Electric Circuits [12 ed.] 9780137648603

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Circuit Variables

Assessment Problems AP 1.1 Use a product of ratios to convert 95% of the speed of light from meters per second to miles per second: (0.95)

1 in 1 ft 1 mile 177,090.79 miles 3 × 108 m 100 cm · · · · = . 1s 1m 2.54 cm 12 in 5280 feet 1s

Now set up a proportion to determine how long it takes this signal to travel 950 miles: 177,090.79 miles 950 miles = . 1s xs Therefore, x=

950 = 0.00536 = 5.36 × 10−3 s = 5.36 ms. 177,090.79

AP 1.2 We begin by expressing $1 trillion in scientific notation: $1 trillion = $1 × 1012 . Divide by 100 = 102 to find the number of $100 bills: $1 trillion =

1012 = 1010 $100 bills. 102

Calculate the height of a stack of 1010 $100 bills: 1010 bills ·

0.11 mm 1m · = 1.1 × 106 m. bill 1000 mm

Now we can convert from meters to miles, again with a product of ratios: 1.1 × 106 m ·

100 cm 1 in 1 ft 1 mi · · · = 683.51 miles. 1m 2.54 cm 12 in 5280 ft 1–1

1–2

CHAPTER 1. Circuit Variables

AP 1.3 [a] First we use Eq. (1.2) to relate current and charge: i=

dq = 0.25te−2000t . dt

Therefore, dq = 0.25te−2000t dt. To find the charge, we can integrate both sides of the last equation. Note that we substitute x for q on the left side of the integral, and y for t on the right side of the integral: Z q(t) q(0)

dx = 0.25

Z t

ye−2000y dy.

0

We solve the integral and make the substitutions for the limits of the integral: e−2000y q(t) − q(0) = 0.25 (−2000y − 1) (−2000)2

t

0

= 62.5 × 10−9 e−2000t (−2000t − 1) + 62.5 × 10−9 = 62.5 × 10−9 (1 − 2000te−2000t − e−2000t ). But q(0) = 0 by hypothesis, so q(t) = 62.5(1 − 2000te−2000t − e−2000t ) nC. [b] q(0.001) = (62.5)[1 − 2000(0.001)e−2000(0.001) − e−2000(0.001) ] = 37.12 nC. AP 1.4 n =

75 × 10−6 C/s = 4.681 × 1014 elec/s. 1.6022 × 10−19 C/elec

AP 1.5 Start by drawing a picture of the circuit described in the problem statement:

Also sketch the four figures from Fig. 1.6:

Problems

1–3

[a] Now we have to match the voltage and current shown in the first figure with the polarities shown in Fig. 1.6. Remember that 250 mA of current entering Terminal 2 is the same as 250 mA of current leaving Terminal 1. We get i = −0.25 A;

(a) v = 50 V, (c) v = −50 V,

i = −0.25 A;

(b) v = 50 V,

i = 0.25 A;

(d) v = −50 V,

i = 0.25 A.

[b] Using the reference system in Fig. 1.6(a) and the passive sign convention, p = vi = (50)(−0.25) = −12.5 W. [c] Since the power is less than 0, the box is delivering power. AP 1.6 p = vi;

w=

Z t

p dx.

0

Since the energy is the area under the power vs. time plot, let us plot p vs. t.

Note that in constructing the plot above, we used the fact that 60 hr = 216,000 s = 216 ks. p(0) = (6)(15 × 10−3 ) = 90 × 10−3 W; p(216 ks) = (4)(15 × 10−3 ) = 60 × 10−3 W; 1 w = (60 × 10−3 )(216 × 103 ) + (90 × 10−3 − 60 × 10−3 )(216 × 103 ) = 16,200 J. 2 AP 1.7 [a] p = vi = (15e−250t )(0.04e−250t ) = 0.6e−500t W; p(0.01) = 0.6e−500(0.01) = 0.6e−5 = 0.00404 = 4.04 mW. [b] wtotal =

Z ∞ 0

p(x) dx =

Z ∞ 0

−500x

0.6e

0.6 −500x ∞ e dx = −500 0

= −0.0012(e∞ − e0 ) = 0.0012 = 1.2 mJ.



1–4

CHAPTER 1. Circuit Variables

Chapter Problems P 1.1

[a] Use a product of ratios to convert 40,075 km to inches: 1000 m 100 cm 1 in · · = 1,577,755,905.5 in. 1 km 1m 2.54 cm Now calculate the number of $20 bills this distance represents:

40,075 km ·

Number of bills =

1,577,755,905.5 in = 256,963,502.5 ≈ 256,963,503. 6.14 in/bill

The total dollar amount represented by this number of $20 bills is Total dollars = (256,963,503)($20) ≈ $5.14 billion. [b] The weight of the number of bills calculated in part (a) is (256,963,503)(1 g) = 256,963.5 kg. Use a product of ratios to convert kg to tons: 1 ton 2.2 lbs · = 282.66 tons. 256,963.5 kg · 1 kg 2000 lbs P 1.2

1s = 10.48 s. 50 × 106 bits 1s = 262 ms. [b] 8(65.5 × 106 ) bits · 2 × 109 bits 1s [c] 8(74 × 1012 ) bits · = 11.84 × 106 s 6 50 × 10 bits [a] 8(65.5 × 106 ) bits ·

11.84 × 106 s ·

1 hr 1 day 1 min · · = 137 days! 60 s 60 min 24 hr

[d] 8(74 × 1012 ) bits · 296 × 103 s · P 1.3

1s = 296 × 103 s. 2 × 109 bits

1 min 1 hr 1 day · · = 3.4 days. 60 s 60 min 24 hr

[a] To begin, we calculate the number of pixels that make up the display: npixels = (1920)(1080) = 2,073,600 pixels. Each pixel requires 24 bits of information. Since 8 bits equal one byte, each pixel requires 3 bytes of information. We can calculate the number of bytes of information required for the display by multiplying the number of pixels in the display by 3 bytes per pixel: nbytes =

2,073,600 pixels 3 bytes · = 6,220,800 bytes/display. 1 display 1 pixel

Problems

1–5

Finally, we use the fact that there are 106 bytes per MB: 1 MB 6,220,800 bytes · 6 = 6.22 MB/display. 1 display 10 bytes 6,220,800 bytes 8 bits · = 4.98 Gbps. 10 ms 1 byte [c] Convert the dimensions of the monitor from inches to mm:

[b]

24 in ·

25.4 mm = 609.6 mm. 1 in

14 in ·

25.4 mm = 355.6 mm. 1 in

Calculate the area of the monitor in mm2 : Screen area = (609.6)(355.6) = 126,773.76 mm2 . Divide the area of the monitor by the number of pixels in the monitor to find the area of an indiviual pixel: 126,773.76 mm2 = 0.1045 mm2 /pixel. 2,073,600 pixels P 1.4

(1024)(600) pixels 3 bytes 60 frames · · = 110,592 × 103 bytes/sec; 1 frame 1 pixel 1 sec (110,592 × 103 bytes/sec)(x secs) = 128 × 109 bytes; x=

128 × 109 = 1157 sec = 19.3 min of video. 110,592 × 103

P 1.5

(0.8)(5.4 × 106 )(1.5 × 103 ) + (0.9)(1.4 × 106 )(45 × 103 ) = 63.18 GWh. 109

P 1.6

[a] (0.6)(64 kWh) · 240 km · [b] 200 mi ·

100 km = 240 km; 16 kWh

0.62 mi = 148.8 mi. 1 km

1 km = 322.58 km; 0.62 mi

100 km 322.58 km = 16 kWh x kWh

so x =

51.61 kWh (100) ≈ 81%. 64 kWh

322.58(16) = 51.61 kWh. 100

1–6 P 1.7

CHAPTER 1. Circuit Variables Remember from Eq. 1.2, current is the time rate of change of charge, or i = dq dt In this problem, we are given the current and asked to find the total charge. To do this, we must integrate Eq. 1.2 to find an expression for charge in terms of current: q(t) =

Z t

i(x) dx.

0

We are given the expression for current, i, which can be substituted into the above expression. To find the total charge, we let t → ∞ in the integral. Thus we have qtotal = = P 1.8

Z ∞

−5000x

20e

0

20 −5000x ∞ 20 dx = e (e−∞ − e0 ) = −5000 −5000 0

20 20 (0 − 1) = = 0.004 C = 4000 µC. −5000 5000

[a] First we use Eq. 1.2 to relate current and charge: i=

dq = 0.025e−1000t . dt

Therefore, dq = 0.025e−1000t dt. To find the charge, we can integrate both sides of the last equation. Note that we substitute x for q on the left side of the integral, and y for t on the right side of the integral: Z q(t) q(0)

dx = 0.025

Z t

e−1000y dy.

0

We solve the integral and make the substitutions for the limits of the integral: e−1000y q(t) − q(0) = 0.025 −1000

t

= 25 × 10−6 (1 − e−1000t ).

0

But q(0) = 0 by hypothesis, so q(t) = 25(1 − e−1000t ) µC. [b] As t → ∞, qT = 25 µC. [c] q(1 × 10−3 ) = (25 × 10−6 )(1 − e(−1000)(0.001) ) = 15.8 µC. P 1.9

First we use Eq. 1.2 to relate current and charge: i=

dq = 0.1 cos 2500t. dt

Therefore, dq = 0.1 cos 2500t dt.

Problems

1–7

To find the charge, we can integrate both sides of the last equation. Note that we substitute x for q on the left side of the integral, and y for t on the right side of the integral: Z q(t)

dx = 0.1

Z t

q(0)

cos 2500y dy.

0

We solve the integral and make the substitutions for the limits of the integral, remembering that sin 0 = 0: 0.1 0.1 0.1 sin 2500y t sin 2500t − sin[2500(0)] = sin 2500t. q(t) − q(0) = 0.1 = 2500 2500 2500 2500 0

But q(0) = 0 by hypothesis, i.e., the current passes through its maximum value at t = 0, so q(t) = 40 × 10−6 sin 2500t C = 40 sin 2500t µC. P 1.10

w = qV = (1.6022 × 10−19 )(9) = 14.42 × 10−19 = 1.442 aJ.

P 1.11

Recall from Eq. 1.2 that current is the time rate of change of charge, or i = dq . In this problem we are given an expression for the charge, and asked to dt find the maximum current. First we will find an expression for the current using Eq. 1.2: d 1 1 t dq i= = − + 2 e−αt 2 dt dt α α α 







d 1 d t −αt d 1 −αt = − − e e 2 dt α dt α dt α2 

= 0− 

= −













1 t 1 −αt e − α e−αt − −α 2 e−αt α α α 





1 1 −αt +t+ e α α 

= te−αt . Now that we have an expression for the current, we can find the maximum value of the current by setting the first derivative of the current to zero and solving for t: di d = (te−αt ) = e−αt + t(−α)e−αt = (1 − αt)e−αt = 0. dt dt Since e−αt never equals 0 for a finite value of t, the expression equals 0 only when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For this value of t, the current is i=

1 −α/α 1 e = e−1 . α α

1–8

CHAPTER 1. Circuit Variables Remember in the problem statement, α = 0.03679. Using this value for α, i=

P 1.12

1 e−1 ∼ = 10 A. 0.03679

[a]

p = vi = −(80)(−4) = 320 W. Power is being absorbed by the box. [b] Entering. [c] Gain. P 1.13

[a] p = −vi = −(80)(4) = −320 W, so power is being delivered by the box. [b] Leaving. [c] Lose.

P 1.14

[a] In Car A, the current i is in the direction of the voltage drop across the 12 V battery(the current i flows into the + terminal of the battery of Car A). Therefore using the passive sign convention, p = vi = (25)(12) = 300 W. Since the power is positive, the battery in Car A is absorbing power, so Car A must have the “dead” battery. Z t 60 s [b] w(t) = p dx; 1 min = 1 · = 60 s; 1 min 0 w(60) =

Z 60

300 dx;

0

w = 300(60 − 0) = 300(60) = 18,000 J = 18 kJ. P 1.15

Assume we are standing at box A looking toward box B. Use the passive sign convention to get p = −vi, since the current i is flowing out of the + terminal of the voltage v. Now we just substitute the values for v and i into the equation for power. Remember that if the power is positive, B is absorbing power, so the power must be flowing from A to B. If the power is negative, B is generating power so the power must be flowing from B to A. [a] p = −(−20)(0.15) = 3 W

3 W from A to B;

[b] p = −(40)(15) = −600 W

600 W from B to A;

[c] p = −(−2000)(−0.05) = −100 W [d] p = −(80)(−3) = 240 W

100 W from B to A;

240 W from A to B.

Problems

P 1.16

3 hr ·

p = (9)(0.15) = 1.35 W;

w(t) =

Z t

p dt;

w(10,800) =

0

P 1.17

1–9

3600 s = 10,800 s; 1 hr

Z 10,800

1.35 dt = 1.35(10,800) = 14.58 kJ.

0

At the Building X end of the line the current is leaving the upper terminal, and thus entering the lower terminal where the polarity marking of the voltage is negative. Thus, using the passive sign convention, p = −vi. Substituting the values of voltage and current given in the figure, p = −(800 × 103 )(1.8 × 103 ) = −1440 × 106 = −1440 MW. Thus, because the power associated with the Building X end of the line is negative, power is being generated at the Building X end of the line and transmitted by the line to be delivered to the Building Y end of the line.

P 1.18

[a] Applying the passive sign convention to the power equation using the voltage and current polarities shown in Fig. 1.5, p = vi. p(t) = (80,000te−500t )(15te−500t ) = 120 × 104 t2 e−1000t ; p(0.01) = 120 × 104 (0.01)2 e−1000(0.01) = 5.45 mW. [b] We know that power is the time rate of change of energy, or p = dw/dt. If we know the power, we can find the energy by integrating Eq. 1.3. To find the total energy, the upper limit of the integral is infinity: wtotal =

Z ∞

120 × 104 x2 e−1000x dx

0

∞

120 × 104 −1000x 2 2 = e [(−1000) x − 2(−1000)x + 2) 3 (−1000) 0

120 × 104 0 = 0− e (0 − 0 + 2) = 2.4 mJ. (−1000)3 P 1.19

[a] p = vi = (100e−500t )(0.02 − 0.02e−500t ) = (2e−500t − 2e−1000t ) W; dp = −1000e−500t + 2000e−1000t = 0 dt 2 = e500t

so

ln 2 = 500t

so thus

2e−1000t = e−500t ; p is maximum at t = 1.4 ms;

pmax = p(1.4 ms) = 0.5 W. [b] w =

Z ∞ 0

=

−500t

[2e

−1000t

− 2e

∞ 2 −500t 2 −1000t ] dt = e − e −500 −1000 0

4 2 − = 2 mJ. 1000 1000





1–10 P 1.20

CHAPTER 1. Circuit Variables [a] We can find the time at which the power is a maximum by writing an expression for p(t) = v(t)i(t), taking the first derivative of p(t) and setting it to zero, then solving for t. The calculations are shown below: p p dp dt dp dt

=

0 t < 0,

= vi = t(3 − t)(6 − 4t) = 18t − 18t2 + 4t3 mW =

18 − 36t + 12t2 = 12(t2 − 3t + 1.5);

=

0

0 ≤ t ≤ 3 s;

when t2 − 3t + 1.5 = 0; √ √ 3± 9−6 3± 3 = ; 2√ 2 √ t2 = 3/2 + 3/2 = 2.366 s. 3/2 − 3/2 = 0.634 s;

t = t1

p = 0 t > 3 s;

=

p(t1 )

= 18(0.634) − 18(0.634)2 + 4(0.634)3 = 5.196 mW;

p(t2 )

= 18(2.366) − 18(2.366)2 + 4(2.366)3 = −5.196 mW.

Therefore, maximum power is being delivered at t = 0.634 s. [b] The maximum power was calculated in part (a) to determine the time at which the power is maximum: pmax = 5.196 mW (delivered). [c] As we saw in part (a), the other “maximum” power is actually a minimum, or the maximum negative power. As we calculated in part (a), maximum power is being extracted at t = 2.366 s. [d] This maximum extracted power was calculated in part (a) to determine the time at which power is maximum: pmax = 5.196 mW (extracted). [e] w =

Z t

pdx =

0

w(0)

Z t

(18x − 18x2 + 4x3 )dx = 9t2 − 6t3 + t4 .

0

=

0 mJ;

w(2)

=

4 mJ;

w(1) = 4 mJ; w(3) = 0 mJ. To give you a feel for the quantities of voltage, current, power, and energy and their relationships among one another, they are plotted below:

Problems

P 1.21

1–11

[a] p = vi = (16,000t + 20)e−800t ][(128t + 0.16)e−800t ] = 2048 × 103 t2 e−1600t + 5120te−1600t + 3.2e−1600t = 3.2e−1600t [640,000t2 + 1600t + 1]; dp = 3.2{e−1600t [1280 × 103 t + 1600] − 1600e−1600t [640,000t2 + 1600t + 1]} dt = −3.2e−1600t [128 × 104 (800t2 + t)] = −409.6 × 104 e−1600t t(800t + 1). dp = 0 when t = 0 so pmax occurs at t = 0. dt = 3.2e−0 [0 + 0 + 1] = 3.2 W.

Therefore, [b] pmax

1–12

CHAPTER 1. Circuit Variables

[c] w =

Z t

pdx;

0

Z t Z t Z t w 2 −1600x −1600x = 640,000x e dx + 1600xe dx + e−1600x dx 3.2 0 0 0 t

640,000e−1600x 4 2 = [256 × 10 x + 3200x + 2] 6 −4096 × 10 0 t

t

1600e−1600x e−1600x + + (−1600x − 1) . 256 × 104 −1600 0 0

When t → ∞ all the upper limits evaluate to zero, hence (640,000)(2) 1600 1 w = + + ; 3.2 4096 × 106 256 × 104 1600 w = 10−3 + 2 × 10−3 + 2 × 10−3 = 5 mJ. P 1.22

[a] p = vi = 30e−500t − 30e−1500t − 40e−1000t + 50e−2000t − 10e−3000t ; p(1 ms) = 3.1 mW. [b]

w(t)

=

Z t

(30e−500x − 30e−1500x − 40e−1000x

0

+ 50e−2000x − 10e−3000x )dx =

21.67 − 60e−500t + 20e−1500t + 40e−1000t − 25e−2000t + 3.33e−3000t µJ;

w(1 ms)

= 1.24 µJ.

[c] wtotal = 21.67 µJ. P 1.23

[a]

p

= vi = (104 t + 5)e−400t ][(40t + 0.05)e−400t ] =

400 × 103 t2 e−800t + 700te−800t + 0.25e−800t

= e−800t [400,000t2 + 700t + 0.25]; dp dt

= {e−800t [800 × 103 t + 700] − 800e−800t [400,000t2 + 700t + 0.25]} =

[−3,200,000t2 + 2400t + 5]100e−800t .

dp Therefore, = 0 when 3,200,000t2 − 2400t − 5 = 0 dt so pmax occurs at t = 1.68 ms. [b] pmax

=

[400,000(.00168)2 + 700(.00168) + 0.25]e−800(.00168)

=

0.67 W.

Problems

[c] w w

= =

Z t Z0t

pdx; 2 −800x

400,000x e

dx +

0

=

Z t

700xe

−800x

dx +

Z t

0.25e−800x dx

0

0 t 400,000e−800x [64 × 104 x2 + 1600x + 2] −512 × 106 0 t t 700e−800x e−800x + (−800x − 1) + 0.25 . 64 × 104 −800 0

0

When t → ∞ all the upper limits evaluate to zero, hence (400,000)(2) 700 0.25 + + = 2.97 mJ. w= 6 4 512 × 10 64 × 10 800 P 1.24

[a] v(10 ms) = 400e−1 sin 2 = 133.8 V; i(10 ms) = 5e−1 sin 2 = 1.67 A; p(10 ms) = v(10 ms)i(10 ms) = 223.79 W. [b]

p

w

= vi = 2000e−200t sin2 200t   1 −200t 1 − cos 400t = 2000e 2 2 = 1000e−200t − 1000e−200t cos 400t; =

Z ∞ 0

=

−200t

1000e

dt −

∞

e−200t 1000 −200 0 (

Z ∞

1000e−200t cos 400t dt

0

) ∞

e−200t −1000 [−200 cos 400t + 400 sin 400t] 2 + (400)2 (200)   200 = 5 − 1000 = 5 − 1 = 4 J. 4 × 104 + 16 × 104 P 1.25

[a] p = vi = 900 sin(200πt) cos(200πt) = 450 sin(400πt) W. Therefore, pmax = 450 W. [b] pmax (extracting) = 450 W. 1 Z 5×10−3 [c] pavg = 450 sin(400πt) dt 0.005 0 5×10−3 225 4 − cos 400πt = 9 × 10 = [1 − cos 2π] = 0. 400π π 0 1 Z 6.25×10−3 [d] pavg = 450 sin(400πt) dt 0.00625 0 =

P 1.26

[a] q

180 180 [1 − cos 2.5π] = = 57.3 W. π π

=

area under i vs. t plot

=

1 (10)(15,000) 2

=

75,000 + 300,000 + 50,000 = 425,000 C.

+ (20)(15,000) + 21 (20)(5000)

0

1–13

1–14

CHAPTER 1. Circuit Variables [b] w

Z

=

p dt =

Z

vi dt;

v = 250 × 10−6 t + 10, 0 ≤ t ≤ 15,000s: i

=

p

=

w1

0 ≤ t ≤ 20 ks.

30 − 666.67 × 10−6 t; 300 + 833.33 × 10−6 t − 166.67 × 10−9 t2 ; Z 15,000

=

(300 + 833.33 × 10−6 t − 166.67 × 10−9 t2 ) dt

0

= (4500 + 93.75 − 187.5)103 = 4406.25 kJ. 15,000 s ≤ t ≤ 20,000 s: i

=

p

=

w2

80 − 4 × 10−3 t; 800 − 20 × 10−3 t − 10−6 t2 ; Z 20,000

=

(4000 − 1750 − 1541.67)103 = 708.33 kJ;

= wT P 1.27

(800 − 20 × 10−3 t − 10−6 t2 ) dt

15,000

= w1 + w2 = 4406.25 + 08.33 = 5114.58 kJ.

[a] q = area under i vs. t plot 1 1 1 = (6)(5000) + (14)(5000) + (6)(10,000) + (8)(10,000) + (8)(5000) 2 2 2 = 15,000 + 70,000 + 30,000 + 80,000 + 20,000 = 215,000 C. [b] w

Z

=

p dt =

Z

vi dt;

v = 250 × 10−6 t + 10, 0 ≤ t ≤ 5000s: i = 20 − 1.2 × 10−3 t; p

=

w1

=

0 ≤ t ≤ 20 ks.

200 − 7 × 10−3 t − 0.3 × 10−6 t2 ; Z 5000

(200 + 7 × 10−3 t − 0.3 × 10−6 t2 ) dt

0

= 106 + 87,500 − 12,500 = 900 kJ. 5000 s ≤ t ≤ 15,000 s: i

=

17 − 0.6 × 10−3 t;

p

=

170 − 1.75 × 10−3 t − 0.15−6 t2 ;

w2

=

Z 15,000

(170 − 1.75 × 10−3 t − 0.15−6 t2 ) dt

5000

=

1.7 × 106 − 175,000 − 162,500 = 1362.5 kJ;

Problems 15,000 s ≤ t ≤ 20,000 s: i

=

p

=

w3

=

32 − 1.6 × 10−3 t; 320 − 8 × 10−3 t − 0.4 × 10−6 t2 ; Z 20,000

(320 − 8 × 10−3 t − 0.4 × 10−6 t2 ) dt

15,000

= 1.6 × 106 − 700,000 − 616,666.67 = 283.33 kJ; wT = w1 + w2 + w3 = 900 + 1362.5 + 283.33 = 2545.83 kJ. P 1.28

[a]

= 5 + 1 × 10−3 t mA,

0 ≤ t ≤ 5 ks;

i(t)

= 10 mA,

5 ks ≤ t ≤ 10 ks;

i(t)

= 20 − 1 × 10−3 t mA,

10 ks ≤ t ≤ 15 ks;

i(t)

= 0,

t > 15 ks.

[b] i(t)

p = vi = 240i so p(t)

= 1200 + 0.24t mW,

0 ≤ t ≤ 5 ks;

p(t)

= 2400 mW,

5 ks ≤ t ≤ 10 ks;

p(t)

= 4800 − 0.24t mW,

10 ks ≤ t ≤ 15 ks;

p(t)

= 0,

t > 15 ks.

[c] To find the energy, calculate the area under the plot of the power: 1 w(5 ks) = (1.2)(5000) + (1.2)(5000) = 9 kJ; 2

1–15

1–16

CHAPTER 1. Circuit Variables w(10 ks) = w(5 ks) + (2.4)(5000) = 21 kJ; 1 w(15 ks) = w(10 ks) + (1.2)(5000) + (1.2)(5000) = 30 kJ. 2

P 1.29

[a] 0 s ≤ t < 10 ms: v = 8 V;

i = 25t A;

p = 200t W.

i = 0.5 − 25t A;

p = 200t − 4 W.

i = −250 mA;

p = 0 W.

10 ms < t ≤ 30 ms: v = −8 V; 30 ms ≤ t < 40 ms: v = 0 V; 40 ms < t ≤ 60 ms: v = 8 V;

i = 25t − 1.25 A; p = 200t − 10 W.

t > 60 ms: v = 0 V;

i = 250 mA;

p = 0 W.

[b] Calculate the area under the curve from zero up to the desired time:

P 1.30

1 (2)(0.01) 2

w(0.01)

=

= 10 mJ;

w(0.03)

= w(0.01) − 21 (2)(0.01) + 21 (2)(0.01) = 10 mJ;

w(0.08)

= w(0.03) − 21 (2)(0.01) + 21 (2)(0.01) = 10 mJ.

pa = (−8)(7) = −56 W; pb = −(−2)(−7) = −14 W; pc = (10)(15) = 150 W; pd = −(10)(5) = −50 W; pe = (−6)(3) = −18 W; pX f = (−4)(3) = −12 W. X Pabs = 150 W; Pdel = 56 + 14 + 50 + 18 + 12 = 150 W.

Problems P 1.31

pa

= va ia = (6)(0.5) = 3 W;

pb

= −vb ib = −(10)(0.1) = −1 W;

pc

= −vc ic = −(−8)(−0.4) = −3.2 W;

pd

= −vd id = −(−2)(0.3) = 0.6 W;

pe

= ve ie = (−2)(0.3) = −0.6 W;

pf

= −vf if = −(4)(−0.2) = 0.8 W;

pg

= −vg ig = −(−6)(0.2) = 1.2 W;

1–17

ph = vh ih = (2)(−0.4) = −0.8 W. Therefore, X

Pabs = 3 + 0.6 + 0.8 + 1.2 = 5.6 W;

X

Pdel = 1 + 3.2 + 0.6 + 0.8 = 5.6 W;

X

Pabs =

X

Pdel .

Thus, the interconnection satisfies the power check. P 1.32

pa

= va ia = (−160)(−10) = 1600 W;

pb

= vb ib = (−100)(−20) = 2000 W;

pc

= −vc ic = −(−60)(6) = 360 W;

pd

= vd id = (800)(−50) = −40,000 W;

pe

= −ve ie = −(800)(−20) = 16,000 W;

pf

= −vf if = −(−700)(14) = 9800 W;

pg

= −vg ig = −(640)(−16) = 10,240 W.

X

Pdel = 40,000 W; Pabs = 1600 + 2000 + 360 + 16,000 + 9800 + 10,240 = 40,000 W; X X Therefore, Pdel = Pabs = 40,000 W.

X

P 1.33

[a] If the power balances, the sum of the power values should be zero: ptotal = −0.918 − 0.810 − 0.012 + 0.400 + 0.224 + 1.116 = 0. Thus, the power balances. [b] When the power is positive, the element is absorbing power. Since elements d, e, and f have positive power, these elements are absorbing power.

1–18

CHAPTER 1. Circuit Variables [c] The voltage can be calculated using v = p/i or v = −p/i, with proper application of the passive sign convention:

P 1.34

va

= −pa /ia = −(−0.918)/(−0.051) = −18 V;

vb

= pb /ib = (−0.81)/(0.045) = −18 V;

vc

= pc /ic = (−0.012)/(−0.006) = 2 V;

vd

= −pd /id = −(0.4)/(−0.02) = 20 V;

ve

= −pe /ie = −(0.224)/(−0.014) = 16 V;

vf

= pf /if = (1.116)/(0.031) = 36 V.

[a] From the diagram and the table we have pa

= −va ia = −(5000)(−0.150) = 750 W;

pb

= vb ib = (2000)(0.250) = 500 W;

pc

= −vc ic = −(3000)(0.200) = −600 W;

pd

= vd id = (−5000)(0.400) = −2000 W;

pe

= −ve ie = −(1000)(−0.050) = 50 W;

pf

= vf if = (4000)(0.350) = 1400 W;

pg

= −vg ig = −(−2000)(0.400) = 800 W;

ph

= −vh ih = −(−6000)(−0.350) = −2100 W.

X X

Pdel

=

600 + 2000 + 2100 = 4700 W;

Pabs

=

750 + 500 + 50 + 1400 + 800 = 3500 W.

Therefore,

X

Pdel 6=

X

Pabs and the subordinate engineer is correct.

[b] The difference between the power delivered to the circuit and the power absorbed by the circuit is −4700 + 3500 = 1200 W. One-half of this difference is 600 W, so it is likely that pc is in error. Either the voltage or the current probably has the wrong sign. (In Chapter 2, we will discover that using KVL the voltage vc should be −3.0 kV, not 3.0 kV!) If the sign of pc is changed from negative to positive, we can recalculate the power delivered and the power absorbed as follows: X Pdel = 2000 + 2100 = 4100 W; X

Pabs = 750 + 500 + 600 + 50 + 1400 + 800 = 4100 W. Now the power delivered equals the power absorbed and the power balances for the circuit.

Problems P 1.35

1–19

[a] We can add the powers supplied together and the powers absorbed together — if the power balances, these power sums should be equal: X Psup = 750 + 400 + 800 = 1950 W; X Pabs = 1600 + 150 + 200 = 1950 W. Thus, the power balances. [b] The current can be calculated using i = p/v or i = −p/v, with proper application of the passive sign convention. Remember that the power supplied is negative and the power absorbed is positive.

P 1.36

ia

= −pa /va = −(−750)/(−3000) = −250 mA;

ib

= −pb /vb = −(1600)/(4000) = −400 mA;

ic

= −pc /vc = −(−400)/(1000) = 400 mA;

id

= pd /vd = (150)/(1000) = 150 mA;

ie

= pe /ve = (−800)/(−4000) = 200 mA;

if

= pf /vf = (200)/(4000) = 50 mA;

pa

= va ia = (120)(−10) = −1200 W;

pb

= −vb ib = −(120)(9) = −1080 W;

pc

= vc ic = (10)(10) = 100 W;

pd

= −vd id = −(10)(−1) = 10 W;

pe

= ve ie = (−10)(−9) = 90 W;

pf

= −vf if = −(−100)(5) = 500 W;

pg

= vg ig = (120)(4) = 480 W;

ph

= vh ih = (−220)(−5) = 1100 W.

X

Pdel = 1200 + 1080 = 2280 W; Pabs = 100 + 10 + 90 + 500 + 480 + 1100 = 2280 W. X X Therefore, Pdel = Pabs = 2280 W. X

Thus, the interconnection now satisfies the power check.

1–20 P 1.37

CHAPTER 1. Circuit Variables [a] The revised circuit model is shown below:

[b] The expression for the total power in this circuit is v a ia − v b ib − v f if + v g ig + v h ih = (120)(−8) − (120)(8) − (−120)(6) + (120)(6) + (−240)ih = 0. Therefore, 240ih = −960 − 960 + 720 + 720 = −480 so −480 = −2 A. 240 Thus, if the power in the modified circuit is balanced the current in component h is −2 A.

ih =

Circuit Elements

Assessment Problems AP 2.1 [a] Because both current sources are in the same branch of the circuit, their values must be the same. Therefore, v1 = 0.4 → v1 = 0.4(50) = 20 V. 50 [b] p = v1 (0.4) = (20)(0.4) = 8 W (absorbed). AP 2.2 [a] The voltage drop from the top node to the bottom node in this circuit must be the same for every path from the top to the bottom. Therefore, the voltages of the two voltage sources are equal: −αi∆ = 6. Also, the current i∆ is in the same branch as the 15 mA current source, but in the opposite direction, so i∆ = −0.015A. Substituting, 6 = 400. 0.015 The interconnection is valid if α = 400 V/A.

−α(−0.015) = 6

→

α=

[b] The voltage across the current source must equal the voltage across the 6 V source, since both are connected between the top and bottom nodes. Using the passive sign convention, p = vi = (6)(0.015) = 0.09 = 90 mW. [c] Since the power is positive, the current source is absorbing power.

2–1

2–2

CHAPTER 2. Circuit Elements

AP 2.3

[a] The resistor and the voltage source are in parallel and the resistor voltage and the voltage source have the same polarities. Therefore these two voltages are the same: vR = vg = 100 V. Note from the circuit that the current through the resistor is ig = 2 A. Use Ohm’s law to calculate the value of the resistor: vR 100 V R= = = 50 Ω. ig 2A Using the passive sign convention to calculate the power in the resistor, pR = (vR )(ig ) = (100 V)(2 A) = 200 W. The resistor is dissipating 200 W of power. [b] Note from part (a) the vR = vg and iR = ig . The power delivered by the source is thus psource −125 W psource = −vg ig so vg = − =− = 500 V. ig 250 mA Since we now have the value of both the voltage and the current for the resistor, we can use Ohm’s law to calculate the resistor value: R=

vg 500 V = = 2 kΩ. ig 250 mA

The power absorbed by the resistor must equal the power generated by the source. Thus, pR = −psource = −(−125 W) = 125 W. [c] Again, note the iR = ig . The power dissipated by the resistor can be determined from the resistor’s current: pR = R(iR )2 = R(ig )2 . Solving for ig , i2g =

pr 20 mW = = 25 × 10−6 R 800 Ω

so

ig =

√

25 × 10−6 = 5 × 10−3 A = 5 mA.

Problems

2–3

Then, since vR = vg vR = RiR = Rig = (800 Ω)(5 mA) = 4 V

so

vg = 4 V.

AP 2.4

[a] Note from the circuit that the current through the conductance G is ig , flowing from top to bottom, because the current source and the conductance are in the same branch of the circuit so must have the same current. The voltage drop across the current source is vg , positive at the top, because the current source and the conductance are also in parallel so must have the same voltage. From a version of Ohm’s law, 20 mA ig = = 4 V. G 5 mS Now that we know the voltage drop across the current source, we can find the power delivered by this source:

vg =

psource = −vg ig = −(4)(0.02) = −80 mW. Thus the current source delivers 80 mW to the circuit. [b] We can find the value of the conductance using the power, and the value of the current using Ohm’s law and the conductance value: pg = Gvg2

so

G=

400 pg = 2 = 0.16 S = 160 mS; 2 vg 50

ig = Gvg = (160 mS)(50 V) = 8 A. [c] We can find the voltage from the power and the conductance, and then use the voltage value in Ohm’s law to find the current: pg = Gvg2 Thus

so vg =

√

vg2 =

pg 50 mW = = 2500. G 20 µS

2500 = 50 V;

ig = Gvg = (20 µS)(50 V) = 0.001 A = 1 mA.

2–4

CHAPTER 2. Circuit Elements

AP 2.5 [a] Redraw the circuit with all of the voltages and currents labeled for every circuit element.

Write a KVL equation clockwise around the circuit, starting below the voltage source: −180 V + v2 + v3 − v1 = 0. Next, use Ohm’s law to find the three unknown voltages in terms of the three currents: v2 = 15i2 ;

v3 = 20i3 ;

v1 = 25i1 .

A KCL equation at the lower right node gives i3 = −i1 ; a KCL equation at the upper right node gives i2 = i3 ; a KCL equation at the lower left node gives is = i1 . Now replace the currents i2 and i3 in the Ohm’s law equations with −i1 : v2 = 15i2 = −15i1 ;

v3 = 20i3 = −20i1 ;

v1 = 25i1 .

Now substitute these expressions for the three voltages into the first equation: 180 = v2 + v3 − v1 = −15i1 − 20i1 − 25i1 = −60i1 . [b] [c] [d] [e]

Therefore i1 = 180/(−60) = −3 A. v1 = 25i1 = 25(−3) = −75 V. v2 = −15i1 = −15(−3) = 45 V. v3 = −20i1 = −20(−3) = 60 V. Since is = i1 , is = −3 A. We can now compute the power associated with the voltage source: p180 = (180)is = (180)(−3) = −540 W. Therefore 180 V source is delivering 540 W.

AP 2.6

i1 = 100/40 = 2.5 A (Ohm’s Law);

Problems 3 = i1 + i2

so

2–5

i2 = 3 − i1 = 3 − 2.5 = 0.5 A (KCL middle top) ;

i2 = i3 (KCL right) ; −100 + v2 + v3 = 0

so

v2 + v3 = 100 (KVL right) ;

v2 = 80i2 = 80(0.5) = 40 V (Ohm’s Law); v3 = Ri3 = Ri2 = 0.5i2 (Ohm’s Law). Substitute the equations for v2 and v3 into the KVL equation and solving for R, 40 + 0.5R = 100

so

R=

100 − 40 = 120 Ω. 0.5

AP 2.7 [a] Plotting a graph of vt versus it gives

Note that when it = 0, vt = 40 V; therefore the voltage source must be 40 V. Since the plot is a straight line, its slope can be used to calculate the value of resistance: 40 − 0 40 ∆v = = = 800 Ω. R= ∆i 0.05 − 0 0.05 A circuit model having the same v − i characteristic is a 40 V source in series with a 800Ω resistor, as shown below:

[b] Draw the circuit model from part (a) and attach a 200 Ω resistor:

2–6

CHAPTER 2. Circuit Elements To find the power delivered to the 200 Ω resistor we must calculate the current through the 200 Ω resistor. Do this by first using KCL to recognize that the current in each of the components is it , flowing in a clockwise direction. Write a KVL equation in the clockwise direction, starting below the voltage source, and using Ohm’s law to express the voltage drop across the resistors in the direction of the current it flowing through the resistors: −40 V + 800it + 200it = 0

so

1000it = 40

so

it =

40 = 40 mA. 1000

Thus, the power delivered to the 200 Ω resistor is p200 = (200)i2t = (200)(0.04)2 = 320 mW. AP 2.8 [a] From the graph in Assessment Problem 2.7(a), we see that when vt = 0, it = 50 mA. Therefore the current source must be 50 mA. Since the plot is a straight line, its slope can be used to calculate the value of resistance: ∆v 40 − 0 40 = = = 800 Ω. ∆i 0.05 − 0 0.05 A circuit model having the same v − i characteristic is a 50 mA current source in parallel with a 800Ω resistor, as shown below:

R=

[b] Draw the circuit model from part (a) and attach a 200 Ω resistor:

Note that by writing a KVL equation around the right loop we see that the voltage drop across both resistors is vt . Write a KCL equation at the top center node, summing the currents leaving the node. Use Ohm’s law to specify the currents through the resistors in terms of the voltage drop across the resistors and the value of the resistors. vt vt −0.05 + + = 0, so 5vt = 40, thus vt = 8 V; 800 200 p200 =

vt2 = 320 mW. 200

Problems

2–7

AP 2.9 Label unknown current:

−20 + 450i + 150i = 0 so

→

600i = 20

(KVL and Ohm’s law); i = 33.33 mA.

vx = 150i = 150(0.0333) = 5 V 

vo = 300

(Ohm’s law);

vx = 300(5/100) = 15 V 100 

(Ohm’s law).

Calculate the power for all components: p20V = −20i = −20(0.0333) = −0.667 W; pd.s. = −vo



vx = −(15)(5/100) = −0.75 W; 100 

p450 = 450i2 = 450(0.033)2 = 0.5 W; p150 = 150i2 = 150(0.033)2 = 0.1667 W; p300 =

vo2 152 = = 0.75 W. 300 300

Thus the total power absorbed is pabs = 0.5 + 0.1667 + 0.75 = 1.4167 W. AP 2.10 [a] io = 0 because no current can exist in a single conductor connecting two parts of a circuit. [b]

2–8

CHAPTER 2. Circuit Elements −200 + 8000ig + 12,000ig = 0 so ig = 200/20,000 = 10 mA; v∆ = (12 × 103 )(10 × 10−3 ) = 120 V; 5 × 10−3 v∆ = 0.6 A; 9000i1 = 3000i2 so i2 = 3i1 ; 0.6 + i1 + i2 = 0 so 0.6 + i1 + 3i1 = 0 thus i1 = −0.15 A. [c] i2 = 3i1 = −0.45 A. [d] pv.s. = −200ig = −200(0.01) = −2 W. v1 = 9000i1 = 9000(−0.15) = −1350 V. pc.s. = 5 × 10−3 v∆ v1 = 5 × 10−3 (120)(−1350) = −810 W. pT = −2 − 810 = −812 W.

Problems

2–9

Problems P 2.1

[a] Yes, independent voltage sources can carry the 8 A current required by the connection; independent current source can support any voltage required by the connection, in this case 20 V, positive at the top. [b] 30 V source:

absorbing;

10 V source:

delivering;

8 A source:

delivering.

[c] P30V P10V P8A X

=

(30)(8) = 240 W (abs);

= −(10)(8) = −80 W (del); = −(20)(8) = −160 W

Pabs =

X

(del).

Pdel = 240 W.

[d] The interconnection is valid, but in this circuit the voltage drop across the 8 A current source is 40 V, positive at the top; 30 V source is absorbing, the 10 V source is absorbing, and the 8 A source is delivering. P30V

=

(30)(8) = 240 W (abs);

P10V

=

(10)(8) = 80 W (abs);

P8A X

P 2.2

= −(40)(8) = −320 W

Pabs =

X

(del).

Pdel = 320 W.

The interconnect is valid since the voltage sources can all carry the 50 mA current supplied by the current source, and the current source can carry the voltage drop required by the interconnection. Note that the voltage drop across the branch with the current source must be the same as the voltage drop across the branch with the three voltage sources. This voltage drop, from left to right, is 12 + 8 − 30 = −10 V. The voltages and currents are summarized in the circuit below:

2–10

CHAPTER 2. Circuit Elements P12V

=

(12)(0.05) = 0.6 W (abs);

P8V

=

(8)(0.05) = 0.4 W

P30V P50mA X

(abs);

= −(30)(0.05) = −1.5 W (dev); = −(−10)(0.05) = 0.5 W

(abs).

Pdev = 1.5 W.

P 2.3

Note that v5 , v20 , and v25 are not specified. The current sources can carry any amount of current and the voltage sources can carry the 5 A of current. Thus, the interconnection is valid. We can now calculate the power developed by the two voltage sources: pv−sources = p60 + p100 = −(60)(5) + (100)(5) = 200 W. Since the power is positive, the sources are absorbing 200 W of power, or developing −200 W of power. P 2.4

The voltage from the top node to the bottom node is 50 V, established by the 50 V source. The voltage from the top node to the bottom node using the right hand branch must also be 50 V, so the voltage drop across the 5 A source must be 20 V.

Now we can calculate the power developed by the current sources. p15A = −(15)(50) = −750 W; p5A = −(5)(20) = −100 W. X

Pdev = 850 W.

Problems P 2.5

2–11

The interconnection is valid, since the voltage sources can carry the 10 A current supplied by the current source, and the current sources can carry whatever voltage drop is required by the interconnection. In particular, note the the voltage drop across the three sources in the right hand branch must be the same as the voltage drop across the 20 A current source in the middle branch, since the middle and right hand branch are connected between the same two terminals. In particular, this means that v1 (the voltage drop across the middle branch) = 100V − 50V − v2 (the voltage drop across the right hand branch). Hence any combination of v1 and v2 such that v1 + v2 = 50 V is a valid solution.

P 2.6

[a] Note from the circuit that vx = −25 V. To find α note that the two current sources are in the same branch of the circuit but their currents flow in opposite directions. Therefore αvx = −15 A. Solve the above equation for α and substitute for vx , α=

−15 A −15 A = = 0.6 A/V. vx −25 V

[b] To find the power associated with the voltage source we need to know the current, ix . Note that this current is in the same branch of the circuit as the dependent current source and these two currents have the same direction. Therefore, the current ix is the same as the current of the dependent source: ix = αvx = (0.6)(−25) = −15 A. Using the passive sign convention, ps = −(ix )(25 V) = −(−15 A)(25 V) = 375 W. Thus the voltage source dissipates 375 W.

2–12

CHAPTER 2. Circuit Elements

P 2.7

[a] Note that the current ib is in the same circuit branch as the 8 A current source; however, ib is defined in the opposite direction of the current source. Therefore, ib = −8 A. Next, note that the dependent voltage source and the independent voltage source are connected between the same two nodes and have the same polarity. Therefore, their voltages are equal, and ib −8 = = −2 V. 4 4 [b] To find the power associated with the 8 A source, we need to find the voltage drop across the source, vi . Note that the two independent sources are connected between the same two nodes, and that the voltages vg and vi have the same polarities, so these voltages are equal: vg =

vi = vg = −2 V. Using the passive sign convention, ps = (8 A)(vi ) = (8 A)(−2 V) = −16 W. Thus the current source generates 16 W of power. P 2.8

The interconnection is invalid. In the middle branch, the value of the current i∆ must be −25 A, since the 25 A current source supplies current in this branch in the direction opposite the direction of the current i∆ . Therefore, the voltage supplied by the dependent voltage source in the left hand branch is 6(−25) = −150 V. This gives a voltage drop from the top terminal to the bottom terminal in the left hand branch of 50 − (−150) = 200 V. But the voltage drop between these same terminals in the right hand branch is 250 V, due to the voltage source in that branch. Therefore, the interconnection is invalid.

Problems

2–13

P 2.9

We assume the interconnection is valid, and therefore the total power in the circuit balances. Note that the voltage drop from the top node to the bottom node must be the same in each branch, so v∆ = 50 + v75

so v75 = v∆ − 50

and v∆ = vdep − 20 so vdep = v∆ + 20. The total power in the circuit is vtotal = 50(75) + 75v75 − 5v∆ vdep + 5v∆ (20) − 25v∆ = 50(75) + 75(v∆ − 50) − 5v∆ (v∆ + 20) + 5v∆ (20) − 25v∆ = 0. Solving for v∆ gives us 10 V, so v75 = −40 V and vdep = 30 V. Use the values for v∆ , v75 , and vdep to calculate the power associated with each source: p50V = (75)(50) = 3750 W p20V = [5(10)](20) = 1000 W

p75A = (75)(−40) = −3000 W; pds = −(50)(30) = −1500 W;

p25A = −(25)(10) = −250 W. X

P 2.10

Pdev = 250 + 3000 + 1500 = 4750 W =

X

Pabs .

[a] Yes, each of the voltage sources can carry the current required by the interconnection, and each of the current sources can carry the voltage drop required by the interconnection. (Note that i1 = 50 mA.)

