2

(1 + j ω C R ) (1 + jω C 1

1

2

R4 )

Comparing to the specified network function gives 1+

R3 R2

(1 + j ω C R ) (1 + jω C 1

1

2

R4 )

=

±40 ω⎞⎛ ω ⎞ ⎛ ⎜1 + j ⎟ ⎜1 + j ⎟ 8 ⎠⎝ 320 ⎠ ⎝

The solution is not unique. For example, we can require 1+

R3 R2

= 40 , C 1 R1 =

1 1 = 0.125 , C 2 R 4 = = 0.00758 8 320

With the given values of capacitance, and choosing R2 = 10 kΩ, we have R1 = 250 kΩ, R2 = 10 kΩ, R3 = 390 kΩ and R4 = 6.25 kΩ (checked using LNAP 10/1/04)

13.3-31

P 13.3-25

R1

The input to the circuit shown in Figure

R2

R3 –

P 13.3-25a is the voltage of the voltage source, vs.

+ –

The output of the circuit is the voltage, vo. The

0.2 μF

vs

+

+

20 kΩ

network function of the circuit is

vo –

V (ω ) H (ω ) = o Vs (ω )

(a) 20 log10|H(ω)| (dB)

Determine the values of the resistances R1, R2, and R3 required to cause the network function of the circuit to correspond to the asymptotic Bode plot shown in Figure P 13.3-25b.

−20

18

dB decade

500 ω (rad/s logarithmic scale)

(b)

Figure P 13.3-25 Solution: From Figure P13.3-25b, H (ω ) has a pole at 500 rad/s and a low frequency gain of 18 dB = 8.

Consequently, the network function corresponding to the Bode plot is H (ω ) =

±8

ω ⎞ ⎛ ⎜1 + j ⎟ 500 ⎠ ⎝

Next, we find the network function corresponding to the circuit. Represent the circuit in the frequency domain.

The node equations are V a − Vs

R1

+

Va Va R2 + = 0 ⇒ Va = Vs 1 R2 R1 + R 2 + j ω C R1 R 2 jω C

13.3-32

Va

and

R2

+

Vo R3

= 0 ⇒ Vo = −

R3 R2

Va

The network function is H=

Vo Vs

− =

R3 R2



R2

R1 + R 2 + j ω C R1 R 2

=

R3

R1 + R 2 R R 1+ jω C 1 2 R1 + R 2

Comparing to the specified network function gives −

R3

R1 + R 2 ±8 = R R ω ⎞ ⎛ 1+ jω C 1 2 ⎜1 + j ⎟ 500 ⎠ R1 + R 2 ⎝ We require R3 R1 + R 2

= 8 and C

R1 R 2 R1 + R 2

=

1 = 0.002 500

The solution is not unique. With the given values of capacitance, and choosing R1 = R2, we have R1 = R2 = 20 kΩ and R3 = 320 kΩ (checked using LNAP 10/1/04)

13.3-33

+ –

The input to the circuit shown in Figure P 13.3-26a is

P 13.3-26

R1

the voltage of the voltage source, vs. The output of the circuit is the vs

voltage, vo. The network function of the circuit is

+ –

V (ω ) H (ω ) = o Vs (ω ) (a)

1 μF



Determine the values of the resistances, R1 and R2, required to cause the network function of the circuit to correspond to Determine the values of the gains K1 and K2 in Figure P 13.3-26b.

(a) 20 log10|H( ω )| (dB)

(b)

vo

R2

the asymptotic Bode plot shown in Figure P 13.3-26b.

+

K1

−20

dB decade

K2 500 20 ω (rad/s logarithmic scale)

(b)

Figure P 13.3-26 Solution: From Figure P13.3-26b, H (ω ) has a pole at 20 rad/s and a zero at 500 rad/s. Consequently, the network

function corresponding to the Bode plot is

ω ⎞ ⎛ ⎜1 + j ⎟ 500 ⎠ H (ω ) = ± K ⎝ . ω⎞ ⎛ ⎜1 + j ⎟ 20 ⎠ ⎝ Next, we find the network function corresponding to the circuit. Represent the circuit in the frequency domain. Apply KCL at the inverting input node of the op amp to get

V o − Vs or

(R

R1

1

+ j ω C 1 ( V o − Vs ) +

Vs R2

=0

+ R 2 + j ω C 1 R1 R 2 ) Vs = ( R 2 + j ω C 1 R1 R 2 ) V o

so H=

Vo Vs

=

R1 + R 2 + j ω C 1 R1 R 2 R 2 + j ω C 1 R1 R 2

=

R1 + R 2 R2

1 + j ω C1

×

R1 R 2 R1 + R 2

1 + j ω C 1 R1

13.3-34

a. Comparing to the specified network function gives

R1 + R 2 R2 C 1 R1 =

We require

×

R1 R 2 R1 + R 2

1 + j ω C 1 R1

=K

1+ j

ω 500

1+ j

ω

20

R1 R 2

1 1 = = 0.002 = .05 and C R1 + R 2 500 20

K=

Notice that

1 + j ω C1

R1 + R 2 R2

1 C 1 R1 × = 20 = 25 R1 R 2 1 C1 R1 + R 2 500

The solution is not unique. For example, choosing C = 1 μF b. The network function is

R1 = 50 kΩ and R2 = 2.083 kΩ

ω ⎞ ⎛ ⎜1 + j ⎟ 500 ⎠ H (ω ) = 25 ⎝ ω⎞ ⎛ ⎜1 + j ⎟ 20 ⎠ ⎝ so 20 ⎞ ⎛ K 1 = 20 log10 ( 25 ) = 28 dB and K 2 = 20 log10 ⎜ 25 × ⎟ = 0 dB 500 ⎠ ⎝ (checked using LNAP 10/1/04)

13.3-35

R2

R1

P 13.3-27

+

The input to the circuit shown in Figure P + –

vs

13.3-27a is the voltage of the voltage source, vs. The

Ria

C

function of the circuit is

vo –

ia

output of the circuit is the voltage, vo. The network

(a)

Vo (ω ) Vs (ω )

20 log10|H( ω )| (dB)

H (ω ) =

+ –

Determine the values of R, C, R1, and R2 required to cause the network function of the circuit to correspond to the asymptotic Bode plot shown in Figure P 13.3-27b.

−12

−20

dB decade

250 ω (rad/s logarithmic scale)

(b)

Figure P 13.3-27 Solution: From Figure P13.3-27b, H (ω ) has a pole at 250 rad/s and a low frequency gain equal to −12 dB = 0.25.

Consequently, the network function corresponding to the Bode plot is H (ω ) =

±0.25 1+ j

ω

.

250

Next, we find the network function corresponding to the circuit. Represent the circuit in the frequency domain. Apply KVL to the left mesh to get

Vs = R1 I a + R I a

⇒ Ia =

Vs R1 + R

Voltage division gives R R1 + R

1 Vo =

R jω C RIa = Ia = Vs 1 1+ j ω C R2 1+ j ω C R2 R2 + jω C

The network function of the circuit is H=

Vo Vs

=

R R1 + R 1 + j ω C R2

Comparing to the specified network function gives 13.3-36

R R1 + R 1+ j ω C R2

=

±0.25 1+ j

ω

250

The solution is not unique. We require

R 1 1 = and C R 2 = = 0.004 R1 + R 4 250 Choosing R = 100 Ω and C = 10 μF we have R1 = 300 Ω and R2 = 400 Ω (checked using LNAP 10/2/04)

13.3-37

P 13.3-28

The input to the circuit shown in Figure P

Gva

13.3-28a is the current of the current source, is. The

io

+

output of the circuit is the current io. The network

va

is

function of the circuit is

R1

R2

C



I o (ω ) I s (ω )

(a) 20 log10|H( ω )| (dB)

H (ω ) =

Determine the values of G, C, R1, and R2 required to cause the network function of the circuit to correspond to the asymptotic Bode plot shown in Figure P 13.328b.

−20

−6

dB decade

200 ω (rad/s logarithmic scale)

(b)

Figure P 13.3-28 Solution: From Figure P13.3-28b, H (ω ) has a pole at 200 rad/s and a low frequency gain equal to −6 dB = 0. 5. Consequently, the network function corresponding to the Bode plot is H (ω ) =

±0.5 1+ j

ω

.

200

Next, we find the network function corresponding to the circuit. Represent the circuit in the frequency domain. Apply KCL at the top node of R1 to get Is =

Va R1

+ G Va

⇒ Va =

R1 1 + G R1

Is

Current division gives 1

Io =

⎛ R1 ⎞ G G jω C Va = Is ⎟ G Va = ⎜⎜ 1 1+ j ω C R2 1 + j ω C R 2 ⎝ 1 + G R1 ⎟⎠ + R2 jω C

The network function of the circuit is G R1 H=

Io Is

=

1 + G R1 1+ j ω C R2

Comparing this network function to the specified network function gives 13.3-38

G R1 1 + G R1

= 0.5 and C R 2 =

1 200

The solution is not unique. Choosing G = 0.01 A/V and C = 10 μF gives R1 = 100 Ω and R2 = 500 Ω (checked using LNAP 10/2/04)

13.3-39

P 13.3-29

A first-order circuit is shown in Figure P

R2

R1

13.3-29. Determine the ratio Vo/Vs and sketch the Bode diagram when RC = 0.1 and R1/R2 = 3. ⎛ R ⎞ 1 Answer: H = ⎜ 1 + 1 ⎟ ⎝ R2 ⎠ 1 + jω RC



R

+

vs

+ –

C

+ vo –

Figure P 13.3-29 Solution: ⎛ R ⎞ Vo (ω ) = ⎜ 1+ 1 ⎟ Vc (ω ) ⎝ R2 ⎠ ⎛ R ⎞⎛ ⎞ 1 = ⎜ 1+ 1 ⎟ ⎜ ⎟ Vs (ω ) ⎝ R2 ⎠ ⎝ 1 + jω C R ⎠

H (ω ) =

⎞ Vo (ω ) ⎛ R1 ⎞⎛ 1 =⎜1+ ⎟⎜ ⎟ Vs (ω ) ⎝ R2 ⎠⎝ 1+ jω C R ⎠

When R C = 0.1 and then H (ω ) =

4 1+j

R1 = 3, R2

ω 10

13.3-40



P 13.3-30 (a) Draw the Bode diagram of the network function Vo/Vs

+

for the circuit of Figure P 13.3-30.



v s +–

vo

30 mF

(b) Determine vo(t) when



Figure P 13.3-30

vs = 10 cos 20t V. Answer: (b) vo = 4.18 cos (20t – 24.3°) V

Solution: a)

Zo = R2 +

1 jω C

ω Vo Zo ω1 jω C = = = ω 1 Vs R1 + Z o R1 + R2 + 1+j ω2 jω C R2 +

where ω1 = and ω 2 =

1 R2 C

1

1+ j

= 16.7 rad/s

1 = 5.56 rad/s ( R1 + R2 )C

vs ( t ) = 10 cos 20 t or Vs = 10∠0°

( (

) )

1+ j 20 Vo 16.7 ∴ = Vs 1+ j 20 5.56 1+ j 1.20 = = 0.418 ∠− 24.3° 1+ j 3.60 b) So Vo = 4.18 ∠ − 24.3° vo (t ) = 4.18cos(20t − 24.3°) V

13.3-41

P 13.3-31

Draw the asymptotic magnitude Bode diagram for

H (ω ) =

10(1 + jω ) jω (1 + j 0.5ω )(1 + j 0.6(ω / 50) + ( jω / 50) 2 )

Hint: At ω = 0.1 rad/s the value of the gain is 40 dB and the slope of the asymptotic Bode plot is –20

dB/decade. There is a zero at 1 rad/s, a pole at 2 rad/s, and a second-order pole at 50 rad/s. The slope of the asymptotic magnitude Bode diagram increases by 20 dB/decade as the frequency increases past the zero, decreases by 20 dB/decade as the frequency increases past the pole, and, finally, decreases by 40 dB/decade as the frequency increases past the second-order pole.

P13.3-31

13.3-42

Section 13-4: Resonant Circuits P 13.4-1

For a parallel RLC circuit with R = 10 kΩ, L = 1/120 H, and C = 1/30 μF, find ω0, Q,

ω1, ω2, and the bandwidth BW. Answer: ω0 = 60 krad/s, Q = 20, ω1 = 58.519 krad/s, ω2 = 61.519 krad/s, and BW = 3 krad/s Solution: For the parallel resonant RLC circuit with R = 10 kΩ, L = 1/120 H, and C = 1/30 µF we have 1 1 = = 60 k rad sec LC ⎛ 1 ⎞⎛ 1 −6 ⎞ x 10 ⎟ ⎜ ⎟⎜ ⎝ 120 ⎠ ⎝ 30 ⎠

ω0 =

1 × 10−6 C 30 Q= R = 10, 000 = 20 1 L 120

ω0

2

2

⎛ω ⎞ ⎛ω ⎞ ω 2 2 ω1 = − + ⎜ 0 ⎟ +ω 0 = 58.52 k rad s and ω 2 = 0 + ⎜ 0 ⎟ +ω 0 = 61.52 k rad s 2Q 2Q ⎝ 2Q ⎠ ⎝ 2Q ⎠ 1 1 = = 3 krad s BW = RC ⎛ 1 −6 ⎞ (10000 )⎜ ×10 ⎟ ⎝ 30 ⎠

Notice that BW = ω 2 − ω1 =

ω0 Q

.

13.4-1

P 13.4-2

A parallel resonant RLC circuit is driven by a current source is = 20 cos ωt mA and

shows a maximum response of 8 V at ω = 1000 rad/s and 4 V at 897.6 rad/s. Find R, L, and C. Answer: R = 400 Ω, L = 50 mH, and C = 20 μF Solution: For the parallel resonant RLC circuit we have

H (ω ) =

k ⎛ω ω ⎞ 1+ Q ⎜ − 0 ⎟ ⎝ ω0 ω ⎠

2

2

so

R = k = H(ω 0 ) =

8 = 400 Ω and ω 0 = 1000 rad s 20⋅10−3

4 = 200, so 20.10−3 400 200 = 2 2 ⎛ 897.6 1000 ⎞ 1+ Q ⎜ − ⎟ ⎝ 1000 897.6 ⎠

At ω = 897.6 rad s , H (ω ) =

⇒ Q =8

Then

1 ⎫ = ω 0 = 1000 ⎪ LC ⎪ C = 20 μ F ⎬ ⇒ L = 50 mH C =Q=8 ⎪ 400 ⎪⎭ L

13.4-2

P 13.4-3

A series resonant RLC circuit has L = 10 mH, C = 0.01 μF, and R = 100 Ω. Determine

ω0, Q, and BW. Answer: ω0 = 105, Q = 10, and BW = 104

Solution: For the series resonant RLC circuit with R = 100 Ω, L = 10 mH, and C = 0.01 µF we have

ω0 =

P 13.4-4

1 1 L R = 105 rad s , Q = = 10, BW = = 104 rad s R C L LC

A quartz crystal exhibits the property that when mechanical stress is applied across its

faces, a potential difference develops across opposite faces. When an alternating voltage is applied, mechanical vibrations occur and electromechanical resonance is exhibited. A crystal can be represented by a series RLC circuit. A specific crystal has a model with L = 1 mH, C = 10 μF, and R = 1 Ω. Find ω0, Q, and the bandwidth. Answer: ω0 = 104 rad/s, Q = 10, and BW = 103 rad/s

Solution: For the series resonant RLC circuit with R = 1 Ω, L = 1 mH, and C = 10 µF we have

ω0 =

1 1 L R = 104 rad s , Q = = 10, BW = = 103 rad s R C L LC

13.4-3

P 13.4-5 Design a parallel resonant circuit to have ω0 = 2500 rad/s, Z(ω0) = 100 Ω, and BW = 500 rad/s. Answer: R = 100 Ω, L = 8 mH, and C = 20 μF Solution: For the parallel resonant RLC circuit we have R = Z (ω 0 ) = 100 Ω

1 = BW = 500 rad/s ⇒ C = 20 μ F 100 C 1 = ω0 = 2500 rad/s ⇒ L = 8 mH −6 ( 20⋅10 ) L

P 13.4-6

Design a series resonant circuit to have ω0 = 2500 rad/s, Y(ω0) = 1/100 Ω, and BW =

500 rad/s. Answer: R = 100 Ω, L = 0.2 H, and C = 0.8 μF

P13.4-6 For the series resonant RLC circuit we have R=

1

Y (ω 0 )

= 100 Ω

100 = BW = 500 rad/s ⇒ L = 0.2 H L 1 = ω 0 = 2500 rad/s ⇒ C = 0.8 μ F ( 0.2 )C

13.4-4

a

P 13.4-7 The circuit shown in Figure P 13.4-7

10 μ H

represents a capacitor, coil, and resistor in parallel.