2–14

CHAPTER 2. Circuit Elements [b] No, because the voltage drop between the top terminal and the bottom terminal cannot be determined. For example, define v1 , v2 , and v3 as shown:

The voltage drop across the left branch, the center branch, and the right branch must be the same, since these branches are connected at the same two terminals. This requires that v1 − 20 = v2 = v3 + 30. But this equation has three unknown voltages, so the individual voltages cannot be determined, and thus the power of the sources cannot be determined. P 2.11

[a] Using the passive sign convention and Ohm’s law, i=

v 120 = = 0.012 = 12 mA. R 10,000

[b] pv.s. = pR = Ri2 = (10,000)(0.012)2 = 1.44 W. [c] Using the passive sign convention with the voltage polarity reversed, i=−

120 v =− = −0.012 = −12 mA; R 10,000

pv.s. = pR = Ri2 = (10,000)(−0.012)2 = 1.44 W. P 2.12

[a] Using the passive sign convention and Ohm’s law, v = −Ri = −(8000)(0.075) = −600 V. (−600)2 v2 = = 45 W. R 8000 [c] Using the passive sign convention with the current direction reversed, [b] pc.s. = pR =

v = Ri = (8000)(0.075) = 600 V; pc.s. = pR =

v2 6002 = = 45 W. R 8000

Problems P 2.13

[a]

[b]

P 2.14

2–15

Vbb

=

no-load voltage of battery;

Rbb

=

internal resistance of battery;

Rx

=

resistance of wire between battery and switch;

Ry

=

resistance of wire between switch and lamp A;

Ra

=

resistance of lamp A;

Rb

=

resistance of lamp B;

Rw

=

resistance of wire between lamp A and lamp B;

Rg1

=

resistance of frame between battery and lamp A;

Rg2

=

resistance of frame between lamp A and lamp B;

S

=

switch.

Since we know the device is a resistor, we can use the power equation. From Fig. P2.14(a), p = vi =

v2 R

so

R=

v2 . p

Using the values in the table of Fig. P2.14(b) R=

(5)2 (10)2 (15)2 (20)2 = = = 16.67 × 10−3 66.67 × 10−3 150 × 10−3 266.67 × 10−3 =

(25)2 (30)2 = = 1.5 kΩ. 416.67 × 10−3 600 × 10−3

Note that this value is found in Appendix H. P 2.15

Since we know the device is a resistor, we can use Ohm’s law to calculate the resistance. From Fig. P2.15(a), v = Ri

so

v R= . i

2–16

CHAPTER 2. Circuit Elements Using the values in the table of Fig. P2.15(b), R=

−22.4 −11.2 11.2 22.4 33.6 = = = = = 560 Ω. −0.04 −0.02 0.02 0.04 0.06

Note that this value is found in Appendix H. P 2.16

The resistor value is the ratio of the power to the square of the current: p R = 2 . Using the values for power and current in Fig. P2.16(b), i 0.072 0.162 0.288 0.018 = = = 2 2 2 (0.001) (0.002) (0.003) (0.004)2 =

0.648 0.45 = = 18 kΩ. 2 (0.005) (0.006)2

Note that this is a value from Appendix H. P 2.17

[a] Write a KVL equation clockwise aroud the right loop, starting below the vertical 10 Ω resistor: −va + vb = 0

so

va = vb .

Using Ohm’s law, va = 10ia

and

vb = 15ib .

Substituting, 2 ib = ia . 3 Write a KCL equation at the top middle node, summing the currents leaving: 10ia = 15ib

so

2 5 ix = ia + ib = ia + ia = ia . 3 3 Write a KVL equation clockwise around the left loop, starting below the voltage source: −ix + ia + ib = 0

−80 V + vx + va = 0.

so

Problems

2–17

From Ohm’s law, vx = 10ix

and

va = 10ia .

Substituting, −80 V + 10ix + 10ia = 0 Substituting for ix : −80 V + 10



50 80 5 ia + 10ia = −80 V + ia + 10ia = −80 V + ia = 0. 3 3 3 

Thus, 80 ia = 80 V 3

so

ia =

80 V = 3 A. 80/3

[b] From part (a), ib = 32 ia = 23 (3 A) = 2 A. [c] From the circuit, vb = 15 Ω(ib ) = 15 Ω(2 A) = 30 V. [d] Use the formula pR = Ri2R to calculate the power absorbed by each resistor: p10Ω(top) =

i2x (10 Ω)

5 = ia 3 

2

5 (10 Ω) = (3) 3 

2

(10 Ω) = (5)2 (10 Ω) = 250 W;

p10Ω(mid) = i2a (10 Ω) = (3)2 (10 Ω) = 90 W; p15Ω = i2b (15 Ω) = (2)2 (15 Ω) = 60 W. [e] Using the passive sign convention, 5 5 = −(80 V)ix = −(80 V) ia = −(80 V) (3 A) 3 3 

psource







= −(80 V)(5 A) = −400 W. Thus the voltage source delivers 400 W of power to the circuit. Check:

P 2.18

X

Pdis = 250 + 90 + 60 = 400 W;

X

Pdel = 400 W.

Label the unknown resistor voltages and currents:

2–18

CHAPTER 2. Circuit Elements

[a] ib =

25 = 0.005 A (Ohm’s law); 5000

i1 = 0.025 − ib = 0.025 − 0.005 = 0.02 A (KCL). [b] v1 = 250i1 = 250(0.02) = 5 V (Ohm’s law); [c] va = 0.025(2000) = 50 V (Ohm’s law); −vg + va + 25 V = 0

so

vg = va + 25 = 50 + 25 = 75 V (KVL).

[d] pg = −vg (0.025) = −75(0.025) = −1.875 W

so 1.875 W delivered.

P 2.19

[a] Write a KCL equation at the top node: −4 + i1 + i2 = 0

so

i1 + i2 = 4.

Write a KVL equation around the right loop: −v80 + v30 + v90 = 0. From Ohm’s law, v80 = 80i1 ,

v30 = 30i2 ,

v90 = 90i2 .

Substituting, −80i1 + 30i2 + 90i2 = 0

so

− 80i1 + 120i2 = 0.

Solving the two equations for i1 and i2 simultaneously, i1 = 2.4 A

and

i2 = 1.6 A.

[b] Write a KVL equation clockwise around the left loop: −vo + v80 = 0 So

but

v80 = 80i1 = 80(2.4) = 192 A.

vo = v80 = 192 V.

[c] Calculate power using p = vi for the source and p = Ri2 for the resistors: psource = −vo (4) = −(192)(4) = −768 W; p80Ω = 2.42 (80) = 460.8 W; p30Ω = 1.62 (30) = 76.8 W; p90Ω = 1.62 (90) = 230.4 W; X

Pdev = 768 W

X

Pabs = 460.8 + 76.8 + 230.4 = 768 W.

Problems P 2.20

2–19

[a] Use KVL for the right loop to calculate the voltage drop across the right-hand branch vo . This is also the voltage drop across the middle branch, so once vo is known, use Ohm’s law to calculate io : vo

=

1000ia + 4000ia + 3000ia = 8000ia = 8000(0.002) = 16 V;

16 =

2000io ; 16 io = = 8 mA. 2000 [b] KCL at the top node: ig = ia + io = 0.002 + 0.008 = 0.010 A = 10 mA. [c] The voltage drop across the source is vo , seen by writing a KVL equation for the left loop. Thus, pg = −vo ig = −(16)(0.01) = −0.160 W = −160 mW. Thus the source delivers 160 mW. P 2.21

Label the unknown resistor currents and voltages:

[a] KCL at the top node: i1 = i2 + 0.0025; KVL around the left loop: −100 + v1 + vo = 0. Use Ohm’s law to write the resistor voltages in the previous equation in terms of the resistor currents: −100 + 5000i1 + 2000i2 = 0

→

5000i1 + 2000i2 = 100.

Use the KCL equation for i1 to eliminate i1 in the KVL equation: 5000(i2 + 0.0025) + 2000i2 = 100

→

7000i2 = 87.5.

Solving, 87.5 = 0.0125 = 12.5 mA. 7000 Therefore,

i2 =

vo = Ri2 = (2000)(0.0125) = 25 V. [b] i1 = i2 + 0.0025 = 0.0125 + 0.0025 = 0.015 A; p100V = −(100)i1 = −(100)(0.015) = −1.5 W; p25mA = (0.0025)vo = (0.0025)(25) = 0.0625 W; p5k = 5000i21 = 5000(0.015)2 = 1.125 W;

2–20

CHAPTER 2. Circuit Elements p2k = 2000i22 = 2000(0.0125)2 = 0.3125 W; ptotal = p25mA + p100V + p5k + p2k = 0.0625 − 1.5 + 1.125 + 0.3125 = 0. Thus the power in the circuit balances.

P 2.22

[a]

iCD = 500/15,000 = 33.33 mA; iBD + iCD = 0.1

so

iBD = 0.1 − 0.033 = 66.67 mA;

4000iBC + 500 − 7500iBD = 0

so

iBC = (500 − 500)/4000 = 0;

iAC = iCD − iBC = 33.33 − 0 = 33.33 mA; 100 = iAB + iAC

so

iAB = 100 − 33.33 = 66.67 mA.

Calculate the power dissipated by the resistors using the equation pR = Ri2R : p5kΩ = (5000)(0.0667)2 = 22.22 W

p7.5kΩ = (7500)(0.0667)2 = 33.33 W;

p10kΩ = (10,000)(0.03333)2 = 11.11 W

p15kΩ = (15,000)(0.0333)2 = 16.67 W;

p4kΩ = (4000)(0)2 = 0 W. [b] Calculate the voltage drop across the current source: vAD = 5000iAB + 7500iBD = 5000(0.0667) + 7500(0.0667) = 833.33 V. Now that we have both the voltage and the current for the source, we can calculate the power supplied by the source: pg = −833.33(0.1) = −83.33 W [c]

X

thus

pg (supplied) = 83.33 W.

Pdis = 22.22 + 33.33 + 11.11 + 16.67 + 0 = 83.33 W. Therefore, X

Psupp =

X

Pdis .

Problems P 2.23

2–21

[a]

v2 = 150 − 50(1) = 100V; i2 =

v2 = 4A; 25

i3 + 1 = i2 ,

i3 = 4 − 1 = 3A;

v1 = 10i3 + 25i2 = 10(3) + 25(4) = 130V; i1 =

v1 130 = = 2A. 65 65

Note also that i4 = i1 + i3 = 2 + 3 = 5 A; ig = i4 + io = 5 + 1 = 6 A. [b]

[c]

p4Ω

=

52 (4) = 100 W;

p50Ω

=

12 (50) = 50 W;

p65Ω

=

22 (65) = 260 W;

p10Ω

=

32 (10) = 90 W;

p25Ω

=

42 (25) = 400 W.

X

Pdis = 100 + 50 + 260 + 90 + 400 = 900 W;

Pdev = 150ig = 150(6) = 900 W. P 2.24

[a] Start by calculating the voltage drops due to the currents i1 and i2 . Then use KVL to calculate the voltage drop across and 100 Ω resistor, and Ohm’s law to find the current in the 100 Ω resistor. Finally, KCL at each of the middle three nodes yields the currents in the two sources and the current in the middle 10 Ω resistor. These calculations are summarized in the figure below:

2–22

CHAPTER 2. Circuit Elements

[b]

p130

= −(130)(15) = −1950 W;

p460

= −(460)(30) = −13,800 W.

X

Pdis = (15)2 (2) + (15)2 (10) + (30)2 (2) + (10)2 (25) + (25)2 (10) + (5)2 (100) = 450 + 2250 + 1800 + 2500 + 6250 + 2500 = 15,750 W.

X

Psup = 1950 + 13,800 = 15,750 W.

Therefore, P 2.25

X

Pdis =

X

Psup = 15,750 W.

[a]

v2 = 80 + 4(12) = 128 V; i1 =

v1 80 = = 5 A; 6 + 10 16

v1 = 128 − (8 + 12 + 4)(2) = 80 V; i3 = i1 − 2 = 5 − 2 = 3 A;

vg = v1 + 24i3 = 80 + 24(3) = 152 V; i4 = 2 + 4 = 6 A; ig = −i4 − i3 = −6 − 3 = −9 A. [b] Calculate power using the formula p = Ri2 : p8 Ω = (8)(2)2 = 32 W;

p12 Ω = (12)(2)2 = 48 W;

p4 Ω(left) = (4)(2)2 = 16 W;

p4 Ω(right) = (4)(6)2 = 144 W;

p24 Ω = (24)(3)2 = 216 W;

p6 Ω = (6)(5)2 = 150 W;

p10 Ω = (10)(5)2 = 250 W;

p12 Ω = (12)(4)2 = 192 W.

Problems

2–23

[c] vg = 152 V. [d] Sum the power dissipated by the resistors: X

pdiss = 32 + 48 + 16 + 144 + 216 + 150 + 250 + 192 = 1048 W.

The power associated with the sources is pvolt−source = (80)(4) = 320 W; pcurr−source = −vg ig = −(152)(9) = −1368 W. Thus the total power dissipated is 1048 + 320 = 1368 W and the total power developed is 1368 W, so the power balances. P 2.26

id = 60/12 = 5 A; therefore, vcd = 60 + 18(5) = 150 V; −240 + vac + vcd = 0; therefore, vac = 240 − 150 = 90 V; ib = vac /45 = 90/45 = 2 A; therefore, ic = id − ib = 5 − 2 = 3 A; vbd = 10ic + vcd = 10(3) + 150 = 180 V; therefore, ia = vbd /180 = 180/180 = 1 A. ie = ia + ic = 1 + 3 = 4 A; −240 + vab + vbd = 0; therefore, vab = 240 − 180 = 60 V. R = vab /ie = 60/4 = 15 Ω. CHECK: ig = ib + ie = 2 + 4 = 6 A; pdev = (240)(6) = 1440 W; X

Pdis =

12 (180) + 42 (15) + 32 (10) + 52 (12) + 52 (18) + 22 (45) = 1440 W (CHECKS).

P 2.27

[a]

2–24

CHAPTER 2. Circuit Elements ib = 60 V/30 Ω = 2 A; va = (30 + 60)(2) = 180 V; −500 + va + vb = 0; so vb = 500 − va = 500 − 180 = 320 V. ie = vb /(60 + 36) = 320/96 = (10/3) A; id = ie − ib = (10/3) − 2 = (4/3) A; vc = 30id + vb = 40 + 320 = 360 V; ic = vc /180 = 360/180 = 2 A; vd = 500 − vc = 500 − 360 = 140 V; ia = id + ic = 4/3 + 2 = (10/3) A; R = vd /ia = 140/(10/3) = 42 Ω. [b] ig = ia + ib = (10/3) + 2 = (16/3) A. pg (supplied) = (500)(16/3) = 2666.67 W.

P 2.28

[a] Plot the v − i characteristic

From the plot: R=

∆v (82 − 50) = = 4 Ω. ∆i (8 − 0)

When it = 0, vt = 50 V; therefore the ideal voltage source has a voltage of 50 V.

[b]

When vt = 0,

it =

−50 = −12.5A. 4

Note that this result can also be obtained by extrapolating the v − i characteristic to vt = 0.

Problems P 2.29

2–25

[a] Plot the v—i characteristic:

From the plot: R=

(420 − 100) ∆v = = 20 Ω. ∆i (16 − 0)

When it = 0, vt = 100 V; therefore the ideal current source must have a current of 100/20 = 5 A.

[b] We attach a 5 Ω resistor to the device model developed in part (a):

Write a KCL equation at the top node: 5 + it = i1 . Write a KVL equation for the right loop, in the direction of the two currents, using Ohm’s law: 20i1 + 5it = 0. Combining the two equations and solving, 20(5 + it ) + 5it = 0

so

25it = −100;

thus

Now calculate the power dissipated by the resistor: p5 Ω = 5i2t = 5(−4)2 = 80 W.

it = −4 A.

2–26 P 2.30

CHAPTER 2. Circuit Elements [a] Begin by constructing a plot of voltage versus current:

[b] Since the plot is linear for 0 ≤ is ≤ 225 mA and since R = ∆v/∆i, we can calculate R from the plotted values as follows: ∆v 75 − 30 45 = = = 200 Ω. ∆i 0.225 − 0 0.225 We can determine the value of the ideal voltage source by considering the value of vs when is = 0. When there is no current, there is no voltage drop across the resistor, so all of the voltage drop at the output is due to the voltage source. Thus the value of the voltage source must be 75 V. The model, valid for 0 ≤ is ≤ 225 mA, is shown below:

R=

[c] The circuit is shown below:

Write a KVL equation in the clockwise direction, starting below the voltage source. Use Ohm’s law to express the voltage drop across the resistors in terms of the current i: −75 V + 200i + 400i = 0 Thus,

i=

so

75 V = 125 mA. 600 Ω

[d] The circuit is shown below:

600i = 75 V.

Problems

2–27

Write a KVL equation in the clockwise direction, starting below the voltage source. Use Ohm’s law to express the voltage drop across the resistors in terms of the current i: −75 V + 200i = 0 Thus,

i=

so

200i = 75 V.

75 V = 375 mA. 200 Ω

[e] The short circuit current can be found in the table of values (or from the plot) as the value of the current is when the voltage vs = 0. Thus, isc = 500 mA

(from table).

[f ] The plot of voltage versus current constructed in part (a) is not linear (it is piecewise linear, but not linear for all values of is ). Since the proposed circuit model is a linear model, it cannot be used to predict the nonlinear behavior exhibited by the plotted data. P 2.31

[a]

[b] ∆v = 25V;

∆i = 2.5 mA;

[c] 10,000i1 = 2500is ,

R=

∆v = 10 kΩ. ∆i

i1 = 0.25is ;

0.02 = i1 + is = 1.25is ,

is = 16 mA.

[d] vs (open circuit) = (20 × 10−3 )(10 × 103 ) = 200 V.

2–28

CHAPTER 2. Circuit Elements [e] The open circuit voltage can be found in the table of values (or from the plot) as the value of the voltage vs when the current is = 0. Thus, vs (open circuit) = 140 V (from the table). [f ] Linear model cannot predict the nonlinear behavior of the practical current source.

P 2.32

Label unknown voltage and current:

io = ix + 3ix = 4ix

(KCL);

−280 + 6ix + 2(4ix ) = 0

(KVL left).

Therefore 14ix = 280 so ix =

280 = 20 A. 14

Thus vo = 2(4ix ) = 8ix = 8(20) = 160 V. The only two circuit elements that could supply power are the two sources, so calculate the power for each source: p280V = −280ix = −280(20) = −5600 W; pd.s. = −(3ix )vo = −3(20)(160) = −9600 W. Both sources are supplying power, so the total power supplied is 5600 + 9600 = 15,200 W. P 2.33

First note that we know the current through all elements in the circuit except the 6 kΩ resistor (the current in the three elements to the left of the 6 kΩ resistor is i1 ; the current in the three elements to the right of the 6 kΩ resistor

Problems

2–29

is 30i1 ). To find the current in the 6 kΩ resistor, write a KCL equation at the top node: i1 + 30i1 = i6k = 31i1 . We can then use Ohm’s law to find the voltages across each resistor in terms of i1 . The results are shown in the figure below:

[a] To find i1 , write a KVL equation around the left-hand loop, summing voltages in a clockwise direction starting below the 5 V source: −5 V + 54,000i1 − 1 V + 186,000i1 = 0. Solving for i1 , 54,000i1 + 186,000i1 = 6 V

so

240,000i1 = 6 V.

Thus, i1 =

6 = 25 µA. 240,000

[b] Now that we have the value of i1 , we can calculate the voltage for each component except the dependent source. Then we can write a KVL equation for the right-hand loop to find the voltage v of the dependent source. Sum the voltages in the clockwise direction, starting to the left of the dependent source: +v − 54,000i1 + 8 V − 186,000i1 = 0. Thus, v = 240,000i1 − 8 V = 240,000(25 × 10−6 ) − 8 V = 6 V − 8 V = −2 V. [c] We now know the values of voltage and current for every circuit element.

2–30

CHAPTER 2. Circuit Elements Let’s construct a power table: Element

Current

Voltage

Power

Power

(µA)

(V)

Equation

(µW)

5V

25

5

p = −vi

−125

54 kΩ

25

1.35

p = Ri2

33.75

1V

25

1

p = −vi

−25

6 kΩ

775

4.65

p = Ri2

3603.75

Dep. source

750

−2

p = −vi

1500

1.8 kΩ

750

1.35

p = Ri2

1012.5

8V

750

8

p = −vi

−6000

The total power generated in the circuit is the sum of the negative power values in the power table: −125 µW + (−25 µW) + (−6000 µW) = −6150 µW. Thus, the total power generated in the circuit is 6150 µW. [d] The total power absorbed in the circuit is the sum of the positive power values in the power table: 33.75 µW + 3603.75 µW + 1500 µW + 1012.5 µW = 6150 µW. Thus, the total power absorbed in the circuit is 6150 µW. P 2.34

[a] The circuit:

v1 = (3)(8) = 24 V v1 = 12io + 4io = 16io 6

(Ohm’s law); (KVL).

Thus, io =

24/6 v1 /6 = = 0.25 = 250 mA. 16 16

Problems

2–31

[b] Calculate the power for all components: p8A = −(8)v1 = −(8)(24) = −192 W; pd.s. = −(v1 /6)io = −(24/6)(0.25) = −1 W; p3 =

242 v12 = = 192 W; 3 3

p12 = 12i2o = 12(0.25)2 = 0.75 W; p4 = 4i2o = 4(0.25)2 = 0.25 W. Therefore, ptotal = −192 − 1 + 192 + 0.75 + 0.25 = 0. Thus the power in the circuit balances. P 2.35

Given that iφ = 2 A, we know the current in the dependent source is 2iφ = 4 A. We can write a KCL equation at the left node to find the current in the 10 Ω resistor. Summing the currents leaving the node, −5 A + 2 A + 4 A + i10Ω = 0

so

i10Ω = 5 A − 2 A − 4 A = −1 A.

Thus, the current in the 10 Ω resistor is 1 A, from right to left, as seen in the circuit below.

[a] To find vs , write a KVL equation, summing the voltages counter-clockwise around the lower right loop. Start below the voltage source. −vs + (1 A)(10 Ω) + (2 A)(30 Ω) = 0

so

vs = 10 V + 60 V = 70 V.

[b] The current in the voltage source can be found by writing a KCL equation at the right-hand node. Sum the currents leaving the node −4 A + 1 A + iv = 0

so

iv = 4 A − 1 A = 3 A.

The current in the voltage source is 3 A, from top to bottom. The power associated with this source is p = vi = (70 V)(3 A) = 210 W. Thus, 210 W are absorbed by the voltage source.

2–32

CHAPTER 2. Circuit Elements [c] The voltage drop across the independent current source can be found by writing a KVL equation around the left loop in a clockwise direction: −v5A + (2 A)(30 Ω) = 0

so

v5A = 60 V.

The power associated with this source is p = −v5A i = −(60 V)(5 A) = −300 W. This source thus delivers 300 W of power to the circuit. [d] The voltage across the controlled current source can be found by writing a KVL equation around the upper right loop in a clockwise direction: +v4A + (10 Ω)(1 A) = 0

v4A = −10 V.

so

The power associated with this source is p = v4A i = (−10 V)(4 A) = −40 W. This source thus delivers 40 W of power to the circuit. [e] The total power dissipated by the resistors is given by (i30Ω )2 (30 Ω) + (i10Ω )2 (10 Ω) = (2)2 (30 Ω) + (1)2 (10 Ω) = 120 + 10 = 130 W. P 2.36

[a] −50 − 20iσ + 18i∆ = 0; −18i∆ + 5iσ + 40iσ = 0,

so

18i∆ = 45iσ .

− 50 − 20iσ + 45iσ = 0,

Therefore,

so

iσ = 2 A.

18i∆ = 45iσ = 90; so i∆ = 5 A; vo = 40iσ = 80 V. [b] ig = current out of the positive terminal of the 50 V source; vd = voltage drop across the 8i∆ source; ig = i∆ + iσ + 8i∆ = 9i∆ + iσ = 47 A; vd = 80 − 20 = 60 V. X X

Pgen

=

50ig + 20iσ ig = 50(47) + 20(2)(47) = 4230 W;

Pdiss

=

18i2∆ + 5iσ (ig − i∆ ) + 40i2σ + 8i∆ vd + 8i∆ (20)

=

(18)(25) + 10(47 − 5) + 4(40) + 40(60) + 40(20)

X

Pgen

=

4230 W; Therefore,

=

X

Pdiss = 4230 W.

Problems

P 2.37

50i2 +

0.250 0.250 + = 0; 50 12.5

i2 = −0.5 mA;

v1 = 100i2 = −50 mV; 20i1 +

(−0.050) + (−0.0005) = 0; 25

i1 = 125 µA;

vg = 10i1 + 40i1 = 50i1 ; Therefore, vg = 6.25 mV. P 2.38

iE − iB − iC = 0; iC = βiB

therefore iE = (1 + β)iB ;

i2 = −iB + i1 ; Vo + iE RE − (i1 − iB )R2 = 0; −i1 R1 + VCC − (i1 − iB )R2 = 0

Vo + iE RE + iB R2 −

or

i1 =

VCC + iB R2 ; R1 + R2

VCC + iB R2 R2 = 0. R1 + R2

Now replace iE by (1 + β)iB and solve for iB . Thus iB = P 2.39

[VCC R2 /(R1 + R2 )] − Vo . (1 + β)RE + R1 R2 /(R1 + R2 )

Here is Equation 2.21: iB =

(VCC R2 )/(R1 + R2 ) − V0 ; (R1 R2 )/(R1 + R2 ) + (1 + β)RE

VCC R2 (15)(80) = = 12 V; R1 + R2 100 R1 R2 (20)(80) = = 16 kΩ; R1 + R2 100 iB =

12 − 0.2 11.8 = = 0.59 mA; 16 + 40(0.1) 20

2–33

2–34

CHAPTER 2. Circuit Elements iC = βiB = (39)(0.59) = 23.01 mA; iE = iC + iB = 23.01 + 0.59 = 23.6 mA; v3d = (23.6)(0.1) = 2.36 V; vbd = Vo + v3d = 2.56 V; i2 =

vbd 2.56 = × 10−3 = 32 µA; R2 80

i1 = i2 + iB = 32 + 590 = 622 µA; vab = 20(0.622) = 12.44 V; iCC = iC + i1 = 23.01 + 0.622 = 23.632 mA; v13 + 23.01(0.5) + 2.36 = 15; v13 = 1.135 V. P 2.40

[a]

[b]

P 2.41

Each radiator is modeled as a 48 Ω resistor:

Problems

2–35

Write a KVL equation for each of the three loops: 240 = 5 A; 48

−240 + 48i1 = 0

→

i1 =

−48i1 + 48i2 = 0

→

i2 = i1 = 5 A;

−48i2 + 48i3 = 0

→

i3 = i2 = 5 A.

Therefore, the current through each radiator is 5 A and the power for each radiator is prad = Ri2 = 48(5)2 = 1200 W. There are three radiators, so the total power for this heating system is ptotal = 3prad = 3(1200) = 3600 W. P 2.42

Each radiator is modeled as a 48 Ω resistor:

Write a KVL equation for this loop: −240 + 48i + 48i + 48i = 0

→

i=

240 = 1.67 A. 3(48)

Calculate the power for each radiator: prad = 48i2 = 48(1.67)2 = 133.33 W. Calculate the total power for this heating system: ptotal = 3prad = 3(133.33) = 400 W. Each radiator has much less power than the radiators in Fig. P2.41, and the total power of this configuration is just 1/9th of the total power in Fig. P2.41.

2–36 P 2.43

CHAPTER 2. Circuit Elements Each radiator is modeled as a 48 Ω resistor:

Write a KVL equation for the left and right loops: −240 + 48i1 = 0

→

−48i1 + 48i2 + 48i2 = 0

i1 =

240 = 5 A; 48

→

i2 =

i1 5 = = 2.5 A. 2 2

The power for the center radiator is pcen = 48i21 = 48(5)2 = 1200 W. The power for each of the radiators on the right is pright = 48i22 = 48(2.5)2 = 300 W. Thus the total power for this heating system is ptotal = pcen + 2pright = 1200 + 2(300) = 1800 W. The center radiator produces 1200 W, just like each of the three radiators in Problem 2.41. But the other two radiators produce only 300 W each, which is 1/4th of the power of the radiators in Problem 2.41. The total power of this configuration is 1/2 of the total power in Fig. P2.41. P 2.44

Each radiator is modeled as a 48 Ω resistor:

Write a KVL equation for the left and right loops: −240 + 48i1 + 48i2 = 0;

Problems −48i2 + 48i3 = 0

→

i2 = i3 .

Write a KCL equation at the top node: i1 = i2 + i3

→

i1 = i2 + i2 = 2i2 .

Substituting into the first KVL equation gives −240 + 48(2i2 ) + 48i2 = 0

→

i2 =

240 = 1.67 A. 3(48)

Solve for the currents i1 and i3 : i3 = i2 = 1.67 A;

i1 = 2i2 = 2(1.67) = 3.33 A.

Calculate the power for each radiator using the current for each radiator: pleft = 48i21 = 48(3.33)2 = 533.33 W; pmiddle = pright = 48i22 = 48(1.67)2 = 133.33 W. Thus the total power for this heating system is ptotal = pleft + pmiddle + pright = 533.33 + 133.33 + 133.33 = 800 W. All radiators in this configuration have much less power than their counterparts in Fig. P2.41. The total power for this configuration is only 22.2% of the total power for the heating system in Fig. P2.41.

2–37

Simple Resistive Circuits

Assessment Problems AP 3.1

Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 35 Ω, the 75 Ω and the 40 Ω resistors in series: 35 Ω + 75 Ω + 40 Ω = 150 Ω. Now combine this 150 Ω resistor in parallel with the 100 Ω resistor: 150 Ωk100 Ω =

(150)(100) 15,000 = = 60 Ω. 150 + 100 250

This equivalent 60 Ω resistor is in series with the 140 Ω resistor: 60 Ω + 140 Ω = 200 Ω. Finally, this equivalent 200 Ω resistor is in parallel with the 300 Ω resistor: 200 Ωk300 Ω =

(200)(300) 60,000 = = 120 Ω. 200 + 300 500

Thus, the simplified circuit is as shown:

3–1

3–2

CHAPTER 3. Simple Resistive Circuits [a] With the simplified circuit we can use Ohm’s law to find the voltage across both the current source and the 120 Ω equivalent resistor: v = (120 Ω)(0.03 A) = 3.6 V. [b] Now that we know the value of the voltage drop across the current source, we can use the formula p = −vi to find the power associated with the source: p = −(3.6 V)(0.03 A) = −108 mW. Thus, the source delivers 108 mW of power to the circuit. [c] We now can return to the original circuit, shown in the first figure. In this circuit, v = 3.6 V, as calculated in part (a). This is also the voltage drop across the 300 Ω resistor, so we can use Ohm’s law to calculate the current through this resistor: 3.6 V = 12 mA. 300 Ω Now write a KCL equation at the upper left node to find the current iB : iA =

−30 mA + iA + iB = 0

so

iB = 30 mA − iA = 30 mA − 12 mA = 18 mA.

Next, write a KVL equation around the outer loop of the circuit, using Ohm’s law to express the voltage drop across the resistors in terms of the current through the resistors: −v + 140iB + 35iC + 75iC + 40iC = 0. So Thus

150iC = v − 140iB = 3.6 V − (140)(0.018) = 1.08 V. iC =

1.08 = 7.2 mA. 150

Now that we have the current through the 75 Ω resistor we can use the formula p = Ri2 to find the power: p75 Ω = (75)(7.2 × 10−3 )2 = 3.888 mW = 3888 µW.

Problems

3–3

AP 3.2

[a] We can use voltage division to calculate the voltage vo across the 4 kΩ resistor: 4000 (60 V) = 20 V. vo (no load) = 4000 + 8000 [b] When we have a load resistance of 4 kΩ then the voltage vo is across the parallel combination of two 4 kΩ resistors. First, calculate the equivalent resistance of the parallel combination: (4000)(4000) = 2000 Ω = 2 kΩ. 4000 + 4000 Now use voltage division to find vo across this equivalent resistance: 4 kΩk4 kΩ =

vo =

2000 (60 V) = 12 V. 2000 + 8000

[c] If the load terminals are short-circuited, the 4 kΩ resistor is effectively removed from the circuit, leaving only the voltage source and the 8 kΩ resistor. We can calculate the current in the resistor using Ohm’s law: i=

60 V = 7.5 mA. 8 kΩ

Now we can use the formula p = Ri2 to find the power dissipated in the 8 kΩ resistor: p8k = (8000)(0.0075)2 = 0.45 W. [d] The power dissipated in the 4 kΩ resistor will be maximum at no load since vo is maximum. In part (a) we determined that the no-load voltage is 20 V, so we can use the formula p = v 2 /R to calculate the power: p4k (max) =

(20)2 = 0.1 W. 4000

3–4

CHAPTER 3. Simple Resistive Circuits

AP 3.3

[a] We will write a current division equation for the current in the branch containing the 5 kΩ resistor and use this equation to solve for R: i5k =

Rk(3000 + 5000 + 7000) (0.03) = 0.012; 3000 + 5000 + 7000

Rk(3000 + 5000 + 7000) =

(3000 + 5000 + 7000)(0.012) = 6000; 0.03

15,000R = 6000; R + 15,000 15,000R = 6000R + 6000(15,000) so 9000R = 6000(15,000); Thus, R =

6000(15,000) = 10,000 = 10 kΩ. 9000

[b] With R = 10 kΩ we can calculate the current through R and then use this current to find the power dissipated by R, using the formula p = Ri2 : iR = 0.03 − i5k = 0.03 − 0.012 = 0.018 = 18 mA, so

pR = (10,000)(0.018)2 = 3.24 W.

[c] The equivalent resistance seen by the current source is Req = 4000 + 10,000k15,000 = 4000 + 6000 = 10,000 = 10 kΩ. The power supplied by the current source equals the power dissipated by the equivalent resistance (whose current is 30 mA) so p = 10,000(0.03)2 = 9 W. Thus, the current source generates 9 W of power. AP 3.4

Problems

3–5

[a] First we need to determine the equivalent resistance of the 80 Ω resistor and the resistors to its right: Req = 80 Ωk(50 Ω + 30 Ω) = 80 Ωk80 Ω = 40 Ω Now we can use voltage division to find the voltage vo : vo =

10 (120 V) = 10 V. 10 + 40 + 70

[b] The current through the 10 Ω resistor can be found using Ohm’s law: 10 V vo = = 1 A. 10 10 Ω Now we use current division to find the current in the 80 Ω branch: 80k(50 + 30) 40 i80Ω = (1 A) = = 0.5 A. 80 80 [c] We can find the power dissipated by the 50 Ω resistor if we can find the current in this resistor. The current in the 10 Ω resistor divides between the 80 Ω branch and the branch with the 50 Ω and 30 Ω resistors. Since 0.5 A is in the 80 Ω branch, the remaining 0.5 A current must be in the other branch. Therefore, i40Ω =

i50Ω = 0.5 A. Thus,

p50Ω = (50)(0.5)2 = 12.5 W.

AP 3.5 [a]

We can find the current i using Ohm’s law: i=

5V = 0.2 A = 200 mA. 25 Ω

[b]

Rm = 50 Ωk5.555 Ω = 5 Ω.

3–6

CHAPTER 3. Simple Resistive Circuits We can use the meter resistance to find the current using Ohm’s law: imeas =

5V = 0.1667 = 166.67 mA. 25 Ω + 5 Ω

AP 3.6 [a]

Use voltage division to find the voltage v: v=

50,000 (150 V) = 100 V. 50,000 + 25,000

[b]

The meter resistance is a series combination of resistances: Rm = 149,950 + 50 = 150,000 Ω. We can use voltage division to find v, but first we must calculate the equivalent resistance of the parallel combination of the 50 kΩ resistor and the voltmeter: (50,000)(150,000) 50,000 Ωk150,000 Ω = = 37.5 kΩ. 50,000 + 150,000 Thus,

vmeas =

37,500 (150 V) = 90 V. 37,500 + 25,000

AP 3.7 [a] Using the condition for a balanced bridge, the products of the opposite resistors must be equal. Therefore, 1000Rx = (2000)(500)

so

Rx =

(2000)(500) = 1000 Ω = 1 kΩ. 1000

Problems

3–7

[b] When the bridge is balanced, there is no current flowing through the meter, so the meter acts like an open circuit. This places the following branches in parallel: The branch with the voltage source, the branch with the series combination R1 and R3 and the branch with the series combination of R2 and Rx . We can find the current in the latter two branches using Ohm’s law: iR1 ,R3 =

30 V = 10 mA; 1000 Ω + 2000 Ω

iR2 ,Rx =

30 V = 20 mA. 500 Ω + 1000 Ω

We can calculate the power dissipated by each resistor using the formula p = Ri2 : pR1 = (1000 Ω)(0.01 A)2 = 100 mW; pR2 = (500 Ω)(0.02 A)2 = 200 mW; pR3 = (2000 Ω)(0.01 A)2 = 200 mW; pRx = (1000 Ω)(0.02 A)2 = 400 mW. Thus, the Rx resistor must dissipate the most power. AP 3.8 Convert the three Y-connected resistors, 1 Ω, 10 Ω, and 40 Ω to three ∆-connected resistors Ra , Rb , and Rc . To assist you the figure below has both the Y-connected resistors and the ∆-connected resistors:

(1)(10) + (1)(40) + (10)(40) = 450 Ω; 1 (1)(10) + (1)(40) + (10)(40) Rb = = 45 Ω; 10 (1)(10) + (1)(40) + (10)(40) Rc = = 11.25 Ω. 40 Ra =

The circuit with these new ∆-connected resistors is shown below:

3–8

CHAPTER 3. Simple Resistive Circuits

From this circuit we see that the 450 Ω resistor is parallel to the 50 Ω resistor: 450 Ωk50 Ω =

(450)(50) = 45 Ω. 450 + 50

Also, the 11.25 Ω resistor is parallel to the 15 Ω resistor: 11.25 Ωk15 Ω =

(11.25)(15) 45 = Ω. 11.25 + 15 7

Once the parallel combinations are made, we can see that the equivalent 45 Ω resistor is in series with the equivalent 45/7 Ω resistor, and that this series combination is in parallel with the voltage source. Therefore, we can use voltage division to find v: v=

45/7 (24) = 3 V. 45/7 + 45

Problems

3–9

Problems P 3.1

[a] The 3 kΩ and 8 kΩ resistors are in series, as are the 5 kΩ and 7 kΩ resistors. The simplified circuit is shown below:

[b] The 180 Ω and 300 Ω resistors are in series, as are the 140 Ω and 200 Ω resistors. The simplified circuit is shown below:

[c] The 40 Ω, 50 Ω, and 60 Ω resistors are in series, as are the 45 Ω and 30 Ω resistors. The simplified circuit is shown below:

[d] The 6 Ω and 1.2 Ω resistors are in series, as are the 6 Ω and 10 Ω resistors. The simplified circuit is shown below:

3–10 P 3.2

CHAPTER 3. Simple Resistive Circuits [a] The 4 kΩ and 12 kΩ resistors are in parallel, as are the 6 kΩ and 3 kΩ resistors. The simplified circuit is shown below:

[b] The 12 Ω and 20 Ω resistors are in parallel, as are the 28 Ω and 21 Ω resistors. The simplified circuit is shown below:

[c] The 30 Ω and 5 Ω resistors are in parallel, as are the 9 Ω and 18 Ω resistors. The simplified circuit is shown below:

[d] The 100 kΩ and 300 kΩ resistors are in parallel, as are the 75 kΩ, 50 kΩ, and 150 kΩ resistors. The simplified circuit is shown below:

P 3.3

Always work from the side of the circuit furthest from the source. Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. [a] Circuit in Fig. P3.1(a): Req = {[(5 k + 7 k)k6 k] + 3 k + 8 k}k10 k = [(12 kk6 k) + 11 k]k10 k = (4 k + 11 k)k10 k = 15 kk10 k = 6 kΩ. Circuit in Fig. P3.1(b): Req = [240k(180 + 300)] + 140 + 200 = (240k480) + 340 = 160 + 340 = 500 Ω.

Problems

3–11

Circuit in Fig. P3.1(c): Req = (40 + 50 + 60)k(30 + 45) = 150k75 = 50 Ω. Circuit in Fig. P3.1(d): Req = ([(6 + 10)k64] + 6 + 1.2)k30 = [(16k64) + 6 + 1.2]k30 = (12.8 + 6 + 1.2)k30 = 20k30 = 12 Ω. [b] Note that in every case, the power delivered by the source must equal the power absorbed by the equivalent resistance in the circuit. For the circuit in Fig. P3.1(a): P = i2s Req = 0.0022 (6000) = 0.024 = 24 mW. For the circuit in Fig. P3.1(b): P =

102 Vs2 = = 0.2 W. Req 500

For the circuit in Fig. P3.1(c): vs2 502 P = = = 50 W. Req 50 For the circuit in Fig. P3.1(d): P = i2s (Req ) = (5)2 (12) = 300 W. P 3.4

Always work from the side of the circuit furthest from the source. Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. [a] Circuit in Fig. P3.2(a): 400 + (6000k3000k(5000 + (4000k12,000))) = 400 + (2000k(5000 + 3000)) = 400 + 1600 = 2000 = 2 kΩ. Circuit in Fig. P3.2(b): Req = 12k20k[18 + (28k21)] = 12k20k(18 + 12) = 12k20k30 = 6 Ω. Circuit in Fig. P3.2(c): Req = 4 + (9k18) + [5k30k(20 + 40)] = 4 + 6 + (5k30k60) = 4 + 6 + 4 = 14 Ω. Circuit in Fig. P3.2(d): Req = (100 kk300 k) + (75 kk50 kk150 k) + 25 k = 75 k + 25 k + 25 k = 125 kΩ.

3–12

CHAPTER 3. Simple Resistive Circuits [b] Note that in every case, the power delivered by the source must equal the power absorbed by the equivalent resistance in the circuit. For the circuit in Fig. P3.2(a): P =

vs2 602 = = 1.8 W. Req 2000

For the circuit in Fig. P3.2(b): P = i2s (Req ) = (0.2)2 (6) = 0.24 W. For the circuit in Fig. P3.2(c): P =

82 vs2 = = 4.57 W. Req 14

For the circuit in Fig. P3.2(d): P = P 3.5

vs2 0.52 = = 2 µW. Req 125,000

[a] Rab = 10 + (5k20) + 6 = 10 + 4 + 6 = 20 Ω. [b] Rab = 30 kk60 kk[20 k + (200 kk50 k)] = 30 kk60 kk(20 k + 40 k) = 30 kk60 kk60 k = 15 kΩ. [c] Rab = 2 k + [(14 k + 10 k)k12 k] + 6 k = (24 kk12 k) + 8 k = 8 k + 8 k = 16 kΩ. [d] Rab = 600k200k300k(250 + 150) = 600k200k300k400 = 80 Ω.

P 3.6

Write an expression for the resistors in series and parallel from the right side of the circuit to the left. Then simplify the resulting expression from left to right to find the equivalent resistance. [a] Rab = [(18 + 12)k30 + 15]k60 = (30k30 + 15)k60 = (15 + 15)k60 = 30k60 = 20 Ω. [b] 60k20 = 1200/80 = 15 Ω; 15 + 8 + 7 = 30 Ω;

12k24 = 288/36 = 8 Ω; 30k120 = 3600/150 = 24 Ω;

Rab = 15 + 24 + 25 = 64 Ω. [c] 35 + 40 = 75 Ω;

75k50 = 3750/125 = 30 Ω;

30 + 20 = 50 Ω;

50k75 = 3750/125 = 30 Ω;

30 + 10 = 40 Ω;

40k60 + 9k18 = 24 + 6 = 30 Ω;

30k30 = 15 Ω;

Rab = 10 + 15 + 5 = 30 Ω.

Problems [d] 50 + 30 = 80 Ω;

P 3.7

3–13

80k20 = 16 Ω;

16 + 14 = 30 Ω;

30 + 24 = 54 Ω;

54k27 = 18 Ω;

18 + 12 = 30 Ω;

30k30 = 15 Ω;

Rab = 3 + 15 + 2 = 20 Ω.

[a] Circuit in Fig. P3.7(a): Req = 15k(18 + 48k16) = 10 Ω. Circuit in Fig. P3.7(b): Req = ([(750 + 250)k1000] + 250)k([(300 + 500)k2400] + 900) = [(1000k1000) + 250]k[(800k2400) + 900] = (500 + 250)k(600 + 900) = 750k1500 = 500 Ω. Circuit in Fig. P3.7(c): 5k10k15k10k(12 + 18) = 2 Ω; 16k(14 + 2) = 8 Ω; Req = 4 + 8 + 12 = 24 Ω. Circuit in Fig. P3.7(d): 144k(4 + 12) = 14.4 Ω; 14.4 + 5.6 = 20 Ω; 20k12 = 7.5 Ω; 7.5 + 2.5 = 10 Ω; 10k15 = 6 Ω; 14 + 6 + 10 = 30 Ω; Req = 30k60 = 20 Ω. [b] Note that in every case, the power delivered by the source must equal the power absorbed by the equivalent resistance in the circuit. For the circuit in Fig. P3.7(a): P =

vs2 202 = = 40 W. Req 10

For the circuit in Fig. P3.7(b): P = i2s (Req ) = (0.05)2 (500) = 1.25 W.

3–14

CHAPTER 3. Simple Resistive Circuits For the circuit in Fig. P3.7(c): P =

vs2 1442 = = 864 W. Req 24

For the circuit in Fig. P3.7(d): P = i2s (Req ) = 52 (20) = 500 W. P 3.8

[a] R + R = 2R. [b] R + R + R + · · · + R = nR. [c] R + R = 2R = 2000 so R = 1000 = 1 kΩ. This is a resistor from Appendix H. [d] nR = 6000; so if n = 4, R = 1.5 kΩ. This is a resistor from Appendix H. So connect four 1.5k resistors in series to get 6 kΩ.

P 3.9

[a] Req = RkR = [b] Req

R R2 = . 2R 2

= RkRkRk · · · kR (n R’s)  −1 1 1 1 = + + ··· R R R  −1 n R = = . R n

R = 6000 so R = 12 kΩ. 2 This is a resistor from Appendix H. R [d] = 900 so R = 900n. n If n = 2 R = 900(2) = 1.8 kΩ. This is a resistor from Appendix H. So connect two 1.8k resistors in parallel to get 900Ω.

[c]

P 3.10

[a] From Ex. 3.1: i1 = 4 A, i2 = 8 A, is = 12 A; at node b: −12 + 4 + 8 = 0, at node d: 12 − 4 − 8 = 0.

Problems [b] v1 = 4is = 48 V;

3–15

v3 = 3i2 = 24 V;

v2 = 18i1 = 72 V;

v4 = 6i2 = 48 V.

loop abda: −120 + 48 + 72 = 0; loop bcdb: −72 + 24 + 48 = 0; loop abcda: −120 + 48 + 24 + 48 = 0. P 3.11

[a] p4Ω = i2s 4 = (12)2 4 = 576 W; p3Ω = (8)2 3 = 192 W;

p18Ω = (4)2 18 = 288 W;

p6Ω = (8)2 6 = 384 W.

[b] p120V (delivered) = 120is = 120(12) = 1440 W. [c] pdiss = 576 + 288 + 192 + 384 = 1440 W. P 3.12

vo (no load) = 25 =

R2 (75), 100 + R2

vo (with load) = 15 =

Req = 25 = 50kRL =

P 3.13

→

Req (75), 100 + Req

50RL , 50 + RL

→

2500 + 25R2 = 75R2 ,

→

→

1500 + 15Req = 75Req ,

1250 + 25RL = 50RL ,

→

R2 = 50 Ω.

→

Req = 25 Ω.

RL = 50 Ω.

160(3300) = 66 V. (4700 + 3300) [b] i = 160/8000 = 20 mA;

[a] vo =

PR1 = (400 × 10−6 )(4.7 × 103 ) = 1.88 W; PR2 = (400 × 10−6 )(3.3 × 103 ) = 1.32 W. [c] Since R1 and R2 carry the same current and R1 > R2 to satisfy the voltage requirement, first pick R1 to meet the 0.5 W specification: iR 1 =

160 − 66 , R1

Thus, R1 ≥

942 0.5



Therefore, or

94 R1

Thus, R2 = 12,408 Ω.

R1 ≤ 0.5;

R1 ≥ 17,672 Ω.

Now use the voltage specification: R2 (160) = 66; R2 + 17,672

2

3–16 P 3.14

CHAPTER 3. Simple Resistive Circuits [a]

40 kΩ + 60 kΩ = 100 kΩ; 25 kΩk100 kΩ = 20 kΩ; vo1 = vo =

20,000 (380) = 80 V; (75,000 + 20,000) 60,000 (vo1 ) = 48 V. (100,000)

[b]

i=

380 = 3.8 mA; 100,000

25,000i = 95 V; vo =

60,000 (95) = 57 V. 100,000

[c] It removes loading effect of second voltage divider on the first voltage divider. Observe that the open circuit voltage of the first divider is 0 vo1 =

25,000 (380) = 95 V. (100,000)

Now note this is the input voltage to the second voltage divider when the current controlled voltage source is used. P 3.15

[a] vo =

100R2 = 20 R1 + R2

Let Re = R2 kRL = vo =

100Re = 16 R1 + Re

so

R1 = 4R2

R2 RL R2 + RL so

R1 = 5.25Re

Problems

Then, 4R2 = 5.25Re = Thus, R2 = 15 kΩ

3–17

5.25(48R2 ) 48 + R2

and

R1 = 4(15k) = 60 kΩ

[b] The resistor that must dissipate the most power is R1 , as it has the largest resistance and carries the same current as the parallel combination of R2 and the load resistor. The power dissipated in R1 will be maximum when the voltage across R1 is maximum. This will occur when the voltage divider has a resistive load. Thus, vR1 = 100 − 16 = 84 V p R1 =

842 = 117.6 mW 60k

Thus the minimum power rating for all resistors should be 1/8 W. P 3.16

Refer to the solution to Problem 3.15. The voltage divider will reach the maximum power it can safely dissipate when the power dissipated in R1 equals 0.15 W. Thus, vR2 1 = 0.15 60,000

so

vR1 = 94.87 V.

vo = 100 − 94.87 = 5.13 V. So,

100Re = 5.13 60,000 + Re

Thus,

and

Re = 3245.55 Ω.

(15,000)RL = 3245.55 15,000 + RL

and

RL = 4141.69 Ω.

The minimum value for RL from Appendix H is 4.7 kΩ. P 3.17

[a] At no load:

vo = kvs =

At full load:

vo = αvs =

Therefore k

=

α

=



Thus

R2 vs . R1 + R2

1−α α



Re vs , R1 + Re

R2 R1 + R2 Re R1 + Re

and and

where Re = (1 − k) R2 . k (1 − α) R1 = Re . α

R1 =

R2 Ro (1 − k) = R2 . Ro + R2 k 

Ro R2 . Ro + R2

3–18

CHAPTER 3. Simple Resistive Circuits

Solving for R2 yields Also, [b] R1 R2

R2 =

(k − α) Ro . α(1 − k)

(1 − k) (k − α) R1 = R2 .·. Ro . k αk   0.25 Ro = 6.667 kΩ; = 0.375   0.25 = Ro = 20 kΩ. 0.125 R1 =

[c]

Maximum dissipation in R2 occurs at no load, therefore, PR2 (max) =

752 = 281.25 mW. 20,000

Maximum dissipation in R1 occurs at full load. PR1 (max) =

[100 − 0.5(100)]2 = 375 mW. 6666.67

[d]

P 3.18

PR1

=

PR2

=

(100)2 = 1.5 W = 1500 mW; 6666.67 (0)2 = 0 W. 20,000

(30)2 = 30, R1 + R2 + R3

Therefore, R1 + R2 + R3 = 30 Ω.

(R1 + R2 )30 = 15; (R1 + R2 + R3 ) Therefore, 2(R1 + R2 ) = R1 + R2 + R3 .

Problems Thus, R1 + R2 = R3 ;

2R3 = 30;

R3 = 15 Ω.

R2 (30) = 5; R1 + R2 + R3 6R2 = R1 + R2 + R3 = 30; Thus, R2 = 5 Ω; P 3.19

i1200Ω =

R1 = 30 − R2 − R3 = 10 Ω.

800k2400k1200 400 (0.32) = (0.32) = 106.667 mA; 1200 1200

P1200Ω = 1200i21200Ω = 1200(106.667 × 10−3 )2 = 13.653 W. P 3.20

[a]

Req = (10 + 40 + 20)k[12 + (20k180)] = 70k30 = 21 Ω; v12A = 12(21) = 252 V; vo = v40Ω = v20Ω = io =

40 (252) = 144 V; 10 + 40 + 20

20k180 18 (252) = (252) = 151.2 V; 12 + (20k180) 30

151.2 = 7.56 A. 20

(252 − 151.2)2 = 846.72 W. 12 [c] p12A = −(252)(12) = −3024 W. Thus the power developed by the current source is 3024 W. [b] p12Ω =

P 3.21

Begin by using the relationships among the branch currents to express all branch currents in terms of i4 : i1 = 2i2 = 2(2i3 ) = 4(2i4 ); i2 = 2i3 = 2(2i4 );

3–19

3–20

CHAPTER 3. Simple Resistive Circuits i3 = 2i4 . Now use KCL at the top node to relate the branch currents to the current supplied by the source. i1 + i2 + i3 + i4 = 1 mA. Express the branch currents in terms of i4 and solve for i4 : 1 mA = 8i4 + 4i4 + 2i4 + i4 = 15i4

so

i4 =

0.001 A. 15

Since the resistors are in parallel, the same voltage, 1 V appears across each of them. We know the current and the voltage for R4 so we can use Ohm’s law to calculate R4 : R4 =

vg 1V = = 15 kΩ. i4 (1/15) mA

Calculate i3 from i4 and use Ohm’s law as above to find R3 : i3 = 2i4 =

0.002 A 15

vg 1V .·. R3 = = = 7.5 kΩ. i3 (2/15) mA

Calculate i2 from i4 and use Ohm’s law as above to find R2 : i2 = 4i4 =

0.004 A 15

vg 1V .·. R2 = = = 3750 Ω. i2 (4/15) mA

Calculate i1 from i4 and use Ohm’s law as above to find R1 : i1 = 8i4 =

0.008 A 15

vg 1V .·. R1 = = = 1875 Ω. i1 (8/15) mA

The resulting circuit is shown below:

P 3.22

[a] Let vo be the voltage across the parallel branches, positive at the upper terminal, then ig = vo G1 + vo G2 + · · · + vo GN = vo (G1 + G2 + · · · + GN ). It follows that

vo =

ig . (G1 + G2 + · · · + GN )

The current in the k th branch is ig G k ik = . [G1 + G2 + · + GN ]

ik = v o G k ;

Thus,

Problems

[b] i6.25 = P 3.23

3–21

1.142(0.16) = 0.032 = 32 mA. [4 + 0.4 + 1 + 0.16 + 0.1 + 0.05]

[a] The equivalent resistance to the right of the 10 kΩ resistor is 3k + 8k + [6kk(5k + 7k)] = 15 kΩ. Therefore, 15kk10k 6k (0.002) = (0.002) = 1.2 mA. 10k 10k [b] The voltage drop across the 10 kΩ resistor can be found using Ohm’s law: i10k =

v10k = (10, 000)i10k = (10, 000)(0.0012) = 12 V. [c] The voltage v10k drops across the 3 kΩ resistor, the 8 kΩ resistor and the equivalent resistance of the 6 kΩ and the parallel branch containing the 5 kΩ and 7 kΩ resistors. Thus, using voltage division, v6k =

6kk(5k + 7k) 4 (12) = (12) = 3.2 V. 3k + 8k + [6kk(5k + 7k)] 15

[d] The voltage v6k drops across the branch containing the 5 kΩ and 7 kΩ resistors. Thus, using voltage division, 5k (3.2) = 1.33 V. v5k = 5k + 7k P 3.24

[a] The equivalent resistance of the resistors to the right of the 30 Ω resistor is Req = 6 + 1.2 + [64k(6 + 10)] = 7.2 + (64k16) = 7.2 + 12.8 = 20 Ω. Using current division, 30k20 (5) = 3 A. 20 [b] The current into the top right node is 3 A and divides to give iReq =

i10 = i10+6 = P 3.25

64k(10 + 6) (3) = 2.4 A. 10 + 6

[a] The equivalent resistance of the circuit to the right of the 60 kΩ resistor is 20 k + (200 kk50 k) = 20 k + 40 k = 60 kΩ. Thus by current division, 30 kk60 kk60 k 15 k (0.02) = (0.02) = 5 mA. 60 k 60 k [b] Using Ohm’s law: i60k =

v60 k = 60,000i60 k = 60,000(0.005) = 300 V. [c] The voltage across the 60 kΩ resistor divides between two resistors – the 20 kΩ resistor and the 200 kk50 k = 40 kΩ equivalent resistance. Using voltage division, 20 k 20 k (v60 k ) = (300) = 100 V. v20 k = 20 k + 40 k 60 k

3–22

CHAPTER 3. Simple Resistive Circuits [d] The current in the 20 kΩ resistor can be found using Ohm’s law: i20 k =

v20 k 100 = = 5 mA. 20 k 20,000

[e] The current i20 k divides between the 200 kΩ branch and the 50 kΩ branch. Using current division, i200 k = P 3.26

200 kk50 k 40 k i20 k = (0.005) = 1 mA. 200 k 200 k

[a] Begin by finding the equivalent resistance of the 120 Ω resistor and all resistors to its right: [(12k24) + (60k20) + 7]k120 = (8 + 15 + 7)k120 = 30k120 = 24 Ω. Now use voltage division to find the voltage across the 15 Ω resistor: v15 =

15 (128) = 30 V. 15 + 24 + 25

[b] Use Ohm’s law to find the current in the 15 Ω resistor: i15 = v15 /15 = 30/15 = 2 A. [c] From part (a), the equivalent resistance of all resistors to the right of the 120 Ω resistor is 30 Ω. Now use current division: 24 120k30 (2) = (2) = 0.4 A. i120 = 120 120 [d] Using Ohm’s law, v120 = 120i120 = 120(0.4) = 48 V. [e] The 48 V divides across the branch to the right of the 120 Ω resistor. This branch has a 12k24 = 8 Ω resistance, a 60k20 = 15 Ω resistance and a 7 Ω resistance. Use voltage division to find the voltage across the 60k20 Ω resistance: 60k20 15 v60k20 = (48) = (48) = 24 V. (12k24) + (60k20) + 7 8 + 15 + 7 [f ] i20 = P 3.27

v60k20 24 = = 1.2 A. 20 20

[a] The equivalent resistance to the right of the 16 Ω resistor is 14 + [5k10k15k10k(12 + 18)] = 14 + 2 = 16 Ω. By voltage division, v4 =

4 4 (144) = (144) = 24 V. 4 + (16k16) + 12 24

Problems

3–23

[b] Using Ohm’s law, v4 24 = = 6 A. 4 4 [c] From part (a), the resistance to the right of the 16 Ω resistor is also 16 Ω. Therefore, the current in the 4 Ω resistor divides equally between the two 16 Ω branches. The current in the 14 Ω resistor is i14 = 6/2 = 3 A. i4 =

[d] Using current division, 5k10k15k10k(12 + 18) 2 i14 = (3) = 0.2 A. 12 + 18 30

i18 = P 3.28

60k30 = 20 Ω; i30Ω =

50 + 25 (25) = 15 A; 30 + 20 + 50 + 75

v2 = (15)(20) = 300 V; v2 + 30i30 = 750 V; v1 − 12(25) = 750; v1 = 1050 V. P 3.29

[a] v6k =

6 (18) = 13.5 V; 6+2

v3k =

3 (18) = 4.5 V; 3+9

vx = v6k − v3k = 13.5 − 4.5 = 9 V. 6 [b] v6k = (Vs ) = 0.75Vs ; 8 v3k =

3 (Vs ) = 0.25Vs ; 12

vx = (0.75Vs ) − (0.25Vs ) = 0.5Vs . P 3.30

Use current division to find the current in the branch containing the 24 kΩ and 16 kΩ resistors, from top to bottom: i24k+16k =

(24 k + 16 k)k(20 k + 40 k) (0.05) = 0.03 = 30 mA. 24 k + 16 k

3–24

CHAPTER 3. Simple Resistive Circuits Use Ohm’s law to find the voltage drop across the 16 kΩ resistor, positive at the top: v16k = (0.03)(16 k) = 480 V. Find the current in the branch containing the 20 kΩ and 40 kΩ resistors, from top to bottom: i20k+40k = 0.05 − i24k+16k = 0.05 − 0.03 = 0.02 = 20 mA. Use Ohm’s law to find the voltage drop across the 40 kΩ resistor, positive at the top: v40k = (40 k)(0.02) = 800 V; vo = v16k − v40k = 480 − 800 = −320 V.

P 3.31

54 Ωk27 Ω = 18 Ω; Therefore, ig =

i2Ω = P 3.32

18 Ω + 2 Ω = 20 Ω;

20k(10 + 15 + 35) = 15 Ω;

675 = 15 A. 30 + 15

20k60 (15) = 11.25 A; 20

io =

27k54 (11.25) = 7.5 A. 27

[a] The equivalent resistance seen by the voltage source is 60k[8 + 30k(4 + 80k20)] = 60k[8 + 30k20] = 60k20 = 15 Ω. Thus, 300 = 20 A. 15 [b] Use current division to find the current in the 8 Ω division: ig =

15 (20) = 15 A. 20 Use current division again to find the current in the 30 Ω resistor: i30 =

12 (15) = 6 A. 30

Thus, p30 = (6)2 (30) = 1080 W.

Problems P 3.33

3–25

At full scale the voltage across the shunt resistor will be 50 mV; therefore the power dissipated will be PA =

(50 × 10−3 )2 ; RA

Therefore RA ≥

(50 × 10−3 )2 = 5 mΩ. 0.5

Otherwise the power dissipated in RA will exceed its power rating of 0.5 W. When RA = 5 mΩ, the shunt current will be iA =

50 × 10−3 = 10 A. 5 × 10−3

The measured current will be imeas = 10 + 0.001 = 10.001 A; .·. Full-scale reading for practical purposes is 10 A. P 3.34

Original meter:

Rs =

Modified meter:

Re =

50 × 10−3 = 0.005 Ω; 10 (0.015)(0.005) = 0.00375 Ω; 0.02

.·. (Ifs )(0.00375) = 50 × 10−3 ; .·. Ifs = 13.33 A. P 3.35

For all full-scale readings the total resistance is Rv + Rmovement =

full-scale reading . 10−3

We can calculate the resistance of the movement as follows: Rmovement = Therefore,

50 mV = 50 Ω. 1 mA Rv = 1000 (full-scale reading) − 50.

3–26

CHAPTER 3. Simple Resistive Circuits [a] Rv = 1000(100) − 50 = 99, 950 Ω; [b] Rv = 1000(5) − 50 = 4950 Ω; [c] Rv = 1000(0.1) − 50 = 50 Ω; 60k5000 (5) = 3.73878 V. 20 + (60k5000) 60 (5) = 3.75 V; [b] vtrue = 20 + 60   3.73878 % error = − 1 × 100 = −0.299%. 3.75

P 3.36

[a] vmeas =

P 3.37

[a] The model of the ammeter is an ideal ammeter in parallel with a resistor whose resistance is given by 100 mV = 50 Ω. 2 mA We can calculate the current through the real meter using current division: (25/12) 25 1 im = (imeas ) = (imeas ) = imeas . 50 + (25/12) 625 25 Rm =

[b] At full scale, imeas = 5 A and im = 2 mA so 5 − 0.002 = 4998 mA flows throught the resistor RA : 100 mV 100 = Ω. 4998 mA 4998

RA = im =

1 (100/4998) (imeas ) = (imeas ). 50 + (100/4998) 2500

[c] Yes P 3.38

The current in the shunt resistor at full-scale deflection is iA = ifullscale − 5 × 10−3 A. The voltage across RA at full-scale deflection is always 200 mV; therefore, RA =

200 × 10−3 200 = . −3 ifullscale − 5 × 10 1000ifullscale − 5

200 = 20.01 mΩ. 10,000 − 5 [b] Let Rm be the equivalent ammeter resistance:

[a] RA =

0.2 = 0.02 = 20 mΩ. 10 200 [c] RA = = 25.0156 mΩ. 8000 − 5 Rm =

Problems

[d] Rm = P 3.39

0.2 = 25 mΩ. 8

Rmeter = Rm + Rmovement =

750 V = 500 kΩ; 1.5 mA

vmeas = (25 kΩk125 kΩk500 kΩ)(30 mA) = (20 kΩ)(30 mA) = 600 V; vtrue = (25 kΩk125 kΩ)(30 mA) = (20.83 kΩ)(30 mA) = 625 V; 600 % error = − 1 100 = −4%. 625 

P 3.40



[a]

40 × 103 i1 + 60 × 103 (i1 − iB ) = 10; 60 × 103 (i1 − iB ) = 0.6 + 50iB (120); .·.

100i1 − 60iB = 10 × 10−3 ;

60i1 − 66iB = 0.6 × 10−3 . Calculator solution yields iB = 180 µA. [b] With the insertion of the ammeter the equations become 100i1 − 60iB = 10 × 10−3

(no change);

60 × 103 (i1 − iB ) = 103 iB + 0.6 + 50iB (120); 60i1 − 67iB = 0.6 × 10−3 ; Calculator solution yields iB = 174.2 µA.   174.2 [c] % error = − 1 100 = −3.22%. 180

3–27

3–28 P 3.41

CHAPTER 3. Simple Resistive Circuits The measured value is ig =

180 = 5.96 A; (20.22 + 10)

The true value is ig =

60k30.5 = 20.22 Ω. imeas =

60 (5.96) = 3.95 A. 90.5

60k30 = 20 Ω.

180 = 6 A; (20 + 10)

itrue =

60 (6) = 4 A; 90

3.95 − 1 × 100 = −1.28%. %error = 4 

P 3.42



Begin by using Ohm’s law to find the actual value of the current io : itrue =

5 = 62.5 mA; 20 + 60

imeas =

5 = 62.112 mA; 20 + 60 + 0.5

62.112 % error = − 1 100 = −0.621% ≈ −0.6%. 62.5 

P 3.43



From the problem statement we have 0.080 =

vs (10 × 106 ) ; 10 × 106 + Rs

0.072 =

vs (5 × 106 ) . 5 × 106 + Rs

In standard form, 10 × 106 vs − 0.08Rs = 0.08(10 × 106 ); 5 × 106 vs − 0.072Rs = 0.072(5 × 106 ). [a] Solving simultaneously on a calculator, vs = 90 mV. [b] Rs = 1250 kΩ.

Problems P 3.44

3–29

[a] vmeter = 150V. [b] Rmeter = (200)(500) = 100 kΩ; 100 kk60 k = 37.5 kΩ; vmeter =

37.5 k (150) = 83.33 V. 30 k + 37.5 k

[c] 30 kk100 k = 23.077 kΩ; vmeter = [d] vmeter

a

23.077 k (150) = 41.67 V. 23.077 k + 60 k

= 150 V;

vmeter b + vmeter c = 125 V. No, because of the loading effect. P 3.45

[a] Since the unknown voltage is greater than either voltmeter’s maximum reading, the only possible way to use the voltmeters would be to connect them in series. [b]

Rm1 = (300)(900) = 270 kΩ;

Rm2 = (150)(1200) = 180 kΩ;

.·. Rm1 + Rm2 = 450 kΩ. i1

max

=

300 × 10−3 = 1.11 mA; 270

i2

max

=

150 × 10−3 = 0.833 mA; 180

.·. imax = 0.833 mA since meters are in series. vmax = (0.833 × 10−3 )(270 + 180)103 = 375 V. Thus the meters can be used to measure the voltage. 320 [c] im = = 0.711 mA; 450 × 103 vm1 = (0.711)(270) = 192 V;

vm2 = (0.711)(180) = 128 V.

3–30 P 3.46

CHAPTER 3. Simple Resistive Circuits The current in the series-connected voltmeters is 205.2 136.8 = = 0.76 mA. 270,000 180,000

im =

v50 kΩ = (0.76 × 10−3 )(50,000) = 38 V; Vpower P 3.47

supply

= 205.2 + 136.8 + 38 = 380 V.

[a] Rmovement = 5 Ω; R1 + Rmovement =

50 = 25 kΩ 2 × 10−3

R2 + R1 + Rmovement =

.·. R1 = 24,995 Ω.

100 = 50 kΩ; 2 × 10−3

R3 + R2 + R1 + Rmovement =

.·. R2 = 25 kΩ.

200 = 100 kΩ; 2 × 10−3

.·. R3 = 50 kΩ. [b]

imove =

188 (2) = 1.88 mA; 200

v1 = (1.88)(50) = 94 V; i1 =

94 = 0.188 mA; 500

i2 = imove + i1 = 1.88 + 0.188 = 2.068 mA; vmeas = vx = 94 + 50i2 = 197.4 V. [c] v1

=

100 V;

i2 = 2 + 0.20 = 2.20 mA;

i1

=

100/500 = 0.20 mA;

vmeas = vx = 100 + 50(2.20) = 210 V.

Problems P 3.48

[a] R1

=

(50)/10−3 = 50 kΩ;

R2

=

(20)/10−3 = 20 kΩ;

R3

=

(2)/10−3 = 2 kΩ.

[b] Let ia

=

actual current in the movement;

id

=

design current in the movement;

ia Then % error = − 1 100. id 



For the 50 V scale: 50 50 ia = = , 50,000 + 50 50,050 ia 50,000 = = 0.9990; id 50,050

P 3.49

50 ; 50,000

% error = (0.9990 − 1)100 = −0.10%.

For the 20 V scale: ia 20,000 = = 0.9975; id 20,050 For the 2 V scale: ia 2000 = = 0.9756; id 2050

id =

% error = (0.9975 − 1.0)100 = −0.25%.

% error = (0.9756 − 1.0)100 = −2.44%.

[a] Rmeter = 300 kΩ + 600 kΩk200 kΩ = 450 kΩ; 450k360 = 200 kΩ; Vmeter =

200 (600) = 500 V. 240

[b] What is the percent error in the measured voltage? True value = 

% error =

360 (600) = 540 V; 400

500 − 1 100 = −7.41%. 540 

3–31

3–32 P 3.50

CHAPTER 3. Simple Resistive Circuits Since the bridge is balanced, we can remove the detector without disturbing the voltages and currents in the circuit.

It follows that i1 =

ig (R2 + Rx ) ig (R2 + Rx ) X ; = R1 + R2 + R3 + Rx R

i2 =

ig (R1 + R3 ) ig (R1 + R3 ) X ; = R1 + R2 + R3 + Rx R

v 3 = R 3 i1 = v x = i2 R x ; R3 ig (R2 + Rx ) Rx ig (R1 + R3 ) X X .·. = ; R R .·. R3 (R2 + Rx ) = Rx (R1 + R3 ). From which Rx = P 3.51

R2 R3 . R1

Note the bridge structure is balanced, that is 15 × 5 = 3 × 25, hence there is no current in the 5 kΩ resistor. It follows that the equivalent resistance of the circuit is Req = 750 + (15,000 + 3000)k(25,000 + 5000) = 750 + 11,250 = 12 kΩ. The source current is 192/12,000 = 16 mA. The current down through the branch containing the 15 kΩ and 3 kΩ resistors is i3k =

11,250 (0.016) = 10 mA; 18,000

.·. p3k = 3000(0.01)2 = 0.3 W.

Problems P 3.52

3–33

[a]

The condition for a balanced bridge is that the product of the opposite resistors must be equal: (1000)(750) = 1500 Ω. 500 [b] The source current is the sum of the two branch currents. Each branch current can be determined using Ohm’s law, since the resistors in each branch are in series and the voltage drop across each branch is 24 V: (500)(Rx ) = (1000)(750)

is =

so

Rx =

24V 24 V + = 28.8 mA. 500 Ω + 750 Ω 1000 Ω + 1500 Ω

[c] We can use Ohm’s law to find the current in each branch: ileft =

24 = 19.2 mA; 500 + 750

iright =

24 = 9.6 mA. 1000 + 1500

Now we can use the formula p = Ri2 to find the power dissipated by each resistor: p500 = (500)(0.0192)2 = 184.32 mW; p1000 = (1000)(0.0096)2 = 92.16 mW;

p750 = (750)(0.0192)2 = 276.18 mW; p1500 = (1500)(0.0096)2 = 138.24 mW.

Thus, the 750 Ω resistor absorbs the most power; it absorbs 276.48 mW of power. [d] From the analysis in part (c), the 1000 Ω resistor absorbs the least power; it absorbs 92.16 mW of power.

3–34 P 3.53

CHAPTER 3. Simple Resistive Circuits Redraw the circuit, replacing the detector branch with a short circuit.

15 kΩk3 kΩ = 2.5 kΩ; 9 kΩk45 kΩ = 7.5 kΩ; is =

50 = 5 mA; 10

v1 = 5(2.5) = 12.5 V; v2 = 5(7.5) = 37.5 V; i1 =

12.5 = 833.3 µA; 15

i2 =

37.5 = 4166.7 µA; 9

id = i1 − i2 = −3333.4 µA. P 3.54

In order that all four decades (1, 10, 100, 1000) that are used to set R3 contribute to the balance of the bridge, the ratio R2 /R1 should be set to 0.001.

P 3.55

Use the figure below to transform the Y to an equivalent ∆:

Ra =

(25)(100) + (25)(40) + (40)(100) 7500 = = 300 Ω; 25 25

Problems

Rb =

(25)(100) + (25)(40) + (40)(100) 7500 = = 187.5 Ω; 40 40

Rc =

(25)(100) + (25)(40) + (40)(100) 7500 = = 75 Ω. 100 100

3–35

Replace the Y with its equivalent ∆ in the circuit to get the figure below:

Find the equivalent resistance to the right of the 5 Ω resistor: 300k[(125k187.5) + (37.5k75)] = 75 Ω. The equivalent resistance seen by the source is thus 5 + 75 = 80 Ω. Use Ohm’s law to find the current provided by the source: is =

40 = 0.5 A. 80

Thus, the power associated with the source is Ps = −(40)(0.5) = −20 W. P 3.56

Use the figure below to transform the Y to an equivalent ∆:

Ra =

(25)(125) + (25)(37.5) + (37.5)(125) 8750 = = 233.33 Ω; 37.5 37.5

Rb =

(25)(125) + (25)(37.5) + (37.5)(125) 8750 = = 350 Ω; 25 25

3–36

CHAPTER 3. Simple Resistive Circuits

Rc =

(25)(125) + (25)(37.5) + (37.5)(125) 8750 = = 70 Ω. 125 125

Replace the Y with its equivalent ∆ in the circuit to get the figure below:

Find the equivalent resistance to the right of the 5 Ω resistor: 350k[(100k233.33) + (40k70)] = 75 Ω. The equivalent resistance seen by the source is thus 5 + 75 = 80 Ω. Use Ohm’s law to find the current provided by the source: is =

40 = 0.5 A. 80

Thus, the power associated with the source is Ps = −(40)(0.5) = −20 W. P 3.57

Use the figure below to transform the ∆ to an equivalent Y:

R1 =

(40)(25) = 9.756 Ω; 40 + 25 + 37.5

R2 =

(25)(37.5) = 9.1463 Ω; 40 + 25 + 37.5

R3 =

(40)(37.5) = 14.634 Ω. 40 + 25 + 37.5

Problems

3–37

Replace the ∆ with its equivalent Y in the circuit to get the figure below:

Find the equivalent resistance to the right of the 5 Ω resistor: (100 + 9.756)k(125 + 9.1463) + 14.634 = 75 Ω. The equivalent resistance seen by the source is thus 5 + 75 = 80 Ω. Use Ohm’s law to find the current provided by the source: is =

40 = 0.5 A. 80

Thus, the power associated with the source is Ps = −(40)(0.5) = −20 W. P 3.58

Begin by transforming the ∆-connected resistors (5 Ω, 10 Ω, 105 Ω) to Y-connected resistors. Both the Y-connected and ∆-connected resistors are shown below to assist in using Eqs. 3.15 – 3.17:

Now use Eqs. 3.15 – 3.17 to calculate the values of the Y-connected resistors: R1 =

(5)(10) = 5/12 Ω; 5 + 10 + 105

R3 =

(10)(105) = 8.75 Ω. 5 + 10 + 105

R2 =

(5)(105) = 4.375 Ω; 5 + 10 + 105

3–38

CHAPTER 3. Simple Resistive Circuits The transformed circuit is shown below:

The equivalent resistance seen by the 2 A source can be calculated by making series and parallel combinations of the resistors to the right of the 2 A source: Req = (28 + 8.75)k(20 + 5/12) + 4.375 = 36.75k(245/12) + 4.375 = 13.125 + 4.375 = 17.5 Ω. Therefore, the voltage v across the 2 A source is given by v = (17.5)(2) = 35 V. P 3.59

[a] After the 20 Ω—100 Ω—50 Ω wye is replaced by its equivalent delta, the circuit reduces to

Now the circuit can be reduced to

i=

96 (1000) = 240 mA; 400

io =

400 (240) = 96 mA. 1000

80 (240) = 48 mA, 400 [c] Now that io and i1 are known return to the original circuit [b] i1 =

Problems

v2 = (50)(0.048) + (600)(0.096) = 60 V; i2 =

v2 60 = = 600 mA. 100 100

[d] vg = v2 + 20(0.6 + 0.048) = 60 + 12.96 = 72.96 V; pg = −(vg )(1) = −72.96 W; Thus the current source delivers 72.96 W. P 3.60

[a] Replace the 60—120—20 Ω delta with a wye equivalent to get

is =

750 750 = = 10 A; 5 + (24 + 36)k(14 + 6) + 12 + 43 75

i1 =

15 (24 + 36)k(14 + 6) (10) = (10) = 2.5 A. 24 + 36 60

[b] io = 10 − 2.5 = 7.5 A; v = 36i1 − 6io = 36(2.5) − 6(7.5) = 45 V. v 45 = 7.5 + = 8.25 A. 60 60 = (750)(10) = 7500 W.

[c] i2 = io + [d] Psupplied P 3.61

[a] Convert the upper delta to a wye. R1 =

(80)(200) = 40 Ω; 400

R2 =

(80)(120) = 24 Ω; 400

3–39

3–40

CHAPTER 3. Simple Resistive Circuits

R3 =

(120)(200) = 60 Ω. 400

Convert the lower delta to a wye. R4 =

(60)(90) = 18 Ω; 300

R5 =

(60)(150) = 30 Ω; 300

R6 =

(90)(150) = 45 Ω. 300

Now redraw the circuit using the wye equivalents.

Rab = 2 + 40 +

(80)(120) + 30 = 42 + 48 + 30 = 120 Ω. 200

[b] When vab = 600 V: 600 = 5 A; ig = 120 (5)(80) i15 = = 2 A; 200 p15Ω = (4)(15) = 60 W. P 3.62

[a] Calculate the values of the Y-connected resistors that are equivalent to the 10 Ω, 40 Ω, and 50 Ω ∆-connected resistors: RX =

(10)(50) = 5 Ω; 10 + 40 + 50

RZ =

(10)(40) = 4 Ω. 10 + 40 + 50

RY =

(50)(40) = 20 Ω; 10 + 40 + 50

Replacing the R2 —R3 —R4 delta with its equivalent Y gives

Problems

3–41

Now calculate the equivalent resistance Rab by making series and parallel combinations of the resistors: Rab = 13 + 5 + [(8 + 4)k(20 + 4)] + 7 = 33 Ω. [b] Calculate the values of the ∆-connected resistors that are equivalent to the 10 Ω, 8 Ω, and 40 Ω Y-connected resistors: (10)(8) + (8)(40) + (10)(40) 800 RX = = = 100 Ω; 8 8 800 (10)(8) + (8)(40) + (10)(40) = = 80 Ω; RY = 10 10 (10)(8) + (8)(40) + (10)(40) 800 RZ = = = 20 Ω 40 40 . Replacing the R2 , R4 , R5 wye with its equivalent ∆ gives

Make series and parallel combinations of the resistors to find the equivalent resistance Rab : 100 Ωk50 Ω = 33.33 Ω;

80 Ωk4 Ω = 3.81 Ω;

.·. 20k(33.33 + 3.81) = 13 Ω; .·. Rab = 13 + 13 + 7 = 33 Ω. [c] Convert the delta connection R4 —R5 —R6 to its equivalent wye. Convert the wye connection R3 —R4 —R6 to its equivalent delta. P 3.63

Replace the upper and lower deltas with the equivalent wyes: R1U =

(10)(50) (50)(40) (10)(40) = 5 Ω; R2U = = 20 Ω; R3U = = 4 Ω; 100 100 100

3–42

CHAPTER 3. Simple Resistive Circuits

R1L =

(10)(60) (60)(30) (10)(30) = 6 Ω; R2L = = 18 Ω; R3L = = 3 Ω. 100 100 100

The resulting circuit is shown below:

Now make series and parallel combinations of the resistors: (4 + 6)k(20 + 32 + 20 + 18) = 10k90 = 9 Ω; Rab = 33 + 5 + 9 + 3 + 40 = 90 Ω. P 3.64

Subtracting Eq. 3.13 from Eq. 3.14 gives R1 − R2 = (Rc Rb − Rc Ra )/(Ra + Rb + Rc ). Adding this expression to Eq. 3.12 and solving for R1 gives R1 = Rc Rb /(Ra + Rb + Rc ). To find R2 , subtract Eq. 3.14 from Eq. 3.12 and add this result to Eq. 3.13. To find R3 , subtract Eq. 3.12 from Eq. 3.13 and add this result to Eq. 3.14. Using the hint, Eq. 3.14 becomes R1 + R3 =

Rb [(R2 /R3 )Rb + (R2 /R1 )Rb ] Rb (R1 + R3 )R2 = . (R2 /R1 )Rb + Rb + (R2 /R3 )Rb (R1 R2 + R2 R3 + R3 R1 )

Solving for Rb gives Rb = (R1 R2 + R2 R3 + R3 R1 )/R2 . To find Ra : First use Eqs. 3.15–3.17 to obtain the ratios (R1 /R3 ) = (Rc /Ra ) or Rc = (R1 /R3 )Ra and (R1 /R2 ) = (Rb /Ra ) or Rb = (R1 /R2 )Ra . Now use these relationships to eliminate Rb and Rc from Eq. 3.13. To find Rc , use Eqs. 3.15–3.17 to obtain the ratios Rb = (R3 /R2 )Rc and Ra = (R3 /R1 )Rc . Now use the relationships to eliminate Rb and Ra from Eq. 3.12. P 3.65

1 R1 = Ra R1 R2 + R2 R3 + R3 R1 1/G1 = (1/G1 )(1/G2 ) + (1/G2 )(1/G3 ) + (1/G3 )(1/G1 ) (1/G1 )(G1 G2 G3 ) G2 G3 = = . G1 + G2 + G3 G1 + G2 + G3 Similar manipulations generate the expressions for Gb and Gc . Ga

=

Problems P 3.66

[a]

va =

vin R4 ; Ro + R4 + ∆R

vb =

R3 vin ; R2 + R3

vo = va − vb =

R4 vin R3 − vin . Ro + R4 + ∆R R2 + R3

When the bridge is balanced, R3 R4 vin = vin ; Ro + R4 R2 + R3 .·.

R4 R3 = . Ro + R4 R2 + R3

Thus,

vo

R4 vin R4 vin − Ro + R4 + ∆R Ro + R4   1 1 − = R4 vin Ro + R4 + ∆R Ro + R4 R4 vin (−∆R) = (Ro + R4 + ∆R)(Ro + R4 ) −(∆R)R4 vin ≈ , since ∆R 0;

32Rf = 6600 − 660(−12)

[b] vo = −12 V; io +

so

Rf = 453.75 kΩ.

A KCL equation at the output node gives

−12 − 10 −12 + = 0; 33,000 453,750

.·. io = 412.12 µA. "

P 5.15

#

Rf Rf Rf vo = − (0.2) + (0.15) + (0.4) ; 4000 5000 20,000 −6 = −0.1 × 10−3 Rf ;

P 5.16

Rf = 60 kΩ;

.·. 0 ≤ Rf ≤ 60 kΩ.

[a]

150,000 =5 Ra

so

150,000 = 10 Rb 150,000 =3 Rc

Ra =

so

150,000 = 30 kΩ; 5

Rb =

so

Rc =

150,000 = 15 kΩ; 10

150,000 = 50 kΩ. 3

[b] −5va − 10(−3) − 3(4) = −5va + 18; −5va + 18 = −12 −5va + 18 = 12

so so

5va = −30

thus

va =

30 = 6 V; 5

− 5va = −6

thus

va =

6 = 1.2 V. 5

5–11

5–12

CHAPTER 5. The Operational Amplifier Thus, 1.2 V ≤ vb ≤ 6 V.

P 5.17

We want the following expression for the output voltage: vo = −(3va + 5vb + 4vc + 2vd ). This is an inverting summing amplifier, so each input voltage is amplified by a gain that is the ratio of the feedback resistance to the resistance in the forward path for the input voltage. Pick a feedback resistor with divisors of 3, 5, 4, and 2 – say 60 kΩ: vo = −



60,000 60,000 60,000 60,000 va + vb + vc + vd . Ra Rb Rc Rd 

Solve for each input resistance value to yield the desired gain: .·. Ra = 60,000/3 = 20 kΩ Rc = 60,000/4 = 15 kΩ; Rb = 60,000/5 = 12 kΩ

Rd = 60,000/2 = 30 kΩ.

Now create the 5 resistor values needed from the realistic resistor values in Appendix H. Note that Rb = 12 kΩ and Rc = 15 kΩ are already values from Appendix H. Create Rf = 60 kΩ by combining 27 kΩ and 33 kΩ in series. Create Ra = 20 kΩ by combining two 10 kΩ resistors in series. Create Rd = 30 kΩ by combining 18 kΩ and 12 kΩ in series. Of course there are many other acceptable possibilities. The final circuit is shown here:

P 5.18

[a] The circuit shown is a non-inverting amplifier. [b] We assume the op amp to be ideal, so vn = vp = 3 V. Write a KCL equation at vn : 3 3 − vo + = 0. 40,000 80,000 Solving, vo = 9 V.

P 5.19

[a] This circuit is an example of the non-inverting amplifier.

Problems

5–13

[b] Use voltage division to calculate vp : vp =

75,000 3vs vs = . 25,000 + 75,000 4

Write a KCL equation at vn = vp = 3vs /4: 3vs /4 3vs /4 − vo + = 0. 8000 32,000 Solving, vo = 12vs /4 + 3vs /4 = 3.75vs . [c] 3.75vs = 15 3.75vs = −9

so

vs = 4 V; vs = −2.4 V;

so

Thus, −2.4 V ≤ vs ≤ 4 V. P 5.20

[a] Write a node voltage equation at vn ; remember that for an ideal op amp, the current into the op amp at the inputs is zero: vn vn − vo + = 0. 4500 63,000 Solve for vo in terms of vn by multiplying both sides by 63,000 and collecting terms: 14vn + vn − vo = 0

so

vo = 15vn .

Now use voltage division to calculate vp . We can use voltage division because the op amp is ideal, so no current flows into the non-inverting input terminal and the 400 mV divides between the 15 kΩ resistor and the Rx resistor: vp =

Rx (0.400). 15,000 + Rx

Now substitute the value Rx = 60 kΩ: 60,000 (0.400) = 0.32 V. vp = 15,000 + 60,000 Finally, remember that for an ideal op amp, vn = vp , so substitute the value of vp into the equation for v0 vo = 15vn = 15vp = 15(0.32) = 4.8 V. [b] Substitute the expression for vp into the equation for vo and set the resulting equation equal to the positive power supply value: 0.4Rx vo = 15 15,000 + Rx

!

= 5;

15(0.4Rx ) = 5(15,000 + Rx ) so Rx = 75 kΩ.

5–14 P 5.21

CHAPTER 5. The Operational Amplifier [a] From Eq. 5.7, vo =

Rs + Rf vg Rs

So,

so

vo Rf =1+ = 5. vg Rs

Rf = 4. Rs

Thus,

Rs =

Rf 100,000 = = 25 kΩ. 4 4

[b] vo = 5vg . When vg = −3 V, vo = 5(−3) = −15 V. When vg = 2 V, vo = 5(2) = 10 V. The power supplies can be set at 10 V and −15 V. P 5.22

[a] From the equation for the non-inverting amplifier, Rs + Rf =4 Rs

so

Rs + Rf = 4Rs

and therefore

Rf = 3Rs .

Choose Rf = 30 kΩ and implement this choice from components in Appendix H by combining two 15 kΩ resistors in series. Choose Rs = Rg = 10 kΩ, which is a component in Appendix H. The resulting non-inverting amplifier circuit is shown here:

[b] vo = 4vg = 12

so

vo = 4vg = −12

vg = 3 V; so

vg = −3 V.

Therefore, −3 V ≤ vg ≤ 3 V. P 5.23

[a] This circuit is an example of a non-inverting summing amplifier.

Problems [b] Write a KCL equation at vp and solve for vp in terms of vs : vp − vs vp + 4 + = 0; 12,000 48,000 4vp − 4vs + vp + 4 = 0

so

vp = 4vs /5 − 4/5.

Now write a KCL equation at vn and solve for vo : vn − vo vn + =0 so vo = 5vn . 10,000 40,000 Since we assume the op amp is ideal, vn = vp . Thus, vo = 5(4vs /5 − 4/5) = 4vs − 4. [c] 4vs − 4 = 10

so

4vs − 4 = −10

vs = 3.5 V; so

vs = −1.5 V;

Thus, −1.5 V ≤ vs ≤ 3.5 V. P 5.24

[a]

vp − va vp − vb vp − vc + + = 0; Ra Rb Rc Rb Rc Ra Rc Ra Rb va + vb + vc , .·. vp = D D D where D = Rb Rc + Ra Rc + Ra Rb . vn vn − vo + = 0; 20,000 Rf !

Rf + 1 vn = vo . 20,000 Let

Rf + 1 = k. 20,000

vo = kvn = kvp ; kRb Rc kRa Rc kRa Rb va + vb + vc ; .·. vo = D D D kRb Rc = 4; D kRa Rc = 1; D Thus,

so Rc =

D ; 1000k

kRb (D/1000k) = 4 and Rb = 4000 = 4 kΩ; D

kRa Rb = 2; D

so Rb =

2D ; 1000k

5–15

5–16

CHAPTER 5. The Operational Amplifier kRc (2D/1000k) = 4 and Rc = 2000 = 2 kΩ; D

Thus,

.·. D = (4000)(2000) + (1000)(2000) + (1000)(4000) = 14 × 106 ; 4D (4)(14) × 106 .·. k = = = 7; Rb Rc (4)(2) × 106 .·.

Rf + 1 = 7, 20,000

Rf = 120 kΩ.

[b] vo = 4(0.75) + 1.0 + 2(1.5) = 7 V; vn = vo /7 = 1 V = vp ;

P 5.25

[a]

ia =

va − vp 0.75 − 1 = = −250 µA; 1000 1000

ib =

1−1 vb − vp = = 0 µA; 4000 4000

ic =

vc − vp 1.5 − 1 = = 250 µA. 2000 2000

vp − va vp − vb vp − vc vp + + + = 0; Ra Rb Rc Rg Rb Rc Rg Ra Rc Rg Ra Rb Rg .·. vp = va + vb + vc , D D D where D = Rb Rc Rg + Ra Rc Rg + Ra Rb Rg + Ra Rb Rc . vn − vo vn + = 0; Rs Rf vn

1 1 + Rs Rf

!

=

vo ; Rf

Rf .·. vo = 1 + vn = kvn , Rs 



Rf where k = 1 + . Rs 



vp = vn ; .·. vo = kvp ; or vo =

kRg Rb Rc kRg Ra Rc kRg Ra Rb va + vb + vc . D D D

Problems kRg Rb Rc =6 D

kRg Ra Rc =3 D

6 Rb = =2 .·. Ra 3 Since

kRg Ra Rb = 4; D

Rc 3 = = 0.75 Rb 4

Ra = 1 kΩ,

5–17

Rb = 2 kΩ

Rc 6 = = 1.5. Ra 4 and

Rc = 1.5 kΩ,

.·. D = [(2)(1.5)(3) + (1)(1.5)(3) + (1)(2)(3) + (1)(2)(1.5)] × 109 = 22.5 × 109 . k(3)(2)(1.5) × 109 = 6; 22.5 × 109 135 × 109 = 15 9 × 109

k=

Rf ; .·. 15 = 1 + Rs Rf = 14; Rs Rf = (14)(15,000) = 210 kΩ. [b] vo = 6(0.5) + 3(2.5) + 4(1) = 14.5 V; vn = vp =

P 5.26

vo 14.5 = = 0.967 V; k 15

is =

vn 0.967 = = 64.44 µA. 15,000 15,000

io =

vn − vo vo − = −4.46 mA. 210,000 3300

[a] This is a difference amplifier circuit. [b] Use Eq. 5.8 with Ra = 10 kΩ, Rb = 20 kΩ, Rc = 15 kΩ, Rd = 25 kΩ, and vb = 8 V: vo = [c]

Rd (Ra + Rb ) Rb 25(10 + 20) 20 vb − (8) − va = 15 − 2va . va = Ra (Rc + Rd ) Ra 10(15 + 25) 10

Rf 25,000(10,000 + Rf ) (8) − (1.625) 10,000(15,000 + 25,000) 10,000 =

5(10,000 + Rf ) 1.625Rf 3.375Rf − =5+ ; 10,000 10,000 10,000

5+

3.375Rf =9 10,000

5+

3.375Rf = −9 10,000

so so

Rf =

(9 − 5)10,000 = 11,852 Ω. 3.375

Rf < 0 which is not a possible solution.

5–18

P 5.27

CHAPTER 5. The Operational Amplifier

[a] vo =

33(100) Rd (Ra + Rb ) Rb va = vb − (0.90) − 4(0.45); Ra (Rc + Rd ) Ra 20(80)

vo = 1.8563 − 1.8 = 56.25 mV. (0.90)(33) = 371.25 mV; 80

[b] vn = vp = ia =

(450 − 371.25)10−3 = 3.9375 µA; 20 × 103

Ra =

va 450 × 10−3 = = 114.3 kΩ. ia 3.9375 × 10−6

[c] Rin b = Rc + Rd = 80 kΩ. P 5.28

vp =

20,000 (−4) = −0.8 V = vn ; 100,000

−0.8 + 4 −0.8 − vo + = 0; 2000 Rf .·. vo = 0.0016Rf − 0.8. vo = 20 V;

Rf = 13 kΩ;

vo = −20 V;

.·. Rf = 13 kΩ.

Rf ≥ 0,

But P 5.29

Rf = −12 kΩ.

vp = 1000ib 1000ib 1000ib − vo + − ia = 0; Ra Rf .·. .·.

1 1 + Ra Rf

1000ib 

1000ib

Rf 1+ Ra



!

− ia =

vo ; Rf

− R f ia = v o .

By hypopthesis, vo = 5000(ib − ia ). Therefore, Rf = 5 kΩ Rf 1000 1 + Ra 

(use two 10 kΩ resistors in parallel). 

= 5000

so

Ra = 1250 Ω.

To construct the 1250 Ω resistor, combine a 1.2 kΩ resistor in series with a parallel combination of two 100 Ω resistors.

Problems

P 5.30

vo =

5–19

Rd (Ra + Rb ) Rb va ; vb − Ra (Rc + Rd ) Ra

By hypothesis: Rb /Ra = 5;

Rc + Rd = 600 kΩ;

Rd (Ra + 5Ra ) =2 .·. Ra 600,000

Rd = 200 kΩ;

so

Rd (Ra + Rb ) = 2. Ra (Rc + Rd )

Rc = 400 kΩ.