22 kΩ

600 pF 1.8 Ω Coil resistance

Calculate the resonant frequency, bandwidth, and Q for the circuit.

b

Figure P 13.4-7 Solution: C = 600 pF L = 10 µH

R1 = 1.8 Ω R 2 = 22 kΩ 1 1 + R1 + jω L R 2

Y (ω ) = jω C +

(R +R = 1

=

2

−ω 2C L R 2 ) + jω ( L +C R1 R 2 ) R1 − jω L × R1 − jω L R 2 ( R1 + jω L )

R1 ( R1 + R 2 −ω 2C L R 2 ) +ω 2 L( L + C R1 R 2 )+ jω R1 ( L + C R1 R 2 ) − jω L( R1 + R 2 −ω 2C L R 2 ) R 2 ( R1 −ω 2 L2 )

ω = ω 0 is the frequency at which the imaginary part of Y (ω ) is zero : R1 ( L +C R1 R 2 ) − L ( R1 + R 2 −ω02 C L R 2 ) = 0

⇒ ω0 =

L R 2 −C R12 R 2 C L2 R 2

= 12.9 M rad sec

13.4-5

P 13.4-8 Consider the simple model of an electric

L

power system as shown in Figure P 13.4-8. The

+

Power line vs

inductance, L = 0.25 H, represents the power line

+ –

C

RL

vo –

and transformer. The customer’s load is RL = 100

Power plant

Ω, and the customer adds C = 25 μF to increase the

Customer load

Figure P 13.4-8

magnitude of Vo.

The source is vs = 1000 cos 400t V, and it is desired that |Vo| also be 1000 V. (a)

Find |V0| for RL = 100 Ω.

(b)

When the customer leaves for the night, he turns off much of his load, making RL = 1 kΩ, at which point sparks and smoke begin to appear in the equipment still connected to the power line. The customer calls you in as a consultant. Why did the sparks appear when RL = 1 kΩ?

Solution: In the frequency domain we have:

(a) Using voltage division yields

(

Vo = 1000∠0°

)

(100 )( − j100 )

100 105 ∠− 45° ∠− 45° 100 − j100 ° 2 2 = 1000∠0 = = 1000∠ − 90° V 100 (100 )( − j100 ) + j100 ∠− 45°+ j100 50 2∠45° 2 100 − j100 ∴|Vo | = 1000 V

(

)

(b) Do a source transformation to obtain

13.4-6

This is a resonant circuit with ω 0 = 1

LC = 400 rad/s. Since this also happens to be the

frequency of the input, so this circuit is being operated at resonance. At resonance the admittances of the capacitor and inductor cancel each other, leaving the impedance of the resistor. Increasing the resistance by a factor of 10 will increase the voltage Vo by a factor of 10. This increased voltage will cause increased currents in both the inductance and the capacitance, causing the sparks and smoke.

P 13.4-9

a

L

b

R1

Consider the circuit in Figure P 13.4-9. R1 = R2 = 1 Ω.

Select C and L to obtain a resonant frequency of ω0 = 100 rad/s.

R2

C

Figure P 13.4-9 Solution:

Let G 2 =

1 . Then R2

Z = R1 + jω L +

(R G = 1

2

1 G 2 + jω C

+ 1 − ω 2 L C ) + j (ω LG 2 + ω C R1 ) G 2 + jω C

At resonance, ∠Z = 0° so tan −1

ω L G2 +ω C R1 ωC = tan −1 2 G2 ( R1G 2 +1−ω L C )

so C − L G 22 ω L G 2 +ω C R1 ωC 2 = ⇒ ω = LC2 ( R1 G 2 +1−ω 2 L C ) G 2

and C > G 22 L

C−L . Then choose C and calculate L: LC2 C = 10 mF ⇒ L = 5 mH 2

With R1 = R 2 = 1 Ω and ω 0 = 100 rad s , ω0 = 104 =

Since C > G 22 L , we are done.

13.4-7

R

P 13.4-10

For the circuit shown in Figure P 13.4-10,

(a) derive an expression for the magnitude response |Zin|

C

L

versus ω, Zin

(b) sketch |Zin| versus ω, and (c) find |Zin| at ω = 1/ LC .

Figure P 13.4-10 Solution: (a) R R −ω 2 R L C ) + jω L ( jω C Z in = jω L + = 1 1+ jω R C R+ jω C Consequently,

( R −ω R L C ) +(ω L ) 2

| Zin | =

2

1+ (ω R C )

2

2

(b)

(c)

ω=

1 LC

⇒ | Zin | =

1 C L

⎛ R2 C ⎞ ⎜1 + ⎟ L ⎠ ⎝

13.4-8

P 13.4-11 The circuit shown in Figure P 13.4-11 shows an experimental setup that could be used to measure the parameters k, Q,

Oscilloscope

and ω0 of this series resonant circuit. These parameters can be determined from a magnitude frequency response plot for Y = I/V. It is more convenient to measure node voltages than currents, so the node voltages V and V2 have been measured. Express |Y| as a

v

+ –

R

i

function of V and V2. Hint: Let V = A and V2 = B∠θ

Then I =

+ v2 –

L C

Figure P 13.4-11

( A − B cos θ ) − jB sin θ R

( A − B cos θ ) 2 + ( B sin θ ) 2 Answer: | Y | = AR Solution: Let V (ω ) = A∠0 and V2 (ω ) = B∠θ . Then I (ω ) = Y (ω ) = V (ω )

V (ω ) − V2 (ω ) A − B∠θ A − B cos θ − j B sin θ R = = AR AR V (ω )

( A − B cosθ ) + ( B sin θ ) 2

| Y (ω ) | =

2

AR

13.4-9

Section 13-6: Plotting Bode Plots Using MATLAB P 13.6-1

The input to the circuit shown in Figure P 13.6-1 is the voltage of the voltage source,

vs. The output of the circuit is the voltage, vo. Use MATLAB to plot the gain and phase shift of this circuit as a function of frequency for frequencies in the range 1 < ω < 1000 rad/s. 10 Ω

20 Ω + –

vs

+

vo



1 mF 0.5 H

Figure P 13.6-1 Solution: Using voltage division twice gives 1 jω L jω C V o (ω ) = V s (ω ) − V s (ω ) 1 R j L + ω 1 R2 + jω C so V o (ω ) 1 jω L H (ω ) = = − Vs (ω ) 1 + j ω C R 2 R1 + j ω L

Modify the MATLAB script given in Section 13.7 of the text: % P13_7_1.m - plot the gain and phase shift of a circuit %--------------------------------------------------------------% Create a list of logarithmically spaced frequencies. %--------------------------------------------------------------wmin=1; wmax=1000;

% starting frequency, rad/s % ending frequency, rad/s

w = logspace(log10(wmin),log10(wmax)); %--------------------------------------------------------------% Enter values of the parameters that describe the circuit. %--------------------------------------------------------------R1 = 10; % Ohms R2 = 20; % Ohms C = 0.001; % Farads

13.6-1

L = 0.5;

% Henries

%--------------------------------------------------------------% Calculate the value of the network function at each frequency. % Calculate the magnitude and angle of the network function. %--------------------------------------------------------------for k=1:length(w) H(k) = 1/(1+j*R2*C*w(k)) - j*L*w(k)/(R1+j*L*w(k)); gain(k) = abs(H(k)); phase(k) = angle(H(k))*180/pi; end %--------------------------------------------------------------% Plot the frequency response. %--------------------------------------------------------------subplot(2,1,1), semilogx(w, gain) xlabel('Frequency, rad/s'), ylabel('Gain, V/V') title('Frequency Response Plots') subplot(2,1,2), semilogx(w, phase) xlabel('Frequency, rad/s'), ylabel('Phase, deg')

Here are the plots produced by MATLAB:

13.6-2

P 13.6-2 The input to the circuit shown in Figure P 13.6-2 is the voltage of the voltage

vs

20 Ω 1 mF

0.5 H 10 Ω

+ –

source, vs. The output of the circuit is the

+ vo –

voltage, vo. Use MATLAB to plot the gain and

Figure P 13.6-2

phase shift of this circuit as a function of frequency for frequencies in the range 1 < ω < 1000 rad/s. Solution: Let Z s = R2 +

1 jω C

and Z p =

R1Z s R1 + Z s

Using voltage division twice gives 1 V a (ω ) =

Zp jω L + Z p

so H (ω ) =

Vs (ω ) and Vo (ω ) =

V o (ω ) V s (ω )

=

jω C V a (ω ) 1 R2 + jω C

Zp

( j ω L + Z ) (1 + j ω C R ) p

2

Modify the MATLAB script given in Section 13.7 of the text: % P13_7_2.m - plot the gain and phase shift of a circuit pi = 3.14159; %--------------------------------------------------------------% Create a list of logarithmically spaced frequencies. %--------------------------------------------------------------wmin=1; wmax=1000;

% starting frequency, rad/s % ending frequency, rad/s

w = logspace(log10(wmin),log10(wmax)); %--------------------------------------------------------------% Enter values of the parameters that describe the circuit. %--------------------------------------------------------------R1 = 10; R2 = 20; C = 0.001; L = 0.5;

% % % %

Ohms Ohms Farads Henries

13.6-3

%--------------------------------------------------------------% Calculate the value of the network function at each frequency. % Calculate the magnitude and angle of the network function. %--------------------------------------------------------------for k=1:length(w) Zs(k) = R2+1/(j*w(k)*C); Zp(k) = R1*Zs(k)/(R1+Zs(k)); H(k) = Zp(k)/((j*w(k)*L+Zp(k))*(1+j*w(k)*C*R2)); gain(k) = abs(H(k)); phase(k) = angle(H(k))*180/pi; end %--------------------------------------------------------------% Plot the frequency response. %--------------------------------------------------------------subplot(2,1,1), semilogx(w, gain) xlabel('Frequency, rad/s'), ylabel('Gain, V/V') title('Frequency Response Plots') subplot(2,1,2), semilogx(w, phase) xlabel('Frequency, rad/s'), ylabel('Phase, deg')

Here are the plots produced by MATLAB:

13.6-4

P 13.6-3

40 Ω

The input to the circuit shown in Figure

P 13.6-3 is the voltage of the voltage source, vs. The output of the circuit is the voltage, vo. Use

20 Ω

25 mF

MATLAB to plot the gain and phase shift of this

+ –

+

25 Ω

vo

vs

circuit as a function of frequency for frequencies in

0.2 H



the range 1 < ω < 1000 rad/s. Figure P 13.6-3 Solution: Let Z1 = R2 +

Using voltage division gives

V a (ω ) =

Z2 Z1 + Z 2

V s (ω ) ⇒ H ( ω ) =

R1 j ω C R1

V o (ω ) V s (ω )

=

and Z 2 = R 3 + j ω L

Z2 Z1 + Z 2

Modify the MATLAB script given in Section 13.7 of the text: % P13_7_3.m - plot the gain and phase shift of a circuit pi = 3.14159; %--------------------------------------------------------------% Create a list of logarithmically spaced frequencies. %--------------------------------------------------------------wmin=1; wmax=1000;

% starting frequency, rad/s % ending frequency, rad/s

w = logspace(log10(wmin),log10(wmax)); %--------------------------------------------------------------% Enter values of the parameters that describe the circuit. %--------------------------------------------------------------R1 = 40; R2 = 20; R3 = 25; C = 0.025; L = 0.2;

% % % % %

Ohms Ohms Ohms Farads Henries

%---------------------------------------------------------------

13.6-5

% Calculate the value of the network function at each frequency. % Calculate the magnitude and angle of the network function. %--------------------------------------------------------------for k=1:length(w) Z1(k) = R2+R1/(j*w(k)*C*R1); Z2(k) = R3+j*w(k)*L; H(k) = Z2(k)/(Z1(k)+Z2(k)); gain(k) = abs(H(k)); phase(k) = angle(H(k))*180/pi; end %--------------------------------------------------------------% Plot the frequency response. %--------------------------------------------------------------subplot(2,1,1), semilogx(w, gain) xlabel('Frequency, rad/s'), ylabel('Gain, V/V') title('Frequency Response Plots') subplot(2,1,2), semilogx(w, phase) xlabel('Frequency, rad/s'), ylabel('Phase, deg')

Here are the plots produced by MATLAB:

13.6-6

Section 13.8 How Can We Check…? P 13.8-1 Circuit analysis contained in a lab report indicates that the network function of a circuit is H (ω ) =

1+ j

ω

630 ω ⎞ ⎛ 10 ⎜1 + j ⎟ 6300 ⎠ ⎝

This lab report contains the following frequency response data from measurements made on the circuit. Do these data seem reasonable? ω, rad/s

200

400

795

1585

3162

|H(ω)|

0.105

0.12

0.16

0.26

0.460

ω, rad/s

6310

12,600

25,100

50,000

100,000

|H(ω)|

0.71

1.0

1.0

1.0

1.0

P13.8-1 When ω < 630 rad/s, H(ω) ≅ 0.1, which agrees with the tabulated values of | H(ω)| corresponding to ω = 200 and 400 rad/s. When ω > 6300 rad/s, H(ω) ≅ 1.0, which agrees with the tabulated values of | H(ω)| corresponding to ω = 12600, 25000, 50000 and 100000 rad/s. At ω = 6300 rad/s, we expect | H(ω)| = −3 dB = 0.707. This agrees with the tabulated value of | H(ω)| corresponding to ω = 6310 rad/s. At ω = 630 rad/s, we expect | H(ω)| = −20 dB = 0.14. This agrees with the tabulated values of | H(ω)| corresponding to ω = 400 and 795 rad/s. This data does seem reasonable.

13.8-1

P 13.8-2

A parallel resonant circuit (see Figure 13.4-2) has Q = 70 and a resonant frequency ω0

= 10,000 rad/s. A report states that the bandwidth of this circuit is 71.43 rad/s. Verify this result. Solution: BW =

P 13.8-3

ω0 Q

=

10,000 = 143 ≠ 71.4 rad s . Consequently, this report is not correct. 70

A series resonant circuit (see Figure 13.4-4) has L = 1 mH, C = 10 μF, and R = 0.5 Ω.