Combine a 180 kΩ resistor and a 220 kΩ resistor in series for Rc ; combine two 100 kΩ resistors in series for Rd . Also, when vo = 0 we have vn − va vn + = 0; Ra Rb Ra .·. vn 1 + Rb 

ia =



= va ;

vn = (5/6)va .

va − (5/6)va 1 va = ; Ra 6 Ra

.·. Ra = 3 kΩ;

Rin =

va = 6Ra = 18 kΩ; ia

Rb = 15 kΩ.

Combine a 1.2 kΩ resistor and a 1.8 kΩ resistor in series for Ra ; use a 15 kΩ resistor for Rb . P 5.31

[a] Assume va is acting alone. Replacing vb with a short circuit yields vp = 0, therefore vn = 0 and we have 0 − va 0 − vo0 + + in = 0, in = 0. Ra Rb Therefore Rb vo0 va =− , vo0 = − va . Rb Ra Ra Assume vb is acting alone. Replace va with a short circuit. Now vb Rd vp = vn = ; Rc + Rd vn − vo00 vn + + in = 0, Ra Rb 

1 1 + Ra Rb

vo00 =



in = 0;

Rd vo00 vb − = 0; Rc + Rd Rb





Rb +1 Ra



vo = vo0 + vo00 =

Rd Rd vb = Rc + Rd Ra

Rd Ra







Ra + Rb vb ; Rc + Rd

Ra + Rb Rb vb − va . Rc + Rd Ra 



5–20

CHAPTER 5. The Operational Amplifier Rd [b] Ra



Ra + Rb Rc + Rd



=

Rb , Ra

Rd Ra = Rb Rc , When

Rd Ra



therefore Rd (Ra + Rb ) = Rb (Rc + Rd );

therefore

Ra + Rb Rc + Rd



=

Eq. 5.8 reduces to vo = P 5.32

Ra Rc = . Rb Rd

Rb . Ra Rb Rb Rb vb − va = (vb − va ). Ra Ra Ra

[a]

vp vp − vc vp − vd + + = 0; 60,000 20,000 30,000 .·. 6vp = 3vc + 2vd = 6vn . vn − va vn − vb vn − vo + + = 0; 10,000 15,000 375,000 .·. vo

=

63.5vn − 37.5va − 25vb

=

63.5[(1/2)vc + (1/3)vd ] − 37.5va − 25vb

=

63.5(0.1 + 0.2) − 37.5(0.4) − 25(0.8) = −15.95 V.

[b] vo = 31.75vc + 21.167(0.6) − 37.5(0.4) − 25(0.8); ±20 = 31.75vc − 22.3; .·. vc = 1332.28 mV

and

vc = 72.44 mV;

.·. 72.44 ≤ vc ≤ 1332.28 mV. P 5.33

Ra = Rc = 1.5 kΩ; Then, Acm =

Rb = 12 kΩ.

1.5Rd − 1.5(12) ; 1.5(1.5 + Rd )

Adm =

Rd (13.5) + 12(1.5 + Rd ) . 2(1.5)(1.5 + Rd )

Problems Set the CMRR equal to 100 and solve for Rd :

Adm CMRR = = Acm

Rd (13.5) + 12(1.5 + Rd ) 25.5Rd + 18 3(1.5 + Rd ) = 100. = 1.5Rd − 18 3Rd − 36 1.5(1.5 + Rd )

Therefore, Rd = 13.18 kΩ. Set the CMRR equal to −100 and solve for Rd : CMRR =

25.5Rd + 18 = −100. 3Rd − 36

Therefore, Rd = 11 kΩ. Thus, 11 kΩ ≤ Rd ≤ 13.18 kΩ. (24)(26) + (25)(25) = 24.98. (2)(1)(25) (1)(24) − 25(1) [b] Acm = = −0.04. 1(25) 24.98 = 624.50. [c] CMRR = 0.04

P 5.34

[a] Adm =

P 5.35

αRg vg [a] vp = αRg + (Rg − αRg )



vo

vn = vp = αvg vn − vg vn − vo + =0 R1 Rf (vn − vg )

Rf + vn − vo = 0 R1

α

vo

0.0

−15 V

α

vo

0.4 −7.8 V

0.1 −13.2 V

0.5

0.2 −11.4 V

0.6 −4.2 V

−9.6 V

0.7 −2.4 V

0.3

−6 V

α

=

Rf Rf 1+ αvg − vg ; R1 R1 

=

(αvg − vg )5 + αvg ;

=

[(α − 1)5 + α]vg ;

=

(6α − 5)vg

=

(6α − 5)(3) = 18α − 15. vo

0.8 −0.6 V 0.9

1.2 V

1.0

3V

5–21

5–22

CHAPTER 5. The Operational Amplifier

[b] Rearranging the equation for vo from (a) gives Rf Rf + 1 vg α + − vg . vo = R1 R1 







Therefore, 

slope =

Rf + 1 vg ; R1 

intercept = −



Rf vg . R1 

[c] Using the equations from (b), −8 =



Rf + 1 vg ; R1 

5=−



Rf vg . R1 

Solving, vg = −3 V; P 5.36

Rf = 5/3. R1

(400 × 10−3 )2 = 16 µW. 104   10 [b] v10 kΩ = (400) = 80 mV; 50 [a] p10 kΩ =

p10 kΩ =

(80 × 10−3 )2 = 0.64 µW. (10 × 103 )

pa 16 = = 25. pb 0.64 [d] Yes, the operational amplifier serves several useful purposes:

[c]

• First, it enables the source to control 25 times as much power delivered to the load resistor. When a small amount of power controls a larger amount of power, we refer to it as power amplification. • Second, it allows the full source voltage to appear across the load resistor, no matter what the source resistance. This is the voltage follower function of the operational amplifier.

Problems

5–23

• Third, it allows the load resistor voltage (and thus its current) to be set without drawing any current from the input voltage source. This is the current amplification function of the circuit. P 5.37

[a] vp = vs ,

vn =

R1 vo , R1 + R2 

Therefore vo =

vn = vp .

R1 + R2 R2 vs = 1 + vs . R1 R1 





[b] vo = vs . [c] Because vo = vs , thus the output voltage follows the signal voltage. P 5.38

[a]

vn − va vn − vo + = 0; R R 2vn − va = vo ; va − vn va − vo va + + = 0; Ra R R 

va

1 2 vn vo + − = ; Ra R R R



va



R 2+ − vn = vo Ra 

vn = vp = va + vg ; .·. 2vn − va = 2va + 2vg − va = va + 2vg ; .·. va − vo = −2vg

(1).

5–24

CHAPTER 5. The Operational Amplifier 

2va + va

R − va − vg = vo ; Ra 

R .·. va 1 + − vo = vg . Ra 



(2)

Now combining equations (1) and (2) yields −va

R = −3vg , Ra

or va = 3vg

Ra . R

Hence ia =

va 3vg = Ra R

Q.E.D.

[b] At saturation vo = ± Vcc , .·. va = ± Vcc − 2vg

(3)

and R .·. va 1 + Ra 



= ± Vcc + vg .

(4)

Dividing Eq (4) by Eq (3) gives 1+

R ± Vcc + vg = ; Ra ± Vcc − 2vg

.·.

± Vcc + vg 3vg R = −1= , Ra ± Vcc − 2vg ± Vcc − 2vg

or Ra = P 5.39

(± Vcc − 2vg ) R 3vg

Q.E.D.

[a] Assume the op-amp is operating within its linear range, then iL =

3 = 2 mA. 1.5

For RL = 2.5 kΩ

vo = (2.5 + 1.5)(2) = 8 V.

Now since vo < 9 V our assumption of linear operation is correct, therefore iL = 2 mA. [b] 9 = 2(1.5 + RL );

RL = 3 kΩ.

Problems

5–25

[c] As long as the op-amp is operating in its linear region iL is independent of RL . From (b) we found the op-amp is operating in its linear region as long as RL ≤ 3 kΩ. Therefore when RL = 6.5 kΩ the op-amp is saturated. We can estimate the value of iL by assuming ip = in  iL . Then iL = 9/(1.5 + 6.5) = 1.125 mA. To justify neglecting the current into the op-amp assume the drop across the 47 kΩ resistor is negligible, and the input resistance to the op-amp is at least 500 kΩ. Then ip = in = (3 − 1.5)/500,000 = 3 µA. But 3 µA  1.125 mA, hence our assumption is reasonable. [d]

P 5.40

vp =

5.6 vg = 0.7vg = 2.8 sin(5π/3)t V; 8.0

vn vn − vo + = 0; 2000 18,000 10vn = vo ;

vn = vp ;

.·. vo = 28 sin(5π/3)t V vo = 0

0 ≤ t ≤ ∞.

t ≤ 0.

At saturation 5π 28 sin t = ±14; 3 



π 5π .·. t= , 3 6 t = 0.10 s,

5π , 6

0.50 s,

sin

5π t = ±0.5; 3

7π , 6

11π , 6

0.70 s,

etc.

etc.;

5–26 P 5.41

CHAPTER 5. The Operational Amplifier It follows directly from the circuit that vo = −5vg From the plot of vg we have vg = 0, t < 0. vg

=

4t

0 ≤ t ≤ 0.5;

vg

=

4 − 4t

0.5 ≤ t ≤ 1.5;

vg

=

4t − 8

1.5 ≤ t ≤ 2.5;

vg

=

12 − 4t 2.5 ≤ t ≤ 3.5;

vg = 4t − 16 3.5 ≤ t ≤ 4.5, etc. Therefore vo = −20t 0 ≤ t ≤ 0.5; vo

=

20t − 20

0.5 ≤ t ≤ 1.5;

vo

=

40 − 20t

1.5 ≤ t ≤ 2.5;

vo

=

20t − 60

2.5 ≤ t ≤ 3.5;

vo = 80 − 20t 3.5 ≤ t ≤ 4.5, etc. These expressions for vo are valid as long as the op amp is not saturated. Since the peak values of vo are ±6, the output is clipped at ±6. The plot is shown below.

Problems

5–27

P 5.42

i1 =

14.7 − 10 = 1 mA; 4700

i2 + i1 + 0 = 5 mA;

i2 = 4 mA;

vo2 = 10 + (1500)(0.004) = 16 V; i3 =

14.7 − 16 = −0.1 mA; 13,000

i4 = i3 + i1 = 0.9 mA; vo1 = 14.7 + 3000(0.0009) = 17.4 V. P 5.43

[a] Let vo1 = output voltage of the amplifier on the left. Let vo2 = output voltage of the amplifier on the right. Then −90 (−0.5) = 3 V; 15 vo2 − vo1 ia = = −4.6 mA. 1000 vo1 =

vo2 =

−120 (0.4) = −1.6 V; 30

[b] ia = 0 when vo1 = v02 so from (a) vo2 = 3 V. Thus −120 (vright ) = 3; 30 90 vright = − = −750 mV. 120

5–28 P 5.44

CHAPTER 5. The Operational Amplifier From Eq. 5.28, 1 vref 1 1 = vn + + R + ∆R R + ∆R R − ∆R Rf

!

−

vo . Rf

Substituting Eq. 5.30 for vp = vn : 



1 1 + R−∆R + R1f vref R+∆R vref  = 1 1 R + ∆R + R−∆R + (R − ∆R) R+∆R

1 Rf



−

vo . Rf

Rearranging, 1 vo 1 = vref − . Rf R − ∆R R + ∆R 



Thus, !

vo = vref P 5.45

2∆R Rf . 2 R − (∆R)2

[a] Replace the op amp with the model from Fig. 5.18:

Write two node voltage equations, one at the left node, the other at the right node: vn − vg vn − vo vn + + = 0; 16,000 24,000 480,000 vo + 105 vn vo − vn vo + + = 0. 2000 24,000 400 Simplify and place in standard form: 51vn − 20vo = 30vg ; (12 × 105 − 1)vn + 73vo = 0. Let vg = 1 V and solve the two simultaneous equations: vo = −1.499767 V;

vn = 91.236 µV.

Thus the voltage gain is vo /vg = −1.499767.

Problems

5–29

[b] From the solution in part (a), vn = 91.236 µV. [c] ig =

vg − vn 1 − 91.236 × 10−6 = ; 16,000 16,000

Rg =

vg 16,000 = = 16,001.46 Ω. ig 1 − 91.236 × 10−6

[d] For an ideal op amp, the voltage gain is the ratio between the feedback resistor and the input resistor: vo 24,000 =− = −1.5. vg 16,000 For an ideal op amp, the difference between the voltages at the input terminals is zero, and the input resistance of the op amp is infinite. Therefore, vn = vp = 0 V; P 5.46

Rg = 16 kΩ.

[a]

vn − vg vn − vo + = 0; 25,000 150,000 .·. vo = 7vn − 6vg . Also

vo = A(vp − vn ) = −Avn ;

−vo .·. vn = ; A 7 .·. vo 1 + = −6vg ; A 

vo =



−6A vg . (7 + A)

−6(150)(0.5) = −2.866 V. (7 + 150) [c] vo = −6(0.5) = −3 V.

[b] vo =

5–30

CHAPTER 5. The Operational Amplifier

[d] −2.94 =

−6(0.5)A 7+A

.·. A = 343. P 5.47

[a]

vn vn − vg vn − vo + + =0 8000 600,000 240,000

or

78.5vn − 2.5vo = vg ;

vo vo − vn vo − 100,000(vp − vn ) + + = 0; 30,000 240,000 5000 57vo − vn − 48 × 105 (vp − vn ) = 0; vp = vg +

(vn − vg )(160) = (11/15)vg + (4/15)vn ; 600

57vo − vn − 48 × 105 [(11/15)vg − (11/15)vn ] = 0; 57vo + 3,520,000vn = 3,520,000vg ; ∆=

78.5 3.52 × 106

−2.5 57

3.52 × 106 vg

No =

78.5 3.52 × 106

vo =

No = 30.98vg ; ∆

[b] N1 =

vn =

= 8,804,474.5;

vg 3.52 × 106 vg

vg

= 272.8 × 106 vg ;

vo = 30.98. vg

−2.5 57

N1 = 0.9995vg ; ∆

= 8,800,057vg ;

vn = 999.5 mV.

vp = (11/15)(1000) + (4/15)(999.5) = 999.87 mV. [c] vp − vn = 367.94 µV. (1000 − 999.87)10−3 = 836.22 pA. 160 × 103 vg vg − vo + = 0, since vn = vp = vg ; [e] 8 240 vo .·. vo = 31vg , = 31. vg

[d] ig =

vn = vp = 1 V;

vp − vn = 0 V;

ig = 0 A.

Problems P 5.48

[a]

vn vn − vTh vn − 0.4 + + = 0; 5000 500,000 100,000 vTh + 3 × 105 vn vTh − vn + = 0. 5000 100,000 Solving, vTh = −7.9994 V. Short-circuit current calculation:

vn − 0.4 vn − 0 vn + + = 0; 500,000 5000 100,000 .·. vn = 0.37736 V. isc =

vn 3 × 105 − vn = −22.64 A; 100,000 5000

RTh =

vTh = 353.33 mΩ. isc

5–31

5–32

CHAPTER 5. The Operational Amplifier [b] The output resistance of the inverting amplifier is the same as the Th´evenin resistance, i.e., Ro = RTh = 353.33 mΩ. [c]



vo =

500 (−7.9994) = −7.99375 V. 500.35333 

vn vn + 7.99375 vn − 0.4 + + = 0; 5000 500,000 100,000 .·. vn = 294.811 µV. ig =

0.4 − 294.811 × 10−6 = 79.94 µA; 5000

Rg = P 5.49

0.4 = 5003.7 Ω. ig

[a] vTh = −

100,000 (0.4) = −8 V; 5000

RTh = 0, since op-amp is ideal.

Problems

5–33

[b] Ro = RTh = 0 Ω. [c] Rg = 5 kΩ P 5.50

since

vn = 0.

[a] Use the approximation for Eq. 5.31 to solve for Rf ; note that since we are using 2% strain gages, δ = ∆R/R = 0.02: Rf =

vo R (4)(150) = = 1250 Ω. 2δvref (2)(0.02)(12)

[b] Now solve for δ given vo = 30 mV: δ=

(0.03)(150) vo R = = 150 × 10−6 . 2Rf vref 2(1250)(12)

The change in strain gage resistance that corresponds to a 30 mV change in output voltage is thus (δ)R = (150 × 10−6 )(150) = 22.5 mΩ. P 5.51

[a]

Let R1 = R + ∆R vp vp − vin vp + + = 0; Rf R R1 .·. vp

"

#

1 1 1 vin + + = ; Rf R R1 R1

.·. vp =

RRf vin = vn . RR1 + Rf R1 + Rf R

vn vn − vin vn − vo + + = 0; R R Rf "

vn

#

1 1 1 vo vin + + − = ; R R Rf Rf R "

#

R + 2Rf vin vo .·. vn − = RRf R Rf ; .·.

"

vo R + 2Rf = Rf RRf

#"

#

RRf vin vin − ; RR1 + Rf R1 + Rf R R

5–34

CHAPTER 5. The Operational Amplifier .·.

"

#

R + 2Rf 1 vo = − vin ; Rf RR1 + Rf R1 + Rf R R

[R2 + 2RRf − R1 (R + Rf ) − RRf ]Rf .·. vo = vin . R[R1 (R + Rf ) + RRf ] Now substitute R1 = R + ∆R and get vo =

−∆R(R + Rf )Rf vin . R[(R + ∆R)(R + Rf ) + RRf ]

If ∆R  R Rf (R + Rf )(−∆R)vin vo ≈ . R2 (R + 2Rf ) [b] vo ≈

350 × 103 (365 × 103 )(−150)12 ≈ −1.42937 V. (15 × 103 )2 (715 × 103 )

[c] vo =

−∆R(R + Rf )Rf vin . R[(R + ∆R)(R + Rf ) + RRf ] =

−(150)(365,000)(350,000)(12) 15,000[(15,150)(365,000) + (15,000)(350,000)]

= −1.422111 V. P 5.52

[a] vo ≈

(R + Rf )Rf (−∆R)vin ; R2 (R + 2Rf )

vo =

(R + Rf )(−∆R)Rf vin ; R[(R + ∆R)(R + Rf ) + RRf ]

R[(R + ∆R)(R + Rf ) + RRf ] approx value .·. = ; true value R2 (R + 2Rf ) R[(R + ∆R)(R + Rf ) + RRf ] − R2 (R + 2Rf ) .·. Error = R2 (R + 2Rf ) =

∆R (R + Rf ) . R (R + 2Rf )

∆R(R + Rf ) .·. % error = × 100. R(R + 2Rf ) [b] % error =

150(365,000) × 100 = 0.51%. 15,000(715,000)

Problems

P 5.53

1=

∆R(365,000) × 100; 15,000(715,000)

.·. ∆R = 293.8356 Ω; 293.8356 × 100 ≈ 1.96%. .·. % change in R = 15,000 P 5.54

[a] It follows directly from the solution to Problem 5.51 that vo =

[R2 + 2RRf − R1 (R + Rf ) − RRf ]Rf vin . R[R1 (R + Rf ) + RRf ]

Now R1 = R − ∆R. Substituting into the expression gives vo =

(R + Rf )Rf (∆R)vin . R[(R − ∆R)(R + Rf ) + RRf ]

Now let ∆R  R and get vo ≈

(R + Rf )Rf ∆Rvin . R2 (R + 2Rf )

[b] It follows directly from the solution to Problem 5.51 that R[(R − ∆R)(R + Rf ) + RRf ] approx value = . .·. true value R2 (R + 2Rf ) (R − ∆R)(R + Rf ) + RRf − R(R + 2Rf ) .·. Error = R(R + 2Rf ) =

−∆R(R + Rf ) . R(R + 2Rf )

−∆R(R + Rf ) .·. % error = × 100. R(R + 2Rf ) [c] R − ∆R = 14,820 Ω

.·. ∆R = 15,000 − 14,820 = 180 Ω;

(365,000)(350,000)(180)(12) .·. vo ≈ ≈ 1.715 V; 15,0002 (715,000) [d] % error =

−180(365,000)(100) = −0.6126%. 15,000(715,000)

5–35

Inductance, Capacitance, and Mutual Inductance

Assessment Problems AP 6.1 [a] v = L

dig ; dt

dig = 18[t(−10e−10t ) + e−10t ] = 18e−10t (1 − 10t); dt v = (50 × 10−6 )(18)e−10t (1 − 10t) = 0.9e−10t (1 − 10t) mV,

t ≥ 0.

[b] p = vig . v(200 ms) = (0.9 × 10−3 )e−2 (1 − 2) = −121.8 µV. ig (200 ms) = 18(0.2)e−2 = 487.2 mA. p(200 ms) = (−121.8 × 10−6 )(487.2 × 10−3 ) = −59.34 µW. [c] delivering. 1 1 [d] w = Li2g = (50 × 10−6 )(487.2 × 10−3 )2 = 5.93 µJ. 2 2 [e] The energy is a maximum where the current is a maximum: dig = 18e−10t (1 − 10t); dt dig = 0 when t = 0.1 s. dt ig(max) = 18(0.1)e−1 = 662.2 mA; 1 wmax = (50 × 10−6 )(662.2 × 10−3 )2 = 10.96 µJ. 2 6–1

6–2

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance

AP 6.2 [a] i = C

dv = (5 × 10−6 )[500t(−2500)e−2500t + 500e−2500t ] dt = 2.5 × 10−3 e−2500t (1 − 2500t) A.

[b] v(100 × 10−6 ) = 500(100 × 10−6 )e−0.25 = 38.94 mV; i(100 × 10−6 ) = (2.5 × 10−3 )e−0.25 (1 − 0.25) = 1.46 mA; p(100 × 10−6 ) = vi = (38.94 × 10−3 )(1.46 × 10−3 ) = 56.86 µW. [c] p > 0, so the capacitor is absorbing power. AP 6.3 [a] v(100 × 10−6 ) = 38.94 mV; 1 1 w = Cv 2 = (5 × 10−6 )(38.94 × 10−3 )2 = 3.79 nJ. 2 2 [b] The energy is maximum when the voltage is maximum: dv = 0 when (1 − 2500t) = 0 or t = 0.4 ms; dt vmax = 500(0.4 × 10−3 )e−1 = 73.58 mV; 1 2 pmax = Cvmax = 13.53 nJ. 2 (24)(48) = 16 mH. 72 [b] i(0+ ) = −0.6 + 0.2 = −0.4 A. 1 Zt [c] i = (0.1e−25x ) dx − 0.4 = (−250e−25t − 150) mA. 0.016 0+ 1 Zt [d] i1 = (0.1e−25x ) dx − 0.6 = (−83.33e−25t − 516.67) mA; + 0.048 0

AP 6.4 [a] Leq =

i2 =

1 Zt (0.1e−25x ) dx + 0.2 = (−166.67e−25t + 366.67) mA; + 0.024 0

i1 + i2 = i. (100)(25) = 20 µF. 125 [b] v(0+ ) = 15 − 10 = 5 V. Z t 1 [c] v(t) = (0.05e−200x ) dx + 5 = (−12.5e−200t + 17.5) V. 20 × 10−6 0+

AP 6.5 [a] Ceq =

Problems Z t 1 [d] v1 = 0.05e−200x dx + 15 = (−2.5e−200t + 17.5) V; −6 + 100 × 10 0 Z t 1 0.05e−200x dx − 10 = −10e−200t V; 25 × 10−6 0+

v2 =

v1 (∞) = 17.5 V,

v2 (∞) = 0 V;

1 w = (100 × 10−6 )(17.5)2 = 15.3125 mJ. 2 AP 6.6 [a] vg

=

5(ig − i1 ) + 20(i2 − i1 ) + 60i2

=

5(16 − 16e−5t − 4 − 64e−5t + 68e−4t ) + 20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t ) + 60(1 − 52e−5t + 51e−4t ) 60 + 5780e−4t − 5840e−5t V.

=

[b] vg (0) = 60 + 5780 − 5840 = 0 V. [c] pdev

= v g ig 960 + 92,480e−4t − 94,400e−5t − 92,480e−9t

=

+ 93,440e−10t W. [d] pdev (∞) = 960 W. [e] i1 (∞) = 4 A;

i2 (∞) = 1 A;

ig (∞) = 16 A;

p5Ω = (16 − 4)2 (5) = 720 W; p20Ω = 32 (20) = 180 W; p60Ω = 12 (60) = 60 W; X

.·.

pabs = 720 + 180 + 60 = 960 W; X

pdev =

X

pabs = 960 W.

22.8 M =√ = 0.95. L1 L2 576 √ = 576 = 24 mH.

AP 6.7 [a] k = √ [b] Mmax

L1 N 2 P1 N1 [c] = 12 = L2 N2 P2 N2 

2

;

N1 2 60 = = 6.25. N2 9.6 N1 √ = 6.25 = 2.5. N2

.·.





6–3

6–4

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance

Problems P 6.1

[a] 0 ≤ t ≤ 1 ms : i=

106 Z t 1Zt vs dx + i(0) = 6 × 10−3 dx + 0 L 0 300 0

= 20x

t =

20t A.

0

1 ms ≤ t ≤ 2 ms : 106 Z t i= (12 × 10−3 − 6x) dx + 20 × 10−3 ; −3 300 10 .·. i = 40t − 10,000t2 − 10 × 10−3 A. 2 ms ≤ t ≤ ∞ : i=

106 Z t (0) dx + 30 × 10−3 = 30 mA. 300 2×10−3

[b]

P 6.2

[a] i = 8e−300t − 8e−1200t A; v=L

di = −9.6e−300t + 38.4e−1200t V, dt

t > 0+ ;

v(0+ ) = −9.6 + 38.4 = 28.8 V. [b] v = 0 when 38.4e−1200t = 9.6e−300t −1500t

or t = (ln 4)/900 = 1.54 ms.

[c] p = vi = 384e − 76.8e − 307.2e−2400t W. dp [d] = 0 when e1800t − 12.5e900t + 16 = 0. dt Let x = e900t x = 1.44766,

−600t

and solve the quadratic x2 − 12.5x + 16 = 0. t=

ln 1.45 = 411.05 µs; 900

ln 11.05 = 2.67 ms; 900 p is maximum at t = 411.05 µs. x = 11.0523,

t=

Problems

6–5

[e] pmax = 384e−1.5(0.41105) − 76.8e−0.6(0.41105) − 307.2e−2.4(0.41105) = 32.72 W. [f ] w is max when i is max, i is max when di/dt is zero. When di/dt = 0, v = 0, therefore t = 1.54 ms. [g] imax = 8[e−0.3(1.54) − e−1.2(1.54) ] = 3.78 A; wmax = (1/2)(4 × 10−3 )(3.78)2 = 28.6 mJ. P 6.3

[a] i(0) = A1 + A2 = 1; di = −2000A1 e−2000t − 8000A2 e−8000t ; dt v = −30A1 e−2000t − 120A2 e−8000t V; v(0) = −30A1 − 120A2 = 60. Solving, A1 = 2

and A2 = −1.

Thus, i = (2e−2000t − e−8000t ) A v = 0.015

t ≥ 0;

di = −60e−2000t + 120e−8000t V, dt

t ≥ 0.

[b] p = vi = 300e−10,000t − 120e−4000t − 120e−16,000t ; p=0

when

Let x = e6000t ;

300e6000t − 120e12,000t − 120 = 0. then

300x − 120x2 − 120 = 0.

Thus x2 − 2.5x + 1 = 0

so

x = 0.5 and x = 2.

If x = e6000t = 0.5, t will be negative. Hence, the solution for t > 0 must be x = 2: e6000t = 2 Thus, t = P 6.4

so

6000t = ln 2.

ln 2 = 115.52 µs. 6000

[a] From Problem 6.3 we have i = A1 e−2000t + A2 e−8000t A; v = −30A1 e−2000t − 120A2 e−8000t V; i(0) = A1 + A2 = 1; v(0) = −30A1 − 120A2 = −300.

6–6

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance A1 = −2;

Solving, Thus,

A2 = 3.

i = −2e−2000t + 3e−8000t A t ≥ 0; v = 0.015

di = 60e−2000t − 360e−8000t V t ≥ 0. dt

[b] i = 0 when 3e−8000t = 2e−2000t ; .·. e6000t = 1.5 so t = (ln 1.5)/6000 = 67.58 µs. Thus, i > 0 for 0 ≤ t ≤ 67.58 µs

i < 0 for 67.58 µs ≤ t < ∞;

and

v = 0 when 60e−2000t = 360e−8000t ; .·. t = (ln 6)/6000 = 298.63 µs. Thus, v < 0 for 0 ≤ t ≤ 298.63 µs

and

v > 0 for 298.63 µs ≤ t < ∞.

Therefore, p < 0 for 0 ≤ t ≤ 67.58 µs

and

298.63 µs ≤ t < ∞

(inductor delivers energy); p > 0 for 67.58 µs ≤ t ≤ 298.63 µs (inductor stores energy). [c] p = vi = 900e−10,000t − 120e−4000t − 1080e−16,000t W; .·. wstored =

Z t2

p dx + w(0).

t1

"

wstored =

t2 −3 −10,000x 10 −90e t1

t2 −4000x +30e t1

t2 # +67.5e−16,000x + 7.5 × 10−3 t1

= 30e−4000t2 + 67.5e−16,000t2 − 90e−10,000t2 + 90e−10,000t1 − 30e−4000t1 − 67.5e−16,000t1 + 7.5 mJ where t1 = 67.58 µs .·. wstored = 5.11 + 7.5 = 12.61 mJ.

and

t2 = 298.63 µs;

Problems Z t1

wextracted = =

Z t1

p dt +

0

Z ∞

p dt

t2

[900e−10,000x − 120e−4000x − 1080e−16,000x ] dx

0

Z ∞

+

[900e−10,000x − 120e−4000x − 1080e−16,000x ] dx

t2

t1 −10,000x −90e

= 10−3

0

−3

−10

−10,000x

90e

t1 −4000x +30e 0

∞

−4000x

+30e

t2

t1 ! +67.5e−16,000x 0

∞

−16,000x

+67.5e

t2

∞ ! t2

= 90e−10,000t2 − 30e−4000t2 − 67.5e−16,000t2 + 30e−4000t1 + 67.5e−16,000t1 − 90e−10,000t1 − 7.5 mJ where t1 = 67.58 µs

and

t2 = 298.63 µs;

.·. wextracted = −12.61 mJ. Thus, the energy stored equals the energy extracted. P 6.5

i = (B1 cos 5t + B2 sin 5t)e−t ; i(0) = B1 = 25 A; di = (B1 cos 5t + B2 sin 5t)(−e−t ) + e−t (−5B1 sin 5t + 5B2 cos 5t) dt = [(5B2 − B1 ) cos 5t − (5B1 + B2 ) sin 5t]e−t ; v=2

di = [(10B2 − 2B1 ) cos 5t − (10B1 + 2B2 ) sin 5t]e−t ; dt

v(0) = 100 = 10B2 − 2B1 = 10B2 − 50

.·. B2 = 150/10 = 15 A.

Thus, i = (25 cos 5t + 15 sin 5t)e−t A, v = (100 cos 5t − 280 sin 5t)e−t V, i(0.5) = −6.70 A;

t ≥ 0; t ≥ 0;

v(0.5) = −150.23 V;

p(0.5) = (−6.70)(−150.23) = 1007.00 W absorbing.

6–7

6–8

P 6.6

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance

[a] vL = L

di = [56 cos 140t + 92 sin 140t]e−20t mV; dt

dvL .·. = [11,760 cos 140t − 9680 sin 140t]e−20t mV/s. dt dvL = 0 when dt .·. t = 6.30 ms.

tan 140t =

Also 140t = 0.8821 + π

11,760 = 1.21; 9680

etc.

Because of the decaying exponential vL will be maximum the first time the derivative is zero. [b] vL (max) = [56 cos 0.8821 + 92 sin 0.8821]e−0.12602 = 93.997 mV; vL max ≈ 94 mV. Note: When P 6.7

t=

0.8821 + π ; 140

vL = −60 mV.

[a] i

=

0

t < 0;

i

=

16t A

0 ≤ t ≤ 25 ms;

i

=

0.8 − 16t A

25 ≤ t ≤ 50 ms;

i

=

0

50 ms < t.

[b] v = L

di = 375 × 10−3 (16) = 6 V dt

v = 375 × 10−3 (−16) = −6 V

0 ≤ t ≤ 25 ms; 25 ≤ t ≤ 50 ms.

v

=

0

t < 0;

v

=

6V

0 < t < 25 ms;

v

= −6 V

25 < t < 50 ms;

v

=

50 ms < t.

0

p = vi. p

=

0

t < 0;

p

=

96t W

0 < t < 25 ms;

p

=

96t − 4.8 W

25 < t < 50 ms;

p

=

0

50 ms < t.

Problems 1 w = Li2 . 2

P 6.8

w

=

0

t < 0;

w

=

48t2 J

0 < t < 25 ms;

w

=

48t2 − 4.8t + 0.12 J

25 < t < 50 ms;

w

=

0

50 ms < t.

[a] i

=

1000 Z t 250 sin 1000x dx − 5 50 0

=

5000

Z t

sin 1000x dx − 5

0

=

− cos 1000x t 5000 −5 1000 0

=

5(1 − cos 1000t) − 5;



i [b]

p



= −5 cos 1000t A. = vi = (250 sin 1000t)(−5 cos 1000t) = −1250 sin 1000t cos 1000t;

p w

w

= −625 sin 2000t W. 1 2 Li = 2 =

1 (50 × 10−3 )25 cos2 1000t 2

=

625 cos2 1000t mJ;

=

[312.5 + 312.5 cos 2000t] mJ.

6–9

6–10

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance

[c] Absorbing power:

P 6.9

Delivering power:

0.5π ≤ t ≤ π ms;

0 ≤ t ≤ 0.5π ms;

1.5π ≤ t ≤ 2π ms;

π ≤ t ≤ 1.5π ms.

0 ≤ t < 0.5 s : t −2.8 1 Zt −20x −20x iL = 2.8e dx + 2 = e +2 0.05 0 (0.05)(20) 0

= −2.8e−20t + 4.8 A. iL (0.5) = 4.8 A. 0.5 s ≤ t < ∞ : t 1 Zt 2.8 −2.8e−20(x−0.5) dx + 4.8 = e−20(x−0.5) +4.8 0.05 0.5 (0.05)(20) 0.5

iL =

Problems = 2.8e−20(t−0.5) + 2 A.

P 6.10

[a] 0 ≤ t ≤ 2 s : v = −25t; i=

1 Zt x 2 t −25x dx + 0 = −10 = −5t2 A. 2.5 0 2 0

2s ≤ t ≤ 6s : v = −100 + 25t; i(2) = −20 A; .·. i

=

1 Zt (25x − 100) dx − 20 2.5 2

=

10

Z t 2

=

Z t

x dx − 40

dx − 20

2

5(t2 − 4) − 40(t − 2) − 20

= 5t2 − 40t + 40 A. 6 s ≤ t ≤ 10 s : v = 200 − 25t;

6–11

6–12

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance i(6) = 5(36) − 240 + 40 = −20 A; i

=

1 Zt (200 − 25x) dx − 20 2.5 6

=

80

Z t

Z t

dx − 10

6

x dx − 20

6

= 80(t − 6) − 10(t2 − 36)/2 − 20 = 80t − 5t2 − 320 A. 10 s ≤ t ≤ 12 s : v = 25t − 300; i(10) = 800 − 500 − 320 = −20 A; i

= =

1 Z 2.5

t

(25x − 300) dx − 20 10

Z t

Z t

x dx − 120

10

10

dx − 20

10

=

5(t2 − 100) − 120(t − 10) − 20

=

5t2 − 120t + 680 A.

t ≥ 12 s : v = 0; i(12) = 5(12)2 − 120(12) + 680 = −40 A; i

=

1 Z 2.5

t

0 dx − 40 12

= −40 A. [b] For 0 ≤ t ≤ 2 s, v=0

v = −25t V;

when

t=0

For 2 ≤ t ≤ 6 s, v=0

when

when

t = 4s

when

For t ≥ 12 s,

so

t = 8s

so

v = 0;

so

i = −5t2 + 80t − 320 A.

i = −5(8)2 + 80(8) − 320 = 0 A.

v = 25t − 300 V;

t = 12 s

i = 5t2 − 40t + 40 A.

i = 5(4)2 − 40(4) + 40 = −40 A.

v = 200 − 25t V;

For 10 ≤ t ≤ 12 s, v=0

i = 0 A.

v = −100 + 25t V;

For 6 ≤ t ≤ 10 s, v=0

so

i = −5t2 A.

i = 5t2 − 120t + 680 A.

i = 5(12)2 − 120(12) + 680 = −40 A.

i = −40 A.

Problems

6–13

[c]

P 6.11

For 0 ≤ t ≤ 1.2 s: 1Zt iL = 14 × 10−3 dx + 0 = 0.7 × 10−3 t; 20 0 iL (1.2 s) = (0.7 × 10−3 )(1.2) = 0.84 mA; Rm = (25)(1000) = 25 kΩ; vm (1.2 s) = (0.84 × 10−3 )(25 × 103 ) = 21 V.

P 6.12

p = vi = 40t[e−10t − 10te−20t − e−20t ]. w=

Z ∞

p dx =

0

Z ∞

40x[e−10x − 10xe−20x − e−20x ] dx = 0.2 J.

0

This is energy stored in the inductor at t = ∞. We can verify this using the energy equation for inductors. Remember that as t → ∞, i = 2 A: 1 1 w = Li2 = (0.1)(2)2 = 0.2 J. (checks) 2 2 P 6.13

[a] v(20 µs)

=

12.5 × 109 (20 × 10−6 )2 = 5 V (end of first interval);

v(20 µs)

=

106 (20 × 10−6 ) − (12.5)(400) × 10−3 − 10

=

5 V (start of second interval);

=

106 (40 × 10−6 ) − (12.5)(1600) × 10−3 − 10

=

10 V (end of second interval).

=

10 V (start of third interval).

v(40 µs)

v(40 µs)

[b] p(10µs) = 62.5 × 1012 (10−5 )3 = 62.5 mW, i(10µs) = 50 mA,

v(10 µs) = 1.25 V,

p(10 µs) = vi = (1.25)(50 m) = 62.5 mW. (checks)

p(30 µs) = 437.50 mW,

v(30 µs) = 8.75 V,

p(30 µs) = vi = (8.75)(0.05) = 62.5 mW. (checks)

i(30 µs) = 0.05 A;

6–14

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [c] w(10 µs) = 15.625 × 1012 (10 × 10−6 )4 = 0.15625 µJ; w = 0.5Cv 2 = 0.5(0.2 × 10−6 )(1.25)2 = 0.15625 µJ; w(30 µs) = 7.65625 µJ; w(30 µs) = 0.5(0.2 × 10−6 )(8.75)2 = 7.65625 µJ.

P 6.14

[a] i = C

d dv = 24 × 10−6 [e−15,000t sin 30,000t] dt dt

= [0.72 cos 30,000t − 0.36 sin 30,000t]e−15,000t A, 

[b] i

π ms = −31.66 mA, 80 



v

i(0+ ) = 0.72 A.

π ms = 20.505 V, 80 

p = vi = −649.23 mW. 

[c] w

π 1 ms = Cv 2 = (0.3 × 10−6 )(20.505)2 = 126.13 µJ. 80 2 



P 6.15

[a] v =

 1 Zt i dx + v(0+ ) C 0+

=

Z t 1 3 cos 50,000x dx = 100 sin 50,000t V. 0.6 × 10−6 0+

[b] p(t) = vi = [300 cos 50,000t] sin 50,000t = 150 sin 100,000t W,

p(max) = 150 W.

1 2 [c] w(max) = Cvmax = (0.3 × 10−6 )(100)2 = 3000 µJ = 3 mJ. 2 P 6.16

[a] v

=

0

t < 0;

v

= −2t V

v

=

2t − 20 V

5 ≤ t ≤ 15 s;

v

=

40 − 2t V

15 ≤ t ≤ 20 s;

v

=

0

20 s < t.

[b] i = C

0 ≤ t ≤ 5 s;

dv : dt

i

=

0

i

= −1 mA

0 < t < 5 s;

i

=

5 < t < 15 s;

i

= −1 mA

15 < t < 20 s;

i

=

20 s < t.

1 mA

0

t < 0;

Problems

6–15

p = vi : p

=

0

t < 0;

p

=

(−2t)(−0.001) = 2t mW

0 < t < 5 s;

p

=

(2t − 20)(0.001) = 2t − 20 mW

5 < t < 15 s;

p

=

(40 − 2t)(−0.001) = 2t − 40 mW

15 < t < 20 s;

p

=

0

20 s < t.

w= w w

Z

= =

p dx : 0 Z t

t < 0; (0.002x) dx =

0

w

=

Z t

t 2 0.001x = t2 mJ

0 < t < 5 s;

0

(0.002x − 0.02) dx + 0.025

5

=

2

(0.001x − 0.02x)

t

+ 0.025

5

= t2 − 20t + 100 mJ w

=

Z t

5 < t < 15 s;

(0.002x − 0.04) dx + 0.025

15

=

2

(0.001x − 0.04x)

t

+ 0.025

15

w

= t2 − 40t + 400 mJ

15 < t < 20 s;

=

20 s < t.

0

[c]

From the plot of power above, it is clear that power is being absorbed for

6–16

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 0 < t < 5 s and for 10 s < t < 15 s, because p > 0. Likewise, power is being delivered for 5 s < t < 10 s and for 15 s < t < 20 s, because p < 0.

P 6.17

1 1 [a] w(0) = C[v(0)]2 = (400 × 10−9 )(25)2 = 125 µJ. 2 2 −1500t [b] v = (A1 t + A2 )e ; v(0) = A2 = 25 V. dv dt

= −1500e−1500t (A1 t + A2 ) + e−1500t (A1 ) (−1500A1 t − 1500A2 + A1 )e−1500t ;

=

dv (0) = A1 − 1500A2 ; dt i=C

dv , dt

i(0) = C

dv(0) ; dt

i(0) 90 × 10−3 dv(0) .·. = = = 225 × 103 ; dt C 0.40 × 10−6 .·. 225 × 103 = A1 − 1500(25). Thus, A1 = 2.25 × 105 + 3.75 × 104 = 262,500

V . s

[c] v = (262,500t + 25)e−1500t ; dv d = 0.40 × 10−6 (262,500t + 25)e−1500t ; dt dt d = [(0.105t + 10 × 10−6 )e−1500t ] dt

i=C i

P 6.18

[a]

v

=

(0.105t + 10 × 10−6 )(−1500)e−1500t + e−1500t (0.105)

=

(−157.5t − 15 × 10−3 + 0.105)e−1500t

=

(0.09 − 157.5t)e−1500t A,

=

(90 − 157,500t)e−1500t mA,

=

5 × 106

Z 250×10−6

t≥0 t ≥ 0.

100 × 10−3 e−1000t dt − 60.6

0

=

w

−1000t 3e

500 × 10

−1000

250×10−6

−60.6

0

=

500(1 − e−0.25 ) − 60.6 = 50 V;

=

1 Cv 2 2

= 12 (0.2)(10−6 )(50)2 = 250 µJ.

Problems [b] v = 500 − 60.6 = 439.40 V; 1 w = (0.2) × 10−6 (439.40)2 = 19.31 mJ = 19,307.24 µJ. 2 P 6.19

[a] 0 ≤ t ≤ 100 µs : 1 = 5 × 106 ; C

C = 0.2 µF 6

v = 5 × 10

Z t

− 0.04 dx + 40;

0

v = −200 × 103 t + 40 V. v(100 µs) = −20 + 40 = 20 V. [b] 100 µs ≤ t ≤ 300 µs : 6

v = 5 × 10

Z t 100×10−6

0.08 dx + 20 = 4 × 105 t − 40 + 20;

v = 4 × 105 t − 20V. v(300 µs) = 4 × 105 (300 × 10−6 ) − 20 = 100 V. [c] 300 µs ≤ t < ∞ : 6

v = 5 × 10

Z t 300×10−6

0 dx + 100 = 100 V.

[d]

P 6.20

dv = 0, t < 0. dt dv = 5e−1000t [cos 3000t + 13 sin 3000t] mA, [b] i = C dt [c] no, v(0− ) = −30 V; v(0+ ) = 10 − 40 = −30 V. [a] i = C

[d] yes,

i(0− ) = 0 A; i(0+ ) = 5 mA.

t ≥ 0.

6–17

6–18

CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [e] v(∞) = 10 V; 1 1 w = Cv 2 = (0.5 × 10−6 )(10)2 = 25 µJ. 2 2

P 6.21

P 6.22

iC = C(dv/dt). 00

τ=

L 0.2 = = 2.5 µs; R 80,000

iL (t) = 48e−400,000t mA,

1 = 400,000; τ t ≥ 0.

p60k = (0.048e−400,000t )2 (60,000) = 138.24e−800,000t W. w60k =

Z t

138.24e−800,000x dx = 172.8 × 10−6 [1 − e−800,000t ] J;

0

1 wL (0) = (.2)(48 × 10−3 )2 = 230.4 µJ; 2 0.25wL (0) = 57.6 µJ; 172.8(1 − e−800,000t ) = 57.6; t=

.·. e800,000t = 1.5;

ln 1.5 = 0.507 µs. 800,000

[b] wdiss (t) = 230.4(1 − e−800,000t ) µJ; wdiss (0.507 µs) = 76.82 µJ; % = (76.82/230.4)(100) = 33.3%.

7–19

7–20 P 7.14

CHAPTER 7. Response of First-Order RL and RC Circuits [a] t < 0 :

ig (0) =

54 = 9 A; 3 + (9k4.5)

iL (0) =

9 ig (0) = 6 A. 9 + 4.5

t > 0:

i∆ =

iT (200) 2 = iT ; 300 3

vT = 50i∆ + iT

(100)(200) 2 200 = 50iT + iT ; 300 3 3

100 200 vT = RTh = + = 100 Ω. iT 3 3

τ=

200 L = × 10−3 R 100

iL = 6e−500t A,

1 = 500; τ

t ≥ 0.

[b] vL = 200 × 10−3 (−3000e−500t ) = −600e−500t V, [c]

vL = 50i∆ + 100i∆ = 150i∆ ; i∆ =

vL = −4e−500t A 150

t ≥ 0+ .

t ≥ 0+ .

Problems

P 7.15

1 w(0) = (200 × 10−3 )(36) = 3.6 J; 2 p50i∆ = −50i∆ iL = −50(−4e−500t )(6e−500t ) = 1200e−1000t W; w50i∆ =

Z ∞

−1000t

1200e

0

% dissipated = P 7.16

e−1000t ∞ dt = 1200 = 1.2 J; −1000 0

1.2 (100) = 33.33%. 3.6

t 0:

Find Th´evenin resistance seen by inductor:

iT = 5vT ;

τ=

vT 1 = RTh = = 0.2 Ω; iT 5

L 50 × 10−3 = = 250 ms; R 0.2

io = 25e−4t A,

t ≥ 0;

1/τ = 4.

7–21

7–22

CHAPTER 7. Response of First-Order RL and RC Circuits

vo = L P 7.17

dio = (50 × 10−3 )(−100e−4t ) = −5e−4t V, dt

t ≥ 0+ .

[a] t < 0:

t = 0+ :

t > 0:

iR = 5e−t/τ mA;

τ=

L = 20 × 10−6 ; R

iR = 5e−50,000t mA.

vR = (2.5 × 103 )(5 × 10−3 )e−50,000t = 12.5e−50,000t V; v1 = 20 × 10−3 [5 × 10−3 (−50,000)e−50,000t ] = −5e−50,000t V; vo = −v1 − vR = −7.5e−50,000t V.

Problems

[b] io = P 7.18

7–23

103 Z t −7.5e−50,000x dx + 0 = 3.125e−50,000t − 3.125 mA. 48 0

[a] From the solution to Problem 7.17, iR = 5 × 10−3 e−50,000t A; pR = (25 × 10−6 e−100,000t )(2.5 × 103 ) = 62.5 × 10−3 e−100,000t W; wdiss =

Z ∞

62.5 × 10−3 e−100,000t dt

0

= 62.5 ×

−100,000t −3 e 10 −105

∞ =

625 nJ.

0

1 1 [b] wtrapped = Leq i2R (0) = (50 × 10−3 )(5 × 10−3 )2 = 625 nJ. 2 2 CHECK: w(0) = 12 (20)(25 × 10−6 ) × 10−3 + 12 (80)(25 × 10−6 ) × 10−3 = 1250 nJ; .·. w(0) = wdiss + wtrapped . P 7.19

[a] t < 0 :

t = 0+ :

220 = iab + (50/3) + (10/3),

iab = 200 A,

t = 0+ .

7–24

CHAPTER 7. Response of First-Order RL and RC Circuits [b] At t = ∞:

t → ∞.

iab = 220/1 = 220 A,

[c] i1 (0) = 50/3,

τ1 =

i2 (0) = 10/3,

2 × 10−3 = 0.167 ms; 12

τ2 =

15 × 10−3 = 0.25 ms; 60

i1 (t) = (50/3)e−6000t A,

t ≥ 0;

i2 (t) = (10/3)e−4000t A,

t ≥ 0;

iab = 220 − (50/3)e−6000t − (10/3)e−4000t A, 220 − (50/3)e−6000t − (10/3)e−4000t = 210; 30 = 50e−6000t + 10e−4000t ; 3 = 5e−6000t + e−4000t ; By trial and error t = 123.1 µs. P 7.20

[a] t < 0:

iL (0− ) =

−2.5(16) = −2 mA. (20)

t ≥ 0;

Problems

7–25

t ≥ 0:

40 × 10−3 = 40 × 10−6 ; 1/τ = 25,000; 103 vo = −1000(−2 × 10−3 )e−25,000t = 2e−25,000t V, τ=

t ≥ 0+ .

1 [b] wdel = (40 × 10−3 )(4 × 10−6 ) = 80 nJ. 2 [c] 0.95wdel = 76 nJ; .·. 76 × 10−9 =

Z to 0

4e−50,000t dt; 1000

to −9 −9 −50,000t −9 −50,000to · . . 76 × 10 = −80 × 10 e ); = 80 × 10 (1 − e 0

.·. e−50,000to = 0.05. 50,000to = ln 20 so to = 59.9 µs; 59.9 to = = 1.498 so to ≈ 1.5τ. .·. τ 40 P 7.21

[a] t < 0 (after source-transforming the parallel resistor and current source):

8 × 10−3 = 0.2 mA; 40 10,000 vC (0− ) = v10k (0− ) = (10 − 2) = 2 V. 20,000 + 5000 + 10,000 + 5000

i1 (0− ) = i2 (0− ) =





[b] t = 0+ :

i1 (0+ ) =

2 × 10−3 = 0.2 mA; 10

7–26

CHAPTER 7. Response of First-Order RL and RC Circuits i2 (0+ ) =

−2 × 10−3 = −0.2 mA. 10

[c] Capacitor voltage cannot change instantaneously, therefore, i1 (0+ ) =

2 = 0.2 mA. 10,000

[d] Switching can cause an instantaneous change in the current in a resistive branch. In this circuit i2 (0− ) = 0.2 mA and i2 (0+ ) = −0.2 mA. [e] vc = 2e−t/τ V,

t ≥ 0;

τ = Re C = 5000(0.4) × 10−6 = 2 × 10−3 ; vc = 2e−500t V, i1 = [f ] i2 = P 7.22

t ≥ 0;

vc = 0.2e−500t mA, 10,000

t ≥ 0.

−vc = −0.2e−500t mA, 5000 + 5000

t ≥ 0+ .

[a] The circuit for t < 0 is shown below. Note that the capacitor behaves like an open circuit.

Find the voltage drop across the open circuit by finding the voltage drop across the 50 kΩ resistor. First use current division to find the current through the 50 kΩ resistor: 80 × 103 (7.5 × 10−3 ) = 4 mA. 80 × 103 + 20 × 103 + 50 × 103 Use Ohm’s law to find the voltage drop: v(0− ) = (50 × 103 )i50k = (50 × 103 )(0.004) = 200 V.

i50k =

[b] To find the time constant, we need to find the equivalent resistance seen by the capacitor for t > 0. When the switch opens, only the 50 kΩ resistor remains connected to the capacitor. Thus, τ = RC = (50 × 103 )(0.4 × 10−6 ) = 20 ms. [c] v(t) = v(0− )e−t/τ = 200e−t/0.02 = 200e−50t V, 1 1 [d] w(0) = Cv 2 = (0.4 × 10−6 )(200)2 = 8 mJ. 2 2

t ≥ 0.

Problems

7–27

1 1 [e] w(t) = Cv 2 (t) = (0.4 × 10−6 )(200e−50t )2 = 8e−100t mJ. 2 2 The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains: 8 × 10−3 e−100t = 2 × 10−3 , P 7.23

e100t = 4,

t = (ln 4)/100 = 13.86 ms.

For t < 0,

V0 =

25,000 (60) = 15 V. 75,000 + 25,000

For t ≥ 0, Req = 25 kΩ,

τ = Req C = (25,000)(5 × 10−9 ) = 125 µs.

so

There is no source in the circuit as t → ∞ so Vf = 0. Thus, vo (t) = Vf + (V0 − Vf )e−t/τ = 0 + (15 − 0)e−t/125µ = 15e−8000t V, P 7.24

t ≥ 0.

This circuit is actually two RC circuits in series, and the requested voltage, vo , is the sum of the voltage drops for the two RC circuits. The circuit for t < 0 is shown below:

Find the current in the loop and use it to find the initial voltage drops across the two RC circuits: i=

15 = 0.2 mA, 75,000

v5 (0− ) = 4 V,

v1 (0− ) = 8 V.

7–28

CHAPTER 7. Response of First-Order RL and RC Circuits There are two time constants in the circuit, one for each RC subcircuit. τ5 is the time constant for the 5 µF – 20 kΩ subcircuit, and τ1 is the time constant for the 1 µF – 40 kΩ subcircuit: τ5 = (20 × 103 )(5 × 10−6 ) = 100 ms; τ1 = (40 × 103 )(1 × 10−6 ) = 40 ms. Therefore, v5 (t) = v5 (0− )e−t/τ5 = 4e−t/0.1 = 4e−10t V, t ≥ 0; v1 (t) = v1 (0− )e−t/τ1 = 8e−t/0.04 = 8e−25t V, t ≥ 0. Finally, vo (t) = v1 (t) + v5 (t) = [8e−25t + 4e−10t ] V, t ≥ 0.

P 7.25

Find the value of the voltage at 60 ms for each subcircuit and use the voltage to find the energy at 60 ms: v1 (60 ms) = 8e−25(0.06) ∼ v5 (60 ms) = 4e−10(0.06) ∼ = 1.79 V, = 2.20 V; 1 1 2 −6 2 ∼ w1 (60 ms) = 2 Cv1 (60 ms) = 2 (1 × 10 )(1.79) = 1.59 µJ; w5 (60 ms) = 12 Cv52 (60 ms) = 21 (5 × 10−6 )(2.20)2 ∼ = 12.05 µJ; w(60 ms) = 1.59 + 12.05 = 13.64 µJ. Find the initial energy from the initial voltage: w(0) = w1 (0) + w2 (0) = 21 (1 × 10−6 )(8)2 + 21 (5 × 10−6 )(4)2 = 72 µJ. Now calculate the energy dissipated at 60 ms and compare it to the initial energy: wdiss = w(0) − w(60 ms) = 72 − 13.64 = 58.36 µJ; % dissipated = (58.36 × 10−6 /72 × 10−6 )(100) = 81.05 %.

P 7.26

[a] Note that there are many different possible correct solutions to this problem. τ R= . C Choose a 10 µF capacitor from Appendix H. Then, 0.004 = 400 Ω. 10 × 10−6 Construct a 400 Ω resistor by combining a 180 Ω resistor and a 220 Ω resistor in series: R=

[b] v(t) = Vo e−t/τ = 120e−250t V, −250t

[c] 120e

= 30

so

ln 4 .·. t = = 5.545 ms. 250

250t

e

t ≥ 0. = 4;

Problems P 7.27

[a] t < 0:

vo (0) =

12,000k68,000 (60) = 51 V. 1800 + (12,000k68,000)

t > 0:

1 τ = (12) × 10−3 = 2 ms; 6 vo = 51e−500t V, p=

1 = 500; τ

t ≥ 0.

vo2 × 10−3 = 216.75 × 10−3 e−1000t W. 12

wdiss =

Z 2×10−3

216.75 × 10−3 e−1000t dt

0

= 216.75 × 10−6 (1 − e−2 ) = 187.42 µJ. 1 2

1 (51)2 × 10−6 = 216.75 µJ; 6

  

[b] w(0) =

0.95w(0) = 205.9125 µJ; Z to

216.75 × 10−3 e−1000x dx = 205.9125 × 10−6 ;

0

Z to

e−1000x dx = 0.95 × 10−3 ;

0

.·. 1 − e−1000to = 0.95; P 7.28

e1000to = 20;

so to = 3 ms.

v = 20 kΩ. i 1 1 1 [b] = = 1000; C= = 0.05 µF. 3 τ RC (10 )(20 × 103 ) 1 [c] τ = = 1 ms. 1000 1 [d] w(0) = (0.05 × 10−6 )(104 ) = 250 µJ. 2 [a] R =

7–29

7–30

CHAPTER 7. Response of First-Order RL and RC Circuits

[e] wdiss =

Z to 0

Z to v2 (104 )e−2000t dt = dt R (20 × 103 ) 0

e−2000t to = 0.5 = 250(1 − e−2000to ) µJ. −2000 0

0.8(250) = 200 = 250(1 − e−2000to ); .·. e−2000to = 0.2; to = P 7.29

1 ln 5; 2000

e2000to = 5; to ∼ = 804.72 µs.

[a] τ = RC = RTh (0.2) × 10−6 = 10−3 ;

1000 .·. RTh = = 5 kΩ. 0.2

vT = 20 × 103 (iT − αv∆ ) + 10 × 103 iT ; v∆ = 10 × 103 iT ; vT = 30 × 103 iT − 20 × 103 α10 × 103 iT ; vT = 30 × 103 − 200 × 106 α = 5 × 103 ; iT .·. 30 − 200,000α = 5;

α = 125 × 10−6 A/V.

[b] vo (0) = (0.018)(5000) = 90 V t > 0:

vo = 90e−1000t V,

t ≥ 0.

t < 0.

Problems

v∆ − vo v∆ + − 125 × 10−6 v∆ = 0; 3 10 × 10 20,000 2v∆ + v∆ − vo − 2500 × 10−3 v∆ = 0; .·. v∆ = 2vo = 180e−1000t V. P 7.30

[a]

pds = (−90e−1000t )(22.5 × 10−3 e−1000t ) = −2025 × 10−3 e−2000t W; Z ∞

wds =

0

pds dt = −1012.5 µJ;

.·. dependent source is delivering 1012.5 µJ. [b] p10k =

(180)2 e−2000t ; 10 × 103

w10k = p20k = w20k =

Z ∞ 0

p10k dt = 1620 µJ;

(90)2 e−2000t ; 20 × 103 Z ∞ 0

p20k dt = 202.5 µJ;

1 wc (0) = (0.2) × 10−6 (90)2 = 810 µJ; 2 X

wdev = 810 + 1012.5 = 1822.5 µJ;

X

wdiss = 202.5 + 1620 = 1822.5 µJ.

7–31

7–32 P 7.31

CHAPTER 7. Response of First-Order RL and RC Circuits t < 0:

t > 0:

vT

=

2 × 104 io + 60,000iT

=

20,000(−iT ) + 60,000iT = 40,000iT ;

vT .·. = RTh = 40 kΩ. iT

τ = RC = 1 ms; vo = 25e−1000t V, io = 25 × 10−9 P 7.32

1 = 1000; τ t ≥ 0;

d [25e−1000t ] = −625e−1000t µA, dt

t ≥ 0+ .

[a] At t = 0− the voltage on each capacitor will be 25 × 10−3 × 200 = 5 V, positive at the upper terminal. Hence at t = 0+ we have

5 5 .·. isd (0+ ) = 0.025 + + = 1.65 A. 5 8

Problems

7–33

At t = ∞, both capacitors will have completely discharged. .·. isd (∞) = 25 mA. [b] isd (t) = 0.025 + i1 (t) + i2 (t); 5

τ1 = (5)(2 × 10−6 ) = 10 µs so v2µ = 5e−10 t V; τ2 = (8)(50 × 10−6 ) = 400 µs so v50µ = 5e−2500t V. d 5 5 .·. i1 (t) = −(2 × 10−6 ) (5e−10 t ) = e−10 t A, dt i2 (t) = −(50 × 10−6 )

d (5e−2500t ) = 0.625e−2500t A, dt

.·. isd = 25 + 1000e−100,000t + 625e−2500t mA, P 7.33

t ≥ 0+ ; t ≥ 0+ ;

t ≥ 0+ .

[a] The equivalent circuit for t > 0:

τ = 0.4 ms;

1/τ = 2500;

vo = 20e−2500t V,

t ≥ 0;

io = 2e−2500t mA,

t ≥ 0+ ;

−2500t

i25kΩ = 0.002e



15 = 0.75e−2500t mA, 40 

t ≥ 0+ ;

p25kΩ = (0.5625 × 10−6 e−5000t )(25,000) = 14,062.5 × 10−6 e−5000t W; w25kΩ =

Z ∞

14,062.5 × 10−6 e−5000t dt = −2.8125 × 10−6 (0 − 1) = 2.8125 µJ;

0

1 1 w(0) = (0.2 × 10−6 )(100) + (0.05 × 10−6 )(900) = 32.5 µJ; 2 2 % diss (25 kΩ) =

2.8125 × 100 = 8.65%. 32.5

[b] p625Ω = 625(2 × 10−3 e−2500t )2 = 2.5 × 10−3 e−5000t W; w625Ω =

Z ∞ 0

p625 dt = 0.50 µJ;

% diss (625Ω) =

0.5 × 100 = 1.54%. 32.5

7–34

CHAPTER 7. Response of First-Order RL and RC Circuits −2500t



i15kΩ = 0.002e

25 = 1.25e−2500t mA, 40 

t ≥ 0+ ;

p15kΩ = (1.25 × 10−3 e−2500t )2 (15,000) = 23.4375 × 10−3 e−5000t W; w15kΩ =

Z ∞

23.4375 × 10−3 e−5000t dt = 4.6875 µJ;

0

% diss (15kΩ) = 14.42%. [c]

X

wdiss = 2.8125 + 0.50 + 4.6875 = 8 µJ;

wtrapped = w(0) − % trapped =

X

wdiss = 32.5 − 8 = 24.5 µJ;

24.5 × 100 = 75.38%. 32.5

Check: 8.65 + 1.54 + 14.42 + 75.38 = 99.99 ≈ 100%. P 7.34

[a] v1 (0− ) = v1 (0+ ) = 75 V

v2 (0+ ) = 0;

Ceq = 2µk8µ = 1.6 µF.

τ = (5)(1.6) × 10−3 = 8ms; i=

75e−125t = 15e−125t mA, 5000

v1 =

1 = 125; τ

vC = 75e−125t V.

t ≥ 0+ .

−106 Z t 15 × 10−3 e−125x dx + 75 = 60e−125t + 15 V, 2 0

106 Z t v2 = 15 × 10−3 e−125x dx + 0 = −15e−125t + 15 V, 8 0 1 [b] w(0) = (2 × 10−6 )(5625) = 5625 µJ. 2

t ≥ 0; t ≥ 0.

Problems

7–35

1 1 [c] wtrapped = (2 × 10−6 )(225) + (8 × 10−6 )225 = 1125 µJ. 2 2 1 wdiss = (1.6 × 10−6 )(5625) = 4500 µJ. 2 Check: wtrapped + wdiss = 1125 + 4500 = 5625 µJ; w(0) = 5625 µJ. P 7.35

t < 0:

iL (0− ) = 6 A. t > 0:

iL (∞) = τ=

32 + 48 = 4 A; 20

L 5 × 10−3 = = 250 µs; R 20

1 = 4000; τ

iL = 4 + (6 − 4)e−4000t = 4 + 2e−4000t A,

t ≥ 0;

vo = −8iL + 48 = −8(4 + 2e−4000t ) + 48 = 16 − 16e−4000t V,

t ≥ 0+ .

diL = 5 × 10−3 [−8000e−4000t ] = −40e−4000t V, dt

t ≥ 0+ ;

[b] vL = 5 × 10−3

vL (0+ ) = −40 V; vo (0+ ) = 0 V. Check: at t = 0+ the circuit is:

vL (0+ ) = 32 − 72 + 0 = −40 V,

vo (0+ ) = 48 − 48 = 0 V.

7–36 P 7.36

CHAPTER 7. Response of First-Order RL and RC Circuits [a] Use the circuit at t < 0, shown below, to calculate the initial current in the inductor:

i(0− ) = 24/2 = 12 A = i(0+ ). Note that i(0− ) = i(0+ ) because the current in an inductor is continuous. [b] Use the circuit at t = 0+ , shown below, to calculate the voltage drop across the inductor at 0+ . Note that this is the same as the voltage drop across the 10 Ω resistor, which has current from two sources — 8 A from the current source and 12 A from the initial current through the inductor.

v(0+ ) = −10(8 + 12) = −200 V. [c] To calculate the time constant we need the equivalent resistance seen by the inductor for t > 0. Only the 10 Ω resistor is connected to the inductor for t > 0. Thus, τ = L/R = (200 × 10−3 /10) = 20 ms. [d] To find i(t), we need to find the final value of the current in the inductor. When the switch has been in position a for a long time, the circuit reduces to the one below:

Note that the inductor behaves as a short circuit and all of the current from the 8 A source flows through the short circuit. Thus, if = −8 A. Now, i(t) = if + [i(0+ ) − if ]e−t/τ = −8 + [12 − (−8)]e−t/0.02 = −8 + 20e−50t A, t ≥ 0. [e] To find v(t), use the relationship between voltage and current for an inductor: di(t) v(t) = L = (200 × 10−3 )(−50)(20e−50t ) = −200e−50t V, t ≥ 0+ . dt

Problems P 7.37

7–37

[a] t < 0:

KVL equation at the top node: vo (0− ) vo (0− ) vo (0− ) + + . 4 20 5 Multiply by 20 and solve: −40 =

−800 = (5 + 1 + 4)vo ;

vo = −80 V;

vo .·. io (0− ) = = −80/5 = −16 A. 5 t > 0:

Use voltage division to find the Th´evenin voltage: 20 VTh = vo = (240) = 60 V. 20 + 60 Remove the voltage source and make series and parallel combinations of resistors to find the equivalent resistance: RTh = 5 + 20k60 = 5 + 15 = 20 Ω. The simplified circuit is:

L 10 × 10−3 1 = = 0.5 ms; = 2000. R 20 τ 60 io (∞) = = 3 A; 20 .·. io = io (∞) + [io (0+ ) − io (∞)]e−t/τ τ=

= 3 + (−16 − 3)e−2000t = 3 − 19e−2000t A,

t ≥ 0.

7–38

CHAPTER 7. Response of First-Order RL and RC Circuits

[b] vo

vo P 7.38

=

dio ; dt 15 − 95e−2000t + 0.01(38,000)(e−2000t );

=

15 − 95e−2000t + 380e−2000t ;

=

15 + 285e−2000t V,

=

5io + (0.01)

t ≥ 0+ .

[a] t < 0:

ig =

240 = 3.75 A; 60 + 20k5

20k5 .·. io (0− ) = (3.75) = 3 A. 5 t > 0:

io (∞) =

4k20k5 (−40) = −16 A; 5

Req = 5 + 4k20 = 8.33 Ω; τ=

10 × 10−3 L = = 1.2 ms; Req 8.33

1 = 833.33; τ

.·. io = io (∞) + [io (0+ ) − io (∞)]e−t/τ = −16 + (3 + 16)e−833.33t = 19e−833.33t − 16 A, [b] vo

dio dt

=

5io + L

=

5(19e−833.33t − 16) + 0.01(−833.33)(19e−833.33t )

= −63.33e−833.33t − 80 V, P 7.39

t ≥ 0.

t ≥ 0+ .

[a] Note that there are many different possible solutions to this problem. R=

L . τ

Problems

7–39

Choose a 10 mH inductor from Appendix H. Then, 0.01 = 80 Ω. 125 × 10−6 Construct the resistance needed by combining 33 Ω, and 47 Ω resistors in series:

R=

[b] i(t) = If + (Io − If )e−t/τ ; Io = 0 A;

If =

Vf 16 = = 200 mA; R 80

.·. i(t) = 200 + (0 − 200)e−8000t mA = 200 − 200e−8000t mA,

t ≥ 0.

[c] i(t) = 0.2 − 0.2e−8000t = (0.5)(0.2) = 0.1; e−8000t = 0.5

e8000t = 2;

so

ln 2 = 86.64 µs. .·. t = 8000 P 7.40

[a] vo (0+ ) = −Ig R2 ;

τ=

L ; R1 + R2

vo (∞) = 0; vo (t) = −Ig R2 e−[(R1 +R2 )/L]t V,

t ≥ 0+ .

[b] vo (0+ ) → ∞, and the duration of vo (t) → zero. L [c] vsw = R2 io ; τ= ; R1 + R2 io (0+ ) = Ig ; Therefore

Therefore [d] |vsw (0+ )| → ∞;

io (∞) = Ig

R1 . R1 + R2

io (t) =

Ig R1 R1 +R2

io (t) =

R1 Ig (R1 +R2 )

vsw =

h

+ Ig − +

R1 Ig (1+R1 /R2 )

duration → 0.

Ig R1 R1 +R2

i

e−[(R1 +R2 )/L]t

R2 Ig e−[(R1 +R2 )/L]t . (R1 +R2 )

+

R2 Ig e−[(R1 +R2 )/L]t , (1+R1 /R2 )

t ≥ 0+ .

7–40

CHAPTER 7. Response of First-Order RL and RC Circuits

P 7.41

Opening the inductive circuit causes a very large voltage to be induced across the inductor L. This voltage also appears across the switch (part [d] of Problem 7.40), causing the switch to arc over. At the same time, the large voltage across L damages the meter movement.

P 7.42

[a] From Eqs. 7.1 and 7.15 Vs Vs −(R/L)t + Io − e ; R R 

i=



v = (Vs − Io R)e−(R/L)t ; Vs = 10; .·. R

Io −

Vs − Io R = 200;

Vs = −10; R

R = 500; L

Vs .·. Io = −10 + = 0 A. R Therefore,

Vs = 200V.

i(∞) = 10 =

200 R

L=

so

R = 20 Ω;

R = 40 mH. 500

[b] i = 10 − 10e−500t ;

i2 = 100 − 200e−500t + 100e−1000t ;

1 1 w = Li2 = (0.04)[100 − 200e−500t + 100e−1000t ] = 2 − 4e−500t + 2e−1000t ; 2 2 w(∞) = 2 J; w(to ) = 2 − 4e−500to + 2e−1000to = 0.25(2); .·. 1 − 2x + x2 = 0.25 Solving,

x = 1.5 and x = 0.5

.·. 0.5 = e−500to P 7.43

and thus

so

to =

but only the second solution is possible;

ln 2 = 1.386 ms. 500

[a]

−

x2 − 2x + 0.75 = 0.

Vs v 1Z t + + v dt + Io = 0. R R L 0

Problems Differentiating both sides, 1 1 dv + v = 0; R dt L dv R + v = 0. dt L

.·. [b]

dv R = − v; dt L dv R dt = − v dt dt L

R dv = − v dt; L

so

R dv = − dt; v L Z v(t) dx

x

Vo

ln

RZ t dy; L 0

v(t) R = − t; Vo L

.·. P 7.44

=−

v(t) = Vo e−(R/L)t = (Vs − RIo )e−(R/L)t .

For t < 0:

vx − 250 vx 10 = 0; + 0.9 (vx − 250) + 20 10 + 40 10 + 40 





vx (vx − 250) + 10 = 0; 20 50 5vx − 5000 + 20vx = 0; io (0− ) = 200/20 = 10 A. t > 0:

vx = 200 V;



7–41

7–42

CHAPTER 7. Response of First-Order RL and RC Circuits After source-transforming to simplify the right-hand side of the circuit we get

Find Th´evenin equivalent with respect to a, b:

VTh − 80 (VTh − 80) +9 =0 18 18



vT = (iT − 0.9vφ )18 = iT − 0.9 vT = 18iT − 9vT

VTh = 80 V.



10vT 18

.·. 10vT = 18iT ;

vT = RTh = 1.8 Ω. iT

io (∞) = 80/21.8 = 3.67 A;



18;

Problems

τ=

87.2 × 10−3 = 4 ms; 21.8

1/τ = 250;

io = 3.67 + (10 − 3.67)e−250t = 3.67 + 6.33e−250t A, P 7.45

calculate io (0− ):

t < 0;

io (0− ) =

t ≥ 0.

5 (0.002) = 0.5 mA; 5 + 15

io (0+ ) = io (0− ) = 0.5 mA. t > 0;

calculate vo (0+ ) using the node voltage method:

−0.002 +

va va − vo + = 0; 5 15

vo − va + 5 × 10−4 + i∆ = 0; 15 i∆ =

vo − 4i∆ − 0.001. 4

Solving, vo (0+ ) = 2 mV. We also know that vo (∞) = 0.

7–43

7–44

CHAPTER 7. Response of First-Order RL and RC Circuits Find the Th´evenin resistance seen by the 2 mH inductor:

iT =

vT vT + − 4i∆ ; 20 4

i∆ =

vT − 4i∆ 4

iT =

vT 4vT vT + − ; 20 4 20

vT ; .·. 5i∆ = 4

i∆ =

vT ; 20

1 1 1 2 iT = + − = = 0.1 S; vT 20 4 5 20 .·. RTh = 10Ω; τ=

2 × 10−3 = 0.2 ms; 10

1/τ = 5000;

.·. vo = 0 + (2 − 0)e−5000t = 2e−5000t mV, P 7.46

t ≥ 0+ .

[a] Let v be the voltage drop across the parallel branches, positive at the top node, then v 1 Zt 1 Zt + v dx + v dx = 0; −Ig + R g L1 0 L2 0   v 1 1 Zt v dx = Ig ; + + Rg L1 L2 0

v 1 Zt + v dx = Ig ; R g Le 0 1 dv v + = 0; Rg dt Le dv Rg + v = 0. dt Le Therefore v = Ig Rg e−t/τ ;

τ = Le /Rg .

Problems Thus i1 = i1 =

1 Zt Ig Rg e−x/τ Ig Rg e−x/τ dx = L1 0 L1 (−1/τ )

0

L2 Ig ; L1 + L2

i2 (∞) =

L1 Ig . L1 + L2

t > 0:

τ=

4 1 = ; 80 20

io = −5e−20t A,

t ≥ 0;

vo = 80io = −400e−20t V, −400e−20t = −80;

t ≥ 0+ ;

e20t = 5;

1 .·. t = ln 5 = 80.47 ms. 20 P 7.48

1 1 [a] wdiss = Le i2 (0) = (4)(25) = 50 J. 2 2 1Zt [b] i12H = (−400)e−20x dx + 5 12 0 −100 e−20x t 5 −20t 10 = + A +5 = e 3 −20 0 3 3

i6H

L1

Ig L2 Ig L1 (1 − e−t/τ ) and i2 = (1 − e−t/τ ). L1 + L2 L1 + L2

[b] i1 (∞) = P 7.47

t I L = g e (1 − e−t/τ );

1Z t (−400)e−20x dx + 0 = 6 0 =

−200 e−20x t 10 −20t 10 e − A. +0 = 3 −20 0 3 3

1 wtrapped = (18)(100/9) = 100 J. 2

7–45

7–46

CHAPTER 7. Response of First-Order RL and RC Circuits 1 [c] w(0) = (12)(25) = 150 J. 2

P 7.49

For t < 0, i40mH (0) = 75/5 = 15 A. For t > 0, after making a Th´evenin equivalent we have

Vs Vs −t/τ i= + Io − e ; R R 



R 4 1 = = = 40; τ Leq 100 × 10−3 Io = 15 A;

Vs 100 = = 25 A; R 4

i = 25 + (15 − 25)e−40t = 25 − 10e−40t A, vo = 0.04 P 7.50

di = 0.04(400e−40t ) = 16e−40t V, dt

t ≥ 0; t > 0+ .

[a] t < 0:

t > 0:

iL (0− ) = iL (0+ ) = 20 A;

τ=

4.8 = 0.1 s; 48

1 = 10; τ

iL (∞) = 10 A; iL = 10 + [20 − 10]e−10t = 10 + 10e−10t A, vo = 4.8[−100e−10t ] = −480e−10t V,

t ≥ 0;

t ≥ 0+ .

Problems 1 Zt [b] i1 = −480e−10x dx + 8 = 4e−10t + 4 A, 12 0 1Z t [c] i2 = −480e−10x dx + 12 = 6e−10t + 6 A, 8 0 P 7.51

7–47

t ≥ 0. t ≥ 0.

[a]

From Example 7.6, vo (t) = −60 + 90e−100t V. Write a KCL equation at the top node and use it to find the relationship between vo and vA : vA − vo vA vA + 75 + + = 0; 8000 160,000 40,000 20vA − 20vo + vA + 4vA + 300 = 0; 25vA = 20vo − 300; vA = 0.8vo − 12. Use the above equation for vA in terms of vo to find the expression for vA : vA (t) = 0.8(−60 + 90e−100t ) − 12 = −60 + 72e−100t V,

t ≥ 0+ .

[b] t ≥ 0+ , since there is no requirement that the voltage be continuous in a resistor. P 7.52

[a] vc (0+ ) = 120 V. Use voltage division to find the final value of voltage: 150 (−200) = −150 V. 150 + 50 Find the Th´evenin equivalent with respect to the terminals of the capacitor: vc (∞) =

VTh = −150 V,

RTh = 12.5 k + 150 kk50 k = 50 kΩ,

Therefore τ = Req C = (50,000)(40 × 10−9 ) = 2 ms.

7–48

CHAPTER 7. Response of First-Order RL and RC Circuits The simplified circuit for t > 0 is:

vc = vc (∞) + [vc (0+ ) − vc (∞)]e−t/τ ; = −150 + [120 − (−150)]e−t/τ = −150 + 270e−500t V,

t ≥ 0.

[b] For t < 0:

150 (−200) = −150 V. 150 + 50 For t ≥ 0: vc (0) =

vc (∞) = 120 V; τ = Req C = 10 × 103 (40 × 10−9 ) = 0.4 ms; vc (t) = vc (∞) + (vc (0) − vc (∞))e−t/τ = 120 + (−150 − 120)e−2500t = 120 − 270e−2500t V. P 7.53

For t < 0:

.·. vo (0− ) = vo (0+ ) = −90 V.

Problems

7–49

t > 0:

vo (∞) = 15 V;

τ = RC = (5 k)(0.05 µ) = 0.25 ms;

1 = 4000; τ

vo = vo (∞) + [vo (0+ ) − vo (∞)]e−t/τ = 15 + [−90 − 15]e−4000t = 15 − 105e−4000t V P 7.54

t ≥ 0.

[a] Find the voltage across the capacitor in the direction of the capacitor current. Call this voltage vc . The initial capacitor charge is zero, so V0 = 0. For t ≥ 0, Req = 20,000 + 30,000 = 50,000 Ω; τ = Req C = (50,000)(0.1 × 10−6 ) = 5 ms. As t → ∞,

Vf = (20,000)(0.0075) = 150 V. Therefore, vC = Vf + (V0 − Vf )e−t/τ = 150 + (0 − 150)e−t/0.005 = 150 − 150e−200t V, Thus, i=C

dvc = (0.1 × 10−6 )[−200(−150e−200t )] dt

= 3e−200t mA, t ≥ 0+ .

t ≥ 0.

7–50

CHAPTER 7. Response of First-Order RL and RC Circuits [b] i20k = (7.5 − 3e−200t ) mA; v = 20,000i20k = 150 − 60e−200t V,

P 7.55

vo (0) =

t ≥ 0+ .

120 (80) = 80 V; 120

vo (∞) = −6(25) = −150 V; 1 = 1000; τ

τ = (25 × 103 )(40 × 10−9 ) = 10−3 s;

vo = −150 + (80 + 150)e−1000t = −150 + 230e−1000t V, P 7.56

t ≥ 0.

[a] Use voltage division to find the initial value of the voltage: 10 k (−75) = −30 V. 10 k + 15 k [b] Use Ohm’s law to find the final value of voltage: vc (0+ ) = v10k =

vc (∞) = v5k = (5 × 10−3 )(5000) = 25 V. [c] Find the Th´evenin equivalent with respect to the terminals of the capacitor: VTh = 25 V,

RTh = 5 k + 20 k = 25 kΩ;

τ = RTh C = 2.5 ms. [d] vc = vc (∞) + [vc (0+ ) − vc (∞)]e−t/τ = 25 + (−30 − 25)e−400t = 25 − 55e−400t V,

t ≥ 0.

We want vc = 25 − 55e−400t = 0: Therefore t = P 7.57

ln(55/25) = 1.97 ms. 400

t < 0; io (0− ) = (0.015)

20 = 6 mA; 50

vo (0− ) = (6)(10) = 60 V.

t = ∞:

20 io (∞) = −0.010 = −4 mA; 50 



vo (∞) = io (∞)(10) = −40 V;

Problems RTh = 10 kΩk40 kΩ = 8 kΩ;

C = 125 nF;

τ = (8000)(125 × 10−9 ) = 1 ms;

1 = 1000; τ

.·. vo (t) = −40 + 100e−1000t V,

t ≥ 0+ .

ic = C i10 =

dvo = −12.5e−1000t mA, dt

t ≥ 0+ ;

vo = −4 + 10e−1000t mA, 0.010

t ≥ 0+ ;

io = ic + i10 = −(4 + 2.5e−1000t ) mA, P 7.58

[a] v = Is R + (Vo − Is R)e−t/RC .·. Is R = 80,

t ≥ 0+ . 

i = Is −

Vo −t/RC e ; R 

Vo − Is R = −80;

.·. Vo = 0 V. Is −

Vo = 50 × 10−3 ; R

.·. Is = 50 mA.

Vo − Is R = 0 − 0.05R = −80 so R = 1 = 2500; RC τ = RC =

C=

80 = 1600 Ω. 0.05

1 1 = = 250 nF; 2500R (2500)(1600)

1 = 400 µs. 2500

[b] v(∞) = 80 V; 1 w(∞) = (250 × 10−9 )(6400) = 800 µJ; 2 0.64w(∞) = 512 µJ; 1 2 512 × 10−6 v (to ) = = 2048; 2 250 × 10−9 80 − 80e−2500to = 64;

v(to ) = 64 V;

e2500to = 5;

.·. to = 643.775 µs.

7–51

7–52 P 7.59

CHAPTER 7. Response of First-Order RL and RC Circuits [a] Note that there are many different possible solutions to this problem. τ R= . C Choose a 10 µH capacitor from Appendix H. Then, 0.002 = 200 Ω. 10 × 10−6 Construct the resistance needed by combining two 100 Ω resistors in series: R=

[b] v(t) = Vf + (V0 − Vf )e−t/τ ; Vf = (If )(R) = (100 × 10−3 )(200) = 20 V;

V0 = −40 V;

.·. v(t) = 20 + (−40 − 20)e−500t V = 20 − 60e−500t V, [c] v(t) = 20 − 60e−500t = 0

so

ln 3 .·. t = = 2.2 ms. 500 P 7.60

[a]

1 Zt Is R = Ri + i dx + Vo ; C 0+ 0=R

di i + + 0; dt C

di i .·. + = 0. dt RC [b]

di i =− ; dt RC Z i(t) i(0+ )

di dt =− ; i RC

dy 1 Zt =− dx; y RC 0+

e−500t =

1 3

t ≥ 0.

Problems

ln

i(t) −t ; = + i(0 ) RC

i(t) = i(0+ )e−t/RC ;

i(0+ ) =

Is R − Vo Vo = Is − ; R R 



Vo −t/RC .·. i(t) = Is − e . R 

P 7.61



For t > 0, VTh = (−15,000)(30)ib = −450 × 103 ib ; ib =

400(12) = 100 µA; 48

VTh = −450 × 103 (100 × 10−6 ) = −45 V; RTh = 15 kΩ.

vo (0+ ) = 0;

vo (∞) = −45 V;

τ = (15, 000)(8)10−6 = 120 ms; vo = −45 + 45e−8.33t V,

1/τ = 8.33;

t ≥ 0;

1 w(t) = (8 × 10−6 )vo2 = 8100(1 − 2e−8.33t + e−16.67t ) µJ; 2 w(∞) = 8100 µJ; .·. 8100(1 − 2e−8.33to + e−16.67to ) = 0.90(8100); .·. 1 − 2x + x2 = 0.90;

x = e−8.33to ;

.·. x2 − 2x + 0.10 = 0; x1 = 1.9487,

x2 = 0.0513;

e−(25/3)to = 0.0513;

(25/3)to = ln 19.4868;

to = 356.4 ms.

7–53

7–54 P 7.62

CHAPTER 7. Response of First-Order RL and RC Circuits For t < 0, t > 0:

vo (0) = 80 V.

3

3

vTh = 30 × 10 i∆ + 0.8(100) = 30 × 10



−100 + 80 = 50 V. 100 × 103 

vT = 30 × 103 i∆ + 16 × 103 iT = 30 × 103 (0.8)iT + 16 × 103 iT = 40 × 103 iT ; RTh =

vT = 40 kΩ. iT

vo = 50 + (80 − 50)e−t/τ ; τ = RC = (40 × 103 )(5 × 10−9 ) = 200 × 10−6 s; vo = 50 + 30e−5000t V. P 7.63

vo (0) = 50 V;

vo (∞) = 80 V;

RTh = 16 kΩ; τ = (16)(5 × 10−6 ) = 80 × 10−6 s;

1 = 12,500; τ

v = 80 + (50 − 80)e−12,500t = 80 − 30e−12,500t V.

1 = 5000; τ

Problems P 7.64

[a] Let i be the current Zin the clockwise direction around the circuit. Then 1 Zt 1 t i dx + i dx Vg = iRg + C1 0 C2 0  1 1 Zt 1 Zt = iRg + + i dx. i dx = iRg + C1 C2 Ceq 0 0 

Now differentiate the equation 0 = Rg

i di + dt Ceq

Therefore i =

or

di 1 + i = 0. dt Rg Ceq

Vg −t/Rg Ceq Vg −t/τ e = e ; Rg Rg

Vg e−x/τ 1 Z t Vg −x/τ e dx = v1 (t) = C1 0 Rg Rg C1 −1/τ

τ = Rg Ceq . t

Vg C2 (1 − e−t/τ ); C1 + C2

τ = Rg Ceq ;

v2 (t) =

Vg C1 (1 − e−t/τ ); C1 + C2

τ = Rg Ceq .

C2 Vg ; C1 + C2

v2 (∞) =

=−

0

v1 (t) =

[b] v1 (∞) = P 7.65

7–55

C1 Vg . C1 + C2

[a] t < 0:

t > 0:

vo (0− ) = vo (0+ ) = 60 V; vo (∞) = 100 V; τ = (0.2)(5) × 10−3 = 1 ms; vo = 100 − 40e−1000t V,

1/τ = 1000; t ≥ 0.

Vg Ceq −t/τ (e − 1); C1

7–56

CHAPTER 7. Response of First-Order RL and RC Circuits

[b] io = −C

dvo = −0.2 × 10−6 [40,000e−1000t ] dt

= −8e−1000t mA; [c] v1 =

t ≥ 0+ .

−106 Z t −8 × 10−3 e−1000x dx + 40 0.3 0

= 66.67 − 26.67e−1000t V, [d] v2 =

t ≥ 0.

−106 Z t −8 × 10−3 e−1000x dx + 20 0.6 0

= 33.33 − 13.33e−1000t V,

t ≥ 0.

1 1 [e] wtrapped = (0.3)10−6 (66.67)2 + (0.6)10−6 (33.33)2 2 2 = 666.67 + 333.33 = 1000 µJ. P 7.66

[a] For t > 0:

τ = RC = (250 × 103 )(8 × 10−9 ) = 2 ms; vo = 50e−500t V, [b] io =

1 = 500; τ

t ≥ 0+ .

50e−500t vo = = 200e−500t µA; 250,000 250,000

Z t −1 200 × 10−6 e−500x dx + 50 = 10e−500t + 40 V, v1 = 40 × 10−9 0

P 7.67

[a] Leq = τ=

(3)(15) = 2.5 H; 3 + 15 Leq 2.5 1 = = s; R 7.5 3

io (0) = 0;

io (∞) =

.·. io = 16 − 16e−3t A,

120 = 16 A; 7.5 t ≥ 0.

vo = 120 − 7.5io = 120e−3t V,

t ≥ 0+ ;

t ≥ 0.

Problems 1Z t 40 40 −3t i1 = 120e−3x dx = − e A, 3 0 3 3 i2 = io − i1 =

8 8 −3t − e A, 3 3

t ≥ 0;

t ≥ 0.

[b] io (0) = i1 (0) = i2 (0) = 0, consistent with initial conditions. vo (0+ ) = 120 V, consistent with io (0) = 0. vo = 3

di1 = 120e−3t V, dt

t ≥ 0+

or di2 = 120e−3t V, t ≥ 0+ . dt The voltage solution is consistent with the current solutions. vo = 15

λ1 = 3i1 = 40 − 40e−3t Wb-turns; λ2 = 15i2 = 40 − 40e−3t Wb-turns; .·. λ1 = λ2 as it must, since vo =

dλ1 dλ2 = . dt dt

λ1 (∞) = λ2 (∞) = 40 Wb-turns; λ1 (∞) = 3i1 (∞) = 3(40/3) = 40 Wb-turns; λ2 (∞) = 15i2 (∞) = 15(8/3) = 40 Wb-turns; .·. i1 (∞) and i2 (∞) are consistent with λ1 (∞) and λ2 (∞). P 7.68

[a] Leq = 4 + 8 − 2(5) = 2 H; τ=

2 1 L = = s; R 50 25

i = 4 − 4e−25t A,

1 = 25; τ

t ≥ 0.

di di di − 5 = − = −(100e−25t ) = −100e−25t V, t ≥ 0+ . dt dt dt di di di [c] v2 (t) = 8 − 5 = 3 = 3(100e−25t ) = 300e−25t V, t ≥ 0+ . dt dt dt [d] i(0) = 4 − 4 = 0, which agrees with initial conditions.

[b] v1 (t) = 4

200 = 50i1 + v1 + v2 = 50(4 − 4e−25t ) − 100e−25t + 300e−25t = 200 V Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, the answers make sense in terms of known circuit behavior.

7–57

7–58 P 7.69

CHAPTER 7. Response of First-Order RL and RC Circuits [a] Leq = 4 + 8 + 2(5) = 22 H; L 22 = s; R 50

1 = 2.273; τ

i = 4 − 4e−2.273t A,

t ≥ 0.

τ=

di di di + 5 = 9 = 9(9.09e−2.273t ) = 81.81e−2.273t V, t ≥ 0+ . dt dt dt di di di [c] v2 (t) = 8 + 5 = 13 = 13(9.09e−2.273t ) = 118.18e−2.273t V, t ≥ 0+ . dt dt dt [d] i(0) = 0, which agrees with initial conditions. [b] v1 (t) = 4

200 = 50i1 + v1 + v2 = 50(4 − 4e−2.273t ) + 81.81e−2.273t + 118.18e−2.273t = 200 V. Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, the answers make sense in terms of known circuit behavior. P 7.70

[a] From Example 7.10, Leq = τ=

0.125 − 0.0625 L1 L2 − M 2 = = 50 mH; L1 + L2 + 2M 0.75 + 0.5

Leq 1 = s; R 5000

1 = 5000; τ

.·. io (t) = 40 − 40e−5000t mA,

t ≥ 0.

[b] vo = 10 − 250io = 10 − 250(0.04 − 0.04e−5000t ) = 10e−5000t V, di1 di2 [c] vo = 0.5 − 0.25 = 10e−5000t V; dt dt io = i1 + i2 ; di1 di2 dio = + = 200e−5000t A/s; dt dt dt di2 di1 .·. = 200e−5000t − ; dt dt di1 di1 .·. 10e−5000t = 0.5 − 50e−5000t + 0.25 ; dt dt di1 .·. 0.75 = 60e−5000t ; dt Z i1 0

i1 =

dx =

Z t

di1 = 80e−5000t dt;

80e−5000y dy;

0

80 −5000y t −5000t e mA, = 16 − 16e −5000 0

t ≥ 0.

t ≥ 0+ .

Problems

7–59

[d] i2 = io − i1 = 0.040 − 0.040e−5000t − 0.016 + 0.016e−5000t = 24 − 24e−5000t mA,

t ≥ 0.

[e] io (0) = i1 (0) = i2 (0) = 0, consistent with zero initial stored energy. vo = Leq

dio = (0.05)(200)e−5000t = 10e−5000t V, dt

t ≥ 0+ (checks).

Also, vo = 0.5

di1 di2 − 0.25 = 10e−5000t V, dt dt

vo = 0.25

di2 di1 − 0.25 = 10e−5000t V, dt dt

t ≥ 0+ (checks). t ≥ 0+ (checks).

vo (0+ ) = 10 V, which agrees with io (0+ ) = 0 A; io (∞) = 40 mA;

io (∞)Leq = (0.04)(0.05) = 2 mWb-turns;

i1 (∞)L1 − i2 (∞)M = (16 × 10−6 )(500) − (24 × 10−6 )(250) = 2 mWb-turns (ok); i2 (∞)L2 − i1 (∞)M = (24 × 10−6 )(250) − (16 × 10−6 )(250) = 2 mWb-turns (ok). Therefore, the final values of io , i1 , and i2 are consistent with conservation of flux linkage. Hence, the answers make sense in terms of known circuit behavior. P 7.71

[a] From Example 7.10, Leq =

(0.008)(0.020) − (0.010)2 L1 L2 − M 2 = = 7.5 mH; L1 + L2 − 2M 0.008 + 0.020 − 2(0.010)

τ=

Leq 7.5 × 10−3 1 = = ; R 75 10,000

io =

15 15 −10,000t − e = 0.2 − 0.2e−10,000t A t ≥ 0. 75 75

[b] vo = 15 − 75io = 15 − 75(0.2 − 0.2e−10,000t ) = 15e−10,000t V t ≥ 0+ . di2 di1 [c] vo = 0.008 + 0.01 ; dt dt io = i1 + i2 ; dio di1 di2 = + dt dt dt; di2 dio di1 di1 = − = 2000e−10,000t − ; dt dt dt dt !

di1 di1 .·. 15e−10,000t = 0.008 + 0.01 2000e−10,000t − ; dt dt

7–60

CHAPTER 7. Response of First-Order RL and RC Circuits di1 .·. = 2500e−10,000t . dt di1 = 2500e−10,000t dt; Z i1

dx = 2500

0

Z t

e−10,000y dy;

0

e−10,000y t .·. i1 = 2500 = 0.25 − 0.25e−10,000t A, −10,000 0

[d] i2

= io − i1 =

0.2 − 0.2e−10,000t − 0.25 + 0.25e−10,000t

= −50 + 50e−10,000t mA, [e] vo

t ≥ 0.