A software program report states that the resonant frequency is f0 = 1.59 kHz and the bandwidth is BW = 79.6 Hz. Are these results correct? Solution: 1 1 L R = 10 k rad s = 1.59 kHz, Q = = 20 and BW = = 500 rad s = 79.6 Hz R C L LC The reported results are correct.

ω0 =

An old lab report contains the approximate

Bode plot shown in Figure P 13.8-4 and concludes that the network function is

ω ⎞ ⎛ 40 ⎜ 1 + j ⎟ 200 ⎠ H (ω ) = ⎝ ω ⎞ ⎛ ⎜1 + j ⎟ 800 ⎠ ⎝ Do you agree?

32

|H ( ω )|, dB

P 13.8-4

20 200 800 ω , rad/sec log scale

Figure P 13.8-4 Solution: The network function indicates a zero at 200 rad/s and a pole at 800 rad/s. In contrast, the Bode plot indicates a pole at 200 rad/s and a zero at 800 rad/s. Consequently, the Bode plot and network function don’t correspond to each other.

13.8-2

PSpice Problems 4 kΩ SP 13-1 The input to the circuit shown in Figure SP 13-1 is the voltage of the voltage + source, vi(t). The output is the voltage, vo(t), + vo(t) vi(t) – 5 μF 1 kΩ across the parallel connection of the capacitor and 1-kΩ resistor. The network function that – represents this circuit is Figure SP 13-1 V (ω ) k H (ω ) = o = Vi (ω ) 1 + j ω p Use PSpice to plot the frequency response of this circuit. Determine the values of the pole, p, and of the dc gain, k.

Solution:

Here are the magnitude and phase frequency response plots:

From the magnitude plot, the low frequency gain is k = 200m = 0.2. From the phase plot, the angle is -45° at p = 2π ( 39.891) = 251 rad/s .

13SP-1

SP 13-2 The input to the circuit shown in Figure SP 13-2 is the voltage of the voltage source, vi(t). The output is the voltage, vo(t), across the series connection of the inductor and 60-Ω resistor. The network function that represents this circuit is

ω

1+ j V (ω ) z H (ω ) = o =k = ω Vi (ω ) 1+ j p Use PSpice to plot the frequency response of this circuit. Determine the values of the pole, p, of the zero, z, and of the dc gain, k. Answer: p = 20 rad/s, z = 12 rad/s, and k = 0.6 V/V

40 Ω +

60 Ω vi(t)

+ –

vo(t)

5H –

Figure SP 13-2

Solution: Here is the magnitude frequency response plot:

The low frequency gain is 0.6 = lim H (ω ) = k ⇒ k = 0.6 . ω →0

The high frequency gain is 1 = lim H (ω ) = k ω →∞

At ω = 2π ( 2.8157 ) = 17.69 rad/s ,

p z

⇒ z = ( 0.6 ) p

2

⎛ 17.69 ⎞ 1+ ⎜ 0.6 p ⎟⎠ ⎝ 0.8 = 0.6 2 ⎛ 17.69 ⎞ 1+ ⎜ ⎟ ⎝ p ⎠



16 p 2 + 869 16 2 = 2 ⇒ p + 313 = p 2 + 869 ⇒ 9 9 p + 313



p = 20 rad/s and z = 12 rad/s

(

)

( 0.77778) p 2 = 312.56

13SP-1

SP 13-3 The input to the circuit shown in Figure SP 13-3 is the voltage of the voltage source, vi(t). The output is the voltage, vo(t), across 30-kΩ resistor. The network function that represents H (ω ) =

this circuit is

Vo (ω ) k = Vi (ω ) 1 + j ω p

Use PSpice to plot the frequency response of this circuit. Determine the values of the pole, p, and of the dc gain, k. Answer: p = 100 rad/s and k = 4 V/V 2 kΩ

15 kΩ +

+ vi(t)

+ –

vC(t)

5 μF 6 vC(t)

+ –

30 kΩ

vo(t) –



Figure SP 13-3 Solution:

From the magnitude plot, the low frequency gain is k = 4.0. Also, the gain is 4/sqrt(2) = 2.828 at 15.914 hertz. From the phase plot, the angle is -45° at p = 2π (15.998 ) = 100.5 rad/s .

13SP-1

SP 13-4 The input to the circuit shown in Figure SP 13-4 is the voltage of the voltage source, vi(t). The output is the voltage, vo(t), across 20-kΩ resistor. The network function that represents this circuit is H (ω ) =

Vo (ω ) k = Vi (ω ) 1 + j ω p

50 kΩ 10 kΩ 2 μF

vi(t)

+ –

– +

20 kΩ

Use PSpice to plot the frequency response of this circuit. Determine the values of the pole, p, and of the dc gain, k. Answer: p = 10 rad/s and k = 5 V/V

+ vo(t) –

Figure SP 13-4

Solution: Here’s the circuit drawn in the PSpice workspace:

Here are the frequency response plots:

13SP-1

From the magnitude plot, the low frequency gain is k = 5.0. From the phase plot, the angle is 180°-45°=135° at p = 2π (1.5849 ) = 9.958 rad/s .

13SP-1

SP 13-5 Figure SP 13-5 shows a circuit and a frequency response. The frequency response plots were made using PSpice and Probe. V(R3:2) and Vp(R3:2) denote the magnitude and angle of the phasor corresponding to vo(t). V(V1:+) and Vp(V1:+) denote the magnitude and angle of the phasor corresponding to vi(t). Hence V(R3:2)/V(V1:+) is the gain of the circuit and Vp(R3:2) – Vp(V1:+) is the phase shift of the circuit. Determine values for R and C required to cause the circuit correspond to the frequency response. Hint: PSpice and Probe use m for milli or 10–3. Hence, the label (159.513, 892.827m) indicates that the gain of the circuit is 892.827*10–3 = 0.892827 at a frequency of 159.513 Hz ≈ 1000rad/sec. Answer: R = 5kΩ and C = 0.2 μF 2.0 (159.513, 892.827 m) (31.878, 1.8565) (318.784, 484.412 m)

1.0

V(R3:2)/V(V1:+) 0

10 kΩ R

175 d C (159.513, 116.515)

150 d (31.878, 158.169) vi(t)

+ –

(318.784, 104.017)



125 d

+

+ 20 kΩ

vo(t) –

100 d

Vp(R3:2)– Vp(V1:+)

10 Hz

30 Hz

100 Hz

300 Hz

1.0 KHz

Frequency

(a) FIGURE

(b)

SP 13-5 (a) A circuit and (b) the corresponding frequency response.

Solution: From the circuit 104 104 R R H (ω ) = − = ∠ − tan −1 (ω C 104 ) 1 + jω C 104 4 2 1 + (ω C 10 )

From the plot, at ω = 200 rad/sec = 31.83 Hertz H(ω) is

1.8565∠158° =

104 R

1 + (ω C 10

)

4 2

∠ − tan −1 (ω C 104 )

13SP-1

Equating phase shifts gives C R 104 ω C 10 = 10 = tan(22°) = 0.404 ⇒ C = 0.2 μ F R + 104 4

3

Equating gains gives 1.8565 =

104 R 1 + (ω C 104 )

2

=

104 R 1 + ( 0.404 )

2

⇒ R = 5 kΩ

13SP-1

SP 13-6 Figure SP 13-6 shows a circuit and a frequency response. The frequency response plots were made using PSpice and Probe. V(R2:2) and Vp(R2:2) denote the magnitude and angle of the phasor corresponding to vo(t). V(V1:+) and Vp(V1:+) denote the magnitude and angle of the phasor corresponding to vi(t). Hence V(R2:2)/V(V1:+) is the gain of the circuit, and Vp(R2:2) – Vp(V1:+) is the phase shift of the circuit. Determine values for R and C required to make the circuit correspond to the frequency response. Hint: PSpice and Probe use m for milli or 10–3. Hence, the label (159.268, 171.408m) indicates that the gain of the circuit is 171.408*10–3 = 0.171408 at a frequency of 159.268 Hz ≈ 1000 rad/sec. Answer: R = 20 kΩ and C = 0.25 μF 400 m (79.239, 256.524 m) (159.268, 171.406 m) 200 m

(316.228, 96.361 m)

V(R2:2)/V(V1:+) 0 0d (79.239, –39.685) (159.268, –59.055) (316.228, –73.197)

–50 d R + vi(t)

+ –

10 kΩ

C

Vp(R2:2)– Vp(V1:+)

vo(t) –100 d 10 Hz –

100 Hz

1.0 KHz

10 KHz

Frequency

(a) FIGURE

(b)

SP 13-6 (a) A circuit and (b) the corresponding frequency response.

Solution: From the circuit 104 1 + jω C R 2

104 104 C R 104 ⎞ R + 104 R + 104 −1 ⎛ H (ω ) = = = ∠ − tan ω ⎜ 4 ⎟ 104 C R 104 4 2 ⎝ R + 10 ⎠ ⎛ ⎞ 10 C R + R 1 + j ω 1 + ⎜ω 1 + jω C 104 R + 104 4 ⎟ ⎝ R + 10 ⎠ From the plot, at ω = 1000 rad/sec = 159.1 Hertz H(ω) is

13SP-1

104 R + 104

⎛ C R 104 ⎞ ∠ − tan ⎜ ω 0.171408∠ − 59° = 4 ⎟ 2 ⎝ R + 10 ⎠ ⎛ C R 104 ⎞ 1+ ⎜ω 4 ⎟ ⎝ R + 10 ⎠ −1

Equating phase shifts gives

ω

4 C R 104 3 C R 10 10 = = tan(59°) = 1.665 R + 104 R + 104

Equating gains gives 104 R + 104

0.171408 =

⎛ C R 10 ⎞ 1+ ⎜ω 4 ⎟ ⎝ R + 10 ⎠ 4

2

=

104 R + 104 1 + (1.665 )

2

⇒ R = 20 kΩ

Substitute this value of R into the equation for phase shift to get: C ( 20 ×103 ) 104 C R 104 3 1.665 = 10 = 10 R + 104 ( 20 ×103 ) + 104 3

⇒ C = 0.25 μ F

13SP-1

Design Problems DP 13-1 Design a circuit that has a low-frequency gain of 2, a high-frequency gain of 5, and makes the transition of H = 2 to H = 5 between the frequencies of 1 kHz and 10 kHz. Solution: Pick the appropriate circuit from Table 13.4-2.

We require 2π × 1000 < z =

R2 1 1 p C , 2π × 10000 > p = , 2=k = and 5 = k = 1 C 1 R1 C2 R 2 z C2 R1

Try z = 2π × 2000. Pick C1 = 0.05 μ F. Then R1 =

Check: p =

1 C C = 1.592 kΩ, R 2 = 2 R1 = 3.183 kΩ and C2 = 1 = 1 = 0.01 μ F p C1 z 2 k z

1 = 31.42 k rad s < 2π ⋅10, 000 rad s. C 2 R2

13DP-1

DP 13-2 Determine L and C for the circuit of Figure DP 13-2 in order to obtain a low-pass filter with a gain of –3 dB at 100 kHz. L + vs

+ –

C

1 kΩ

vo –

Figure DP 13-2 Solution:

1

1 V (ω ) jω C 1+ jω C R LC H (ω ) = o = = = 1 1 R Vs (ω ) ⎛ 1 ⎞ −ω 2 + jω + jω L + ⎜ || R ⎟ jω L + 1+ jω C R RC LC ⎝ jω C ⎠

Pick

|| R

R

1 = ω 0 = 2π (100 ⋅103 ) rad s . When ω = ω 0 LC

1 LC H 0 (ω ) = 1 1 1 1 − +j + LC LC RC LC So H (ω 0 ) = R

C . We require L −3 dB = 0.707 = H (ω 0 ) = R

C C = 1000 L L

Finally 1 ⎫ = 2π (100⋅103 ) ⎪ LC C =1.13 nF ⎪ ⎬⇒ C ⎪ L = 2.26 mH 0.707 =1000 ⎪⎭ L

13DP-2

0.47 μ F DP 13-3 British Rail has constructed an instrumented 8.06 kΩ railcar that can be pulled over 2.37 MΩ its tracks at speeds up to 180 10 kΩ 1 MΩ km/hr and will measure the – Circuit – B track-grade geometry. Using + v + s – Circuit A + + such a railcar, British Rail can vo 0.1 μ F monitor and track gradual – degradation of the rail grade, 866 kΩ 499 kΩ especially the banking of – curves, and permit preventive + maintenance to be scheduled as Circuit C needed well in advance of track-grade failure. Figure DP 13-3 The instrumented railcar has numerous sensors, such as angular-rate sensors (devices that output a signal proportional to rate of rotation) and accelerometers (devices that output a signal proportional to acceleration), whose signals are filtered and combined in a fashion to create a composite sensor called a compensated accelerometer (Lewis, 1988). A component of this composite sensor signal is obtained by integrating and high-pass filtering an accelerometer signal. A first-order low-pass filter will approximate an integrator at frequencies well above the break frequency. This can be seen by computing the phase shift of the filter-transfer function at various frequencies. At sufficiently high frequencies, the phase shift will approach 90°, the phase characteristic of an integrator. A circuit has been proposed to filter the accelerometer signal, as shown in Figure DP 133. The circuit is comprised of three sections, labeled A, B, and C. For each section, find an expression for and name the function performed by that section. Then find an expression for the gain function of the entire circuit, Vo/Vs. For the component values, evaluate the magnitude and phase of the circuit response at 0.01, 0.02, 0.05, 0.1, 0.2, 0.5, 1.0, 2.0, 5.0, and 10.0 Hz. Draw a Bode diagram. At what frequency is the phase response approximately equal to 0°? What is the significance of this frequency?

13DP-3

Solution: R1 = 10 kΩ R 2 = 866 kΩ R 3 = 8.06 kΩ R 4 = 1 MΩ R 5 = 2.37 MΩ R 6 = 499 kΩ C 1 = 0.47 μ F C 2 = 0.1 μ F

Va = −

Circuit A

R3 R2

Vc −

R3 R1

Vs = −H 1 Vc − H 2 Vs

R5 Circuit B

Vo = −

Circuit C

Vc = −

R4 1 + jω C 1 R 5

Va = − H 3 Va

1 Vo = −H 4 Vo jω C 2 R 6

Then Vc = H 3 H 4 Va

Va = −H 2 Vs − H1 H 3 H 4 Va

Vo = −H 3 Va =

⇒ Va =

−H 2 Vs 1 + H1 H 3 H 4

H 2 H3 Vs 1 + H1 H 3 H 4

After some algebra jω

Vo =

R3 R1 R 4 C 1

R3 R 2 R 4 R 6 C1 C 2

−ω + j 2

ω

Vs

R5 C1

This MATLAB program plots the Bode plot: R1=10; R2=866; R3=8.060; R4=1000;

% units: kOhms and mF so RC has units of sec

13DP-4

R5=2370; R6=449; C1=0.00047; C2=0.0001; pi=3.14159; fmin=5*10^5; fmax=2*10^6; f=logspace(log10(fmin),log10(fmax),200); w=2*pi*f; b1=R3/R1/R4/C1; a0=R3/R2/R4/R6/C1/C2; a1=R5/C1; for k=1:length(w) H(k)=(j*w(k)*b1)/(a0-w(k)*w(k)+j+w(k)*a1); gain(k)=abs(H(k)); phase(k)=angle(H(k)); end subplot(2,1,1), semilogx(f, 20*log10(gain)) xlabel('Frequency, Hz'), ylabel('Gain, dB') title('Bode Plot') subplot(2,1,2), semilogx(f, phase*180/pi) xlabel('Frequency, Hz'), ylabel('Phase, deg')

13DP-5

DP 13-4 Design a circuit that has the network function H (ω ) =

jω ω ⎞⎛ ω ⎞ ⎛ ⎜1 + j ⎟ ⎜1 + j ⎟ 200 ⎠ ⎝ 500 ⎠ ⎝

Hint: Use two circuits from Table 13.4-1. Connect the circuits in cascade. That means that the

output of one circuit is used as the input to the next circuit. H(ω) will be the product of the network functions of the two circuits from Table 13.3-2. Solution: Pick the appropriate circuits from Table 13.4-2.