= L2 =

t ≥ 0.

di1 di2 +M dt dt

0.02(−500e−10,000t ) + 0.01(2500e−10,000t )

= 15e−10,000t V, t ≥ 0+ (checks). i1 (0) = 0.25 − 0.25 = 0; agrees with initial conditions; i2 (0) = −0.05 + 0.05 = 0; agrees with initial conditions; The final values of io , i1 , and i2 can be checked via the conservation of Wb-turns: io (∞)Leq = 0.2 × (7.5 × 10−3 ) = 1.5 mWb-turns; i1 (∞)L1 + i2 (∞)M = 0.25(0.008) − 0.05(0.01) = 1.5 mWb-turns; i2 (∞)L2 + i1 (∞)M = −0.05(0.02) + 0.25(0.01) = 1.5 mWb-turns. Thus our solutions make sense in terms of known circuit behavior. P 7.72

For t < 0:

i(0) =

0.08(150) = 60 mA. 200

Problems 0 ≤ t ≤ 250 µs:

i = 60e−2000t mA; i(250µs) = 60e−0.5 = 36.39 mA. 250 µs ≤ t ≤ 650 µs:

Req =

(50)(75) = 30 Ω; 125

R 30 1 = = × 103 = 1200; τ L 25 i = 36.39e−1200(t−250×10

−6 )

mA.

650 µs ≤ t < ∞: i(650µs) = 36.39e−0.48 = 22.52 mA; i = 22.52e−2000(t−650×10 v=L

di ; dt

−6 )

mA;

L = 25 mH;

di −6 −6 = (22.52 × 10−3 )(−2000)e−2000(t−650×10 ) = −45.035e−2000(t−650×10 ) ; dt v = (25 × 10−3 )(−45.035)e−2000(t−650×10 = −1.126e−2000(t−650×10

−6 )

V,

v(1ms) = −1.126e−2000(350)×10

−6

−6 )

t > 650+ µs. = −559.1 mV.

7–61

7–62 P 7.73

CHAPTER 7. Response of First-Order RL and RC Circuits From the solution to Problem 7.72, the initial energy is 1 w(0) = (25 mH)(60 mA)2 = 45 µJ. 2 For 650 µs ≤ t < ∞: 1 −6 w(t) = (25 mH)(22.52e−2000(t−650×10 ) mA)2 = (0.04)(45 µJ). 2 Solving, t = 964.72 µs.

P 7.74

[a] Prior to switch a closing at t = 0, there are no sources connected to the inductor; thus, i(0− ) = 0. At the instant switch a is closed, i(0+ ) = 0. For 0 ≤ t ≤ 1 s,

The equivalent resistance seen by the 10 V source is 2 + (3k0.8). The current leaving the 10 V source is 10 = 3.8 A. 2 + (3k0.8) The final current in the inductor, which is equal to the current in the 0.8 Ω resistor is 3 If = (3.8) = 3 A. 3 + 0.8 The resistance seen by the inductor is calculated to find the time constant: L 2 [(2k3) + 0.8]k3k6 = 1 Ω τ= = = 2 s. R 1 Therefore, i = if + [i(0+ ) − if ]e−t/τ = 3 − 3e−0.5t A,

0 ≤ t ≤ 1 s.

For part (b) we need the value of i(t) at t = 1 s: i(1) = 3 − 3e−0.5 = 1.18 A. .

Problems

7–63

[b] For t > 1 s

Use current division to find the final value of the current: 9 i= (−8) = −4.8 A 9+6 The equivalent resistance seen by the inductor is used to calculate the time constant: 2 L = = 0.8 s. 3k(9 + 6) = 2.5 Ω τ= R 2.5 Therefore, i = if + [i(1+ ) − if ]e−(t−1)/τ = −4.8 + 5.98e−1.25(t−1) A, P 7.75

t ≥ 1 s.

[a] t < 0:

ig =

500 = 5 A; 4 + (480k120)

i(0− ) =

5(480) = 4 A = i(0+ ). 600

[b] 0 ≤ t ≤ 100 µs: i = 4e−t/τ ; 1 R 120 + 96k480 = = = 10,000; τ L 20 × 10−3 i = 4e−10,000t ; i(25µs) = 4e−10

4 (25×10−6 )

= 4e−0.25 = 3.12 A.

7–64

CHAPTER 7. Response of First-Order RL and RC Circuits [c] i(100µs) = 4e−1 = 1.47 A. 100 µs ≤ t < ∞:

R 120 1 = = = 6000; τ L 0.02 i = 1.47e−6000(t−100×10

−6 )

A;

i(200µs) = 1.47e−6000(100)×10

−6

= 1.47e−0.6 = 807.59 mA.

[d] 0 ≤ t ≤ 100 µs: i = 4e−10,000t ; v=L

di 4 4 = (20 × 10−3 )(4)(−104 )e−10 t = −800e−10 t V; dt

v(100− µs) = −800e−10

4 (100×10−6 )

= −800e−1 = −294.30 V.

[e] 100 µs ≤ t < ∞: i = 1.47e−6000(t−100×10

−6 )

;

v = (20 × 10−3 )(1.47)(−6000)e−6000(t−100×10 = −176.58e−6000(t−100×10

−6 )

−6 )

V;

v(100+ µs) = −176.58 V. P 7.76

0 ≤ t ≤ 50 µs:

5 τ= × 10−9 (180,000) = 100 µs; 9 

Re = 720,000k240,000 = 180 kΩ; vc = 494.6e−10,000t mV; vc (50 µs) = 494.6e−0.5 = 300 mV.



Problems 50 µs ≤ t < ∞:

Re = 120,000k40,000 + 600,000k200,000 = 30,000 + 150,000 = 180 kΩ; 5 τ= × 10−9 (180,000) = 100 µs; 9 



1 = 10,000; τ

vc = 300e−10,000(t − 50 µs) mV; v1 =

30 vc = 50e−10,000(t − 50 µs) mV; 180

v2 =

150 vc = 250e−10,000(t − 50 µs) mV; 180

i1 =

v1 = 416.7e−10,000(t − 50 µs) nA; 120 × 103

i2 =

v2 = 416.7e−10,000(t − 50 µs) nA; 600 × 103

isw = i1 − i2 = 0 A; isw (100 µs) = 0 A. P 7.77

0 ≤ t ≤ 3 µs:

τ = RC = (2 × 103 )(5 × 10−9 ) = 10 µs; vo (0) = 0 V;

vo (∞) = 6 V;

vo = 6 − 6e−100,000t V

0 ≤ t ≤ 3 µs.

1/τ = 100,000;

7–65

7–66

CHAPTER 7. Response of First-Order RL and RC Circuits vo (3 µs) = 6 − 6e−0.3 = 1.555 V; 3 µs ≤ t < ∞:

t = ∞:

i=

6 − (−12) = 1.8 mA; 10,000

vo (∞) = 6 − 2000i = 2.4 V; vo = 2.4 + (1.555 − 2.4)e−(t − 3 µs)/τ ; RTh = 2 kΩk8 kΩ = 1.6 kΩ; τ = (1600)(5 × 10−9 ) = 8 µs; vo = 2.4 − 0.845e−125,000(t − 3 µs) P 7.78

1/τ = 125,000; 3 µs ≤ t < ∞.

Note that for t > 0, vo = (35/40)vc , where vc is the voltage across the 50 nF capacitor. Thus we will find vc first. t 0:

vT = −75 × 103 i∆ + 20 × 103 iT ; i∆ =

100 iT = 0.8iT ; 125

.·. vT = −60 × 103 iT + 20 × 103 iT . RTh =

vT = −40 kΩ; iT

Problems τ = RC = −40 × 103 (0.025) × 10−6 = −10−3; s vc = 25e1000t V; 1000t = ln 2000 P 7.89

25e1000t = 50,000; .·.

t = 7.6 ms.

[a]

vT = 2000iσ ; 6 iσ = (iT + βiσ ) = 0.75iT + 0.75βiσ ; 8 iσ (1 − 0.75β) = 0.75iT ; iσ =

0.75iT ; 1 − 0.75β

RTh =

1500iT = vT ; (1 − 0.75β)

1500 vT = = −3000; iT 1 − 0.75β

1 − 0.75β = −0.5 [b] Find

2000iσ =

.·. β = 2.

VTh :

VTh − 120 VTh (VTh − 120) + −2 = 0; 2000 6000 2000 VTh = 180 V.

i(0) = 0;

i(∞) = −60 mA;

τ=

L 0.3 = = −0.1 ms. R −3000

7–77

7–78

CHAPTER 7. Response of First-Order RL and RC Circuits .·. i = −60 + 60e10,000t mA. v=L

di = (0.3)(10,000)(0.06e10,000t ) = 180e10,000t = 36,000; dt

ln 200 .·. t = = 529.83 µs. 10,000 P 7.90

0 ≤ t ≤ 32 ms:

t 1 Zt 1 1 vo = − −10 dx + 0 = − (−10x) = (10t); RCf 0 RCf RCf 0

RCf = (200 × 103 )(0.2 × 10−6 ) = 40 × 10−3 .·.

so

1 = 25; RCf

vo (t) = 250t so vo (0.032) = (250)(0.032) = 8 V.

t ≥ 32 ms:

t 1 1 Zt 1 vo = − 5 dy + 8 = − (5y) +8 = − 5(t − 32 × 10−3 ) + 8; RCf 32×10−3 RCf RCf 32×10−3

RCf = (250 × 103 )(0.2 × 10−6 ) = 50 × 10−3 .·.

so

1 = 20; RCf

vo (t) = −20(5)(t − 32 × 10−3 ) + 8 = −100t + 11.2 V.

The output will saturate at the negative power supply value: −15 = −100t + 11.2

.·.

t = 262 ms.

Problems P 7.91

Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: vp = Vf + [Vo − Vf ]e−t/τ = −2 + (0 + 2)e−t/τ ; τ = (160 × 103 )(10 × 10−9 ) = 1.6 × 10−3 s; vp = −2 + 2e−625t V;

1/τ = 625;

vn = vp .

Write a KCL equation at the inverting input, and use it to determine vo : vn − vo vn + = 0; 10,000 40,000 .·. vo = 5vn = 5vp = −10 + 10e−625t V. The output will saturate at the negative power supply value: −10 + 10e−625t = −5; P 7.92

e−625t = 1/2;

t = (ln 2)/625 = 1.11 ms.

Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: vp = Vf + [Vo − Vf ]e−t/τ = −2 + (1 + 2)e−625t = −2 + 3e−625t V. The analysis for vo is the same as in Problem 7.91: vo = 5vp = −10 + 15e−625t V. The output will saturate at the negative power supply value: −10 + 15e−625t = −5;

P 7.93

e−625t = 1/3;

t = (ln 3)/625 = 1.76 ms.

[a] RC = (200 × 103 )(25 × 10−9 ) = 5 × 10−3 ; 0 ≤ t ≤ 5 µs:

1 = 200. RC

vg = 0.6 × 106 t V; vo = −200

Z t

0.6 × 106 x dx + 0

0 7x

= −12 × 10

2

2

t =

−6 × 107 t2 V;

0

vo (5 µs) = −6 × 10 (5 × 10−6 )2 = −1.5 × 10−3 V. 7

7–79

7–80

CHAPTER 7. Response of First-Order RL and RC Circuits 5 µs ≤ t ≤ 15 µs: vg = 6 − 0.6 × 106 t V; vo = −200

Z t 5×10−6

(6 − 0.6 × 106 x) dx − 1.5 × 10−3

x2 t −1.5 × 10−3 −6 −6 2 5×10 5×10 = −1200t + 6 × 10−3 + 6 × 107 t2 − 1.5 × 10−3 − 1.5 × 10−3 = 6 × 107 t2 − 1200t + 3 × 10−3 V. t



+12 × 107

= −1200x

vo (15 µs) = 6 × 107 (15 × 10−6 )2 − 1200(15 × 10−6 ) + 3 × 10−3 = −1.5 × 10−3 V. 15 µs ≤ t ≤ 20 µs: vg = −12 + 0.6 × 106 t V; vo = −200

Z t

= 2400x

15×10−6 t



(−12 + 0.6 × 106 x) dx − 1.5 × 10−3 7x

2

t

−12 × 10

−1.5 × 10−3

2 = 2400t − 36 × 10 − 6 × 107 t2 + 13.5 × 10−3 − 1.5 × 10−3 = −6 × 107 t2 + 2400t − 24 × 10−3 V. 15×10−6

15×10−6

−3

vo (20 µs) = −6 × 107 (20 × 10−6 )2 + 2400(20 × 10−6 ) − 24 × 10−3 = 0. [b]

[c] The output voltage will also repeat. This follows from the observation that at t = 20 µs the output voltage is zero, hence there is no energy stored in the capacitor. This means the circuit is in the same state at t = 20 µs as it was at t = 0, thus as vg repeats itself, so will vo . P 7.94

[a] RC = 40(50) × 10−6 = 2 ms;

1 = 500; RC

[b] 0 ≤ t ≤ 50 ms : vo = −500

Z t 0

−0.50 dx = 250t V.

vo = 0,

t < 0.

Problems

7–81

[c] 50 ms ≤ t ≤ 100 ms; vo (0.05) = 250(0.05) = 12.5 V; vo (t) = −500

Z t

0.50 dx + 12.5 = −250(t − 0.05) + 12.5 = −250t + 25 V.

0.05

[d] 100 ms ≤ t ≤ ∞ : vo (0.1) = −25 + 25 = 0 V; vo (t) = 0 V.

P 7.95

Write a KCL equation at the inverting input to the op amp, where the voltage is 0: 0 − vg 0 − vo d + + Cf (0 − vo ) = 0; Ri Rf dt .·.

dvo 1 vg + vo = − . dt Rf Cf Ri Cf

Note that this first-order differential equation is in the same form as Eq. 7.18 if Is = −vg /Ri . Therefore, its solution is the same as Eq. 7.19: −vg Rf −vg Rf −t/Rf Cf vo = + Vo − e . Ri Ri 

[a] vo = 0,



t < 0.

[b] Rf Cf = (4 × 106 )(50 × 10−9 ) = 0.2;

1 = 5; Rf Cf

−vg Rf −(−0.5)(4 × 106 ) = = 50; Ri 40,000 Vo = vo (0) = 0; .·.

vo = 50 + (0 − 50)e−5t = 50(1 − e−5t ) V,

0 ≤ t ≤ 50 ms.

7–82

CHAPTER 7. Response of First-Order RL and RC Circuits

[c]

1 = 5; Rf Cf −vg Rf −(0.5)(4 × 106 ) = = −50; Ri 40,000 Vo = vo (0.05) = 50(1 − e−0.25 ) ∼ = 11.06 V; .·.

vo = −50 + [11.06 − (−50)]e−5(t−0.05) ; = 61.06e−5(t−0.05) − 50 V,

[d]

50 ms ≤ t ≤ 100 ms.

1 = 5; Rf Cf −vg Rf = 0; Ri Vo = vo (0.10) = 61.06e−0.25 − 50 ∼ = −2.45 V; vo = 0 + (−2.45 − 0)e−5(t−0.1) = −2.45e−5(t−0.1) V,

P 7.96

[a]

Cdvp vp − vb + = 0; dt R

therefore

100 ms ≤ t ≤ ∞.

dvp 1 vb + vp = ; dt RC RC

vn − va d(vn − vo ) +C = 0; R dt therefore

dvo dvn vn va = + − . dt dt RC RC

But vn = vp . So

dvn vn dvp vp vb + = + = . dt RC dt RC RC

Therefore

dvo 1 = (vb − va ); dt RC

vo =

1 Zt (vb − va ) dy. RC 0

Problems

7–83

[b] The output is the integral of the difference between vb and va and then scaled by a factor of 1/RC. 1 Zt [c] vo = (vb − va ) dx; RC 0 RC = (40 × 103 )(25 × 10−9 ) = 1 ms; vb − va = 50 mV; vo = 1000 P 7.97

Z t 0

0.05 dx = 50t;

50tsat = 12;

tsat = 240 ms.

[a] While T2 has been ON, C2 is charged to VCC , positive on the left terminal. At the instant T1 turns ON the capacitor C2 is connected across b2 − e2 , thus vbe2 = −VCC . This negative voltage snaps T2 OFF. Now the polarity of the voltage on C2 starts to reverse, that is, the right-hand terminal of C2 starts to charge toward +VCC . At the same time, C1 is charging toward VCC , positive on the right. At the instant the charge on C2 reaches zero, vbe2 is zero, T2 turns ON. This makes vbe1 = −VCC and T1 snaps OFF. Now the capacitors C1 and C2 start to charge with the polarities to turn T1 ON and T2 OFF. This switching action repeats itself over and over as long as the circuit is energized. At the instant T1 turns ON, the voltage controlling the state of T2 is governed by the following circuit:

It follows that vbe2 = VCC − 2VCC e−t/R2 C2 . [b] While T2 is OFF and T1 is ON, the output voltage vce2 is the same as the voltage across C1 , thus

It follows that vce2 = VCC − VCC e−t/RL C1 . [c] T2 will be OFF until vbe2 reaches zero. As soon as vbe2 is zero, ib2 will become positive and turn T2 ON. vbe2 = 0 when VCC − 2VCC e−t/R2 C2 = 0, or when t = R2 C2 ln 2.

7–84

CHAPTER 7. Response of First-Order RL and RC Circuits [d] When t = R2 C2 ln 2,

we have

vce2 = VCC − VCC e−[(R2 C2 ln 2)/(RL C1 )] = VCC − VCC e−10 ln 2 ∼ = VCC . [e] Before T1 turns ON, ib1 is zero. At the instant T1 turns ON, we have

ib1 =

VCC VCC −t/RL C1 + e . R1 RL

[f ] At the instant T2 turns back ON, t = R2 C2 ln 2; therefore ib1 =

VCC VCC −10 ln 2 ∼ VCC + e . = R1 RL R1

[g]

[h]

P 7.98

[a] tOFF2 = R2 C2 ln 2 = 18 × 103 (2 × 10−9 ) ln 2 ∼ = 25 µs [b] tON2 = R1 C1 ln 2 ∼ = 25 µs [c] tOFF1 = R1 C1 ln 2 ∼ = 25 µs [d] tON1 = R2 C2 ln 2 ∼ = 25 µs 9 9 + = 3.5 mA 3000 18,000 9 9 −25/6 ∼ + e = = 0.5465 mA 18,000 3000

[e] ib1 = [f ] ib1

Problems

7–85

[g] vce2 = 9 − 9e−25/6 ∼ = 8.86 V P 7.99

[a] tOFF2 = R2 C2 ln 2 = (18 × 103 )(2.8 × 10−9 ) ln 2 ∼ = 35 µs [b] tON2 = R1 C1 ln 2 ∼ = 37.4 µs [c] tOFF1 = R1 C1 ln 2 ∼ = 37.4 µs [d] tON1 = R2 C2 ln 2 = 35 µs [e] ib1 = 3.5 mA 9 [f ] ib1 = × 10−3 + 3 × 10−3 e−35/9 ∼ = 0.561 mA 18 [g] vce2 = 9 − 9e−35/9 ∼ = 8.81 V Note in this circuit T2 is OFF 35 µs and ON 37.4 µs of every cycle, whereas T1 is ON 35 µs and OFF 37.4 µs every cycle.

P 7.100 If R1 = R2 = 20RL = 100 kΩ,

then

50 × 10−6 C1 = = 721.35 pF; 100 × 103 ln 2 If R1 = R2 = 2RL = 10 kΩ, C1 =

50 × 10−6 = 7.21 nF; 10 × 103 ln 2

75 × 10−6 C2 = = 1.08 nF. 100 × 103 ln 2

then C2 =

75 × 10−6 = 10.82 nF. 10 × 103 ln 2

Therefore 721.35 pF ≤ C1 ≤ 7.21 nF and 1.08 nF ≤ C2 ≤ 10.82 nF. P 7.101 [a] T2 is normally ON since its base current ib2 is greater than zero, i.e., ib2 = VCC /R when T2 is ON. When T2 is ON, vce2 = 0, therefore ib1 = 0. When ib1 = 0, T1 is OFF. When T1 is OFF and T2 is ON, the capacitor C is charged to VCC , positive at the left terminal. This is a stable state; there is nothing to disturb this condition if the circuit is left to itself. [b] When S is closed momentarily, vbe2 is changed to −VCC and T2 snaps OFF. The instant T2 turns OFF, vce2 jumps to VCC R1 /(R1 + RL ) and ib1 jumps to VCC /(R1 + RL ), which turns T1 ON. [c] As soon as T1 turns ON, the charge on C starts to reverse polarity. Since vbe2 is the same as the voltage across C, it starts to increase from −VCC toward +VCC . However, T2 turns ON as soon as vbe2 = 0. The equation for vbe2 is vbe2 = VCC − 2VCC e−t/RC . So vbe2 = 0 when t = RC ln 2, therefore T2 stays OFF for RC ln 2 seconds. P 7.102 [a] For t < 0, vce2 = 0. When the switch is momentarily closed, vce2 jumps to VCC 12 R1 = (8000) = 2 V. R1 + RL 8000 + 40, 000 T2 remains open for (36,067)(100 × 10−12 ) ln 2 ∼ = 2.5 µs. 

vce2 =



7–86

CHAPTER 7. Response of First-Order RL and RC Circuits

[b] ib2 =

VCC = 332.71 µA, R

ib2 = 0, ib2

−2 ≤ t ≤ 0 µs;

0 < t < RC ln 2(= 2.5 µs);

=

VCC VCC −(t−RC ln 2)/RL C + e R RL

=

332.71 + 300e−250,000(t−2.5×10

−6 )

µA,

2.5 µs < t;

P 7.103 [a] 0 ≤ t ≤ 0.5: i = If + (I0 − If )e−t/τ =

9 12 9 + − e−t/τ 30 30 30 



i = 0.3 + 0.1e−30t/L ; i(0.1) = 0.3 + 0.1e−3/L = 0.35; .·. e3/L = 2;

L=

3 = 4.33 H. ln 2

[b] 0 ≤ t ≤ tr , where tr is the time the relay releases: 12 i=0+ − 0 e−30t/L = 0.4e−30t/L ; 30 



.·. 0.35 = 0.4e−30tr /L ;

e30tr /L = 1.143;

where τ = L/R.

Problems

tr =

7–87

4.33 ln 1.143 ∼ = 19.3 ms. 30

P 7.104 From the Practical Perspective, vC (t) = 0.75VS = VS (1 − e−t/RC ). 0.25 = e−t/RC

t = −RC ln 0.25.

so

In the above equation, t is the number of seconds it takes to charge the capacitor to 0.75VS , so it is a period. We want to calculate the heart rate, which is a frequency in beats per minute, so H = 60/t. Thus, H=

60 . −RC ln 0.25

P 7.105 In this problem, Vmax = 0.8VS , so the equation for heart rate in beats per minute is H=

60 . −RC ln 0.2

Given R = 100 kΩ and C = 20 µF, H=

60 = 74.56. −(100,000)(5 × 10−6 ) ln 0.2

Therefore, the heart rate is about 75 beats per minute. P 7.106 From the Practical Perspective, vC (t) = Vmax = VS (1 − e−t/RC ). Solve this equation for the resistance R: Vmax = 1 − e−t/RC VS

so

−t Vmax = ln 1 − ; RC VS 

Then,

.·.

e−t/RC = 1 −

R=

−t Vmax C ln 1 − VS 



.

Vmax . VS

7–88

CHAPTER 7. Response of First-Order RL and RC Circuits In the above equation, t is the time it takes to charge the capacitor to a voltage of Vmax . But t and the heart rate H are related as follows: H=

60 . t

Therefore, R=

−60 Vmax HC ln 1 − VS 

.

P 7.107 From Problem 7.106, R=

−60 Vmax HC ln 1 − VS 

.

Note that from the problem statement, Vmax = 0.7. VS Therefore, R=

−60 = 69.215 kΩ. (72)(10 × 10−6 ) ln (1 − 0.7)

Natural and Step Responses of RLC Circuits

Assessment Problems AP 8.1 [a] α =

1 1 = = 250; 2RC 2(1000)(2 × 10−6 )

ω02 =

1 1 = = 40,000; LC (12.5)(2 × 10−6 )

s1,2 = −250 ±

q

2502 − 40,000 = −250 ± 150;

s1 = −100 rad/s;

s2 = −400 rad/s.

[b] Overdamped. 1 1 = = 160; [c] α = 2RC 2(1562.5)(2 × 10−6 ) ω02 =

1 1 = = 40,000; LC (12.5)(2 × 10−6 )

s1,2 = −160 ±

q

1602 − 40,000 = −160 ± j120;

s1 = −160 + j120 rad/s; [d] α =

q

40,000 =

1 ; 2RC

s2 = −160 − j120 rad/s. .·. R =

1 = 1250 Ω. 2(200)(2 × 10−6 )

AP 8.2 From the given values of R, L, and C, α=

1 1 = = 2500 rad/s; 2RC 2(5000)(40 × 10−9 )

1 1 = = 20002 rad2 /s2 ; LC 6.25(40 × 10−9 ) √ s1 , s2 = −2500 ± 25002 − 20002 so s1 = −1000 rad/s and s2 = −4000 rad/s. ω02 =

8–1

8–2

CHAPTER 8. Natural and Step Responses of RLC Circuits −10 = −2 mA. 5000 [b] iC (0+ ) = −(iL (0+ ) + iR (0+ )) = −(8 − 2) = −6 mA. dvc (0+ ) −0.006 dvc (0+ ) = ic (0+ ) = −0.006, therefore = = −150,000 V/s. [c] C dt dt C [d] v = [A1 e−1000t + A2 e−4000t ] V, t ≥ 0+ ; [a] v(0− ) = v(0+ ) = −10,

v(0+ ) = A1 + A2 ,

therefore iR (0+ ) =

dv(0+ ) = −1000A1 − 4000A2 . dt

Therefore A1 + A2 = −10,

−A1 − 4A2 = −150;

A1 = −63.33 V.

[e] A2 = 53.33 V. [f ] v = [53.33e−4000t − 63.33e−1000t ] V,

t ≥ 0.

v(t) v(t) = = 10.67e−4000t − 12.67e−1000t mA. R 5000 dv(t) [b] iC (t) = C = 40 × 10−9 [(−4000)53.33e−4000t + (1000)63.33e−1000t ] dt

AP 8.3 [a] iR (t) =

= 2.533e−1000t − 8.533e−4000t mA. [c] iL (t) = −[iR (t) + iC (t)] = 2.133e−4000t − 10.133e−1000t mA. AP 8.4 [a] α = 400; ωd =

ωd = 300;

q

ω02 − α2 ;

.·. ω02 = ωd2 + α2 = 9 × 104 + 16 × 104 = 25 × 104 . 1 = 25 × 104 ; LC 1 L= = 16 mH. 4 (25 × 10 )(250 × 10−6 ) ω02 =

[b] α =

1 ; 2RC

.·. R =

1 1 = = 5 Ω. 2αC (800)(250 × 10−6 )

[c] V0 = v(0) = 120 V. [d] I0 = iL (0) = −iR (0) − iC (0); iR (0) =

120 = 24 A; 5

iC (0) = C

dv (0) = 250 × 10−6 [−400(120) + 300(80)] = −6 A; dt

.·. I0 = −24 + 6 = −18 A.

Problems [e] iC (t) = 250 × 10−6

dv(t) = e−400t (−17 sin 300t − 6 cos 300t) A; dt

v(t) = e−400t (24 cos 300t + 16 sin 300t) A; 5

iR (t) =

iL (t) = −iR (t) − iC (t) = e−400t (−18 cos 300t + sin 300t) A,

t ≥ 0.

Check: diL = 16 × 10−3 e−400t [7500 cos 300t + 5000 sin 300t]; L dt v(t) = e−400t [120 cos 300t + 80 sin 300t] V. 

AP 8.5 [a]

1 2RC

2

=

.·. C = [b]

1 = (500)2 ; LC

1 = 1 µF. (500)2 (4)

1 = 500; 2RC .·. R =

1 = 1 kΩ. 2(500)(10−6 )

[c] v(0) = D2 = 8 V; 8 = 8 mA; [d] iR (0) = 1000 iC (0) = −8 + 10 = 2 mA; dv 2 × 10−3 = 2000 V/s; (0) = D1 − 500D2 = dt 10−6 .·. D1 = 2000 + 500(8) = 6000 V/s. [e] v = 6000te−500t + 8e−500t V,

t ≥ 0;

dv = [−3 × 106 t + 2000]e−500t ; dt iC = C

dv = (−3000t + 2)e−500t mA, dt

t ≥ 0+ .

V0 50 = = 4 A. R 12.5 [b] iC (0+ ) = I − iR (0+ ) − iL (0+ ) = 2 − 4 − 1 = −3 A. Vo 50 diL (0+ ) [c] = = = 2000 A/s. dt L 0.025

AP 8.6 [a] iR (0+ ) =

8–3

8–4

CHAPTER 8. Natural and Step Responses of RLC Circuits 1 1 = 640; = 640,000; s1,2 = −640 ± j480 rad/s. 2RC LC [e] iL = If + B10 e−αt cos ωd t + B20 e−αt sin ωd t, If = I = 2 A; [d] α =

iL (0+ ) = 1 = If + B10 ,

therefore B10 = −1 A;

diL (0+ ) = 2000 = −αB10 + ωd B20 , dt

therefore B20 = 2.833 A;

Therefore iL (t) = 2 + e−640t [− cos 480t + 2.833 sin 480t] A, [f ] v(t) = AP 8.7 α =

LdiL = e−640t [50 cos 480t − 33.33 sin 480t] V dt

t ≥ 0.

R 200 = = 400 rad/s; 2L 2(0.25) s

ω0 =

1 = LC

α2 < ω02 : ωd =

√

s

1 (250 ×

10−3 )(16

× 10−6 )

= 500 rad/s;

underdamped.

5002 − 4002 = 300 rad/s;

vo = Vf + B10 e−400t cos 300t + B20 e−400t sin 300t; vo (∞) = 200(0.08) = 16 V; vo (0) = 0 = Vf + B10 = 0

so

dvo (0) = 0 = −400B10 + 300B20 dt

B10 = −16 V; so

B20 = −21.33 V;

.·. vo (t) = 16 − 16e−400t cos 300t − 21.33e−400t sin 300t V, AP 8.8 α =

R 250 = = 500 rad/s; 2L 2(0.25) s

ω0 =

1 = LC

α2 = ω02 :

s

1 (250 ×

10−3 )(16

critically damped.

vo = Vf + D10 te−500t + D20 e−500t ;

× 10−6 )

= 500 rad/s;

t ≥ 0.

t ≥ 0.

Problems vo (0) = 0 = Vf + D20 ; .·. D20 = −20 V;

vo (∞) = (250)(0.08) = 20 V; dvo (0) = 0 = D10 − αD20 dt .·. AP 8.9 α =

D10 = (500)(−20) = −10,000 V/s;

vo (t) = 20 − 10,000te−500t − 20e−500t V,

t ≥ 0.

R 312.5 = = 625 rad/s; 2L 2(0.25) s

ω0 =

so

1 = LC

α2 > ω02 :

s

1 (250 ×

10−3 )(16

× 10−6 )

= 500 rad/s;

overdamped.

s1,2 = −625 ±

√

6252 − 5002 = −250, −1000 rad/s;

vo = Vf + A01 e−250t + A02 e−1000t ; vo (0) = 0 = Vf + A01 + A02 ; vo (∞) = (312.5)(08) = 25 V;

.·. A01 + A02 = −25 V;

dvo (0) = 0 = −250A01 − 1000A02 ; dt Solving,

A01 = −33.33 V;

A02 = 8.33 V;

vo (t) = 25 − 33.33e−250t + 8.33e−1000t V,

t ≥ 0.

8–5

8–6

CHAPTER 8. Natural and Step Responses of RLC Circuits

Problems P 8.1

1 1 = , therefore C = 500 nF. 2 (2RC) LC 1 [b] α = 5000 = , therefore C = 1 µF; 2RC [a]

s

25 × 106 −

s1,2 = −5000 ± [c] √

1 = 20,000, LC 

s1,2 = −40 ±

therefore C = 125 nF;

q

(40)2

−

s1 = −5.36 krad/s, P 8.2

s1 = −α +

q

α2 − ω02 ;

s1 + s2 = −2α R=

→

(103 )(106 ) = (−5000 ± j5000) rad/s. 20

202



103 ,

s2 = −74.64 krad/s. s2 = −α +

q

α2 − ω02 ;

α = −(s1 + s2 )/2 = −(−200 − 1800)/2 = 1000 rad/s.

1 106 = = 27.78 Ω; 2αC (2000)(18)

v(0+ ) = −24 V; iR (0+ ) =

−24 = −864 mA; 27.78

dv = 2400e−200t + 21,600e−1800t ; dt dv(0+ ) = 2400 + 21,600 = 24,000 V/s; dt iC (0+ ) = 18 × 10−6 (24,000) = 432 mA; iL (0+ ) = −[iR (0+ ) + iC (0+ )] = −[−864 + 432] = 432 mA.

Problems P 8.3

[a] −α +

q

α2 − ω02 = −5000;

−α −

q

α2 − ω02 = −20,000;

.·. −2α = −25,000; α = 12,500 rad/s; 106 1 = = 12,500; 2RC 2R(0.05) R = 800 Ω; q

2 α2 − ω02 = 15,000; 4(α2 − ω02 ) = 225 × 106 ; .·. ω0 = 10,000 rad/s. ωo2 = 108 = .·. L = [b] iR =

1 ; LC

1 = 200 mH. 108 C

v(t) = −6.25e−5000t + 25e−20,000t mA, R

iC = C

dv(t) = 1.25e−5000t − 20e−20,000t mA, dt

iL = −(iR + iC ) = 5e−5000t − 5e−20,000t mA, P 8.4

[a] ω02 =

109 1 = = 25 × 106 ; LC 40

ω0 = 5000 rad/s; 1 = 5000; 2RC R=

R=

1 ; 10,000C

109 = 12.5 kΩ. 8 × 104

[b] v(t) = D1 te−5000t + D2 e−5000t ; v(0) = −25 V = D2 ; dv = (D1 t − 25)(−5000e−5000t ) + D1 e−5000t ; dt iC (0) dv (0) = 125 × 103 + D1 = ; dt C

t ≥ 0+ ; t ≥ 0+ ; t ≥ 0+ .

8–7

8–8

CHAPTER 8. Natural and Step Responses of RLC Circuits iC (0) = −iR (0) − iL (0); iR (0) =

−25 = −2 mA; 12,500

.·. iC (0) = 2 − (−1) = 3 mA; 3 × 10−3 dv (0) = = 0.375 × 106 = 3.75 × 105 ; .·. dt 8 × 10−9 .·. 1.25 × 105 + D1 = 3.75 × 105 . D1 = 2.5 × 105 = 25 × 104 V/s; .·. v(t) = (25 × 104 t − 25)e−5000t V, [c] iC (t) = 0 when

t ≥ 0.

dv (t) = 0; dt

dv = (25 × 104 t − 25)(−5000)e−5000t + e−5000t (25 × 104 ) dt = (375,000 − 125 × 107 t)e−5000t ; dv = 0 when 125 × 107 t1 = 375,000; dt

.·. t1 = 300 µs;

v(300µs) = 50e−1.5 = 11.1565 V. [d] iL (300µs) = −iR (300µs) =

11.1565 = 0.89252 mA; 12.5

wC (300µs) = 4 × 10−9 (11.1565)2 = 497.87 nJ; wL (300µs) = (2.5)(0.89252)2 × 10−6 = 1991.48 nJ; w(300µs) = ωC + ωL = 2489.35 nJ; w(0) = 4 × 10−9 (625) + 2.5(10−6 ) = 5000 nJ. % remaining = P 8.5

[a] iR (0) =

2489.35 (100) = 49.79%. 5000

90 = 45 mA; 2000

iL (0) = −30 mA; iC (0) = −iL (0) − iR (0) = 30 − 45 = −15 mA.

Problems

[b] α =

1 1 = = 25,000; 2RC 2(2000)(10 × 10−9 )

ω02 =

1 1 = = 4 × 108 ; LC (0.25)(10 × 10−9 )

s1,2 = −25,000 ±

q

6.25 × 108 − 108 (4) = −25,000 ± 15,000;

s1 = −10,000 rad/s;

s2 = −40,000 rad/s;

v = A1 e−10,000t + A2 e−40,000t . v(0) = A1 + A2 = 90; dv −15 × 10−3 (0) = −104 A1 − 4A2 × 104 = = −1.5 × 106 V/s; −9 dt 10 × 10 −A1 − 4A2 = −150; .·. −3A2 = −60;

A2 = 20;

v = 70e−10,000t + 20e−40,000t V, [c] iC

= C =

A1 = 70. t ≥ 0.

dv dt

10 × 10−9 [−70 × 104 e−10,000t − 80 × 104 e−40,000t ]

= −7e−10,000t − 8e−40,000t mA; iR = 35e−10,000t + 10e−40,000t mA; iL = −iC − iR = −28e−10,000t − 2e−40,000t mA, P 8.6

α=

t ≥ 0.

1 1 = = 20,000; 2RC 2(2500)(10 × 10−9 )

α2 = 4 × 108 ;

.·. α2 = ωo2 .

Critical damping: v = D1 te−αt + D2 e−αt ; iR (0+ ) =

90 = 36 mA; 2500

iC (0+ ) = −[iL (0+ ) + iR (0+ )] = −[−30 + 36] = −6 mA; v(0) = D2 = 90;

8–9

8–10

CHAPTER 8. Natural and Step Responses of RLC Circuits dv = D1 [t(−αe−αt ) + e−αt ] − αD2 e−αt ; dt iC (0) −6 × 10−3 dv (0) = D1 − αD2 = = = −6 × 105 ; −9 dt C 10 × 10 D1 = αD2 − 6 × 105 = (2 × 104 )(90) − 6 × 105 = 120 × 104 ; v = (120 × 104 t + 90)e−20,000t V,

P 8.7

t ≥ 0.

1 1 = = 12,000; 2RC 2(12,500/3)(10 × 10−9 ) 1 = 4 × 108 ; LC s1,2 = −12,000 ± j16,000 rad/s; .·. response is underdamped. v(t) = B1 e−12,000t cos 16,000t + B2 e−12,000t sin 16,000t; v(0+ ) = 90 V = B1 ;

iR (0+ ) =

90 = 21.6 mA; (12,500/3)

iC (0+ ) = −[iL (0+ ) + iR (0+ )] = −[−30 + 21.6] = 8.4 mA; dv(0+ ) 8.4 × 10−3 = = 840,000 V/s; dt 10 × 10−9 dv(0+ ) = −12,000B1 + 16,000B2 = 840,000, dt or − 3B1 + 4B2 = 210;

.·. B2 = 120 V;

v(t) = 90e−12,000t cos 16,000t + 120e−12,000t sin 16,000t V, P 8.8

√ 1 = 1000 2, ω0 = 103 , 2RC s1 = −414.21, s2 = −2414.21;

[a] α =

t ≥ 0.

therefore overdamped.

therefore v = A1 e−414.21t + A2 e−2414.21t ; dv(0+ ) iC (0+ ) = = 98,000 V/s; dt C − 414.21A1 − 2414.21A2 = 98,000;

v(0+ ) = 0 = A1 + A2 ; Therefore A1 = 49,

A2 = −49;

v(t) = 49[e−414.21t − e−2414.21t ] V,

t ≥ 0.

Problems [b]

Example 8.4: vmax ∼ = 74.1 V at 1.4 ms; Example 8.5: vmax ∼ = 36.1 V at 1.0 ms; Problem 8.8: vmax ∼ = 28.2 V at 0.9 ms. P 8.9

1 = 8000, therefore R = 62.5 Ω. 2RC 10 V [b] iR (0+ ) = = 160 mA; 62.5 Ω [a]

iC (0+ ) = −(iL (0+ ) + iR (0+ )) = −80 − 160 = −240 mA = C Therefore

dv(0+ ) −0.240 = = −240 kV/s. dt C

[c] B1 = v(0+ ) = 10 V,

dvc (0+ ) = ωd B2 − αB1 . dt

Therefore 6000B2 − 8000B1 = −240,000, [d] iL = −(iR + iC );

iR = v/R;

v = e−8000t [10 cos 6000t −

iC = C

iC = e−8000t [−240 cos 6000t + iL = 10e−8000t [8 cos 6000t + 

[a]

1 2RC

2

=

B2 = (−80/3) V. dv ; dt

80 sin 6000t] V. 3

Therefore iR = e−8000t [160 cos 6000t −

P 8.10

dv(0+ ) . dt

1280 sin 6000t] mA. 3

460 sin 6000t] mA; 3

82 sin 6000t] mA, 3

t ≥ 0.

1 1 1 = , → = 500, LC (0.4)(10 × 10−6 ) 2RC

R = 100 Ω.

8–11

8–12

CHAPTER 8. Natural and Step Responses of RLC Circuits [b] 0.5CV02 = 12.5 × 10−3 , [c]

0.5LI02

−3

= 12.5 × 10 ,

[d] D2 = v(0+ ) = 50, iR (0+ ) =

therefore V0 = 50 V. I0 = 250 mA.

dv(0+ ) = D1 − αD2 ; dt

50 = 500 mA. 100

Therefore iC (0+ ) = −(500 + 250) = −750 mA; Therefore

−0.75 dv(0+ ) = = −75,000 V/s; dt C

Therefore D1 − αD2 = −75,000;

α=

1 = 500, 2RC

D1 = −50,000 V/s.

[e] v = [50e−500t − 50,000te−500t ] V; iR = P 8.11

v = [0.5e−500t − 500te−500t ] A, R

t ≥ 0+ .

From the form of the solution we have v(0) = A1 + A2 ; dv(0+ ) = −α(A1 + A2 ) + jωd (A1 − A2 ). dt We know both v(0) and dv(0+ )/dt will be real numbers. To facilitate the algebra we let these numbers be K1 and K2 , respectively. Then our two simultaneous equations are K 1 = A1 + A2 ; K2 = (−α + jωd )A1 + (−α − jωd )A2 . The characteristic determinant is

∆=

1 (−α + jωd )

1 (−α − jωd )

= −j2ωd .

The numerator determinants are

N1 =

K1 K2

1 (−α − jωd )

= −(α + jωd )K1 − K2

Problems

and N2 =

1 (−α + jωd )

It follows that A1 = and A2 =

K1 K2

= K2 + (α − jωd )K1 .

ωd K1 − j(αK1 + K2 ) N1 = ; ∆ 2ωd

N2 ωd K1 + j(αK1 + K2 ) . = ∆ 2ωd

We see from these expressions that P 8.12

8–13

A1 = A∗2 .

By definition, B1 = A1 + A2 . From the solution to Problem 8.11 we have A1 + A2 =

2ωd K1 = K1 . 2ωd

But K1 is v(0), therefore, B1 = v(0), which is identical to Eq. 8.16. By definition, B2 = j(A1 − A2 ). From Problem 8.11 we have B2 = j(A1 − A2 ) =

αK1 + K2 j[−2j(αK1 + K2 )] = . 2ωd ωd

It follows that K2 = −αK1 + ωd B2 ,

dv(0+ ) but K2 = dt

and K1 = B1 .

Thus we have dv + (0 ) = −αB1 + ωd B2 , dt which is identical to Eq. 8.17. P 8.13

1 = 1 rad/s; 2RC 1 = 10; ω02 = LC √ ωd = 10 − 1 = 3 rad/s;

[a] α =

.·. v = B1 e−t cos 3t + B2 e−t sin 3t. v(0) = B1 = 0; iR (0+ ) = 0 A;

v = B2 e−t sin 3t; iC (0+ ) = −iL (0+ ) = 3 A;

12 = −αB1 + ωd B2 = −1(0) + 3B2 ; .·. B2 = 4; .·. v = 4e−t sin 3t V,

t ≥ 0.

dv + 3 (0 ) = = 12 V/s; dt 0.25

8–14

CHAPTER 8. Natural and Step Responses of RLC Circuits

[b]

dv = 4e−t (3 cos 3t − sin 3t); dt dv = 0 when 3 cos 3t = sin 3t or dt .·. 3t1 = 1.25, t1 = 416.35 ms; 3t2 = 1.25 + π,

tan 3t = 3;

t2 = 1463.55 ms;

3t3 = 1.25 + 2π,

t3 = 2510.74 ms.

2π 2π = = 2094.40 ms. ωd 3 Td 2094.40 [d] t2 − t1 = 1047.20 ms; = = 1047.20 ms. 2 2 [e] v(t1 ) = 4e−(0.41635) sin 3(0.41635) = 2.50 V; [c] t3 − t1 = 2094.40 ms;

Td =

v(t2 ) = 4e−(1.46355) sin 3(1.46355) = −0.88 V; v(t3 ) = 4e−(2.51074) sin 3(2.51074) = 0.31 V. [f ]

P 8.14

[a] α = 0;

ωd = ω0 =

√

10 = 3.16 rad/s;

v = B1 cos ω0 t + B2 sin ω0 t; C

v(0) = B1 = 0;

dv (0) = −iL (0) = 3; dt

12 = −αB1 + ωd B2 = −0 + √ .·. B2 = 12/ 10 = 3.79 V; v = 3.79 sin 3.16t V, [b] ωd = 3.16 rad/s. [c] 3.79 V.

√

10B2 ;

t ≥ 0.

v = B2 sin ω0 t;

Problems P 8.15

[a] s1 = −α +

q

α2 − ω02 = −40;

s2 = −α −

.·. s1 + s2 = −2α = −200;

q

α2 − ω02 = −160;

α = 100 rad/s;

q

s1 − s2 = 2 α2 − ω02 = 120;

ω0 = 80 rad/s;

C=

1 1 = = 25 µF; 2αR 200(200)

L=

1 106 = = 6.25 H; ω02 C (80)2 (25)

iC (0+ ) = A1 + A2 = 15 mA; diL (0) diR (0) diC (0) =− − ; dt dt dt diL (0) 0 = = 0 A/s; dt 6.25 1 dv(0) 1 iC (0) 15 × 10−3 diR (0) = = = = 3 A/s; dt R dt R C (200)(25 × 10−6 ) diC (0) .·. = −3 A/s; dt .·. 160A1 + 40A2 = 3. 4A1 + A2 = 75 × 10−3 ;

.·. A1 = 20 mA;

.·. iC = 20e−160t − 5e−40t mA,

A2 = −5 mA;

t ≥ 0.

[b] By hypothesis v = A3 e−160t + A4 e−40t ,

t ≥ 0;

v(0) = A3 + A4 = 0; 15 × 10−3 dv(0) = = 600 V/s dt 25 × 10−6 −160A3 − 40A4 = 600; v = −5e−160t + 5e−40t V,

.·. A3 = −5 V; t ≥ 0.

v = −25e−160t + 25e−40t mA, 200 [d] iL = −iR − iC ; [c] iR (t) =

iL = 5e−160t − 20e−40t mA,

A4 = 5 V

t ≥ 0.

t ≥ 0+ .

8–15

8–16 P 8.16

CHAPTER 8. Natural and Step Responses of RLC Circuits t 0:

iR (0) =

60 = 37.5 mA; 1600

iL (0) = 45 mA;

iC (0) = −37.5 − 45 = −82.5 mA; α=

1 109 = = 5000 rad/s; 2RC 3200(62.5)

ω02 =

109 1 = = 16 × 106 ; LC 62.5

s1,2 = −5000 ±

√

25 × 106 − 16 × 106 = −5000 ± 3000;

s1 = −2000 rad/s;

s2 = −8000 rad/s;

.·. vo = A1 e−2000t + A2 e−8000t . A1 + A2 = vo (0) = 60; dvo −82.5 × 10−3 (0) = −2000A1 − 8000A2 = = −1320 × 103 ; dt 62.5 × 10−9 Solving,

A1 = −140 V,

A2 = 200 V;

.·. vo = −140e−2000t + 200e−8000t V,

t ≥ 0.