We require 10 = − k1k2 = R 2C1

R4 R3

, 200 = p1 =

1 1 and 500 = p2 = R1 C 1 C 2 R4

Pick C 1 = 1 μ F. Then R1 =

1 1 = 5 kΩ. Pick C 2 = 0.1 μF. Then R 4 = = 20 kΩ. p1 C1 p2C2

Next

10 =

R2 R3

(10−6 )(20 ⋅103 ) ⇒

R2 R3

= 500

Let R 2 = 500 kΩ and R 3 = 1 kΩ.

13DP-6

DP 13-5 Strain-sensing instruments can be used to measure orientation and magnitude of strains running in more than one direction. The search for a way to predict earthquakes focuses on identifying precursors, or changes, in the ground that reliably warn of an impending event. Because so few earthquakes have occurred precisely at instrumented locations, it has been a slow and frustrating quest. Laboratory studies show that before rock actually ruptures—precipitating an earthquake—its rate of internal strain increases. The material starts to fail before it actually breaks. This prelude to outright fracture is called “tertiary creep” (Brown, 1989). The frequency of strain signals varies from 0.1 to 100 rad/s. A circuit called a band-pass filter is used to pass these frequencies. The network function of the band-pass filter is Kjω H (ω ) = ⎛ ω ⎞⎛ ω⎞ ⎜ 1 + j ⎟⎜ 1 + j ⎟ ω1 ⎠⎝ ω2 ⎠ ⎝ Specify ω1, ω2, and K so that the following are the case: 1.

The gain is at least 17 dB over the range 0.1 to 100 rad/s.

2.

The gain is less than 17 dB outside the range 0.1 to 100 rad/s.

3.

The maximum gain is 20 dB.

Solution: Pick the appropriate circuits from Table 13.4-2.

13DP-7

We require 20 dB = 10 = − k1k2 = R 2C1

Pick C 1 = 20 μ F. Then R1 =

R4 R3

, 0.1 = p1 =

1 1 and 100 = p2 = R1 C 1 C 2 R4

1 1 = 500 kΩ. Pick C 2 = 1 μF. Then R 4 = = 10 kΩ. p1 C 1 p2C2

Next 10 =

R2 R3

(20 ⋅10−6 )(10 ⋅103 ) ⇒

R2 R3

= 50

Let R 2 = 200 kΩ and R 3 = 4 kΩ.

13DP-8

DP 13-6 Is it possible to design the circuit shown in Figure DP 13-6 to have a phase shift of – 45°and a gain of 2 V/V both at a frequency of 1000 radians/second using a 0.1 microfarad capacitor and resistors from the range 1 k ohm to 200 k ohm? R3

R2 –

R1

+

A cos (ω t + θ) +–

+ 100 kΩ

C

vo(t) –

Figure DP 13-6

Solution:

1+ The network function of this circuit is H (ω ) =

R2 R3

1+ jω R1C

The phase shift of this network function is θ = − tan −1 ω R1C 1+ The gain of this network function is G = H (ω ) =

R3 R2

1+ (ω R1C ) 2

1+ =

R3 R2

1+ ( tan θ )

2

Design of this circuit proceeds as follows. Since the frequency and capacitance are known, R1 is tan(−θ ) . Next pick R2 = 10kΩ (a convenient value) and calculated R3 using calculated from R1 = ωC R 3 = (G ⋅ 1+ (tan θ ) 2 − 1) ⋅ R 2 . Finally

θ = −45 deg, G = 2, ω = 1000 rad s ⇒ R1 = 10 kΩ, R 2 = 10 kΩ, R 3 = 18.284 kΩ, C = 0.1 μF

13DP-9

DP 13-7 Design the circuit shown in Figure DP 13-7a to have the asymptotic Bode plot shown in Figure DP 13-7b. R2

0.5 μ F

C

vin(t)



32 | H(ω ) |, dB

+ –

R1

+ 100 kΩ

vout(t)

+



20 200 800 ω , rad/sec log scale

(a)

(b)

Figure DP 13-7

Solution: From Table 13.4-2 and the Bode plot: 800 = z =

1 ⇒ R1 = 2.5 kΩ R1 (0.5×10−6 )

32 dB = 40 = 200 = p =

R2 R1

⇒ R 2 = 100 kΩ

1 1 ⇒ C = = 0.05μ F R2C (200)(100×103 )

(Check: 20 dB = 10 = k

p 0.5×10−6 0.5×10−6 = = ) z C 0.05×10−6

13DP-10

DP 13-8 For the circuit of Figure DP 13.-8, select R1 and R2 so that the gain at high frequencies is 10 V/V and the phase shift is 195° at ω = 1000 rad/s. Determine the gain at ω = 10 rad/s. R1

R2 0.1 μ F

vs

+ –

– +

+ vo –

Figure DP 13.-8 Solution: −R2 jω C R 2 = − 1 1+ jω C R1 1+ jω C tan(270°−195°) 195° = 180 + 90 − tan −1 ω C R1 ⇒ R1 = = 37.3 kΩ (1000)(0.1×10−6 ) R 10 = lim H (ω ) = 2 ⇒ R 2 = 10 R1 = 373 kΩ ω →∞ R1 H (ω ) =

13DP-11

Chapter 14: The Laplace Transform Exercises Exercise 14.7-1 Determine the voltage vC(t) and the current iC(t) for t ≥ 0 for the circuit of Figure 14.7-1. Hint: vC(0) = 4 V Answer:

vC(t) = (6–2e–0.67t) u(t)

V and

iC(t) =

2 –0.67t e u(t) A 3

Figure 14.7-1

Solution:

KCL at top node:

VC ( s ) 3

s 2 + VC ( s ) = + 2 2 s

VC ( s ) =

(

6 2 − s s+ 2 3

v C (t ) = 6 − 2 e

2 VC ( s ) I C (s) = −2= 3 2 2 s+ 3 s

⇒ iC (t ) =

− ( 2 / 3) t

) u(t ) V

2 − ( 2 / 3) t e u (t ) A 3

14-1

Exercise 14.8-1 The transfer function of a circuit is H(s) =

–5s . Determine the s + 15s + 50 2

impulse response and step response of this circuit. Answer:

5 10 ⎤ (a) impulse response = ℒ –1 ⎡ = (5e–5t–10e–10t)u(t) ⎢⎣ s + 5 – s + 10 ⎥⎦

1 ⎤ ⎡ 1 (b) step response = ℒ –1 ⎢⎣ s + 10 – s + 5 ⎥⎦ = (e–10t – e–5t)u(t) Solution:

10 ⎤ ⎡ 5 (a) impulse response = L−1 ⎢ − = ( 5 e − 5 t − 10 e −10 t ) u (t ) ⎥ 5 10 s s + + ⎣ ⎦ 1 ⎤ ⎡ 1 (b) step response = L−1 ⎢ − = ( e−10 t − e − 5 t ) u (t ) ⎥ ⎣ s + 10 s + 5 ⎦

Exercise 14.8-2 The impulse response of a circuit is h(t) = 5e–2t sin (4t)u(t). Determine the step

response of this circuit. Hint:

H(s) = ℒ [5e–2t sin (4t)u(t)] =

Answer:

5(4) 20 = 2 2 2 ( s + 2) + 4 s + 4s + 20

s+4 ⎤ ⎡1 ⎡ H (s) ⎤ step response = ℒ –1 ⎢⎣ s ⎥⎦ = ℒ –1 ⎢⎣ s – s 2 + 4s + 20 ⎥⎦ 1 ⎛ ⎞⎞ –2 t ⎛ = ⎜1 – e ⎜⎝ cos 4t + 2 sin 4t ⎟⎠ ⎟ u (t ) ⎝ ⎠

Solution:

H ( s ) = L ⎡⎣5 e − 2 t sin ( 4 t ) u ( t ) ⎤⎦ =

5 ( 4)

( s + 2)

2

+4

2

=

20 s + 4 s + 20 2

⎡ H (s) ⎤ ⎤ s+4 1 -1 ⎡ 1 step response = L-1 ⎢ = (1 − e − 2 t (cos 4t − sin 4t )) u (t ) ⎥=L ⎢ − 2 ⎥ 2 ⎣ s s + 4 s + 20 ⎦ ⎣ s ⎦

14-2

Exercise 14.10-1 The input to a circuit is the voltage vi(t). The output is the voltage vo(t). The

transfer function of this circuit is H ( s) =

Vo ( s ) ks = 2 Vi ( s ) s + (3 − k ) s + 2

Determine the following: (a)

The steady-state response when vi(t) = 5 cos 2t V and the gain of the VCVS is k = 2 V/V

(b)

The impulse response when k = 3– 2 2 = 0.17 V/V

(c)

The impulse response when k = 3 + 2 2 = 5.83 V/V

Answer: (a) vo(t) = 7.07 cos (2t–45°) V

(b) h(t) = 0.17e − (c) h(t) = 5.83e

(1 − 2t ) u(t ) (1 + 2t ) u(t )

2t

2t

Solution:

The poles of the transfer function are p1,2 = a.) When k = 2 V/V, the poles are p1,2 =

− (3 − k ) ±

(3 − k ) 2

2

−8

.

−1 ± −7 so the circuit is stable. The transfer function is 2

H (s) =

Vo ( s ) Vi ( s )

=

2s s +s+2 2

The circuit is stable when k =2 V/V so we can determine the network function from the transfer function by letting s = jω. V o (ω ) V i (ω )

= H (ω ) = H ( s ) s = j ω =

2s 2 jω = s + s + 2 s = j ω ( 2 − ω 2 ) + jω 2

The input is v i ( t ) = 5cos 2 t V . The phasor of the steady state response is determine by multiplying the phasor of the input by the network function evaluated at ω = 2 rad/s. ⎛ 2 jω V o (ω ) = H (ω ) ω = 2 × V i (ω ) = ⎜ ⎜ ( 2 − ω 2 ) + jω ⎝

⎞ ⎟ ( 5∠0° ) = ⎛⎜ j 4 ⎞⎟ ( 5∠0° ) = 7.07∠ − 45° ⎟ ⎝ −2 + j 2 ⎠ ω =2 ⎠

14-3

The steady state response is vo ( t ) = 7.07 cos ( 2 t − 45° ) V . b. When k = 3 − 2 2 , the poles are p1,2 =

−2 2 ± 0 = − 2, − 2 so the circuit is stable. The 2

transfer function is H (s) =

0.17 s

=

0.17



0.17 2

(s + 2) (s + 2) (s + 2) 2

The impulse response is h ( t ) = L -1⎡⎣ H ( s ) ⎤⎦ = 0.17 e −

2t

(1 −

2

)

2 t u (t )

We see that when k = 3 − 2 2 the circuit is stable and lim h(t ) = 0 . t →∞

c. When, k = 3 + 2 2 the poles are p1,2 =

2 2± 0 = 2, 2 so the circuit is not stable. The 2

transfer function is H (s) =

5.83s

=

5.83

+

5.83 2

(s − 2) (s − 2) (s − 2) 2

The impulse response is h ( t ) = L -1⎡⎣ H ( s ) ⎤⎦ = 5.83 e

2t

(1 +

2

)

2 t u (t )

We see that when k = 3 + 2 2 the circuit is unstable and lim h(t ) = ∞ . t →∞

14-4

Exercise 14.12-1

A circuit is specified to have a transfer function of H(s) =

25 s + 10s + 125 2

and a unit step response of vo(t) = 0.1(2 – e–5t(2 cos 10t + 3 sin 10t))u(t) Verify that these specifications are consistent.

Solution: For the poles to be in the left half of the s-plane, the s-term needs to be positive.

⎡2 ⎛ ⎞⎤ 10 2s + 20 s +5 ⎡2 ⎤ V0 ( s ) = 0.1 ⎢ −⎜ 2 + ⎟ ⎥ = 0.1 ⎢ − 2 2 2 2 2 ⎟ ⎜ ⎣ s s +10 s + 125 ⎥⎦ ⎢⎣ s ⎝ ( s + 5 ) +10 ( s +5) +10 ⎠ ⎥⎦ = 0.1 = 0.1

2( s 2 +10s +125 )−( 2s + 20 ) s s ( s 2 +10s +125 )

250 25 = 2 s ( s +10s +125 ) s ( s +10s +125 ) 2

Then H (s) =

V0 ( s )

L ⎡⎣u ( t ) ⎤⎦

=

(

25 2

s s +10 s +125

1 s

)=

25 s 2 +10 s +125

These specifications are consistent.

14-5

Section 14.2 Laplace Transforms  P14.2‐1 Determine the Laplace Transform of  v ( t ) = (17 e − 4 t − 147 e− 5 t ) u ( t )  V   Answer:  V ( s ) =

3 s + 29   s + 9 s + 20 2

Solution:  

L ⎡⎣17 e− 4 t − 14 e − 5 t ⎤⎦ =

17 14 17 ( s + 5 ) − 14 ( s + 4 ) 3 s + 29 − = = 2 s+4 s+5 s + 9 s + 20 ( s + 4 )( s + 5)

  P14.2‐2 Determine the Laplace Transform of    v ( t ) = 13cos ( 6 t − 22.62° )  V.  Answer:  V ( s ) =

12 s + 30   s 2 + 36

Solution:  

v ( t ) = 13cos ( 6 t − 22.62° ) = 13cos ( −22.62° ) cos ( 6 t ) − 13sin ( −22.62° ) sin ( 6 t ) = 12 cos ( 6 t ) + 5sin ( 6 t ) V V ( s ) = L ⎡⎣12 cos ( 6 t ) + 5sin ( 6 t ) ⎤⎦ = L ⎡⎣12 cos ( 6 t ) ⎤⎦ + L ⎡⎣ +5sin ( 6 t ) ⎤⎦ = (12 )

s 6 12 s + 30 + ( 5) 2 = 2 s + 36 s + 36 s + 36 2

  P14.2‐3 Determine the Laplace Transform of  v ( t ) = 10 e −5 t cos ( 4 t + 36.86° ) u ( t )  V.  Answer:  V ( s ) =

8 s + 16   s + 25 s + 41 2

Solution:  

10 cos ( 4 t + 36.86° ) = 10 cos ( 36.86° ) cos ( 4 t ) − 10sin ( 36.86° ) sin ( 4 t ) = 8cos ( 4 t ) − 6sin ( 4 t ) V L ⎡⎣10 cos ( 4 t + 36.86° ) ⎤⎦ = ( 8 )

s 4 8s − 4 − ( 6) 2 = 2 s + 16 s + 16 s + 16 2

V ( s ) = L ⎡⎣10 e −5 t cos ( 4 t + 36.86° ) ⎤⎦ = L ⎡⎣10 cos ( 4 t + 36.86° ) ⎤⎦ =

s ← s +5

8 ( s + 5) − 4 8s − 4 8 s + 16   = = 2 s 2 + 16 s ← s +5 ( s + 5 ) + 16 s 2 + 25 s + 41  

P14.2‐4 Determine the Laplace Transform of  v ( t ) = 3 t e − 2 t u ( t )  V   Answer:  V ( s ) =

3   s + 4s + 4 2

Solution:   L ⎡⎣3 t e − 2 t ⎤⎦ = L [3 t ] s ←s + 2 =

3 s2

= s←s + 2

3

( s + 2)

2

=

3 s + 4s + 4 2

  P14.2‐5 Determine the Laplace Transform of  v ( t ) = 16 (1 − 2 t ) e − 4 t u ( t )  V.  Answer:  V ( s ) =

16 ( s + 2 )   s 2 + 8 s + 16

Solution:  

 

16 32 ⎛ 16 32 ⎞ L ⎡⎣16 (1 − 2 t ) e − 4 t ⎤⎦ = L [16 − 32 t ] s ←s + 4 = ⎜ − 2 ⎟ = − 2 ⎝ s s ⎠ s ←s + 4 s + 4 ( s + 4 ) 16 ( s + 4 ) − 32 16 ( s + 2 ) = 2 = 2 s + 8 s + 16 s + 8 s + 16

 

Section 14-3: Pulse Inputs P 14.3-1 Determine the Laplace transform of f(t) shown in Figure P 14.3-1.