Problems

P 8.17

ω02 =

α=

1 1 = = 25 × 106 ; −9 LC 0.64(62.5 × 10 )

1 1 = = 4000 rad/s; 2RC 2(2000)(62.5 × 10−9 )

ωd =

q

α2 = 16 × 103 ;

(25 − 16) × 106 = 3000 rad/s;

s1,2 = −4000 ± j3000rad/s; vo (t) = B1 e−4000t cos 3000t + B2 e−4000t sin 3000t. vo (0) = B1 = 60 V; iR (0) =

60 = 30 mA; 2000

iL (0) = 45 mA; iC (0) = −iR (0) − iL (0) = −75 mA; iC (0) −75 × 10−3 = = −12 × 105 ; C 62.5 × 10−9 dvo (0) = −4000B1 + 3000B2 = −12 × 105 ; dt .·. 3B2 = 4B1 − 1200 = 240 − 1200 = −960;

.·. B2 = −320 V;

vo (t) = 60e−4000t cos 3000t − 320e−4000t sin 3000t V, P 8.18

ω02 =

α=

1 1 = = 108 ; LC (0.16)(62.5 × 10−9 )

1 1 = = 104 ; −9 2RC 2(800)(62.5 × 10 )

.·. α2 = ω02 (critical damping). vo (t) = D1 te−10,000t + D2 e−10,000t ; vo (0) = D2 = 60 V;

ω0 = 104 ;

t ≥ 0.

8–17

8–18

CHAPTER 8. Natural and Step Responses of RLC Circuits

iR (0) =

60 = 75 mA; 800

iL (0) = 45 mA; iC (0) = −(75 + 45) = −120 mA; dvo iC (0) −120 × 10−3 (0) = −10,000D2 + D1 = = = −1920 × 103 ; dt C 62.5 × 10−9 .·.

D1 = −1320 × 103 V/s;

vo (t) = (60 − 132 × 104 t)e−10,000t V,

t > 0.

P 8.19

vT = 104

iT (150 × 103 ) (150)(60)106 + iT ; 210 × 103 210 × 103

1500 × 103 9000 × 103 10,500 vT = + = × 103 = 50 kΩ; iT 210 210 210 75 (6) = 45 V; 10

Vo =

Io = 0;

iC (0) = −iR (0) − iL (0) = −

45 = −0.9 mA; 50,000

iC (0) −0.9 × 10−3 = = −720 × 103 ; C 1.25 × 10−9 ω02 =

α=

1 1 = = 108 ; LC (8)(1.25 × 10−9 )

ω0 = 104 rad/s;

1 1 = = 8000 rad/s; 2RC (2)(50,000)(1.25 × 10−9 )

ωd =

q

(100 − 64) × 106 = 6000 rad/s;

Problems vo = B1 e−8000t cos 6000t + B2 e−8000t sin 6000t. vo (0) = B1 = 45 V; dvo (0) = 6000B2 − 8000B1 = −720 × 103 ; dt .·. 6000B2 = 8000(45) − 720 × 103 ;

.·. B2 = −60 V;

vo = 45e−8000t cos 6000t − 60e−8000t sin 6000t V,

t ≥ 0.

P 8.20

× 103 ) (150)(60)106 + iT ; 210 × 103 210 × 103

4 iT (150

vT = 10

vT 1500 × 103 9000 × 103 10,500 = + = × 103 = 50 kΩ; iT 210 210 210 75 (6) = 45 V; 10

Vo =

Io = 0;

iC (0) = −iR (0) − iL (0) = −

45 = −0.9 mA; 50,000

iC (0) −0.9 × 10−3 = = −900 × 103 ; −9 C 10 ω02 =

α=

1 1 = = 108 ; LC (10)(10−9 )

ω0 = 10,000 rad/s;

1 1 = = 10,000 rad/s; 2RC (2)(50,000)(10−9 )

α2 = ω02

so the response is critically damped.

vo = D1 te−10,000t + D2 e−10,000t ;

8–19

8–20

CHAPTER 8. Natural and Step Responses of RLC Circuits vo (0) = D2 = 45 V; dvo (0) = D1 − αD2 = −900 × 103 ; dt .·. D1 = −900 × 103 + (10,000)(45);

.·. D1 = −450,000 V/s;

vo = −450,000te−10,000t + 45e−10,000t V,

t ≥ 0.

P 8.21

vT = 104

iT (150 × 103 ) (150)(60)106 + iT ; 210 × 103 210 × 103

1500 × 103 9000 × 103 10,500 vT = + = × 103 = 50 kΩ; iT 210 210 210 75 (6) = 45 V; 10

Vo =

Io = 0;

iC (0) = −iR (0) − iL (0) = −

45 = −0.9 mA; 50,000

−0.9 × 10−3 iC (0) = = −1125 × 103 ; C 0.8 × 10−9 ω02 =

α=

1 1 = = 108 ; LC (12.5)(0.8 × 10−9 )

ω0 = 10,000 rad/s;

1 1 = = 12,500 rad/s; 2RC (2)(50,000)(0.8 × 10−9 )

α2 > ω02

so the response is overdamped.

vo = A1 es1 t + A2 es2 t ; s1,2 = −α ±

q

α2 − ω02 = −12,500 ±

q

(12,500)2 − 108 = −12,500 ± 7500;

Problems

8–21

.·. s1,2 = −5000 rad/s, −20,000 rad/s; A1 + A2 = Vo = 45 .·. A1 = −15,

− 5000A1 − 20,000A2 = −1125 × 103 ;

and

A2 = 60;

vo = −15e−5000t + 60e−20,000t V, P 8.22

[a]

t ≥ 0.

1 = 20,0002 . LC There are many possible solutions. This one begins by choosing L = 1 mH. Then, C=

1 = 2.5 µF. (10−3 )(20,000)2

We can achieve this capacitor value using components from Appendix H by combining a 2.2 µF and three 0.1 µF capacitors in parallel. Critically damped: .·. R =

α = ω0 = 20,000

so

1 = 20,000; 2RC

1 = 10 Ω. 2(2.5 × 10−6 )(20,000)

Appendix H has a 10 Ω resistor. The final circuit:

q

[b] s1,2 = −α ± α2 − ω02 = −20,000 ± 0; Therefore there are two repeated real roots at −20,000 rad/s. P 8.23

[a] Underdamped response: α < ω0

so

α < 20,000.

Therefore we choose a larger resistor value than the one used in Problem 8.22. Choose R = 15 Ω: 1 α= = 13,333.33; 2(15)(2.5 × 10−6 ) s1,2 = −13,333.33 ±

q

13,333.332 − 20,0002 = −13,333.33 ± j14,907.12 rad/s.

8–22

CHAPTER 8. Natural and Step Responses of RLC Circuits [b] Overdamped response: α > ω0

so

α > 20,000.

Therefore we choose a smaller resistor value than the one used in Problem 8.22. Choose R = 5 Ω, which we can construct using two parallel-connected 10 Ω resistors: α=

1 = 40,000; 2(5)(2.5 × 10−6 )

s1,2 = −40,000 ±

q

40,0002 − 20,0002 = −40,000 ± 34,641;

= −5359 rad/s; P 8.24

− 74,641 rad/s.

and

For t < 0 : iL (0− ) = iL (0+ ) =

12 = 30 mA; 400

vo (0− ) = vo (0+ ) = 0.

For t > 0

α=

1 = 1000 rad/s; 2RC

s1 = −400 rad/s

ω02 =

1 = 64 × 104 ; LC

s2 = −1600 rad/s;

vo (∞) = 0 = Vf ; vo = A01 e−400t + A02 e−1600t ; iC (0+ ) = −30 + 30 + 0 = 0; dvo iC (0+ ) · .. = = 0 = −400A01 − 1600A02 ; dt C .·. A01 + 400A02 = 0; .·. A01 = 0;

A02 = 0.

.·. vo = 0 for t ≥ 0.

vo (0) = A01 + A02 = 0.

Problems

Note:

vo (0) = 0;

vo (∞) = 0;

8–23

dvo (0) = 0. dt

Hence the 30 mA current circulates between the current source and the ideal inductor in the equivalent circuit. In the original circuit the 12 V source sustains a current of 30 mA in the inductor. This is an example of a circuit going directly into steady state when the switch is closed. There is no transient period, or interval. P 8.25

For

α=

t>0

1 = 100; 2RC

1 = 6400; LC

s1,2 = −100 ± 60; s1 = −40 rad/s;

s2 = −160 rad/s;

vo = Vf + A01 e−40t + A02 e−160t ; Vf = 0;

vo (0+ ) = 0;

iC (0+ ) = 37.5 mA;

.·. A01 + A02 = 0; iC (0+ ) dvo (0+ ) = = 6000 V/s; dt 6.25 × 10−6 dvo (0+ ) = −40A01 − 160A02 ; dt −40A01 − 160A02 = 6000; A01 + 4A02 = −150; A01 + A02 = 0; .·. A01 = 50 V;

A02 = −50 V.

vo = 50e−40t − 50e−160t V,

t ≥ 0.

8–24 P 8.26

CHAPTER 8. Natural and Step Responses of RLC Circuits [a] From the solution to Prob. 8.25 s1 = −40 rad/s and s2 = −160 rad/s, therefore io = If + A01 e−40t + A02 e−160t . If = 37.5 mA;

dio (0+ ) = 0; dt

io (0+ ) = 0;

.·. 0 = 37.5 + A01 + A02 ;

−40A01 − 160A02 = 0.

It follows that A01 = −50 mA;

A02 = 12.5 mA;

.·. io = 37.5 − 50e−40t + 12.5e−160t mA, [b]

dio = 2e−40t − 2e−160t ; dt vo = L

dio = 25[2e−40t − 2e−160t ]; dt

vo = 50e−40t − 50e−160t V, P 8.27

α=

t ≥ 0.

t ≥ 0.

106 1 = = 100; 2RC (1600)(6.25)

ω02 =

α2 = 104 ;

1 106 = = 6400; LC (25)(6.25)

s1,2 = −100 ±

√

104 − 6400 = −100 ± 60;

s1 = −40 rad/s;

s2 = −160 rad/s;

io = If + A01 e−40t + A02 e−160t ; If =

30 = 37.5 mA; 800

io (0) = 0;

0 = 37.5 × 10−3 + A01 + A02 ,

.·. A01 + A02 = −37.5 × 10−3 ;

dio 30 (0) = = −40A01 − 160A02 . dt 25 Solving,

A01 = −40 mA;

A02 = 2.5 mA.

io = 37.5 − 40e−40t + 2.5e−160t mA,

t ≥ 0.

Problems

P 8.28

[a] α =

1 106 = = 100; 2RC (1600)(6.25)

8–25

α2 = 104 ;

1 106 = = 6400; LC (25)(6.25) √ = −100 ± 104 − 6400 = −100 ± 60;

ω02 = s1,2

s1 = −40 rad/s;

s2 = −160 rad/s;

vo (∞) = 0 = Vf ; .·. vo = A01 e−40t + A02 e−160t . vo (0) = 30 = A01 + A02 ; iC (0+ ) = 0;

Note:

dvo (0) = 0 = −40A01 − 160A02 . .·. dt A01 = 40 V,

Solving,

A02 = −10 V.

vo (t) = 40e−40t − 10e−160t V, [b] io (t) =

t > 0+ .

1Zt (40e−40x − 10e−160x ) dx + I0 25 0 40e−40t 10e−160x − −40 −160

1 = 25

! t

0

t

= (−0.04e−40x + 0.0025e−160x )

0

= −40e−40t + 2.5e−160t + 37.5 mA,

t ≥ 0.

!

P 8.29

P 8.30

diL [a] v = L = 16[e−20,000t − e−80,000t ] V, t ≥ 0. dt v [b] iR = = 40[e−20,000t − e−80,000t ] mA, t ≥ 0+ . R [c] iC = I − iL − iR = [−8e−20,000t + 32e−80,000t ] mA, t ≥ 0+ . diL [a] v = L dt

!

= 40e−32,000t sin 24,000t V,

[b] iC (t) = I − iR − iL = 24 × 10−3 −

t ≥ 0.

v − iL 625

= [24e−32,000t cos 24,000t − 32e−32,000t sin 24,000t] mA, P 8.31

diL v=L dt

!

= 960,000te−40,000t kV,

t ≥ 0.

t ≥ 0+ .

8–26

P 8.32

CHAPTER 8. Natural and Step Responses of RLC Circuits 1 vC (0+ ) = (240) = 120 V; 2 iL (0+ ) = 60 mA; α=

iL (∞) =

240 = 48 mA; 5000

1 1 = = 40; 2RC 2(5000k5000)(5 × 10−6 )

ω02 =

1 1 = = 2500; LC 80(5 × 10−6 )

α2 = 1600;

α2 < ω02 ;

.·.

underdamped;

√ s1,2 = −40 ± j 2500 − 1600 = −40 ± j30 rad/s; iL

= If + B10 e−αt cos ωd t + B20 e−αt sin ωd t =

0.048 + B10 e−40t cos 30t + B20 e−40t sin 30t.

iL (0) = 0.048 + B10 ;

B10 = 0.060 − 0.048 = 12 mA;

120 diL (0) = 30B20 − 40B10 = = 1.5 = 1500 × 10−3 ; dt 80 .·. 30B20 = 40(12 × 10−3 ) + 1500 × 10−3 ;

B20 = 66 mA;

.·. iL = 48 + 12e−40t cos 30t + 66e−40t sin 30t mA, P 8.33

t 0:

iL (0− ) = I0 = 3/150 = 20 mA;

300k150 = 100 Ω;

iL (0) = 20 mA,

120 − 20 = 100 mA.

iL (∞) = −100 mA;

t ≥ 0.

vC (0− ) = V0 = 0.

Problems ω02 = α=

1 1 = = 64 × 106 ; −9 LC (0.03125)(500 × 10 )

1 1 = = 104 ; 2RC 2(100)(500 × 10−9 )

ω0 = 8000 rad/s;

α2 = 100 × 106 ;

α2 − ω02 = (100 − 64)106 = 36 × 106 ; s1,2 = −10,000 ± 6000; s1 = −4000 rad/s;

s2 = −16,000 rad/s;

iL = If + A01 e−4000t + A02 e−16,000t ; iL (∞) = If = −100 mA; iL (0) = A01 + A02 + If = 20 mA; .·. A01 + A02 − 100 = 20 so A01 + A02 = 120 mA; diL (0) = 0 = −4000A01 − 16,000A02 ; dt A01 = 160 mA,

Solving,

A02 = −40 mA.

iL = −100 + 160e−4000t − 40e−16,000t mA, s

P 8.34

ω0 =

1 = LC

s

t ≥ 0.

1 = 1250 rad/s; (640 × 10−3 )(10−6 )

1 1 = = 1000 rad/s 2RC 2(500)(10−6 ) √ ωd = 12502 − 10002 = 750;

α=

.·. underdamped.

If = −1 A; iL = −1 + B10 e−1000t cos 750t + B20 e−1000t sin 750t; iL (0) = −1 + B10 = 0.5

so

B10 = 1.5;

diL V0 (0) = −αB10 + ωd B20 = ; dt L .·.

−1000(1.5) + 750B20 =

40 , 640 × 10−3

so

iL (t) = −1 + e−1000t [1.5 cos 750t + (25/12) sin 750t] A,

B20 = 25/12; t ≥ 0.

8–27

8–28

P 8.35

CHAPTER 8. Natural and Step Responses of RLC Circuits

α=

1 1 = = 1000 rad/s; 2RC 2(500)(10−6 ) s

ω0 =

1 = LC

s

1 = 800 rad/s; (1.5625)(10−6 ) s1,2 = −1000 ±

Overdamped:

√

10002 − 8002 = −400, −1600 rad/s;

If = −1 A; iL = −1 + A01 e−400t + A02 e−1600t; iL (0) = −1 + A01 + A02 = 0.5

so

A01 + A02 = 1.5;

40 V0 diL (0) = −400A01 − 1600A02 = = = 25.6; dt L 1.5625 A01 = 2.02133,

Solving,

A02 = −0.52133;

iL (t) = −1 + 2.02133e−400t − 0.52133e−1600t A, P 8.36

α=

1 1 = = 1000; 2RC 2(500)(10−6 ) s

ω0 =

t ≥ 0.

1 = LC

α2 = ω02

s

1 = 1000 rad/s; (1)(10−6 )

Critically damped.

If = −1 A; iL = −1 + D10 te−1000t + D20 e−1000t ; iL (0) = −1 + D20 = 0.5;

.·. D20 = 1.5 A;

V0 diL (0) = D10 − αD20 = ; dt L .·.

D10 − 1000(1.5) =

40 1

so

iL = −1 + 1540te−1000t + 1.5e−1000t A,

D10 = 1540; t ≥ 0.

Problems P 8.37

t < 0: vo (0− ) = vo (0+ ) =

1000 (25) = 20 V; 1250

iL (0− ) = iL (0+ ) = 0. t>0

−0.1 +

20 + iC (0+ ) + 0 = 0; 200

.·. iC (0+ ) = 0;

106 1 = = 250 rad/s; 2RC (400)(10) ω02 =

1 106 = = 62,500; LC 10(1.6)

.·. α2 = ω02

critically damped.

[a] iL = If + D10 te−250t + D20 e−250t ; +

iL (0 ) = 0;

If = 100 mA;

.·. 0 = 0.1 + D20 ;

20 diL (0+ ) = = 12.5 A/s; dt 1.6

D20 = −100 mA;

−250D20 + D10 = 12.5;

D10 = −12.5 A/s;

.·. iL = 100 − 12,500te−250t − 100e−250t mA [b] vo = L

dio = 1.6(−12.5e−250t + 3125te−250t + 25e−250t ) dt

= 5000te−250t + 20e−250t V, P 8.38

[a] wL =

t ≥ 0.

Z ∞ 0

pdt =

Z ∞ 0

t ≥ 0+ .

vo iL dt;

vo = 5000te−250t + 20e−250t V; iL = 0.1 − 12.5te−250t − 0.1e−250t A; p = 2e−250t + 500te−250t − 750te−500t − 62,500t2 e−500t − 2e−500t W.

8–29

8–30

CHAPTER 8. Natural and Step Responses of RLC Circuits Z wL = 2

∞

Z ∞

−250t

e

dt + 250

0

−250t

te

dt − 375

0

Z ∞

31,250

t2 e−500t dt −

∞ e−250t −250 0

te−500t −

0

Z ∞

0

=

Z ∞

e−500t dt

0

∞

250 −250t e (−250t − 1) + − (250)2 0 ∞

375 −500t e (−500t − 1) − (500)2 0 ∞

31,250 −500t 2 2 e (500 t + 1000t + 2) − 3 (−500) 0 ∞

e−500t . (−500) 0 All the upper limits evaluate to zero hence 1 250 375 (31,250)(2) 1 wL = + − − − ; 4 3 6 2 250 62,500 25 × 10 (5) 10 500 wL = 8 + 8 − 3 − 1 − 4 = 8 mJ. Note this value corresponds to the final energy stored in the inductor, i.e. 1 wL (∞) = (1.6)(0.1)2 = 8 mJ. 2 −250t [b] v = 5000te + 20e−250t V; iR =

v = 25te−250t + 0.1e−250t A; 200

pR = viR = 2e−500t [62,500t2 + 500t + 1]; wR =

Z ∞ 0

pR dt;

Z wR = 62,500 2

∞

Z ∞

t2 e−500t dt + 500

0

te−500t dt +

Z ∞

0

0

∞ 62,500e−500t 4 2 + = [25 × 10 t + 1000t + 2] −125 × 106 0

∞ e−500t ∞ 500e−500t + (−500t − 1) . 25 × 104 (−500) 0 0

Since all the upper limits evaluate to zero we have wR 62,500(2) 500 1 = + + . 2 125 × 106 25 × 104 500 wR = 2 + 4 + 4 = 10 mJ.

e−500t dt

Problems [c] 100 = iR + iC + iL

8–31

(mA);

iR + iL = 25,000te−250t + 100e−250t + 100 − 12,500te−250t − 100e−250t mA = 100 + 12,500te−250t mA; .·. iC = 100 − (iR + iL ) = −12,500te−250t mA = −12.5te−250t A; pC = viC = [5000te−250t + 20e−250t ][−12.5te−250t ] = −250[250t2 e−500t + te−500t ]. Z ∞ Z ∞ wC 2 −500t = 250 te dt + te−500t dt; −250 0 0 ∞ ∞ 250e−500t e−500t wC 4 2 + . = [25 × 10 t + 1000t + 2] (−500t − 1) −250 −125 × 106 25 × 104 0 0



Since all upper limits evaluate to zero we have 250(1) −250(250)(2) − = −1000 × 10−6 − 10 × 10−4 = −2 mJ. 125 × 106 25 × 104 Note this 2 mJ corresponds to the initial energy stored in the capacitor, i.e., 1 wC (0) = (10 × 10−6 )(20)2 = 2 mJ. 2 Thus wC (∞) = 0 mJ which agrees with the final value of v = 0.

wC =

[d] The energy delivered by the voltage source is the same as the energy delivered by the equivalent current source is = 100 mA. ps (del) = 0.1vo W = 0.1[5000te−250t + 20e−250t ] = 2e−250t + 500te−250t W; Z ws = 2

= =

∞

−250t

e

dt +

0

250te−250t dt

0

e−250t ∞

−250

Z ∞

0

∞ 250e−250t + (−250t − 1) 62,500 0

1 1 + ; 250 250

2(2) 4 = = 16 mJ. 250 250 [e] wL = 8 mJ (absorbed); wR = 10 mJ (absorbed); ws =

wC = 2 mJ (delivered); X

wdel = wabs = 18 mJ.

ws = 16 mJ (delivered);

8–32 P 8.39

CHAPTER 8. Natural and Step Responses of RLC Circuits α = 800 rad/s;

ωd = 600 rad/s;

ωo2 − α2 = 36 × 104 ; α=

R = 800; 2L

ωo2 = 100 × 104 ;

wo = 1000 rad/s;

R = 1600L;

1 = 100 × 104 ; LC

L=

106 = 2 mH; (100 × 104 )(500)

.·. R = 3.2 Ω; i(0+ ) = B1 = 0 A;

at t = 0+ .

12 + 0 + vL (0+ ) = 0;

vL (0+ ) = −12 V;

di(0+ ) −12 = = −6000 A/s; dt 0.002 di(0+ ) .·. = 600B2 − 800B1 = −6000; dt .·. 600B2 = 800B1 − 6000; .·. i = −10e−800t sin 600t A, P 8.40

.·. B2 = −10 A; t ≥ 0.

From Prob. 8.39 we know vc will be of the form vc = B3 e−800t cos 600t + B4 e−800t sin 600t. From Prob. 8.39 we have vc (0) = 12 V = B3 , and iC (0) dvc (0) = = 0; dt C dvc (0) = 600B4 − 800B3 ; dt .·. 600B4 = 800B3 + 0;

B4 = 16 V;

vc (t) = 12e−800t cos 600t + 16e−800t sin 600t V

t ≥ 0.

Problems P 8.41

q

α2 − ω02 = −4000;

[a] −α +

.·. α = 10,000 rad/s, α=

R = 10,000; 2L

ω02 =

−α −

q

α2 − ω02 = −16,000;

ω02 = 64 × 106 ; R = 20,000L;

1 = 64 × 106 ; LC

L=

109 = 0.5 H; 64 × 106 (31.25)

R = 10,000 Ω. [b] i(0) = 0; L

di(0) = vc (0); dt

.·. vc2 (0) = 576;

1 (31.25) × 10−9 vc2 (0) = 9 × 10−6; 2 vc (0) = 24 V;

24 di(0) = = 48 A/s. dt 0.5 [c] i(t) = A1 e−4000t + A2 e−16,000t ; i(0) = A1 + A2 = 0; di(0) = −4000A1 − 16,000A2 = 48. dt Solving, .·. A1 = 4 mA;

A2 = −4 mA;

i(t) = 4e−4000t − 4e−16,000t mA, [d]

t ≥ 0.

di(t) = −16e−4000t + 64e−16,000t ; dt di = 0 when 64e−16,000t = 16e−4000t , dt or e12,000t = 4; .·. t =

ln 4 = 115.52 µs. 12,000

[e] imax = 4e−0.4621 − 4e−1.8484 = 1.89 mA. di [f ] vL (t) = 0.5 = [−8e−4000t + 32e−16,000t ] V, dt

t ≥ 0+ .

8–33

8–34

P 8.42

CHAPTER 8. Natural and Step Responses of RLC Circuits [a] ω02 = α=

1 109 108 = = = 25 × 106 ; LC (0.25)(160) 4 R = ω0 = 5000 rad/s; 2L

.·. R = (5000)(2)L = 2500 Ω. [b] i(0) = iL (0) = 24 mA; vL (0) = 90 − (0.024)(2500) = 30 V; 30 di (0) = = 120 A/s. dt 0.25 [c] vC = D1 te−5000t + D2 e−5000t ; vC (0) = D2 = 90 V; dvC iC (0) −iL (0) (0) = D1 − 5000D2 = = ; dt C C D1 − 450,000 = −

24 × 10−3 = −150,000; 160 × 10−9

.·. D1 = 300,000 V/s; vC = 300,000te−5000t + 90e−5000t V, P 8.43

t ≥ 0+ .

[a] For t > 0:

Since i(0− ) = i(0+ ) = 0, va (0+ ) = 300 V. 4

[b] va = 200i + 5 × 10

Z t

i dx + 300;

0

dva di = 200 + 5 × 104 i dt dt dva (0+ ) di(0+ ) di(0+ ) = 200 + 5 × 104 i(0+ ) = 200 ; dt dt dt −L

di(0+ ) = 300; dt

Problems di(0+ ) = −0.2(300) = −60 A/s; dt dva (0+ ) .·. = −12,000 V/s. dt R 800 [c] α = = = 80 rad/s; 2L 10 1 106 = = 104 ; LC (5)(20) √ = −80 ± 6400 − 104 = −80 ± j60 rad/s.

ω02 = s1,2

Underdamped: va = B1 e−80t cos 60t + B2 e−80t sin 60t; va (0) = B1 = 300 V; dva (0) = −80B1 + 60B2 = −12,000; dt

.·. B2 = 200 V;

va = 300e−80t cos 60t + 200e−80t sin 60t V, P 8.44

iL (0− ) = iL (0+ ) =

t ≥ 0+ .

70 = 280 mA; 50 + 200

vc (0− ) = vc (0+ ) = 200(0.280) = 56 V; ω02 = α=

1 1 = = 50 × 106 ; LC (0.100)(200 × 10−9 )

R 200 = = 1000; 2L 2(0.100)

α2 < ω02

.·.

α2 = 106 ;

underdamped.

s1,2 = −1000 ± j7000 rad/s; i = B1 e−1000t cos 7000t + B2 e−1000t sin 7000t; i(0) = B1 = 280 mA; di 1 (0) = 7000B2 − 1000B1 = [−200(0.28) − (−56)] = 0; dt 0.1 1 .·. B2 = B1 = 40 mA; 7 i = 280e−1000t cos 7000t + 40e−1000t sin 7000t mA,

t ≥ 0+ .

8–35

8–36 P 8.45

CHAPTER 8. Natural and Step Responses of RLC Circuits [a] t < 0: 120 = 15 mA; vo = (5000)(0.015) = 75 V. 8000 t > 0: R 5000 α= = = 2500 rad/s; 2L 2(1) io =

ω02 =

109 1 = = 4 × 106 = 400 × 104 ; LC (1)(250)

α2 − ω02 = 625 × 104 − 400 × 104 = 225 × 104 ; .·. s1,2 = −2500 ± 1500. s1 = −1000 rad/s

s2 = −4000 rad/s;

.·. io (t) = A1 e−1000t + A2 e−4000t . io (0) = A1 + A2 = 15 × 10−3 ; dio 1 (0) = −1000A1 − 4000A2 = [−5000(0.015) − (−75)] = 0; dt 1 Solving,

A2 = −5 mA.

A1 = 20 mA;

io (t) = 20e−1000t − 5e−4000t mA,

t ≥ 0+ .

[b] vo (t) = A1 e−1000t + A2 e−4000t ; vo (0) = A1 + A2 = 75; −15 × 10−3 dvo (0) = −1000A1 − 4000A2 = ; dt 250 × 10−9 Solving,

A1 = 80 V;

A2 = −5 V.

vo (t) = 80e−1000t − 5e−4000t V,

t ≥ 0+ .

Check: 5000io + 1

dio = vo ; dt

5000io = 100e−1000t − 25e−4000t ; dio = −20e−1000t + 20e−4000t ; dt dio .·. 5000io + = 80e−1000t − 5e−4000t V dt

(checks).

Problems P 8.46

t < 0:

i(0) =

100 100 = = 5 A; 4 + [10k(10 + 30)] + 8 20

10 = 70 V. vo (0) = 100 − 5(4) − 10(5) 50 



t > 0:

α=

20 R = = 5, 2L 4

ω02 =

α2 = 25;

1 100 = = 50; LC 2

ω02 > α2

underdamped.

vo = B1 e−αt cos ωd t + B2 e−αt sin ωd t;

ωd =

√

50 − 25 = 5;

vo = B1 e−5t cos 5t + B2 e−5t sin 5t; vo (0) = B1 = 70 V; C

dvo (0) = −5, dt

dvo −5 = × 103 = −500 V/s; dt 10

dvo (0) = −5B1 + 5B2 = −500; dt 5B2 = −500 + 5B1 = −500 + 350; .·. vo = 70e−5t cos 5t − 30e−5t sin 5t V,

B2 = −150/5 = −30 V; t ≥ 0.

8–37

8–38

P 8.47

CHAPTER 8. Natural and Step Responses of RLC Circuits

[a]

1 = 50002 . LC There are many possible solutions. This one begins by choosing L = 10 mH. Then, C=

1 (10 ×

10−3 )(5000)2

= 4 µF.

We can achieve this capacitor value using components from Appendix H by combining four 1 µF capacitors in parallel. Critically damped:

α = ω0 = 5000

so

R = 5000; 2L

.·. R = 2(0.01)(5000) = 100 Ω. Appendix H has a 100 Ω resistor. The final circuit:

q

[b] s1,2 = −α ± α2 − ω02 = −5000 ± 0. Therefore there are two repeated real roots at −5000 rad/s. P 8.48

[a] Underdamped response: α < ω0

so

α < 5000.

Therefore we choose a smaller resistor value than the one used in Problem 8.47 to give a smaller value of α. For convenience, pick α = 4000 rad/s: R = 4000 so R = 2(4000)(0.01) = 80 Ω. 2L We can create an 80 Ω resistance by combining a 33 Ω resistor and a 47 Ω resistor in series. √ s1,2 = −4000 ± 40002 − 50002 = −4000 ± j3000 rad/s.

α=

[b] Overdamped response: α > ω0

so

α > 5000.

Therefore we choose a larger resistor value than the one used in Problem 8.47. Choose R = 125 Ω, which can be created by combining a 100 Ω resistor, a 15 Ω resistor and a 10 Ω resistor in series: α= s1,2

R = 6250; 2L √ = −6250 ± 62502 − 50002 = −6250 ± 3750 = −2500 rad/s;

and

− 10,000 rad/s.

Problems P 8.49

8–39

i(0+ ) = 0, since there is no source connected to L for t < 0. !

15,000 (80) = 50 V; 15,000 + 9000

−

+

vC (0 ) = vC (0 ) =

s

R α= = 8000; 2L ωd =

q

ω0 =

1 = 10,000; LC

Vf = 100 V.

the response is underdamped.

10,0002 − 80002 = 6000;

vC = Vf + B10 e−8000t cos 6000t + B20 e−8000t sin 6000t; Vf + B10 = V0

so

B10 = V0 − Vf = 50 − 100 = −50;

−αB10 + ωd B20 =

I0 , C

so B20

1 = ωd



1 I0 + αB10 = [0 + 8000(−50)] = −66.67; C 6000 

Therefore, vC = 100 − e−8000t (50 cos 6000t + 66.67 sin 6000t) V, P 8.50

i=C

t ≥ 0.

dvC dt

= 2 × 10−6 [8000e−8000t (50 cos 6000t + 66.67 sin 6000t) − e−8000t (6000)(−50 sin 6000t + 66.67 cos 6000t)] = 1.67e−8000t sin 6000t A, P 8.51

t > 0.

[a] From Problem 8.49, I0 = 0, V0 = 50 V, and Vf = 100 V. Then, R α= = 10,000; 2L

s

ω0 =

1 = 10,000; LC

the response is critically damped.

vC = Vf + D10 te−10,000t + D20 e−10,000t ; Vf + D20 = V0 D10 − αD20 =

so

D20 = V0 − Vf = 50 − 100 = −50;

I0 , C

so I0 + αD20 = 0 + 10,000(−50)] = −5 × 105 ; C Therefore, D10 =

vC = 100 − 500,000te−10,000t − 50e−10,000t V,

t ≥ 0.

8–40

CHAPTER 8. Natural and Step Responses of RLC Circuits

[b] i = C

dvC dt

= 2 × 10−6 [−500,000e−10,000t + 5 × 109 te−10,000t + 500,000e−10,000t ] = 10,000te−10,000t A, P 8.52

t > 0.

t < 0:

io (0− ) =

48 = 3 mA; 16,000

vC (0− ) = 20 − (12,000)(0.003) = −16 V. t = 0+ :

12 kΩk24 kΩ = 8 kΩ; .·. vo (0+ ) = (0.003)(8000) − 16 = 24 − 16 = 8 V, and vL (0+ ) = 20 − 8 = 12 V. vo (t) = 8000io + vC ; dvo dio dvC (t) = 8000 + ; dt dt dt dvo + dio dvC + (0 ) = 8000 (0+ ) + (0 ); dt dt dt vL (0+ ) = L

dio + (0 ); dt

Problems

8–41

vL (0+ ) 12 dio + (0 ) = = = 60 A/s; dt L 0.2 C

dvc + (0 ) = io (0+ ); dt

3 × 10−3 dvc + .·. (0 ) = = 375,000; dt 8 × 10−9 dvo + .·. (0 ) = 8000(60) + 375,000 = 855,000 V/s. dt ω02 =

α=

1 109 = = 625 × 106 ; LC 1.6

ω0 = 25,000 rad/s;

R 8000 = = 20,000 rad/s; 2L 0.4

α2 < ω02

α2 = 400 × 106 ;

underdamped.

s1,2 = −20,000 ± j15,000 rad/s; vo (t) = Vf + B10 e−20,000t cos 15,000t + B20 e−20,000t sin 15,000t; Vf = vo (∞) = 20 V; 8 = 20 + B10 ;

B10 = −12 V;

−20,000B10 + 15,000B20 = 855,000. Solving,

B20 = 41 V.

.·. vo (t) = 20 − 12e−20,000t cos 15,000t + 41e−20,000t sin 15,000t V, P 8.53

iC (0) = 0; α=

ω02

vo (0) = 200 V;

R 4 = = 50 rad/s; 2L 2(0.04)

1 103 = = = 2500; LC 0.4

.·. α2 = ωo2 ;

critical damping.

t ≥ 0+ .

8–42

CHAPTER 8. Natural and Step Responses of RLC Circuits vo (t) = Vf + D10 te−50t + D20 e−50t ; Vf = 100 V; vo (0) = 100 + D20 = 200;

D20 = 100 V;

dvo (0) = −50D20 + D10 = 0; dt D10 = 50D20 = 5000 V/s; vo = 100 + 5000te−50t + 100e−50t V, P 8.54

t ≥ 0.

t < 0:

iL (0) =

−160 = −100 mA; 1600

vC (0) = 1000iL (0) = −100 V. t > 0:

α=

R 1000 = × 103 = 5000 rad/s; 2L 200

ω02 =

1 (109 )(103 ) 108 = = = 25 × 106 ; LC (100)(400) 4

ω0 = 5000 rad/s

.·.

critical damping.

vC (t) = Vf + D10 te−5000t + D20 e−5000t ; vC (0) = −100 V;

Vf = −60 V;

Problems .·. −100 = −60 + D20 ; C

8–43

D20 = −40 V;

dvC (0) = iL (0) = −100 × 10−3 ; dt

dvC −100 × 10−3 (0) = = −250,000 V/s; dt 400 × 10−9 .·. D10 = 5000(−40) − 250,000 = −450,000. vC (t) = −60 − 450,000te−5000t − 40e−5000t V, P 8.55

t ≥ 0.

[a] vC = V + [B10 cos ωd t + B20 sin ωd t] e−αt ; dvC = [(ωd B20 − αB10 ) cos ωd t − (αB20 + ωd B10 ) sin ωd t]e−αt . dt Since the initial stored energy is zero, dvC (0+ ) = 0; dt

vC (0+ ) = 0 and

It follows that B10 = −V

and B20 =

αB10 . ωd

When these values are substituted into the expression for [dvC /dt], we get dvC = dt But

α2 + ωd V e−αt sin ωd t. ωd !

α2 + ωd2 ω2 α2 + ωd = = 0. ωd ωd ωd

Therefore [b]

ω02 V e−αt sin ωd t. ωd !

dvC = dt

dvC = 0 when dt

sin ωd t = 0,

or ωd t = nπ,

where n = 0, 1, 2, 3, . . . . Therefore t = [c] When tn = and

nπ , ωd

nπ . ωd cos ωd tn = cos nπ = (−1)n ,

sin ωd tn = sin nπ = 0.

Therefore vc (tn ) = V [1 − (−1)n e−αnπ/ωd ].

8–44

CHAPTER 8. Natural and Step Responses of RLC Circuits [d] It follows from [c] that v(t1 ) = V + V e−(απ/ωd ) Therefore But

vC (t1 ) − V e−(απ/ωd ) = −(3απ/ω ) = e(2απ/ωd ) . d vC (t3 ) − V e

2π = t3 − t1 = Td , ωd (

P 8.56

and vc (t3 ) = V + V e−(3απ/ωd ) .

vc (t1 ) − V 1 ln α= Td vc (t3 ) − V

thus α =

)

;

Td = t3 − t1 =

13.505 12,000 ln = 5000; α= 2π 0.985 

1 [vC (t1 ) − V ] ln . Td [vC (t3 ) − V ]



ωd =

3π π 2π − = ms. 12 12 12

2π = 12,000 rad/s. Td

ω02 = ωd2 + α2 = 144 × 106 + 25 × 106 = 169 × 106 ; L= P 8.57

1 = 29.6 mH; (169)(0.2)

R = 2αL = 295.86 Ω.

[a] Let i be the current in the direction of the voltage drop vo (t). Then by hypothesis i = If + B10 e−αt cos ωd t + B20 e−αt sin ωd t . If = i(∞) = 0,

i(0) =

Vg = B10 . R

Therefore i = B10 e−αt cos ωd t + B20 e−αt sin ωd t. L

di(0) = 0, dt

therefore

di(0) = 0; dt

di = [(ωd B20 − αB10 ) cos ωd t − (αB20 + ωd B10 ) sin ωd t] e−αt . dt Therefore ωd B20 − αB10 = 0; Therefore

B20 =

α 0 α Vg B1 = . ωd ωd R

Problems α2 Vg ωd Vg di + sin ωd t e−αt vo = L = − L dt ωd R R (

!

LVg =− R

)

Vg L =− R

α2 + ωd2 −αt e sin ωd t ωd

Vg L =− R

ω02 −αt e sin ωd t ωd

Vg L =− Rωd

!

!



1 e−αt sin ωd t; LC 

Vg −αt e sin ωd t V, RCωd

vo = −

)

α2 + ωd sin ωd t e−αt ωd

(

[b]

!

t ≥ 0.

dvo Vg =− {ωd cos ωd t − α sin ωd t}e−αt ; dt ωd RC dvo = 0 when dt

tan ωd t =

ωd . α

Therefore ωd t = tan−1 (ωd /α) (smallest t). 1 ωd tan−1 . ωd α 

t= P 8.58



[a] From Problem 8.57 we have vo =

−Vg −αt e sin ωd t; RCωd

α=

R 120 = = 12,000 rad/s; 2L 0.01

ω02 = ωd =

1012 1 = = 400 × 106 ; LC 2500 q

ω02 − α2 = 16 krad/s;

−Vg −(−600)109 = = 625; RCωd (120)(500)(16) × 103 .·. vo = 625e−12,000t sin 16,000t V. [b] From Problem 8.57 1 ωd td = tan−1 ωd α 

td = 57.96 µs.



1 16,000 = tan−1 16,000 12,000

!

8–45

8–46

CHAPTER 8. Natural and Step Responses of RLC Circuits [c] vmax = 625e−0.012(57.96) sin[(0.016)(57.96)] = 249.42 V. [d] R = 12 Ω;

α = 1200 rad/s;

ωd = 19,963.97 rad/s; vo = 5009.02e−1200t sin 19,963.97t V,

t ≥ 0;

td = 75.67 µs; vmax = 4565.96 V. P 8.59

[a]

d2 vo 1 = vg ; 2 dt R1 C1 R2 C2 1 1 = = 125; R1 C1 R2 C2 (50,000)(20,000)(2 × 10−6 )(4 × 10−6 ) d2 vo .·. = 125vg . dt2 0 ≤ t ≤ 0.2− : vg = 400 mV; d2 vo = 50. dt2 Let g(t) = Z g(t)

dvo , dt

dx = 50

Z t

g(0)

then

dg = 50 or dg = 50 dt. dt

dy;

0

g(t) − g(0) = 50t, g(t) =

g(0) =

dvo (0) = 0; dt

dvo = 50t; dt

dvo = 50t dt; Z vo (t)

dx = 50

Z t

vo (0)

x dx;

0

vo (t) = 25t2 V,

vo (t) − vo (0) = 25t2 ,

0 ≤ t ≤ 0.2− ;

dvo1 1 =− vg = −10vg = −4; dt R1 C1 dvo1 = −4 dt; Z vo1 (t) vo1 (0)

dx = −4

Z t 0

dy;

vo (0) = 0;

Problems vo1 (t) − vo1 (0) = −4t,

vo1 (0) = 0;

0 ≤ t ≤ 0.2− .

vo1 (t) = −4t V, 0.2+ ≤ t ≤ tsat : d2 vo = −12.5, dt2 dg(t) = −12.5; dt Z g(t) g(0.2+ )

dvo ; dt

let g(t) =

dg(t) = −12.5 dt;

dx = −12.5

Z t

dy;

0.2

g(t) − g(0.2+ ) = −12.5(t − 0.2) = −12.5t + 2.5; g(0.2+ ) = C

dvo (0.2+ ) ; dt

0 − vo1 (0.2+ ) dvo ; (0.2+ ) = dt 20 × 103

vo1 (0.2+ ) = vo1 (0.2− ) = −4(0.2) = −0.80 V; dvo (0.2+ ) 0.80 .·. C = 40 µA. = dt 20 × 103 40 × 10−6 dvo (0.2+ ) = = 10 V/s; dt 4 × 10−6 dvo .·. g(t) = −12.5t + 2.5 + 10 = −12.5t + 12.5 = ; dt .·. dvo = −12.5t dt + 12.5 dt. Z vo (t) vo (0.2+ )

dx =

Z t 0.2+

−12.5y dy +

+

vo (t) − vo (0.2 ) = −6.25y

2

t

Z t 0.2+

12.5 dy;

+ 12.5y

0.2

t

;

0.2

vo (t) = vo (0.2+ ) − 6.25t2 + 0.25 + 12.5t − 2.5; vo (0.2+ ) = vo (0.2− ) = 1 V; .·. vo (t) = −6.25t2 + 12.5t − 1.25 V, dvo1 = −10(−0.1) = 1, dt dvo1 = dt;

Z vo1 (t) vo1 (0.2+ )

0.2+ ≤ t ≤ tsat .

0.2+ ≤ t ≤ tsat ; dx =

Z t 0.2+

dy;

8–47

8–48

CHAPTER 8. Natural and Step Responses of RLC Circuits vo1 (0.2+ ) = vo1 (0.2− ) = −0.8 V

vo1 (t) − vo1 (0.2+ ) = t − 0.2; .·. vo1 (t) = t − 1 V,

0.2+ ≤ t ≤ tsat .

Summary: 0 ≤ t ≤ 0.2− s :

vo1 = −4t V,

0.2+ s ≤ t ≤ tsat :

vo = 25t2 V;

vo1 = t − 1 V,

vo = −6.25t2 + 12.5t − 1.25 V.

[b] −10 = −6.25t2sat + 12.5tsat − 1.25; .·. 6.25t2sat − 12.5tsat − 8.75 = 0. t2sat − 2tsat − 1.4 = 0; √ tsat = 1 ± 2 + 1.4 = 1 ± 1.844; .·. tsat = 2.844 sec. vo1 (tsat ) = 1.844 − 1 = 0.844 V. P 8.60

τ1 = (0.25 × 106 )(2 × 10−6 ) = 0.50 s; 1 = 2; τ1

τ2 = (0.25 × 106 )(4 × 10−6 ) = 1 s;

.·.

1 = 1; τ2

dvo d2 vo +3 + 2vo = 50. .·. 2 dt dt s2 + 3s + 2 = 0; (s + 1)(s + 2) = 0;

s1 = −1,

vo = Vf + A01 e−t + A02 e−2t ;

s2 = −2;

Vf = (250/50)(250/20)(0.4) = 25 V;

vo = 25 + A01 e−t + A02 e−2t ; vo (0) = 0 = 25 + A01 + A02 ; .·. A01 = −50,

A02 = 25 V.

vo (t) = 25 − 50e−t + 25e−2t V, dvo1 + 2vo1 = −4; dt

dvo (0) = 0 = −A01 − 2A02 ; dt

0 ≤ t ≤ 0.2 s;

.·. vo1 = −2 + 2e−2t V,

0 ≤ t ≤ 0.2 s.

Problems

8–49

vo (0.2) = 25 − 50e−0.2 + 25e−0.4 = 0.8215 V; vo1 (0.2) = −2 + 2e−0.4 = −0.6594 V. At t = 0.2 s

iC =

C

0 + 0.6594 = 32.97 µA; 20 × 103

dvo = 32.97 µA; dt

dvo 32.97 = = 8.24 V/s. dt 4

0.2 s ≤ t < ∞: d2 vo dvo +3 + 2 = −12.5; 2 dt dt vo (∞) = −6.25; .·. vo = −6.25 + A01 e−(t−0.2) + A02 e−2(t−0.2) . 0.8215 = −6.25 + A01 + A02 ; dvo (0.2) = 8.24 = −A01 − 2A02 ; dt .·. A01 + A02 = 7.07; A01 = 22.38;

−A01 − 2A02 = 8.24.

A02 = −15.31;

.·. vo = −6.25 + 22.38e−(t−0.2) − 15.31e−2(t−0.2) V,

0.2 ≤ t < ∞.

dvo1 + 2vo1 = 1; dt .·. vo1 = 0.5 + (−0.6594 − 1)e−2(t−0.2) = 0.5 − 1.66e−2(t−0.2) V,

0.2 ≤ t < ∞.