5 ⎞ 5 ⎛ 21 ⎞ ⎛ 21 ⎞ ⎛ Hint: f ( t ) = ⎜ 5 − t ⎟ u ( t ) + ⎜ t − ⎟ u ⎜ t − ⎟ 3 ⎠ 3⎝ 5⎠ ⎝ 5⎠ ⎝

5 f (t) 0 –2

5e−4.2 s + 15s − 5 Answer: F ( s ) = 3s 2

3

t

Figure P 14.3-1

Solution:

⎛ 5 ⎞ ⎛ 5 ⎞ f ( t ) = ⎜ − t + 5 ⎟ u ( t ) − ⎜ − ( t − 4.2 ) ⎟ u ( t − 4.2 ) ⎝ 3 ⎠ ⎝ 3 ⎠

⎛ 5 5⎞ ⎛ 5 ⎞ 15 s + 5 ( e F ( s ) = ⎜ − 2 + ⎟ − e −4.2 s ⎜ − 2 ⎟ = s⎠ 3 s2 ⎝ 3s ⎝ 3s ⎠

−4.2 s

P 14.3-2 Use the Laplace transform to obtain the transform of the signal f(t) shown in Figure P14.3-2. Answer: F ( s ) =

− 1)

3 f (t)

3 (1 − e −2 s ) s

0

2

t

Figure P14.3-2 Solution:



F (s) = ∫0

3 e − st f (t ) e dt = ∫ 0 3 e dt = −s − st

2

2

− st

= 0

3(1−e −2 s ) s

1

P 14.3-3 Determine the Laplace transform of f(t) shown in Figure P 14.3-3. 5 Answer: F ( s ) = 2 (1 − e −2 s − 2 se −2 s ) 2s

5 f (t)

0

1

2

t

Figure P 14.3-3 Solution:

⎧5 2 t 0 0 since pi = 1 ± 3 j . No final value of f ( t ) exists.

14-2

P14.5-5

as+b where v ( t ) is the voltage shown in Figure P14.5-5, determine the Given that L ⎡⎣v ( t ) ⎤⎦ = 2 s +8s values of a and b.

Figure P14.5-5 Solution: From the plot, v(0) = 4 V and lim v ( t ) = 12 V . From the final value theorem, t →∞

lim v ( t ) = lim sV ( s ) = lim s

t →∞

Consequently, 12 =

s →0

s →0

as+b as+b b = lim = . 2 s + 8 s s →0 s + 8 8

b ⇒ b = 96 . From the initial value theorem 8 as+b as+b lim v ( t ) = lim sV ( s ) = lim s 2 = lim =a t →0 s →∞ s →∞ s + 8 s s →∞ s + 8

Consequently a = 4.

14-3

P14.5-6

as+b Given that L ⎡⎣v ( t ) ⎤⎦ = 2 where v ( t ) is the voltage shown in Figure P14.5-6, determine 2 s + 40 s the values of a and b.

Figure P14.5-6 Solution:

From the plot, v(0+) = 10 V and lim v ( t ) = 2 V . From the final value theorem, t →∞

lim v ( t ) = lim sV ( s ) = lim s

t →∞

s →0

s →0

as+b as+b b = lim = . 2 2 s + 40 s s→0 2 s + 40 40

b ⇒ b = 80 . From the initial value theorem 40 as+b as+b a lim v ( t ) = lim sV ( s ) = lim s 2 = lim = t →0 + s →∞ s →∞ 2 s + 40 s s →∞ 2 s + 40 2 a ⇒ a = 20 . Consequently 10 = 2

Consequently, 2 =

14-4

Section 14.6 Solution of Differential Equations Describing a Circuit

Figure P14.6-1 P14.6-1 The circuit shown in Figure P14.6-1 is at steady state before the switch closes at time t = 0. Determine the inductor current, i(t), after the switch closes. Solution: The initial inductor current is

i ( 0) =

12 =2A 6

d i (t ) + 4 i (t ) dt Take the Laplace Transform of both sides of this equation:

Apply KVL after the switch closes

12 = 2

12 = 2 ⎡⎣ s I ( s ) − 2 ⎤⎦ + 4 I ( s ) s Solve for I(s): Taking the Inverse Laplace Transform:

I (s) =

2s +6 3 1 = − s ( s + 2) s s + 2

i ( t ) = 3 − e −2t A

Figure P14.6-2 P14.6-2 The circuit shown in Figure P14.6-1 is represented by the differential equation

d 2v ( t ) d v (t ) +7 + 10 v ( t ) = 120 2 dt dt after time t = 0. The initial conditions are i(0) = 0 and v(0) = 4 V Determine the capacitor, v(t), after time t = 0. Solution: Let’s take the Laplace Transform of both sides of the differential equation.

⎡ d v (t ) ⎤ £⎢ ⎥ = sV ( s ) − 4 ⎣ dt ⎦

First, using Table 14.2-2

Next, notice that Using Table 14.2-2 again Now we have: or:

Solve for V(s):

d v (t ) d v ( 0) i ( 0) ⇒ = =0 dt dt 0.1 ⎡ d 2v ( t ) ⎤ ⎡ d v (t ) ⎤ d v ( 0) £⎢ = s 2 V (s) − 4 s ⎥− 2 ⎥ = s£ ⎢ dt ⎣ dt ⎦ ⎣ dt ⎦ i ( t ) = 0.1

s 2 V ( s ) − 4 s + 7 ( sV ( s ) − 4 ) + 10V ( s ) =

120 s

120 + 28 + 4 s s 40 16 − 2 4 s + 28 s + 120 3 1 12 V (s) = = − = + 3 + 3 s ( s + 2 )( s + 5 ) s s + 2 s s + 2 s + 5

(s

2

+ 7 s + 10 ) V ( s ) =

Taking the inverse Laplace transform: v ( t ) = 12 −

40 − 2 t 16 −5t e + e A 3 3

Figure P14.6-3 P14.6-3 The circuit shown in Figure P14.6-3 is at steady state before time t = 0. The input to the circuit is v s ( t ) = 2.4 u ( t ) V

Consequently, the initial conditions are i1(0)=0 and i2(0)=0. Determine the inductor current, i2(t), after time t = 0. Solution: First, let’s simplify the circuit by replacing the 3 12-Ω resistors at the right of the circuit by an equivalent resistor:

vs (t ) = 2

Write the mesh equations

and

0=2

d i 2 (t ) dt

d i1 ( t ) dt

+ 12 ( i1 ( t ) − i 2 ( t ) )

+ 8 i 2 ( t ) − 12 ( i1 ( t ) − i 2 ( t ) ) = 2

d i2 (t ) dt

+ 20 i 2 ( t ) − 12 i1 ( t )

Taking the Laplace Transforms of these equations: 2.4 = 2 s I 1 ( s ) + 12 ( I 1 ( s ) − I 2 ( s ) ) s 0 = 2 s I 2 ( s ) + 20 I 2 ( s ) − 12 I 1 ( s ) and

1 5 5⎞ ⎛1 s I 2 (s) + I 2 (s) = ⎜ s + ⎟ I 2 (s) 6 3 3⎠ ⎝6 2.4 5⎞ ⎛1 = ( 2 s + 12 ) I 1 ( s ) − 12 I 2 ( s ) = ( 2 s + 12 ) ⎜ s + ⎟ I 2 ( s ) − 12 I 2 ( s ) s 3⎠ ⎝6 7.2 0.3 0.03794 0.33974 = + − I 2 (s) = 2 s s + 14.3 s + 1.68 2 s + 16 s + 24

Next, some algebra:

Solving for I2(s):

I1 (s) =

(

)

Taking the Inverse Laplace Transform gives i2(t) for t ≥ 0: i 2 ( t ) = 300 + 39.74 e −14.3t − 339.74 e −1.68t A

Figure P14.6-4 P14.6-4 The circuit shown in Figure P14.6-4 is at steady state before the switch opens at time t = 0. Determine the capacitor voltage, v(t), after the switch opens. Solution:

v (0) 3 4 =− A (12 ) = 4 V and i ( 0 ) = − 3 3 3+9 2 1 d v (t ) 3 d v (t ) + + v (t ) Apply KVL after the switch opens 0 = 2 dt 2 2 dt The initial conditions are

v (0) =

d 2 v (t ) d v (t ) That is 0= +3 + 2 v (t ) 2 dt dt Let’s take the Laplace Transform of both sides of the differential equation.

⎡ d v (t ) ⎤ £⎢ ⎥ = sV ( s ) − 4 ⎣ dt ⎦

First, using Table 14.2-2

d v (t ) d v ( 0) i ( 0) 8 ⇒ =− = A dt dt 0.5 3 ⎡ d 2v ( t ) ⎤ ⎡ d v (t ) ⎤ d v ( 0) 2 Using Table 14.2-2 again £ ⎢ = s 2 V (s) − 4 s − ⎥− 2 ⎥ = s£ ⎢ dt 3 ⎣ dt ⎦ ⎣ dt ⎦ Next, notice that

Now we have:

Solve for V(s):

i ( t ) = −0.5

8⎞ ⎛ 2 ⎜ s V ( s ) − 4 s − ⎟ + 3 ( sV ( s ) − 4 ) + 2V ( s ) = 0 3⎠ ⎝ 44 32 20 4s+ 3 = 3 − 3 V (s) = ( s + 1)( s + 2 ) s + 1 s + 2

Taking the Inverse Laplace Transform: v ( t ) =

32 − t 20 − 2 t e − e V 3 3

Figure P14.6-5 P14.6-5 The circuit shown in Figure P14.6-5 is at steady state before the switch closes at time t = 0. Determine the capacitor voltage, v(t), after the switch closes. Solution:

v (0) =

The initial capacitor voltage is

40 (12 ) = 0.98 V 10 + 40

Write a node equation after the switch closes: vs (t ) − v (t ) 10 × 10

3

= ( 2 × 10− 6 )

1200 − 100 v ( t ) = 2 600 =

v (t ) d v (t ) + dt 10 × 10 3

d v ( t ) + 100 v ( t ) dt

d v ( t ) + 100 v ( t ) dt

Take the Laplace Transform of both sides of this equation:

600 = ⎡⎣ sV ( s ) − 0.98⎤⎦ + 100V ( s ) s Solve for V(s):

V (s) =

Taking the Inverse Laplace Transform:

600 s + 0.98 6 5.02 = − s ( s + 100 ) s s + 100

v ( t ) = 6 − 5.02e −100 t V

Section 14.7 Circuit Analysis Using Impedance and Initial Conditions P14.7-1 Figure P14.7-1a shows a circuit represented in the time domain. Figure P14.7-1b shows the same circuit, now represented in the complex frequency domain. Figure P14.7-1c shows a plot of the inductor current.

(b)

(a)

(c) Figure P14-7-1 Determine the values of D and E used to represent the circuit in the complex frequency domain. Determine the values of the resistance R 2 and the inductance L. Solution: First, we find the values of D and E used to represent the circuit in the complex frequency domain:

L ⎡⎣12 − 6 u ( t ) ⎤⎦ = L ⎡⎣6 u ( t ) ⎤⎦ =

6 ⇒ D = 6 A. s

E is the initial inductor current = 8 A from the plot. Next, we find the values of the resistance R 2 and the inductance L: The circuit is at steady state before t = 0, so the inductor acts like a short circuit. Using current division, ⎛ 30 ⎞ 8=⎜ 12 ⇒ R 2 = 15 Ω . Similarly, the circuit will at steady state for t → ∞. Again, the inductor ⎜ 30 + R 2 ⎟⎟ ⎝ ⎠ acts like a short circuit. Using current division, 14.7-1

⎛ 30 ⎞ 4=⎜ 6 ⇒ R 2 = 15 Ω . ⎜ 30 + R 2 ⎟⎟ ⎝ ⎠

The inductor current can be represented as v ( t ) = 4 + 4 e − at for t ≥ 0. From the plot,

then

⎛ 4.54 − 4 ⎞ ln ⎜ ⎟ 4 ⎠ 4.54 = 4 + 4 e − a ( 0.5) so a = ⎝ = 4.005 ≅ 4 1/s . −0.5 1 45 L =τ = ⇒ L= = 11.25 H . 4 15 + 30 4

As a check, apply KVL to the center mesh in the complex frequency domain to get R D E D L s + R1 Es+ 1 E D ⎛ ⎞ ⎛ ⎞ s s = L R 2 I ( s ) + L s ⎜ I ( s ) − ⎟ + R1 ⎜ I ( s ) − ⎟ = 0 ⇒ I ( s ) = R + R2 ⎞ s⎠ s⎠ L s + R1 + R 2 ⎛ ⎝ ⎝ s⎜s + 1 ⎟ L ⎠ ⎝ 8 s + 16 4 4 = + s ( s + 4) s s + 4 Taking the inverse Laplace transform gives

Substituting values gives

I (s) =

4 ⎤ ⎡4 i ( t ) = L−1 ⎡⎣ I ( s ) ⎤⎦ = L−1 ⎢ + = 4 + 4 e−4 t A for t ≥ 0 ⎥ ⎣ s s + 4⎦

14.7-2

P14.7-2 Figure P14.7-2a shows a circuit represented in the time domain. Figure P14.7-2b shows the same circuit, now represented in the complex frequency domain. Figure P14.7-2c shows a plot of the inductor current.

(a)

(b)

(c) Figure P14-7-2

Determine the values of D and E used to represent the circuit in the complex frequency domain. Determine the values of the resistance R 1 and the capacitance C. Solution:

First, we find the values of D and E used to represent the circuit in the complex frequency domain: L ⎡⎣6 + 12 u ( t ) ⎤⎦ = L ⎡⎣18 u ( t ) ⎤⎦ =

18 ⇒ D = 18 V. s

E is the initial capacitor voltage = 4 V from the plot. Next, we determine the values of the resistance R 1 and the capacitance C: The circuit is at steady state before t = 0, so the capacitor acts like an open circuit. Using voltage ⎛ 30 ⎞ 6 ⇒ R1 = 15 Ω . Similarly, the circuit will at steady state for t → ∞. Again, the division, 4 = ⎜ ⎜ R1 + 30 ⎟⎟ ⎝ ⎠ capacitor acts like an open circuit. Using voltage division, 14.7-3

⎛ 30 ⎞ 12 = ⎜ 18 ⇒ R1 = 15 Ω . ⎜ R1 + 30 ⎟⎟ ⎝ ⎠ The capacitor voltage can be represented as v ( t ) = 12 − 8 e − at for t ≥ 0. From the plot, ⎛ 11.6 − 12 ⎞ ln ⎜ ⎟ −8 ⎠ = 7.9886 ≅ 8 1/s . 11.6 = 12 − 8 e − a ( 0.375) so a = ⎝ −0.375 1 1 then = τ = (15 || 30 ) C = 10 C ⇒ C = = 12.5 mF . 8 80 As a check, apply KCL at the top node of the 30 Ω resistor, R 2 to get D D Es+ R1 C s + V ( s ) + C s ⎛V ( s ) − E ⎞ = 0 ⇒ V ( s ) = ⎜ ⎟ R1 R2 s⎠ ⎛ R + R2 ⎞ ⎝ s⎜s + 1 ⎟ ⎜ R1 R 2 C ⎟⎠ ⎝ Substituting values and performing a partial fraction expansion gives V (s) −

V (s) =

4 s + 96 12 8 = − s ( s + 8) s s + 8

Taking the inverse Laplace transform gives

8 ⎤ ⎡12 = 12 − 8 e− 8t V for t ≥ 0 v ( t ) = L−1 ⎡⎣V ( s ) ⎤⎦ = L−1 ⎢ − ⎥ ⎣ s s + 8⎦

14.7-4

P14.7-3

Figure P14.7-3a shows a circuit represented in the time domain. Figure P14.7-3b shows the same circuit, now represented in the complex frequency domain. Determine the values of a, b and d used to represent the circuit in the complex frequency domain.

(a)

(b) Figure P14.7-3

Solution:

24 − 36 u ( t ) = −12 for t > 0

L [ −12] =

−12 ⇒ a = −12 V. s

The circuit is at steady state before t = 0, and the input is constant, so the capacitor acts like an open circuit and the inductor acts like a short circuit. 24 ⎛ 8 ⎞ v (0) = ⎜ =2A ⎟ 24 = 16 V and i ( 0 ) = 4+8 ⎝ 4+8⎠ Consequently, b = v ( 0 ) = 16 V and d = L i ( 0 ) = ( 6 )( 2 ) = 12 .

14.7-5

P 14.7-4

The input to the circuit shown in

t=0

Figure P 14.7-4 is the voltage of the voltage source, 12 V. The output of this circuit is the voltage, vo(t), across the capacitor. Determine

12 V

vo(t) for t > 0.

– +

6Ω 6Ω

+ 0.5 F

6Ω –t/2

Answer: vo (t) = – (4 + 2e

vo(t) –

)V for t > 0 Figure P 14.7-4

Solution: t < 0

time domain

frequency domain

Mesh equations in the frequency domain: 6 I 1 ( s ) + 6 ( I 1 ( s ) − I 2 ( s )) + 6 I 1 ( s ) +

12 2 2 = 0 ⇒ I1 (s) = I 2 (s) − s 3 3s

2 6 2⎞ 6 ⎛ I 2 ( s ) − − 6 ( I 1 ( s ) − I 2 ( s )) = 0 ⇒ ⎜ 6 + ⎟ I 2 ( s ) − 6 I 1 ( s ) = s s s⎠ s ⎝ 1 ⎛2 2⎞ 2 ⎞ 6 ⎛ ⇒ I 2 (s) = 2 Solving for I2(s): ⎜ 6 + ⎟ I 2 (s) − 6⎜ I 2 (s) − ⎟ = 1 s⎠ 3s ⎠ s ⎝ ⎝3 s+ 2 ⎛ 1 ⎞ −2 1 6 1⎜ 2 ⎟ 6 4 Calculate for Vo(s): Vo ( s ) = I 2 ( s ) − = ⎜ − = − ⎟ s 2⎜ s+ 1 ⎟ s s+ 1 s 2 ⎝ 2⎠ 2 Take the Inverse Laplace transform:

(

vo ( t ) = − 4 + 2 e −t /2

)

V for t > 0

(Checked using LNAP, 12/29/02)

14.7-6

P 14.7-5

The input to the circuit shown in Figure

t=0

P 14.7-5 is the voltage of the voltage source, 12 V. 2Ω

The output of this circuit is the current, i(t), in the 12 V

inductor. Determine i(t) for t > 0.

– +

2Ω 5H

Answer: i(t) = –3(1 + e–0.8t) A for t > 0

i(t)

Figure P 14.7-5 Solution: t 0 (Checked using LNAP, 12/29/02)

14.7-7

P 14.7-6 The input to the circuit shown in Figure P 14.7-6 is the voltage of the voltage source, 18 V. The output of this circuit, the voltage across the capacitor, is given by

vo (t) = 6 + 12e–2t V when t > 0 Determine the value of the capacitance, C, and the value of the resistance, R.

Figure P 14.7-6 Solution:

Steady-state for t < 0:

Steady-state for t > 0:

From the equation for vo(t): vo ( ∞ ) = 6 + 12 e From the circuit: Therefore:

6=

− 2 (∞)

vo ( ∞ ) =

=6 V

3 (18) R+3

3 (18) ⇒ R = 6 Ω R+3

⎛ 1 ⎞ 18 6 −6 I (s)⎜ 2 + ⎟ + − = 0 ⇒ I (s) = 1 Cs⎠ s s ⎝ s+ 2C

14.7-8

⎛ 1 18 1 ⎜ −6 Vo ( s ) = I (s) + = ⎜ Cs s Cs ⎜ s + 1 ⎜ 2C ⎝

Taking the inverse Laplace transform:

⎞ ⎟ 18 −12 12 18 12 6 + + = + ⎟+ = 1 s s s+ 1 s ⎟⎟ s s+ 2C 2C ⎠

vo ( t ) = 6 + 12 e − t / 2C V for t > 0

Comparing this to the given equation for vo(t), we see that 2 =

1 2C

⇒ C = 0.25 F . (Checked using LNAP, 12/29/02)

14.7-9

P 14.7-7

The input to the circuit shown in Figure P 14.7-7 is the voltage source voltage vs(t) = 3 – u(t)

V

The output is the voltage vo(t) = 10 + 5e–100t V for t ≥ 0 Determine the values of R1 and R2.

Figure P 14.7-7 Solution:

We will determine Vo ( s ) , the Laplace transform of the output, twice, once from the given equation and once from the circuit. From the given equation for the output, we have Vo ( s ) =

10 5 + s s + 100

Next, we determine Vo ( s ) from the circuit. For t ≥ 0 , we represent the circuit in the frequency domain using the Laplace transform. To do so we need to determine the initial condition for the capacitor. When t < 0 and the circuit is at steady state, the capacitor acts like an open circuit. Apply KCL at the noninverting input of the op amp to get 3 − v (0 −) = 0 ⇒ v (0 −) = 3 V R1 The initial condition is v (0 +) = v (0 −) = 3 V

Now we can represent the circuit in the frequency domain, using Laplace transforms.

Apply KCL at the noninverting input of the op am to get 2 3 −V ( s) V ( s) − s s = 6 R1 10 s Solving gives 106 3s + 2 R1 2 1 V (s) = = + ⎛ 106 ⎞ s ⎛ 106 ⎞ s ⎜s+ ⎟ ⎜s+ ⎟ ⎜ ⎜ R1 ⎟⎠ R1 ⎟⎠ ⎝ ⎝ Apply KCL at the inverting input of the op amp to get ⎛ ⎞ ⎜ ⎟ Vo ( s) −V ( s) V ( s) R2 ⎞ R2 ⎞⎜ 2 ⎛ ⎛ ⎟ 1 = ⇒ V o ( s ) = ⎜1 + ⎟V ( s ) = ⎜1 + ⎟⎜ + ⎟ R2 1000 ⎝ 1000 ⎠ ⎝ 1000 ⎠ ⎜ s ⎛ 106 ⎞ ⎟ ⎜s+ ⎟⎟ ⎜ ⎜ ⎟ R 1 ⎝ ⎠⎠ ⎝ The expressions for Vo(s) must be equal, so ⎛ ⎞ ⎜ ⎟ R ⎛ ⎞⎜ ⎟ 10 5 2 1 2 + = ⎜1 + ⎟⎜ + ⎟ s s + 100 ⎝ 1000 ⎠ ⎜ s ⎛ 106 ⎞ ⎟ ⎜s+ ⎟ ⎜ ⎜ R1 ⎟⎠ ⎟ ⎝ ⎝ ⎠ Equating coefficients gives 106 1+ = 5 ⇒ R 2 = 4 kΩ and = 100 ⇒ R1 = 10 kΩ 1000 R1 R2

(checked using LNAPTR 7/31/04)

P 14.7-8 Determine the inductor current, iL(t), in the circuit shown in Figure P 14.7-8 for each of the following cases:

(a)

R = 2 Ω, L = 4.5 H, C = 1/9 F, A = 5 mA, B = – 2 mA

(b)

R = 1 Ω, L = 0.4 H, C = 0.1 F, A = 1 mA, B = – 2 mA

(c)

R = 1 Ω, L = 0.08 H, C = 0.1 F, A = 0.2 mA, B = – 2 mA

Figure P 14.7-8 Solution:

For t < 0, The input is constant. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. The circuit is at steady state at time t = 0 − so vC ( 0 − ) = 0 and i L ( 0 − ) = B

The capacitor voltage and inductor current are continuous so vC ( 0 + ) = vC ( 0 − ) and iL (0 +) = iL (0 −) .

For t < 0, represent the circuit in the frequency domain using the Laplace transform as shown. V C ( s ) is the node voltage at the top node of the circuit. Writing a node equation gives

A + B VC ( s ) B VC ( s ) = + + + C sV C ( s ) s R s Ls so

A L s + R + R L C s2 = VC ( s ) s RLs Then

A AR L C = VC ( s ) = R L C s2 + L s + R s2 + 1 s + 1 RC LC

A B B LC and I L (s) = + = + Ls s ⎛ 1 1 ⎞ s s ⎜ s2 + s+ ⎟ RC LC ⎠ ⎝ 1 a.) When R = 2 Ω, L = 4.5 H, C = F, A = 5 mA and B = −2 mA , then 9 5 40 −2 3 10 + = + 7 − 7 I L (s) = 2 s s+4 s+ 1 s ( s + 4.5 s + 2 ) s 2 Taking the inverse Laplace transform gives VC ( s )

5 5 i L ( t ) = 3 + e− 4 t − e − 0.5 t mA for t ≥ 0 7 7 b.) When R = 1 Ω, L = 0.4 H, C = 0.1 F, A = 1 mA and B = −2 mA , then

I L (s) =

⎛1 25 25 5 1 ⎞ −2 −2 + = + = − + + ⎜ ⎟ 2 ⎜ s ( s + 5)2 s + 5 ⎟ s s ( s 2 + 10 s + 25 ) s s ( s + 5) ⎝ ⎠

Taking the inverse Laplace transform gives

i L ( t ) = − (1 + 5 t e − 5 t − e− 5 t ) mA for t ≥ 0 c.) When R = 1 Ω, L = 0.08 H, C = 0.1 F, A = 0.2 mA and B = −2 mA , then

I L (s) =

25 10 s+5 −2 −1.8 −0.2 s − 2 −1.8 0.2 0.1 + = + = − − 2 2 2 s s ( s + 10 s + 125 ) s ( s + 5) + 102 s ( s + 5) + 102 ( s + 5) + 102 2

Taking the inverse Laplace transform gives i L ( t ) = −1.8 − e − 5 t ( 0.2 cos (10 t ) + 0.1sin (10 t ) ) mA for t ≥ 0

P 14.7-9 Determine the capacitor current, ic(t), in the circuit shown in Figure P 14.7-9 for each of the following cases:

(a)

R = 3 Ω, L = 2 H, C = 1/24 F, A = 12 V

(b)

R = 2 Ω, L = 2 H, C = 1/8 F, A = 12 V

(c)

R = 10 Ω, L = 2 H, C = 1/40 F, A = 12 V

Figure P 14.7-9 Solution:

For t < 0, the switch is open and the circuit is at steady state. and the circuit is at steady state. At steady state, the capacitor acts like an open circuit. A A and vC ( t ) = i (t ) = 2R 2 Consequently, A A and vC ( 0 − ) = i (0 −) = 2R 2 Also iC (0 −) = 0 The capacitor voltage and inductor current are continuous so vC ( 0 + ) = vC ( 0 − ) and iL (0 +) = iL (0 −) .

For t > 0, the voltage source voltage is 12 V. Represent the circuit in the frequency domain using the Laplace transform as shown.

I L ( s ) and I C ( s ) are mesh currents. Writing a mesh equations gives

AL A + R ( I L ( s ) − I C ( s )) − = 0 2R s 1 A + R ( I L ( s ) − I C ( s )) = 0 I C (s) − Cs 2s ⎛ AL A⎞ −R ⎞ ⎛Ls+ R ⎜ 2R + s ⎟ I s ⎛ ⎞ ( ) L ⎜ ⎟ ⎟ 1 ⎟ ⎜⎜ ⎟⎟ = ⎜ ⎜ −R R + I s ⎜ A ⎟ ( ) ⎜ ⎠ C s ⎟⎠ ⎝ C ⎝ ⎜ − 2s ⎟ ⎝ ⎠

L s I L (s) −

Or, in matrix form



I C (s) =

⎛ AL A⎞ A⎞ A + ⎟ ⎟ + R⎜ 2L ⎝ 2s ⎠ ⎝ 2R s ⎠ = 1 1 ⎛ 1 ⎞ s2 + s+ ( L s + R ) ⎜ R + ⎟ − R2 RC LC Cs⎠ ⎝

( L s + R)⎜ −

a.) When R = 3 Ω, L = 2 H, C =

1 F and A = 12 V , 24

3 3 3 2 I C (s) = 2 = = 4 − 4 . s + 8 s + 12 ( s + 2 )( s + 6 ) s + 2 s + 8 Taking the inverse Laplace transform gives 3 ⎛3 ⎞ i C ( t ) = ⎜ e− 2t − e− 6t ⎟ u ( t ) A 4 ⎝4 ⎠

b.) R = 2 Ω, L = 2 H, C =

1 F and A = 12 V , 8 3 3 I C (s) = 2 = s + 4 s + 4 ( s + 2 )2

Taking the inverse Laplace transform gives i C ( t ) = 3 t e− 2t u ( t ) A

1 F and A = 12 V 40 3 3 3 4 I C (s) = 2 = = × 2 s + 4 s + 20 ( s + 2 ) + 16 4 ( s + 2 )2 + 16

c.) R = 10 Ω, L = 2 H, C =

Taking the inverse Laplace transform gives i C (t ) =

3 − 2t e sin ( 4 t ) u ( t ) A 4

(checked using LNAP 4/11/01)

P 14.7-10

The voltage source voltage in the circuit shown in Figure P 14.7-10 is

Determine v(t) for t ≥ 0.

vs(t) = 12 – 6u(t) V

Figure P 14.7-10 Solution: For t < 0, The input is 12 V. At steady state, the capacitor acts like an open circuit.

Notice that v(t) is a node voltage. Express the controlling voltage of the dependent source as a function of the node voltage: va = −v(t)

Writing a node equation: ⎛ 12 − v ( t ) ⎞ v ( t ) ⎛ 3 ⎞ −⎜ + ⎜ − v (t ) ⎟ = 0 ⎟+ 8 4 ⎝ 4 ⎠ ⎝ ⎠ −12 + v ( t ) + 2 v ( t ) − 6 v ( t ) = 0 ⇒ v ( t ) = −4 V

v ( 0 + ) = v ( 0 − ) = −4 V

For t < 0, represent the circuit in the frequency domain using the Laplace transform as shown. V ( s ) is a node voltage. Express the controlling

voltage of the dependent source in terms of the node voltages V a ( s ) = −V ( s ) Writing a node equation gives 6 s + V ( s ) + 3 s ⎛ V ( s ) + 4 ⎞ = 0.75 V ( s ) ⎜ ⎟ 8 4 40 ⎝ s⎠

V (s) −

Solving gives −2 10 10 4 2 4 1 ⎞ ⎛1 − 4 ⇒ V (s) = − = + − = −2 ⎜ + ⎟ s s ( s − 5) s − 5 s s − 5 s − 5 ⎝ s s −5⎠ Taking the inverse Laplace transform gives

( s − 5)V ( s ) =

v ( t ) = −2 (1 + e5 t ) V for t ≥ 0

This voltage becomes very large as time goes on.

P 14.7-11

Determine the output voltage, vo(t), in the circuit shown in Figure P 14.7-11.

Figure P 14.7-11 Solution:

For t < 0, the voltage source voltage is 2 V and the circuit is at steady state. At steady state, the capacitor acts like an open circuit. i (0 −) =

2−0 = 0.04 mA 10 × 103 + 40 × 103

and vC ( 0 − ) = ( 40 × 103 )( 0.04 × 10−3 ) = 1.6 V

The capacitor voltage is continuous so vC ( 0 + ) = vC ( 0 − ) . For t > 0, the voltage source voltage is 12 V. Represent the circuit in the frequency domain using the Laplace transform as shown.

V C ( s ) and V o ( s ) are node voltages. Writing a node equation gives

12 1.6 VC ( s ) − s + s + V C ( s ) = 0 ⇒ 4 ⎛ V ( s ) − 12 ⎞ + 0.08 s ⎛ V ( s ) − 1.6 ⎞ + V ( s ) = 0 ⎜ C ⎟ ⎜ C ⎟ C 6 3 0.5 × 10 10 × 10 40 × 103 s ⎠ s ⎠ ⎝ ⎝ s 48 80 s + 48 1.6 s + 600 9.6 −8 V C ( s )( 0.08 s + 5 ) = + 0.128 ⇒ V C ( s ) = = = + s s ( 0.08 s + 5 ) s ( s + 62.5 ) s s + 62.5

VC ( s ) −

Taking the inverse Laplace transform gives

v C ( t ) = 9.6 − 8 e − 62.5 t V for t ≥ 0

The 40 kΩ resistor, 50 kΩ resistor and op amp comprise an inverting amplifier so v o (t ) = −

50 50 v C ( t ) = − ( 9.6 − 8 e − 62.5 t ) = −12 + 10 e − 62.5 t V for t ≥ 0 40 40

so ⎧ −2 V for t ≤ 0 v o (t ) = ⎨ − 62.5 t V for t ≥ 0 ⎩ −12 + 10 e

(checked using LNAP 10/11/04)

P 14.7-12

Determine the capacitor voltage, v(t), in the circuit shown in Figure P 14.7-12.

Figure P 14.7-12 Solution: For t < 0, the voltage source voltage is 5 V and the circuit is at steady state. At steady state, the capacitor acts like an open circuit. Using voltage division twice v (0 −) =

and

32 30 5− 5 = 0.25 V 32 + 96 120 + 30

v ( 0 + ) = v ( 0 − ) = 0.25 V

For t > 0, the voltage source voltage is 20 V. Represent the circuit in the frequency domain using the Laplace transform as shown. We could write mesh or node equations, but finding a Thevenin equivalent of the part of the circuit to the left of terminals a-b seems promising.

Using voltage division twice ⎛ 32 ⎞ 20 ⎛ 30 ⎞ 20 5 − 4 1 V oc ( s ) = ⎜ = = V ⎟ −⎜ ⎟ s s ⎝ 32 + 96 ⎠ s ⎝ 120 + 30 ⎠ s

Z t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48 Ω

After replacing the part of the circuit to the left of terminals a-b by its Thevenin equivalent circuit as shown 1 0.25 − s s = 0.75 I (s) = 80 48 s + 80 48 + s

V (s) =

80 0.25 ⎛ 80 ⎞ 0.75 0.25 I (s) + =⎜ ⎟ + s s s ⎝ s ⎠ 48 s + 80

V (s) =

60 0.25 1.25 0.25 0.75 −0.75 0.25 1 −0.75 + = + = + + = + s ( 48 s + 80 ) s s ( s + 1.67 ) s s s + 1.67 s s s + 1.67

Taking the inverse Laplace transform gives v ( t ) = 1 − 0.75e −1.67 t V for t ≥ 0

Then

⎧0.25 V for t ≤ 0 v (t ) = ⎨ −1.67 t V for t ≥ 0 ⎩ 1 − 0.75e

(checked using LNAP 7/1/04)

P 14.7-13

Determine the voltage vo(t) for t ≥ 0 for the circuit of Figure P 14.7-13.

Hint: vC(0) = 4 V Answer: vo(t) = 24e0.75t u(t) V (This circuit is unstable.)

Figure P 14.7-13 Solution:

Mesh Equations: 4 1 4 ⎛ 1 ⎞ − − I C (s) − 6 I (s) + I C (s) = 0 ⇒ − = ⎜ 6 + ⎟ I C (s) + 6 I (s) 2s ⎠ s 2s s ⎝ 10 6 ( I ( s ) − I C ( s )) + 3 I ( s ) + 4 I C ( s ) = 0 ⇒ I ( s ) = − I C ( s ) 9 Solving for I C(s): 4 ⎛ 2 1 ⎞ 6 − = ⎜ − + ⎟ I C (s) ⇒ I C (s) = 3 s ⎝ 3 2s ⎠ s− 4

(

So Vo(s) is Back in the time domain:

)

Vo ( s ) = 4 I C ( s ) =

24 3 s− 4

v o ( t ) = 24 e0.75t u (t ) V for t ≥ 0

P 14.7-14

Determine the current iL(t) for t ≥ 0 for the circuit of Figure P 14.7-14.

Hint: vC(0) = 8 V and iL(0) = 1 A

1 ⎛ ⎞ Answer: iL ( t ) = ⎜ e − t cos 2t + e− t sin 2t ⎟ u ( t ) A 2 ⎝ ⎠

Figure P 14.7-14 Solution:

KVL: 8 ⎛ 20 ⎞ + 4 = ⎜ + 8 + 4s ⎟ I L ( s ) s ⎝ s ⎠ so I L ( s) =

( s + 1) + 1 2+ s = s + 2 s + 5 ( s + 1) 2 + 4 2

Taking the inverse Laplace transform: 1 ⎛ ⎞ i L ( t ) = ⎜ e − t cos 2 t + e − t sin 2 t ⎟ u ( t ) A 2 ⎝ ⎠

P14.7-15

The circuit shown in Figure P14.7-23 is at steady state before the switch opens at time t = 0. Determine the inductor voltage v(t) for t > 0.

Figure P14.7-15 Solution: The circuit shown in Figure P14.7-23 is at steady state before the switch opens at time t = 0. Determine the inductor voltage v(t) for t > 0.

Determine the initial conditions, i.e. the inductor current and capacitor voltage at t = 0, as shown in the circuit on the left below. Use those initial conditions to represent the circuit in the s-domain as s shown in the circuit on the left below.

Figure P14.7-15

Analysis of the s-domain circuit shows that ⎛ ⎞ ⎜ ⎟⎛ 4 16 ( s + 2 ) 4 ⎞ 16 ( s + 2 ) V (s) = ⎜ + 2⎟ = 2 = ⎟ ⎜ 2 8 ⎠ s + 8 s + 16 ( s + 4 ) ⎜⎜ 0.5 s + 4 + ⎟⎟ ⎝ s s⎠ ⎝

Performing the partial fraction expansion, we get 16 ( s + 2 )

( s + 4) Finally

2

=

k −32 + s + 4 ( s + 4 )2

⇒ 16 ( s + 2 ) = k ( s + 2 ) − 32 ⇒ k = 16

V (s) =

16 −32 + s + 4 ( s + 4 )2

Now use linearity, e − at f ( t ) ↔ F ( s + a ) and t ↔

1 to find the inverse Laplace transform s2

1 2 v ( t ) = 16 e −4 t ℒ -1[ − 2 ] = 16 (1 − 2 t ) e − 4 t for t > 0 s s

P14.7-16 The circuit shown in Figure P14.7-16 is at steady state before time t = 0. Determine the resistor voltage v(t) for t > 0.

Figure P14.7-16 Solution: The circuit shown in Figure 14.7-16 is at steady state before time t = 0. Determine the resistor voltage v(t) for t > 0.

Determine the initial conditions, i.e. the inductor current and capacitor voltage at t = 0, as shown in the circuit on the left below. Use those initial conditions to represent the circuit in the s-domain as s shown in the circuit on the left below.

Figure P14.7-16

Analysis of the s-domain circuit shows that V (s) V (s) 2 ⎛ s ⎞ + + + ⎜ ⎟V ( s ) = 0 ⇒ 5 6s s ⎝ 30 ⎠

( 6 s + 5 + s )V ( s ) = −60 2

Performing the partial fraction expansion, we get V (s) =

−60 −15 15 = + s + 6 s + 5 s +1 s + 5 2

The inverse Laplace transform is

v ( t ) = 15 ( e −5t − e − t ) V for t > 0

⇒ V (s) =

−60 s +6s +5 2

P14.7.17 The input to the circuit shown in Figure P14.7-17 is the voltage source voltage

⎧10 V when t < 0 v i ( t ) = 10 + 5 u ( t ) V = ⎨ ⎩15 V when t > 0 Determine the response, vo(t). Assume that the circuit is at steady state when t < 0. Sketch vo(t) as a function of t.

Figure P14.7-17

Solution:

⎧10 V when t < 0 v i ( t ) = 10 + 5 u ( t ) V = ⎨ ⎩15 V when t > 0

The circuit is at steady state when t < 0 so the capacitors act like open circuits. Since the current in the resistor is 0 A, vo(0−) = 0 V. Then KVL gives v1(0−) = 10 V. Use the initial conditions to represent the circuit for t>0 in the s-domain: Calculate 5 ||

500 500 = s s + 100

and 500 0.8 s s = 125 500 s + 20 + 5 || s s 5 ||

Use voltage division to write 500 4 ⎛ 15 10 ⎞ 0.8 s ⎛ 5 ⎞ s Vo ( s ) = − ⎟= = ⎜ ⎜ ⎟ 125 500 ⎝ s s ⎠ s + 20 ⎝ s ⎠ s + 20 + 5 || s s Taking the inverse Laplace transforms gives v o ( t ) = 4 e − 20 t V when t > 0 . 5 ||

In summary when t < 0 ⎧ 0V v o ( t ) = ⎨ − 20 t V when t > 0 ⎩4 e Let t = 0.05 s and calculate v o ( 0.05 ) = 4 e − 20 ( 0.05) =1.47 V

P14.7-18 The input to the circuit shown in Figure P14.7-18 is the current source current

⎧25 mA when t < 0 i ( t ) = 25 − 15 u ( t ) mA = ⎨ ⎩10 mA when t > 0 Determine the response, i2(t). Assume that the circuit is at steady state when t < 0. Sketch i2(t) as a function of t.

Figure P14.7-18

Solution:

⎧25 mA when t < 0 i ( t ) = 25 − 15 u ( t ) mA = ⎨ ⎩10 mA when t > 0

The circuit is at steady state when t < 0 so the inductors act like short circuits. Since the voltage across the resistor is 0 V, i2(0−) = 0 V. Then KCL gives i1(0−) = 25 mA. Use the initial conditions to represent the circuit for t>0 in the s-domain:

Use current division to write I2 ( s ) =

1.25 s ⎛ 0.01 0.025 ⎞ 0.2 s ⎛ 0.015 ⎞ −0.003 − ⎜ ⎟= ⎜− ⎟= 1.25 s + ( 25 + 5 s ) ⎝ s s ⎠ s+4⎝ s ⎠ s + 20

Taking the inverse Laplace transforms gives i 2 ( t ) = 3 e − 4 t mA when t > 0 .

In summary ⎧ 0 mA when t < 0 i 2 ( t ) = ⎨ − 4t mA when t > 0 ⎩3 e Let t = 0.1 s and calculate i 2 ( 0.1) = 3 e − 4 ( 0.1) = 2.01 mA

P 14.7-19 All new homes are required to install a device called a ground fault circuit interrupter (GFCI) that will provide protection from shock. By monitoring the current going to and returning from a receptacle, a GFCI senses when normal flow is interrupted and switches off the power in 1/40 second. This is particularly important if you are holding an appliance shorted through your body to ground. A circuit model of the GFCI acting to interrupt a short is shown in Figure P 14.7-19. Find the current flowing through the person and the appliance, i(t), for t ≥ 0 when the short is initiated at t = 0. Assume v = 160 cos 400t and the capacitor is intially uncharged.

FIGURE

P 14.7-19

Circuit model of person and appliance shorted to ground

Solution:

We are given v ( t ) = 160 cos 400 t . The capacitor is initially uncharged, so v C ( 0 ) = 0 V . Then i ( 0) = 10−3

KCL yields

dvC dt

+

vC 100

160 cos ( 400 × 0 ) − 0 = 160 A 1

=i

Apply Ohm’s law to the 1 Ω resistor to get v −v C i= ⇒ vC = v− i 1 di + 1010 i = 1600 cos 400t − ( 6.4 × 104 ) sin 400t Solving yields dt Taking the Laplace transform yields s I ( s ) − i (0) + (1010 ) I ( s ) =

so Next

I (s) =

s 2 + ( 400 )

( 6.4×10 ) ( 400 ) − 2

1600s 2

s 2 + ( 400 )

160 1600s − 2.5×107 + s + 1010 ( s + 1010 ) ⎡⎣ s 2 + (400) 2 ⎤⎦

2

1600s − 2.5×107 A B B* = + + ( s + 1010 ) ⎡⎣ s 2 + (400)2 ⎤⎦ s + 1010 s + j 400 s − j 400 where A =

B =

1600 s − 2.5 x 107 ( s +1010 ) ( s − j 400 )

1600 s − 2.5×107 s 2 + ( 400 ) =

s = − j 400

Then I (s) =

= − 23.1 ,

2 s = −1010

2.56 × 107 ∠1.4° 8.69 × 10 ∠68.4 5

°

= 11.5 − j 27.2 and B* = 11.5 + j 27.2

136.9 11.5− j 27.2 11.5 + j 27.2 + + s + 1010 s + j 400 s − j 400

Finally i ( t ) = 136.9e−1010t + 2 (11.5 ) cos 400t − 2 ( 27.2 ) sin 400t for t > 0 = 136.9e −1010t + 23.0 cos 400t − 54.4sin 400t for t > 0

P 14.7-20 Using the Laplace transform, find vc(t) for t > 0 for the circuit shown in Figure P 14.720. The initial conditions are zero. Hint:Use a source transformation to obtain a single mesh circuit. Answer: vc =–5e–2t + 5 (cos 2t + sin 2t) V

Figure P 14.7-20. Solution:

vC (0) = 0

vc +15×103 i = 10 cos 2t ⎫ ⎪ ⎬ ⇒ ⎛ 1 −3 ⎞ d vc i = ⎜ ×10 ⎟ ⎪ ⎝ 30 ⎠ dt ⎭

d vc + 2 vc = 20 cos 2t dt

Taking the Laplace Transform yields: sVC ( s ) − vC ( 0 ) + 2VC ( s ) = 20

20s s A B B* ⇒ V s = = + + ( ) C s2+ 4 ( s + 2 )( s 2 + 4 ) s + 2 s + j 2 s − j 2

where A=

20 s s2 +4

= s = −2

−40 20s = −5, B = 8 ( s + 2 )( s − j 2 )

= s = − j2

5 5 5 5 5 = + j and B* = − j 1− j 2 2 2 2

Then

5 5 5 5 +j −j −5 2 2 2 2 VC ( s ) = + + s+2 s+ j2 s− j2

⇒ vC ( t ) = −5e−2t + 5 ( cos 2t + sin 2t ) V

P 14.7-21

Determine the inductor current, i(t), in the circuit shown in Figure P 14.7-21.

Figure P 14.7-21 Solution: After the switch opens, apply KCL and KVL to get d ⎛ ⎞ R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs dt ⎝ ⎠

Apply KVL to get d i (t ) + R2 i (t ) dt Substituting v ( t ) into the first equation gives v (t ) = L

d⎛ d d ⎛ ⎞⎞ R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs dt ⎝ dt dt ⎠⎠ ⎝ 2 d d then R 1 C L 2 i ( t ) + R1 C R 2 + L i ( t ) + R1 + R 2 i ( t ) = Vs dt dt Dividing by R1 C L :

(

)

(

)

⎛ R1 C R 2 + L ⎞ d ⎛ R1 + R 2 ⎞ Vs i t i t + + ⎜ ⎟ ⎜ ⎟ i (t ) = ( ) ( ) 2 ⎜ ⎟ ⎜ ⎟ R1 C L dt ⎝ R1 C L ⎠ dt ⎝ R1 C L ⎠ d2

With the given values:

d2 dt

Taking the Laplace transform:

2

i ( t ) + 25

d i ( t ) + 156.25 i ( t ) = 125 dt

125 ⎡ 2 ⎛d ⎞⎤ ⎢ s I ( s ) − ⎜ dt i ( 0 + ) + s i ( 0 + ) ⎟ ⎥ + 25 ⎡⎣ s I ( s ) − i ( 0 + ) ⎤⎦ + 156.25 I ( s ) = s ⎝ ⎠⎦ ⎣ We need the initial conditions. For t < 0, the switch is closed and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. Using voltage division v (0 −) =

9 20 = 14.754 V 9 + (16 || 4 )

Then, using current division ⎛ 4 ⎞ v (0 −) i (0 −) = ⎜ = 0.328 A ⎟ ⎝ 16 + 4 ⎠ 9 The capacitor voltage and inductor current are continuous so v ( 0 + ) = v ( 0 − ) and i ( 0 + ) = i ( 0 − ) .

After the switch opens v (t ) = L

d i (t ) + R2 i (t ) ⇒ dt

v ( 0 + ) 9 i ( 0 + ) 14.754 9 ( 0.328 ) d i (0 +) = + = + = 29.508 dt 0.4 0.4 0.4 0.4

Substituting these initial conditions into the Laplace transformed differential equation gives ⎡ s 2 I ( s ) − ( 29.508 + 0.328 s ) ⎤ + 25 ⎡ s I ( s ) − 0.328⎤ + 156.25 I ( s ) = 125 ⎣ ⎦ ⎣ ⎦ s

( s2 + 25 s + 156.25) I ( s ) = 125s + ( 29.508 + 0.328 s ) + 25 ( 0.328) so I (s) =

=

0.328 s 2 + ( 29.508 + 25 ( 0.328 ) ) + 125

(

s s 2 + 25 s + 156.25

)

0.328 s 2 + ( 29.508 + 25 ( 0.328 ) ) + 125 s ( s + 12.5 )

2

=

−0.471 23.6 0.8 + + 2 s + 12.5 ( s + 12.5 ) s

Taking the inverse Laplace transform i ( t ) = 0.8 + e −12.5 t ( 23.6 t − 0.471) A for t ≥ 0

So

⎧0.328 A for t ≤ 0 i (t ) = ⎨ −12.5 t ( 23.6 t − 0.471) A for t ≥ 0 ⎩ 0.8 + e

(checked using LNAP 10/11/04)

P 14.7-22

Find v2(t) for the circuit of Figure P 14.7-22 for t ≥ 0.

Hint: Write the node equations at a and b in terms of v1 and v2. The initial conditions are v1(0) =

10 V and v2(0) = 25 V. The source is vs = 50 cos 2t u(t) V. Answer: v2 ( t ) =

23 − t 16 −4t e + e + 12 cos 2t + 12 sin 2t V t ≥ 0 3 3

Figure P 14.7-22 Solution:

Apply KCL at node a to get 1 d v1 v 2 − v1 = 48 dt 24

⇒ 2 v1 +

d v1 dt

= 2 v2

Apply KCL at node b to get v 2 − 50 cos 2 t 20

+

v 2 − v1 24

+

v2 30

+

d v2 1 d v2 = 0 ⇒ − v1 + 3 v 2 + = 60 cos 2 t 24 dt dt

Take the Laplace transforms of these equations, using v1 (0) = 10 V and v2 (0) = 25 V , to get

( 2+ s ) V1 ( s) − 2V2 ( s) = 10 and − V1 ( s) + ( 3+ s ) V2 ( s) =

25s 2 + 60s +100 s2 + 4

Solve these equations using Cramer’s rule to get ⎛ 25s 2 + 60s +100 ⎞ ( 2+ s ) ⎜ ⎟ +10 ( 2+ s ) ( 25s 2 + 60 s +100 )+10 ( s 2 + 4 ) s2 +4 ⎝ ⎠ V2 ( s ) = = ( 2+ s ) (3+ s)− 2 ( s 2 + 4 ) ( s +1)( s + 4 ) = Next, partial fraction expansion gives

25s 3 +120s 2 + 220s + 240 ( s 2 + 4 ) ( s +1)( s + 4 )

V2 ( s ) =

A A* B C + + + s + j 2 s − j 2 s +1 s + 4

where A =

25s 3 +120 s 2 + 220 s + 240 ( s +1) ( s + 4 ) ( s − j 2 )

s =− j 2

=

−240− j 240 = 6 + j6 −40

A* = 6 − j 6 25s 3 +120 s 2 + 220 s + 240 B = ( s2 +4) ( s+4) 25s 3 +120 s 2 + 220 s + 240 C = ( s 2 + 4 ) ( s +1)

Then V2 ( s ) =

s =−1

s =−4

=

115 23 = 15 3

=

−320 16 = −60 3

6+ j 6 6− j 6 23 3 16 3 + + + s + j 2 s − j 2 s +1 s + 4

Finally v2 (t ) = 12 cos 2 t + 12sin 2 t +

23 − t 16 −4t e + e V t≥0 3 3

P 14.7-23 The motor circuit for driving the snorkel shown in Figure P 14.7-23a is shown in Figure P 14.7-23b. Find the motor current I2(s) when the initial conditions are i1(0 –) = 2 A and i2(0 –) = 3 A. Determine i2(t) and sketch it for 10 s. Does the motor current smoothly drive the snorkel?

Solution: Here are the equations describing the coupled coils:

di1 di +M 2 dt dt di di v2 (t ) = L2 2 + M 1 dt dt

v1 (t ) = L1

⇒ V1 ( s) = 3 ( s I1 ( s) − 2 ) + ( sI 2 ( s) − 3) = 3s I1 ( s) + sI 2 ( s ) − 9 ⇒ V2 ( s) = s( I1 ( s ) − 2 ) + 2( sI 2 ( s) −3) = sI1 ( s) + 2sI 2 ( s ) − 8

Writing mesh equations: 5 5 = 2 ( I1 ( s ) + I 2 ( s ) ) + V1 = 2 ( I1 ( s ) + I 2 ( s ) ) + 3s I1 ( s ) + sI 2 ( s ) − 9 ⇒ ( 3s + 2 ) I1 + ( s + 2 ) I 2 = 9 + s s V1 ( s ) = V2 ( s ) + 1I 2 ( s ) ⇒ 3s I1 ( s ) + sI 2 ( s ) − 9 = sI1 ( s ) + 2 sI 2 ( s ) − 8+ I 2 ( s ) ⇒ 2s I1 − ( s +1) I 2 =1

Solving the mesh equations for I2(s):

I2 ( s ) =

15s + 8 3s + 1.6 0.64 2.36 = + = 2 s + 0.26 s + 1.54 5s + 9s + 2 ( s + 0.26 )( s +1.54 )

Taking the inverse Laplace transform: i2 (t ) = 0.64e −0.26t + 2.36e −1.54t A for t > 0

P14.7-24

Using Laplace transforms, find vo(t) for t > 0 for the circuit shown in Figure 14.7-24.

Figure 14.7-24 Solution: Find vo(t) for t > 0 for this circuit:

For t < 0, with the circuit at steady state, we have

v ( 0 ) = 2 V and i ( 0 ) =

For t > 0 we have

Apply KVL to the left mesh to get ⎛ 1 d ⎞ 10 ⎜ v C (t ) ⎟ + v C (t ) − 8 = 0 ⇒ ⎝ 20 dt ⎠ Apply KVL to the right mesh to get

1 d vC (t ) + vC (t ) = 8 2 dt

1 A 2

4

d d i L ( t ) + 12 i L ( t ) − 3 v C ( t ) = 0 ⇒ 4 i L ( t ) + 12 i L ( t ) = 3 v C ( t ) dt dt

Take the Laplace transform of these differential equations to get 1 8 ⎡⎣ sVC ( s ) − v C ( 0 ) ⎤⎦ + VC ( s ) = ⇒ 2 s

1 8 2 s + 16 ⎡⎣ sVC ( s ) − 2 ⎤⎦ + VC ( s ) = ⇒ VC ( s ) = 2 s s ( s + 2)

and 1⎤ ⎡ 4 ⎡⎣ s I L ( s ) − i L ( 0 ) ⎤⎦ + 12 I L ( s ) = 3VC ( s ) ⇒ 4 ⎢ s I L ( s ) − ⎥ + 12 I L ( s ) = 3VC ( s ) 2⎦ ⎣ 3 1 ⇒ ( s + 3) I L ( s ) = VC ( s ) + 4 2 3 ⎛ 2 s + 16 ⎞ 1 ⇒ ( s + 3) I L ( s ) = ⎜⎜ ⎟+ 4 ⎝ s ( s + 2 ) ⎟⎠ 2 1 2 5 s + s + 12 2 ⇒ IL ( s ) = 2 s ( s + 2 )( s + 3)

Taking the inverse Laplace transform gives

Finally

9 − 2 3 ⇒ IL ( s ) = + 2 + s s+2 s+3

9 ⎛ ⎞ i L ( t ) = ⎜ 2 − e − 2 t + 3 e − 3t ⎟ u ( t ) A 2 ⎝ ⎠ −2t v o ( t ) = 12 i L ( t ) = 24 − 54 e + 36 e − 3t u ( t ) V

(

)

(checked with LNAP 2/28/05)

P14.7-25

The circuit shown in Figure P14.7-25 is at steady state before the switch opens at time t = 0. Determine the inductor voltage v(t) for t > 0.

Figure P14.7-25 Solution: Determine the initial conditions, i.e. the inductor current and capacitor voltage at t = 0, as shown in the circuit on the left below. Use those initial conditions to represent the circuit in the sdomain as shown in the circuit on the left below.

Analysis of the s-domain circuit shows that ⎛

1000 ⎞ 1000 ⎞ ⎛ ⎟ (1.5 ) ⎜ 40 s + ⎟ 3.846 s ⎠ 12 s + 78 3.846 ⎠ ⎝ ⎝ V (s) = − = = 2 1000 1000 s + 8 s + 52 5 s + 40 + 5 s 2 + 40 s + 3.846 s 3.846 The denominator does not factor any further in the real numbers. Let’s complete the square in the denominator 12 ( s + 4 ) + 30 12 ( s + 4 ) 5 ( 6) 12 s + 78 12 s + 78 12 s + 78 V (s) = 2 = 2 = = = + 2 2 2 2 s + 8 s + 52 ( s + 8 s + 16 ) + 36 ( s + 4 ) + 36 ( s + 4 ) + 36 ( s + 4 ) + 6 ( s + 4 )2 + 62

( −1.5) ⎜ 40 +

Now use the property e − at f ( t ) ↔ F ( s + a ) and the Laplace transform pairs sin ωt for t ≥ 0 ↔

ω s + ω2 2

and cos ωt for t ≥ 0 ↔

s s + ω2 2

to find the inverse Laplace transform v ( t ) = e − 4 t ℒ-1[

5 ( 6) 12 s + 2 ] = e − 4t [ 12 cos(6t) + 5 sin (6t) ] for t > 0 2 2 s +6 s +6 2

P14.7-26

The circuit shown in Figure P14.7-26 is at steady state before the switch opens at time t = 0. Determine the inductor voltage v(t) for t > 0.

Figure P14.7-26 Solution

The circuit shown in Figure P14.7-22 is at steady state before the switch opens at time t = 0. Determine the inductor voltage v(t) for t > 0. Determine the initial conditions, i.e. the inductor current and capacitor voltage at t = 0, as shown in the circuit on the left below. Use those initial conditions to represent the circuit in the s-domain as s shown in the circuit on the left below.

Figure P14.7-26

Analysis of the s-domain circuit shows that ⎛ ⎞ ⎜ ⎟ ⎛ 12 ⎞ −12 −144 −60 ⎟⎜ ⎟ = V (s) = ⎜ = 2 ⎜ 2.4 s + 12 + 1000 ⎟ ⎝ s ⎠ 2.4 s 2 + 12 s + 1000 s + 5 s + 48.5 ⎜ 8.59 s ⎟⎠ 8.59 ⎝ The denominator does not factor any further in the real numbers. Let’s complete the square in the denominator

V (s) =

−9.23 ( 6.5 ) −60 −60 −60 = = = 2 2 2 2 s + 5 s + 48.5 ( s + 2.5 ) + 42.25 ( s + 2.5 ) + 6.5 ( s + 2.5) + 6.5 2 2

Now use e − at f ( t ) ↔ F ( s + a ) and sin ωt for t > 0 ↔ transform v ( t ) = e − 2.5t ℒ-1[

−9.23 ( 6.5 )

( s + 2.5)

2

+ 6.5

2

ω to find the inverse Laplace s + ω2 2

] = −9.23 e − 2.5t sin (6.5 t) for t > 0

Section 14.8 Transfer Functions    P14.8‐1 The input to the circuit shown in  Figure P14.8‐1 is the voltage  v i ( t )  and 

   

the output is the voltage  v o ( t ) . 

Determine the values of L, C, k, R1 and R2  that cause the step response of this  circuit to be: 

(

)

v o ( t ) = 5 + 20 e−5000 t − 25 e−4000t u ( t ) V

Figure 14.8‐1 

Answer: One solution is R1 = 400 Ω, L = 0.1 H, k = 5 V/V, C = 0.1 μF, R1 = 2 kΩ.  Solution: First, determine the transfer function from the circuit. To do so, represent the circuit in the s‐ domain as shown below. 

  Applying voltage division twice  k R 2C R1 k Va ( s ) = V (s)   Va ( s ) = V i ( s ) = L V i ( s )    and    V o ( s ) = 1 1 a R1 R1 + L s R2 + s+ s+ Cs R 2C L 1 Cs

R1

Substitu