8–50 P 8.61

CHAPTER 8. Natural and Step Responses of RLC Circuits [a]

2C

dva va − vg va + + = 0. dt R R

(1) Therefore

va vg dva + = . dt RC 2RC

d(0 − vb ) 0 − va +C =0 R dt (2) Therefore

dvb va + = 0, dt RC

va = −RC

dvb . dt

dvb d(vb − vo ) 2vb +C +C = 0. R dt dt (3) Therefore

dvb vb 1 dvo + = . dt RC 2 dt

From (2) we have

d2 vb dva = −RC 2 dt dt

and va = −RC

When these are substituted into (1) we get (4) − RC

d2 vb dvb vg − = . 2 dt dt 2RC

Now differentiate (3) to get (5)

1 dvb 1 d2 vo d2 vb + = . dt2 RC dt 2 dt2

But from (4) we have (6)

d2 vb 1 dvb vg + = − . dt2 RC dt 2R2 C 2

Now substitute (6) into (5) d2 vo vg = − 2 2. 2 dt R C

dvb . dt

Problems

8–51

d2 vo vg = 2 2. 2 dt R C The two equations are the same except for a reversal in algebraic sign.

[b] When R1 C1 = R2 C2 = RC :

[c] Two integrations of the input signal with one operational amplifier. P 8.62

At t = 0 the voltage across each capacitor is zero. It follows that since the operational amplifiers are ideal, the current in the 500 kΩ is zero. Therefore there cannot be an instantaneous change in the current in the 1 µF capacitor. Since the capacitor current equals C(dvo /dt), the derivative must be zero.

P 8.63

Problem 8.63(a) — Example 8.14, with R1 and R2 removed: [a] Ra = 100 kΩ; d2 vo 1 = 2 dt Ra C1 

C1 = 0.1 µF; 

Rb = 25 kΩ; 1 = 100 Ra C1

1 vg ; Rb C2 

vg = 250 × 10−3 ;

therefore

C2 = 1 µF; 1 = 40; Rb C2

d2 vo = 1000. dt2

dvo (0) , our solution is vo = 500t2 . dt The second op-amp will saturate when

[b] Since vo (0) = 0 =

vo = 6 V,

or tsat =

q

6/500 ∼ = 0.1095 s.

dvo1 1 =− vg = −25. dt Ra C1 [d] Since vo1 (0) = 0, vo1 = −25t V;

[c]

At t = 0.1095 s,

vo1 ∼ = −2.74 V.

Therefore the second amplifier saturates before the first amplifier saturates. Our expressions are valid for 0 ≤ t ≤ 0.1095 s. Once the second op-amp saturates, our linear model is no longer valid. Problem 8.63(b) — Example 8.14 with vo1 (0) = 5 V and vo (0) = −10 V: [a] Initial conditions will not change the differential equation; hence the equation is the same as Example 8.14. [b] vo = 5 + A01 e−10t + A02 e−20t

(from Example 8.14);

vo (0) = −10 = 5 + A01 + A02 .

8–52

CHAPTER 8. Natural and Step Responses of RLC Circuits

iC (0+ ) = 0.1 m − 0.2 m = −0.1 mA; iC (0+ ) = C

dvo (0+ ) = −0.1 mA; dt

−0.1 × 10−3 dvo (0+ ) = = −100 V/s; dt 10−6 dvo = −10A01 e−10t − 20A02 e−20t ; dt dvo + (0 ) = −10A01 − 20A02 = −100. dt Therefore −A01 − 2A02 = −10 and A01 + A02 = −15. Thus, A01 = −40 and A02 = 25. vo = 5 − 40e−10t + 25e−20t V. [c] Same as Example 8.14: dvo1 + 20vo1 = −25. dt [d] From Example 8.14: vo1 (∞) = −1.25 V;

v1 (0) = 5 V (given).

Therefore vo1 = −1.25 + (5 + 1.25)e−20t = −1.25 + 6.25e−20t V. P 8.64

[a] From Example 8.13 therefore

d2 vo = 2; dt2

dg(t) = 2, dt

g(t) − g(0) = 2t;

g(t) =

dvo ; dt

g(t) = 2t + g(0);

g(0) =

dvo (0) . dt

Problems

iR =

8–53

dvo (0) 12 × 10−3 = 24 µA = iC = −C ; 500 dt

dvo (0) −24 × 10−6 = = −24 = g(0); dt 1 × 10−6 dvo = 2t − 24; dt dvo = 2t dt − 24 dt; vo − vo (0) = t2 − 24t; vo = t2 − 24t − 6,

vo (0) = −6 V; 0 ≤ t ≤ tsat .

[b] t2 − 24t − 6 = −9; t2 − 24t + 3 = 0; t∼ = 125.66 ms. Check the output from the first stage to make sure it doesn’t saturate at an earlier time. From Eq. 8.38 dvo1 = −40(25 × 10−3 ) = −1; dt dvo1 = −dt,

so

If saturation occurs,

vo1 − vo1 (0) = −t; vo1 = −t + 12 = 5.

Therefore, the first stage would saturate at 7 s, which is later than the time the second stage saturates. s

P 8.65

[a] ω0 = .·. [b] I0 =

1 = LC f0 =

s

1 (8 ×

10−9 )(5

×

10−12 )

= 5 × 109 rad/sec;

ω0 5 × 109 = = 0.796 × 109 Hz = 0.796 GHz. 2π 2π

12 = 0.24 A; 50

1 1 w(0) = LI02 = (8 × 10−9 )(0.24)2 = 2.304 × 10−10 J = 0.23 nJ. 2 2

8–54

CHAPTER 8. Natural and Step Responses of RLC Circuits [c] Because the inductor and capacitor are assumed to be ideal, none of the initial energy will ever be dissipated, so for all t ≥ 0 the 0.23 nJ will be continually exchanged between the inductor and capacitor.

P 8.66

[a] ω0 = 2πf0 = 2π(5 × 109 ) = 10π × 109 rad/s; ω02 =

1 ; LC

.·.

C=

1 1 = = 2.03 × 10−12 = 2.03 pF. 2 −9 9 2 Lω0 (0.5 × 10 )(10π × 10 )

V 12 sin ω0 t = sin 10π × 109 t 9 ω0 RC (10π × 10 )(25)(2.03 × 10−12 )

[b] vo (t) =

= 7.5 sin 10π × 109 t V, P 8.67

[a] α =

0.024 R = = 1.5 × 106 rad/s; −9 2L 2(8 × 10 ) s

ω0 =

1 = LC

[b] ω02 > α2 [c] ωd =

t ≥ 0.

q

f0 =

s

1 (8 ×

10−9 )(5

× 10−12 )

= 5 × 109 rad/s.

so the response is underdamped.

ω02 − α2 =

q

(5 × 109 )2 − (1.5 × 106 )2 ≈ 5 × 109 rad/s;

ω0 5 × 109 = = 0.796 × 109 Hz = 0.796 GHz. 2π 2π

Therefore the addition of the 24 mΩ resistance does not change the frequency of oscillation. [d] Because of the added resistance, the oscillation now occurs within a decaying exponential envelope (see Fig. 8.9). The form of the 6 exponential envelope is e−αt = e−1.5×10 t . Let’s assume that the 6 oscillation is described by the function Ke−1.5×10 t cos 5 × 109 t, where the maximum magnitude, K, exists at t = 0. How long does it take before the magnitude of the oscillation has decayed to 0.01K? 6

Ke−1.5×10 t = 0.01K .·.

t=

so

6

e−1.5×10 t = 0.01;

ln 0.01 = 3.07 × 10−6 s = 3.07 µs. −1.5 × 106

Therefore, the oscillations will persist for only 3.07 µs, due to the presence of a small amount of resistance in the circuit. This is why an LC oscillator is not used in the clock generator circuit.

Sinusoidal Steady State Analysis

Assessment Problems AP 9.1 [a] I = 25/60◦ mA. [b] 45 sin(50t − 30◦ ) = 45 cos(50t − 120◦ ); .·.

V = 45/−120◦ V.

[c] V = 10/53.13◦ + 4.47/−116.565◦ = 5.66/45◦ V. [d] sin(10πt − 45◦ ) = cos(10πt − 135◦ ). Thus, I = −150/−135◦ + 150/45◦ = 300/45◦ A. AP 9.2 [a] i = 400 cos(ωt + 38◦ ) mA. [b] V = 50/−50◦ − 80/60◦ = 107.87/−94.18◦ . Therefore v = 107.87 cos(ωt − 94.18◦ ) V. [c] V = 80 − j40 + 25/−75◦ = 107.67/−36.57◦ . v = 107.67 cos(ωt − 36.57◦ ) V. AP 9.3 [a] 400 Hz. [b] θv = 0◦ ; 100 100/0◦ /−90◦ ; = I= jωL ωL

θi = −90◦ .

100 = 20; ωL = 5 Ω. ωL 5 [d] L = = 1.99 mH. 800π

[c]

9–1

9–2

CHAPTER 9. Sinusoidal Steady State Analysis [e] ZL = jωL = j5 Ω.

AP 9.4 [a] ω = 2πf = 314,159.27 rad/s. 10 × 10−3 /0◦ V = [b] I = = jωC(10 × 10−3 )/0◦ = 10 × 10−3 ωC /90◦ ; ZC 1/jωC .·. θi = 90◦ . [c] 628.32 × 10−6 = 10 × 10−3 ωC; 1 10 × 10−3 = = 15.92 Ω, ωC 628.32 × 10−6 [d] C =

.·. XC = −15.92 Ω.

1 1 = ; 15.92(ω) (15.92)(100π × 103 )

C = 0.2 µF. −1 [e] Zc = j ωC 



= −j15.92 Ω.

AP 9.5 Convert the three currents to phasors and add the phasors: I4 = −(80/30◦ − 100/−225◦ + 50/−90◦ ) = −161.59/−29.965◦ = 161.59/150.035◦ . i4 = 161.59 cos(ωt + 150.035◦ ) A. AP 9.6 [a] V =

(40/−90◦ )(100/45◦ ) ; |Z|/θZ

But −45 − θZ = −90◦

.·. θZ = 45◦ ;

Z = 90 − j40 + jXL . To make the phase angle of Z equal 45◦ , the real part of Z must equal the imaginary part of Z, or −40 + XL = 90. .·. XL = 130 Ω;

XL = jωL = 130;

130 .·. L = = 26 mH. 5000 [b] V =

−j40(100/45◦ ) = 31.43/−90◦ V; (90 + j130 − j40)

.·. |V| = 31.43 V.

Problems AP 9.7 [a]

ω = 8000 rad/s; ωL = 64 Ω,

−1 = −100 Ω; ωC

Zxy = 100k − j100 + 30 + j64 = 50 − j50 + 30 + j64 = (80 + j14) Ω. [b] ω = 4000 rad/s so ωL = 32 Ω,

−1 = −200 Ω; ωC

Zxy = 100k − j200 + 30 + j32 = 80 − j40 + 30 + j32 = (110 − j8) Ω. [c] Zxy = 100k(−j/ωC) + 30 + ωL =

−100(j/ωC) + 30 + jωL 100 − (j/ωC)

100 − j1002 ωC + 30 + jωL. = (100ωC)2 + 1 The impedance will be purely resistive when the j terms cancel, i.e., (1002 C/L) − 1 1002 ωC 2 = ωL so ω = = 36 × 106 . (100ωC)2 + 1 1002 C 2 Solving for ω yields ω = 6000 rad/s. [d] Zxy = 100k − j133.33 + 30 + j48 = 64 − j48 + 30 + j48 = 94 Ω. AP 9.8 The frequency 6000 rad/s was found to give Zxy = 94 Ω in Assessment Problem 9.7. Thus, V = 470/0◦ ,

Is =

V 470/0◦ = = 5/0◦ A. Zxy 94

Using current division, IC =

100k − j133.33 (5) = 1.8 + j2.4 = 3/53.13◦ A; −j133.33

iC = 3 cos(6000t + 53.13◦ ) A,

Im = 3 A.

9–3

9–4

CHAPTER 9. Sinusoidal Steady State Analysis

AP 9.9

Convert the Y connection consisting of the 60 Ω, 75 Ω and j30 Ω impedances into the equivalent Delta connection. Za =

60(75) + 75(j30) + 60(j30) = 75 + j67.5 Ω, 60

Zb =

60(75) + 75(j30) + 60(j30) = 60 + j54 Ω, 75

Zc =

60(75) + 75(j30) + 60(j30) = 135 − j150 Ω. j30

Now make series and parallel combination of impedances to find the equivalent impedance seen by the voltage source. Zeq = 25 + [(15kZc − j45kZb )kZa "

= 25 +

!

#

15(135 − j150) −j45(60 + j54) + k(75 + j67.5) 150 − j150 60 + j9

= 25 + [14.25 − j0.75 + 33.007 − j49.95]k(75 + j67.5) = 25 + 55.25 − j12.6 = 80.25 − j12.6 Ω. Therefore, I=

−j50 = 0.62/−81.07◦ A. 80.25 − j12.6

AP 9.10 Calculate impedances: ZL = j(500)(.12) = j60 Ω;

ZC =

−j = −j100 Ω. (500)(20 × 10−6 )

Step 1 to Step 2: source transform to get two voltage sources, (4)(50) = 200 V;

(−j2)(−j100) = −200 V.

Problems

Now use Ohm’s law to calculate the phasor current: I=

400 200 − (−200 = = 4 + j2 = 4.47/26.57◦ A. 30 + 50 + j60 − j100 80 − j40

Therefore, i(t) = 4.47 cos(500t + 26.57◦ ) A AP 9.11 Short circuit current

Isc =

5Vx −40 + j40 + 5Vx + ; −j10 10

Vx = −40 + j40 + 5Vx .·.

Isc =

so

Vx = 10 − j10;

50 − j50 −40 + j40 + 50 − j50 + = 6 + j4 A. −j10 10

9–5

9–6

CHAPTER 9. Sinusoidal Steady State Analysis Open circuit voltage

I=

−40 + j40 = −4 A; 10 − j10

Vx = 10I = −40 V; Voc = 5Vx − j10I = −200 + j40 V; ZrmN =

−200 + j40 = −20 + j20 Ω. 6 + j4

AP 9.12 Calculate impedances: jωL = j(2500)(1.6 × 10−3 ) = j4 Ω; 1 −j = = −j4 Ω; jωCtop (2500)(100 × 10−6 ) 1 −j = = −j8 Ω; jωCmid (2500)(50 × 10−6 ) Ig = 5/0◦ A;

−5 +

Vg = 20/90◦ V.

V1 − V2 V1 − j20 + = 0; −j8 −j4

Problems V2 − V1 V2 V2 − j20 + + = 0. −j8 j4 12 Solving, V2 = −8 + j4 V;

Io =

V2 = 1 + j2 = 2.24/63.43◦ A; j4

io = 2.24 cos(2500t + 63.43◦ ) A. AP 9.13 Calculate impedances: jωL = j5000(60 × 10−3 ) = j300 Ω; 1 −j = = −j100 Ω. jωC (5000)(2 × 10−6 )

−400/0◦ + (50 + j300)Ia − 50Ib − 150(Ia − Ib ) = 0; (150 − j100)Ib − 50Ia + 150(Ia − Ib ) = 0. Solving, Ia = −0.8 − 1.6 A;

Ib = −1.6 + j0.8 A;

Vo = 100Ib = −160 + j80 = 178.89/153.43◦ ; vo = 178.89 cos(5000t + 153.43◦ ) V.

9–7

9–8

CHAPTER 9. Sinusoidal Steady State Analysis

AP 9.14 [a] jωL1 = j(5000)(2 × 10−3 ) = j10 Ω; jωL2 = j(5000)(8 × 10−3 ) = j40 Ω; jωM = j10 Ω.

70 = (10 + j10)Ig + j10IL ; 0 = j10Ig + (30 + j40)IL . Solving, Ig = 4 − j3 A;

IL = −1 A;

ig = 5 cos(5000t − 36.87◦ ) A; iL = 1 cos(5000t − 180◦ ) A. 2 M = √ = 0.5. L1 L2 16 [c] When t = 100π µs, [b] k = √

5000t = (5000)(100π) × 10−6 = 0.5π = π/2 rad = 90◦ ; ig (100πµs) = 5 cos(53.13◦ ) = 3 A; iL (100πµs) = 1 cos(−90◦ ) = 0 A; 1 1 1 w = L1 i21 + L2 i22 + M i1 i2 = (2 × 10−3 )(9) + 0 + 0 = 9 mJ. 2 2 2 When t = 200π µs, 5000t = π rad = 180◦ ; ig (200πµs) = 5 cos(180 − 53.13) = −4 A; iL (200πµs) = 1 cos(180 − 180) = 1 A; 1 1 w = (2 × 10−3 )(16) + (8 × 10−3 )(1) + 2 × 10−3 (−4)(1) = 12 mJ. 2 2

Problems AP 9.15

Mesh current equations: −200 + (10 + j5)I1 + V1 = 0; (4000 − j8000)I2 − V2 = 0. Ideal transformer equations: V1 V2 =− ; 1 20 I1 = −20I2 . Place in standard form and solve using a calculator: V2 = 3577.71/153.43◦ V I2 = 0.4/−143.13◦ A

9–9

9–10

CHAPTER 9. Sinusoidal Steady State Analysis

Problems P 9.1

[a] Right as φ becomes more negative. [b] Left. P 9.2

[a]

T 1250 250 = + = 250 µs; 2 6 6 f=

T = 500 µs;

106 1 = = 2000Hz. T 500

[b] v = Vm sin(ωt + θ; ) ω = 2πf = 4000π rad/s; −250 4000π × 10−6 + θ = 0; 6 



π .·. θ = rad = 30◦ ; 6

v = Vm sin[4000πt + 30◦ ]; 75 = Vm sin 30◦ ;

Vm = 150 V;

v = 150 sin[4000πt + 30◦ ] = 150 cos[4000πt − 60◦ ] V. P 9.3

[a] By hypothesis i = 10 cos(ωt + θ); di = −10ω sin(ωt + θ); dt .·. 10ω = 20,000π; ω = 2000π rad/s.

Problems

[b] f =

ω = 1000 Hz; 2π

T =

1 = 1 ms = 1000 µs; f

150 3 3 .·. θ = −90 − (360) = −144◦ ; = , 1000 20 20 .·. i = 10 cos(2000πt − 144◦ ) A. √

P 9.4

Vm =

P 9.5

[a] 170 V.

2Vrms =

√

2(230) = 325.27 V.

[b] 2πf = 120π;

f = 60Hz.

[c] ω = 120π = 376.99 rad/s. −π −π (60) = = −1.05 rad. [d] θ(rad) = 180 3 [e] θ = −60◦ . 1 1 = 16.67 ms. [f ] T = = f 60 π 1 [g] 120πt − = 0; .·. t = = 2.78 ms. 3 360     0.125 π [h] v = 170 cos 120π t + − 18 3 = 170 cos[120πt + (15π/18) − (π/3)] = 170 cos[120πt + (π/2)] = −170 sin 120πt V. [i] 120π(t − to ) − (π/3) = 120πt − (π/2); π .·. 120πto = ; 6 P 9.6

to =

[a] ω = 2πf = 240π rad/s,

25 ms. 18 f=

ω = 120 Hz. 2π

[b] T = 1/f = 8.33 ms. [c] Vm = 100 V. [d] v(0) = 100 cos(45◦ ) = 70.71 V. 45◦ (2π) π [e] φ = 45◦ ; φ= = = 0.7854 rad. 360◦ 4 [f ] V = 0 when 240πt + 45◦ = 90◦ . Now resolve the units: (240π rad/s)t =

45◦ π = rad, ◦ 57.3 /rad 4

t = 1.042 ms.

9–11

9–12

CHAPTER 9. Sinusoidal Steady State Analysis [g] (dv/dt) = (−100)240π sin(240πt + 45◦ ); (dv/dt) = 0 when 240πt + 45◦ = 180◦ 135◦ 3π 240πt = = rad. ◦ 57.3 /rad 4

or

Therefore t = 3.125 ms. s

P 9.7

Vrms = Z T /2 0

1 Z T /2 2 2 2π t dt; Vm sin T 0 T

Vm2 sin2



   2π V 2 Z T /2 V 2T 4π t dt = m 1 − cos t dt = m ; T 2 0 T 4 s

Therefore Vrms =

P 9.8

Z to +T to

Vm2

1 Vm2 T Vm = . T 4 2

2

Z to +T 1

cos (ωt + φ) dt =

Vm2

=

Vm2

2 Vm2 = 2 Vm2 = 2

2

t

(o R to +T to

+

1 cos(2ωt + 2φ) dt 2

dt +

Z to +T

)

cos(2ωt + 2φ) dt

to

i 1 h sin(2ωt + 2φ) |ttoo +T T+ 2ω   1 T+ [sin(2ωto + 4π + 2φ) − sin(2ωto + 2φ)] 2ω     1 2 T 2 T + (0) = Vm . = Vm 2 2ω 2

P 9.9





[a] The numerical values of the terms in Eq. 9.7 are Vm = 100, R/L = 533.33, √ R2 + ω 2 L2 = 50;

ωL = 30;

φ = 60◦ ,

θ = 36.87◦ ;

θ = tan−1 30/40,

h

i

i = −1.84e−533.33t + 2 cos(400t + 23.13◦ ) A,

t ≥ 0.

[b] Transient component = −1.84e−533.33t A; Steady-state component = 2 cos(400t + 23.13◦ ) A. [c] By direct substitution into the equation from part (a), i(1.875 ms) = 133.61 mA. [d] 2 A,

400 rad/s,

23.13◦ .

[e] The current lags the voltage by 36.87◦ .

Problems P 9.10

[a] From Eq. 9.7 we have L

di Vm R cos(φ − θ) −(R/L)t ωLVm sin(ωt + φ − θ) √ e − ; = √ 2 dt R + ω 2 L2 R 2 + ω 2 L2

Ri =

−Vm R cos(φ − θ)e−(R/L)t Vm R cos(ωt + φ − θ) √ √ + ; R 2 + ω 2 L2 R 2 + ω 2 L2 #

"

di R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ) √ L + Ri = Vm . dt R 2 + ω 2 L2 But √

R = cos θ R 2 + ω 2 L2

and

√

ωL = sin θ. R 2 + ω 2 L2

Therefore the right-hand side reduces to Vm cos(ωt + φ). At t = 0, Eq. 9.7 reduces to i(0) =

−Vm cos(φ − θ) Vm cos(φ − θ) √ + √ 2 = 0. R 2 + ω 2 L2 R + ω 2 L2

Vm cos(ωt + φ − θ). + ω 2 L2 Therefore diss −ωLVm L sin(ωt + φ − θ) =√ 2 dt R + ω 2 L2

[b] iss = √

R2

and Riss = √

Vm R cos(ωt + φ − θ). + ω 2 L2

R2

"

diss R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ) √ L + Riss = Vm dt R 2 + ω 2 L2 = Vm cos(ωt + φ). P 9.11

[a] Y = 100/45◦ + 500/ − 60◦ = 483.86/ − 48.48◦ ; y = 483.86 cos(300t − 48.48◦ ). [b] Y = 250/30◦ − 150/50◦ = 120.51/4.8◦ ; y = 120.51 cos(377t + 4.8◦ ). [c] Y = 60/60◦ − 120/ − 215◦ + 100/90◦ = 152.88/32.94◦ ; y = 152.88 cos(100t + 32.94◦ ).

#

9–13

9–14

CHAPTER 9. Sinusoidal Steady State Analysis [d] Y = 100/40◦ + 100/160◦ + 100/ − 80◦ = 0; y = 0.

P 9.12

[a] ωL = (104 )(20 × 10−3 ) = 200 Ω. [b] ZL = jωL = j200 Ω. [c] VL = IZL = (10/30◦ )(200/90◦ ) = 2000/120◦ V. [d] vL = 2 cos(10,000t + 120◦ ) kV.

P 9.13

[a] XC =

−1 −1 = = −50 Ω. ωC 4000(5 × 10−6 )

[b] ZC = jXC = −j50 Ω. 30/25◦ V = = 0.6/115◦ A. [c] I = ZC 50/−90◦ [d] i = 0.6 cos(4000t + 115◦ ) A. P 9.14

[a] Vg = 150/20◦ ;

Ig = 30/ − 52◦ ;

Vg = 5/72◦ Ω. .·. Z = Ig [b] ig lags vg by 72◦ : 2πf = 8000π;

f = 4000 Hz;

T = 1/f = 250 µs;

72 .·. ig lags vg by (250) = 50 µs. 360 P 9.15

[a] Z1 = R1 + jωL1 ; Z2 =

R2 (jωL2 ) ω 2 L22 R2 + jωL2 R22 = ; R2 + jωL2 R22 + ω 2 L22

Z1 = Z2 [b] R1 =

when R1 =

ω 2 L22 R2 R22 + ω 2 L22

(4 × 108 )(6.25)(5 × 104 ) = 2.5 × 104 ; 25 × 108 + (4 × 108 )(6.25)

.·. R1 = 25 kΩ; L1 =

and L1 =

(25 × 108 )2.5 = 1.25 H. 50 × 108

R22 L2 . R22 + ω 2 L22

Problems

P 9.16

[a] Y2 =

1 j − ; R2 ωL2 R1 − jωL1 1 = 2 . R1 + jωL1 R1 + ω 2 L21

Y1 =

Therefore R2 =

Y2 = Y1

R12 + ω 2 L21 R1

when

and L2 =

R12 + ω 2 L21 . ω 2 L1

25 × 106 + 108 (0.25) = 10 × 103 ; 5 × 103

[b] R2 =

.·. R2 = 10 kΩ. 50 × 106 = 1 H. 108 (0.5)

L2 = P 9.17

[a] Z1 = R1 − j

1 ; ωC1

R2 R2 − jωR22 C2 R2 /jωC2 = = ; R2 + (1/jωC2 ) 1 + jωR2 C2 1 + ω 2 R22 C22

Z2 =

Z1 = Z2

when R1 =

ωR22 C2 1 = ωC1 1 + ω 2 R22 C22 [b] R1 =

1 + (64 ×

C1 = P 9.18

[a] Y2 =

(64 ×

R2 1 + ω 2 R22 C22

or C1 =

and

1 + ω 2 R22 C22 . ω 2 R22 C2

500 = 250 Ω; × 104 )(625 × 10−18 )

108 )(25

108 )(25

2 = 50 nF. × 104 )(25 × 10−9 )

1 + jωC2 ; R2 1 jωC1 ω 2 R1 C12 + jωC1 = = . R1 + (1/jωC1 ) 1 + jωR1 C1 1 + ω 2 R12 C12

Y1 =

Therefore R2 = [b] R2 = C2 =

Y1 = Y2

1 + ω 2 R12 C12 ω 2 R1 C12

when

and C2 =

C1 . 1 + ω 2 R12 C12

1 + (4 × 108 )(4 × 106 )(2500 × 10−18 ) = 2500 = 2.5kΩ; (4 × 108 )(2 × 103 )(2500 × 10−18 ) 50 × 10−9 = 10 nF. 5

9–15

9–16

P 9.19

CHAPTER 9. Sinusoidal Steady State Analysis

[a] Y =

1 1 1 + + 4 − j3 16 + j12 −j100

= 0.16 + j0.12 + 0.04 − j0.03 + j0.01 = 0.2 + j0.1 = 223.6/26.57◦ mS. [b] G = 200 mS. [c] B = 100 mS. [d] I = 50/0◦ A,

V=

I 50 = 223.61/−26.57◦ V; = Y 0.223/26.57◦

V 223.6/−26.57◦ IC = = 2.24/63.43◦ A; = ◦ / ZC 100 −90 iC = 2.24 cos(ωt + 63.43◦ ) A, P 9.20

Im = 2.24 A.

[a]

500/60◦ = 1/23.13◦ A. 400 + j700 − j400 [c] i = 1 cos(8000t + 23.13◦ ) A. [b] I =

P 9.21

[a] jωL = j(5 × 104 )(40 × 10−6 ) = j2 Ω; 106 1 = −j = −j20 Ω; jωC 5 × 104

Ig = 20/−20◦ A.

[b] Vo = 20/−20◦ Ze ; Ze =

1 ; Ye

Ye =

1 1 1 +j + ; 20 20 1 + j2

Ye = 0.05 + j0.05 + 0.20 − j0.40 = 0.25 − j0.35 S; Ze =

1 = 2.32/54.46◦ Ω; 0.25 − j0.35

Vo = (20/−20◦ )(2.32/54.46◦ ) = 46.4/34.46◦ V.

Problems

9–17

[c] vo = 46.4 cos(5 × 104 t + 34.46◦ ) V. P 9.22

[a] Using the notation and results from Problem 9.16: RkL = 20 + j40 so R1 = 20,

L1 =

R2 =

202 + 10002 (0.04)2 = 100 Ω; 20

L2 =

202 + 10002 (0.04)2 = 50 mH; 10002 (0.04)

40 = 40 mH; 1000

R2 kjωL2 = 100kj50 = 20 + j40 Ω. (checks) The circuit, using combinations of components from Appendix H, is shown here:

[b] Using the notation and results from Problem 9.18: RkC = 20 − j40 so R1 = 20, R2 = C2 =

C1 = 25 µF;

1 + 10002 (20)2 (25 µ)2 = 100 Ω; 10002 (20)(25 µ)2 25 µ 1+

10002 (20)2 (25 µ)2

= 20 µF;

R2 k(−j/ωC2 ) = 100k(−j50) = 20 − j40 Ω. (checks) The circuit, using combinations of components from Appendix H, is shown here:

P 9.23

[a] (20 + j40)k(−j/ωC) = 100kj50k(−j/ωC). To cancel out the j50 Ω impedance, the capacitive impedance must be −j50 Ω: −j 1 = −j50 so C = = 20 µF. 1000C (50)(1000)

9–18

CHAPTER 9. Sinusoidal Steady State Analysis Check: RkjωLk(−j/ωC) = 100kj50k(−j50) = 100 Ω. Create the equivalent of a 20 µF capacitor from components in Appendix H by combining two 10 µF capacitors in parallel. [b] (20 − j40)k(jωL) = 100k(−j50)k(jωL). To cancel out the −j50 Ω impedance, the inductive impedance must be j50 Ω: j1000L = j50 so L =

50 = 50 mH. 1000

Check: RkjωLk(−j/ωC) = 100kj50k(−j50) = 100 Ω. Create the equivalent of a 50 mH inductor from components in Appendix H by combining five 10 mH inductors in series. P 9.24

[a] R = 800 Ω = 330 Ω + 470 Ω; ωL −

1 1 = −600 so 5000L − = −600. ωC 5000C

Choose L = 40 mH by combining four 10 mH inductors in series. Then, 1 1 = 200 + 600 so C = = 0.25 µF. 5000C 5000(800) We can achieve the desired capacitance by combining a 0.22 µF capacitor in parallel with three 0.01 µF capacitors. 1 1 2 so ω = = 108 ; [b] 0.04ω = ω(0.25 × 10−6 ) 0.04(0.25 × 10−6 ) .·. ω = 10,000 rad/s. P 9.25

Z = 400 + j(5)(40) − j

1000 = 500/ − 36.87◦ Ω; (5)(0.4)

750/0◦ × 10−3 = 1.5/36.87◦ mA; Io = ◦ / 500 − 36.87 io (t) = 1.5 cos(5000t + 36.87◦ ) mA. P 9.26

ZL = j(5000)(48 × 10−3 ) = j240 Ω; ZC =

−j = −j80 Ω. (5000)(2.5 × 10−6 )

Problems Construct the phasor domain equivalent circuit:

Using current division: I=

(80 + j240) (0.2) = 0.1 + j0.1 A; 240 − j80 + 80 + j240

Vo = 240I = 24 + j24 = 33.94/45◦ ; vo = 33.94 cos(5000t + 45◦ ) V. P 9.27

1 109 = = −j4000 Ω; jωC (31.25)(8000) jωL = j8000(500)10−3 = j4000 Ω; Vg = 64/0◦ V.

Ze =

(2000)(j4000) = 1600 + j800 Ω; 2000 + j4000

ZT = 1600 + j800 − j4000 = 1600 − j3200 Ω; Ig =

64/0◦ = 8 + j16 mA; 1600 − j3200

Vo = Ze Ig = (1600 + j800)(0.008 + j0.016) = j32 = 32/90◦ V; vo = 32 cos(8000t + 90◦ ) V.

9–19

9–20 P 9.28

CHAPTER 9. Sinusoidal Steady State Analysis Vs = 250/ − 90◦ V; 1 = −j400 Ω; jωC jωL = j125 Ω.

Zeq = 62.5 + j125k500k(300 − j400) = 125 + j125 Ω; Vo 250/ − 90◦ Io = = = −1 − j1 = 1.414/ − 135◦ A; Zeq 125 + j125 io = 1.414 cos(2500t − 135◦ ) A. P 9.29

Z1 = 10 − j40 Ω; Z2 =

(5 − j10)(10 + j30) = 10 − j10 Ω; 15 + j20

Z3 =

20(j20) = 10 + j10 Ω; 20 + j20

.·. Zab = Z1 + Z2 + Z3 = 30 − j40 Ω = 50/ − 53.13◦ Ω. P 9.30

First find the admittance of the parallel branches: Yp =

1 1 1 1 + + + = 0.375 − j0.125 S; 6 − j2 4 + j12 5 j10

Zp =

1 1 = = 2.4 + j0.8 Ω; Yp 0.375 − j0.125

Zab = −j12.8 + 2.4 + j0.8 + 13.6 = 16 − j12 Ω; Yab =

1 1 = = 0.04 + j0.03 S Zab 16 − j12

= 40 + j30 mS = 50/36.87◦ mS.

Problems

P 9.31

[a] Zab = j5ω +

(4000)(109 /jω625) 4000 + (109 /j625ω)

= j5ω +

4 × 1012 25 × 105 jω + 109

= j5ω +

4 × 107 104 + j25ω

= j5ω +

4 × 1011 100 × 107 ω − j ; 108 + 625ω 2 108 + 625ω 2

.·. 5 =

109 ; 108 + 625ω 2

5 × 108 + 3125ω 2 = 109 ; ω = 4 × 102 = 400 rad/s. [b] Zab (400) = j2000 + P 9.32

[a]

(4000)(−j4000) = 2 kΩ. 4000 − j4000

109 1 = = −j10 Ω; jωC j8 × 105 (125) jωL = j8 × 105 (25 × 10−6 ) = j20 Ω; Ig = 5/0◦ ;

−5 +

V V V − Vo + + = 0; −j10 20 12

Vo − V Vo + = 0. 12 j20 Solving, Vo = 44.72/ − 10.30◦ V; vo = 44.72 cos(8 × 105 t − 10.30◦ ) V. [b] ω = 2πf = 8 × 105 ; T =

f=

1 π = = 2.5π µs; f 4 × 105

4 × 105 ; π

9–21

9–22

CHAPTER 9. Sinusoidal Steady State Analysis 10.30 .·. (2.5π) = 224.82 ns; 360 .·. vo lags ig by 224.82 ns.

P 9.33

[a]

Va = (120 + j40)(0.04/0◦ ) = 4.8 + j1.6 V; Ib =

4.8 + j1.6 = 20 + j20 mA; 160 − j80

Ic = 40/0◦ + (20 + j20) + (40 + j80) mA = 100 + j100 mA; Vg = 25Ic + Va = 25(0.100 + j0.100) + 4.8 + j1.6 = 7.3 + j4.1 V. [b] ib = 28.28 cos(800t + 45◦ ) mA; ic = 141.42 cos(800t + 45◦ ) mA; vg = 8.37 cos(800t + 29.32◦ ) V. P 9.34

[a] Yp = =

1 + j4 × 10−3 ω 10 + j2ω 10 − j2ω + j4 × 10−3 ω 100 + 4ω 2

10 j2ω − + j4 × 10−3 ω. 2 100 + 4ω 100 + 4ω 2 Yp is real when =

4 × 10−3 ω = or

2ω , 100 + 4ω 2

ω 2 = 100;

[b] Yp (10 rad/s) =

ω = 10 rad/s;

10 = 20 mS; 500

Zp (10 rad/s) =

103 = 50 Ω; 20

Z(10 rad/s) = 50 + 150 = 200 Ω;

f = 5/π = 1.59Hz.

Problems

9–23

Vg 10/0◦ A= = 50/0◦ mA; 200 200

Io =

io = 50 cos 10t mA. P 9.35

[a] Y1 =

1 = 0.2 × 10−3 S; 5000

Y2 = =

1 1200 + j0.2ω 1200 0.2ω −j ; 6 2 1.44 × 10 + 0.04ω 1.44 × 106 + 0.04ω 2

Y3 = jω50 × 10−9 ; YT = Y1 + Y2 + Y3 . For ig and vo to be in phase the j component of YT must be zero; thus, ω50 × 10−9 =

0.2ω , 1.44 × 106 + 0.04ω 2

or 0.04ω 2 + 1.44 × 106 =

0.2 × 109 = 4 × 106 ; 50

.·. 0.04ω 2 = 2.56 × 106 [b] YT = 0.2 × 10−3 +

1.44 ×

.·. ω = 8000 rad/s = 8 krad/s. 106

1200 = 0.5 × 10−3 S; + 0.04(64) × 106

.·. ZT = 2000 Ω. Vo = (2.5 × 10−3 /0◦ )(2000) = 5/0◦ ; vo = 5 cos 8000t V. P 9.36

109 104 (j2ω) + [a] Zg = 4000 − j 25ω 104 + j2ω

.·.

= 4000 − j

109 2 × 104 jω(104 − j2ω) + 25ω 108 + 4ω 2

= 4000 − j

109 4 × 104 ω 2 2 × 108 ω + 8 + j ; 25ω 10 + 4ω 2 108 + 4ω 2

0.2 × 109 ω 109 = . 25ω 108 + 4ω 2

108 + 4ω 2 = 5ω 2 ; ω 2 = 108 ;

ω = 10,000 rad/s.

9–24

CHAPTER 9. Sinusoidal Steady State Analysis [b] When ω = 10,000 rad/s Zg = 4000 +

4 × 104 (104 )2 = 12,000 Ω; 108 + 4(104 )2

45/0◦ .·. Ig = = 3.75/0◦ mA. 12,000 Vo = Vg − Ig Z1 ; Z1 = 4000 − j

109 = 4000 − j4000 Ω; 25 × 104

Vo = 45/0◦ − (3.75 × 10−3 )(4000 − j4000) = 45 − (15 − j15) = 30 + j15 = 33.54/26.57◦ V; vo = 33.54 cos(10,000t + 26.57◦ ) V. P 9.37

Va = j2Ia = j2(−j5) = 10/0◦ V; Vc = 60/0◦ − Va = 50/0◦ V; Ic =

50/0◦ Vc = = 5/53.13◦ = 3 + j4 A; 6 − j8 10/ − 53.13◦

Ib = Ic − Ia = 3 + j4 − (−j5) = 3 + j9 A = 9.49/71.57◦ A; Vb = Ib (j5) = (3 + j9)(j5) = −45 + j15 V; Vz = Vb + Vc = −45 + j15 + 50 + j0 = 5 + j15 V; Vd + Vz = 60/0◦ ;

.·. Vd = 60 − 5 − j15 = 55 − j15 V.

Problems

Id =

9–25

Vd = 3 + j11 A; −j5

Iz = Id − Ib = 3 + j11 − 3 − j9 = j2 A; Z= P 9.38

5 + j15 Vz = = 7.5 − j2.5 Ω. Iz j2

V2 is the voltage across the −j10 Ω impedance. V1 − Vg V1 V1 − V2 + + = 0; 20 j5 Z (40 + j30) − (100 − j50) 40 + j30 (40 + j30) − V2 + + = 0; 20 j5 Z .·. V2 = 40 + j30 + (3 − j4)Z. V2 V2 − Vg V2 − V1 + − Ig + = 0; Z −j10 3 + j1 V2 − (40 + j30) V2 V2 − (100 − j50) + − (20 + j30) + = 0. Z −j10 3 + j1 Substituting the expression for V2 found at the start and simplifying yields Z = 12 + j16 Ω.

P 9.39

Ig = 100/ − 8.13◦ mA;

Vg = 50/ − 45◦ V; Z=

Vg = 500/ − 36.87◦ Ω = 400 − j300 Ω; Ig

2.5 × 106 Z = 400 + j 0.04ω − ; ω !

2.5 × 106 .·. 0.04ω − = −300; ω .·. ω 2 + 7500ω − 62.5 × 106 = 0; .·. ω = −3750 ± ω > 0,

q

(3750)2 + 62.5 × 106 = −3750 ± 8750.

.·. ω = 5000 rad/s.

9–26

P 9.40

CHAPTER 9. Sinusoidal Steady State Analysis

[a] Z1 = 1600 − j Z2 =

109 = 1600 − j1600 Ω; 104 (62.5)

4000(j104 L) 4 × 105 L2 + j16 × 104 L = ; 4000 + j104 L 16 + 100L2

ZT = Z1 + Z2 = 1600 +

16 × 104 L 4 × 105 L2 − j1600 + j . 16 + 100L2 16 + 100L2

ZT is resistive when 16 × 104 L = 1600, 16 + 100L2

or

L2 − L + 0.16 = 0. Solving, L1 = 0.8 H and L2 = 0.2 H. [b] When L = 0.8 H: ZT = 1600 + Ig =

4 × 105 (0.64) = 4800 Ω; 16 + 64

96/0◦ × 10−3 = 20/0◦ mA; 4.8

ig = 20 cos 10,000t mA. When L = 0.2 H: ZT = 1600 +

4 × 105 (0.04) = 2400 Ω; 16 + 4

ig = 40 cos 10,000t mA.

P 9.41

R R jωC [a] Zp = = R + (1/jωC) 1 + jωRC =

12,500 12,500 = 1 + j(1000)(12,500)C 1 + j12.5 × 106 C

=

12,500(1 − j12.5 × 106 C) 1 + 156.25 × 1012 C 2

=

12,500 156.25 × 109 C − j ; 1 + 156.25 × 1012 C 2 1 + 156.25 × 1012 C 2

jωL = j1000(5) = j5000; .·. 5000 =

156.25 × 109 C ; 1 + 156.25 × 1012 C 2

Problems .·. 781.25 × 1015 C 2 − 156.25 × 109 C + 5000 = 0; .·. C 2 − 20 × 10−8 C + 64 × 10−16 = 0; √ .·. C1,2 = 10 × 10−8 ± 100 × 10−16 − 64 × 10−16 . C1 = 10 × 10−8 + 6 × 10−8 = 16 × 10−8 = 0.16 µF; C2 = 10 × 10−8 − 6 × 10−8 = 4 × 10−8 = 0.04 µF. 12,500 . 1 + 156.25 × 1012 C 2 When C = 160 nF Re = 2500 Ω;

[b] Re =

250/0◦ = 0.1/0◦ A; ig = 100 cos 1000t mA. 2500 When C = 40 nF Re = 10,000 Ω;

Ig =

Ig = P 9.42

250/0◦ = 0.025/0◦ A; 10,000

ig = 25 cos 1000t mA.

Convert the Y on the left side to a delta.

The numerator for the three delta-connected impedances is the same: N = −j4(20 + j60) − j40 + 10(20 + j60) = 440 + j480. Then, Za =

440 + j480 = 44 + j48 Ω; 10

Zb =

440 + j480 = −120 + j110 Ω; −j4

Zc =

440 + j480 = 9.4 − j4.2 Ω. 20 + j60

9–27

9–28

CHAPTER 9. Sinusoidal Steady State Analysis Substitute the delta-connected impedances for the Y-connected impedances to give

Now combine the impedances in series and in parallel to give the equivalent impedance seen by the voltage source: Zeq = (44 + j48)k{[(9.4 − j4.2)k(63.2 + j2.4)] + [(−120 + j110)k(−j20)]} = 18 − j24 Ω. Therefore, I0 = P 9.43

120/0◦ = 2.4 + j3.2 = 4/53.13◦ A. 18 − j24

[a] jωL = j(5000)(50) × 10−3 = j250 Ω; 1 1 = −j = −j500 Ω. jωC (5000)(400 × 10−9 )

Using voltage division, Vab =

(250 + j250)k(−j500) (23.36/26.565◦ ) = 20/0◦ ; j250 + (250 + j250)k(−j500)

VTh = Vab = 20/0◦ V. [b] Remove the voltage source and combine impedances in parallel to find ZTh = Zab : 1 1 1 Yab = + + = 2 − j4 mS; j250 250 + j250 −j500 ZTh = Zab =

1 = 100 + j200 Ω. Yab

Problems [c]

P 9.44

Step 1 to Step 2: 75/0◦ = −j4.167 = 4.167/ − 90◦ A. j18 Step 2 to Step 3: (j18)k24 =

(j18)(24) = 8.64 + j11.52 Ω. 24 + j18

Step 3 to Step 4: (4.167/ − 90◦ )(8.64 + j11.52) = 60/ − 36.87◦ V.

P 9.45

V1 = 240/53.13◦ = 144 + j192 V; V2 = 96/−90◦ = −j96 V; jωL = j(4000)(15 × 10−3 ) = j60 Ω; 1 6 × 106 = −j = −j60 Ω. jωC (4000)(25) Perform a source transformation: V1 144 + j192 = = 3.2 − j2.4 A; j60 j60

9–29

9–30

CHAPTER 9. Sinusoidal Steady State Analysis V2 96 = −j = −j4.8 A. 20 20

Combine the parallel impedances: Y =

1 1 1 j5 1 1 + + + = = ; j60 30 −j60 20 j60 12

Z=

1 = 12 Ω. Y

Vo = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48/36.87◦ V; vo = 48 cos(4000t + 36.87◦ ) V. P 9.46

IN =

0.1/120◦ + 1/30◦ mA, ZN

IN =

0.3/ − 60◦ + (−3/210◦ ) mA, ZN

ZN in kΩ.

ZN in kΩ;

Problems 0.3/ − 60◦ 0.1/120◦ + 1/30◦ = + (−3/210◦ ; ) ZN ZN 0.3/ − 60◦ − 0.1/120◦ = 1/30◦ + 3/210◦ ; ZN ZN =

0.3/ − 60◦ − 0.1/120◦ = 0.2/90◦ = j0.2 kΩ; ◦ ◦ / / 1 30 + 3 210

IN =

0.1/120◦ + 1/30◦ = 1.5/30◦ mA. 0.2/90◦

P 9.47

(27 + j12)Ia − 3Ib = −87/0◦ ; −3Ia + (27 − j12)Ib = 87/0◦ . Solving, Ia = −2.4167 + j1.21;

Ib = 2.4167 + j1.21;

VTh = 12Ia + (12 − j12)Ib = 14.5/0◦ V.

9–31

9–32

CHAPTER 9. Sinusoidal Steady State Analysis Short Circuit Test:

(27 + j12)Ia − 3Ib − 12Isc = −87; −3Ia + (27 − j12)Ib − (12 − j12)Isc = 87; −12Ia − (12 − j12)Ib + (24 − j12)Isc = 0. Solving, Isc = 1/0◦ ; .·.

IN = Isc = 1/0◦ ;

ZN =

VTh 14.5/0◦ = 14.5 Ω. = Isc 1/0◦

Alternate calculation for ZN :

X

Z = 12 + 3 + 12 − j12 = 27 − j12;

Z1 =

36 12 = 27 − j12 9 − j4;

Z2 =

36 − j36 12 − j12 = ; 27 − j12 9 − j4

Problems

Z3 =

12(12 − j12) 48 − j48 = . 27 − j12 9 − j4

Za = 12 + j12 +

Zb = 12 +

Za kZb =

12 12(14 + j5) = ; 9 − j4 9 − j4

12 − j12 12(10 − j5) = ; 9 − j4 9 − j4

165 − j20 18 − j8;

ZN = Z3 + Za kZb =

48 − j48 165 − j20 + = 14.5 Ω. 9 − j4 18 − j8

P 9.48

V1 − 75 0.02V1 (40) V1 − + = 0; 150(4 + j1) 40 − j150 40 − j150 75(4 − j15) .·. V1 = . 16 − j12 VTh = =

40V1 4 75(4 − j15) = · 40 − j150 4 − j15 16 − j12 75 = 15/36.87◦ V. 4 − j3

9–33

9–34

CHAPTER 9. Sinusoidal Steady State Analysis

Isc =

75 1 = A; 600 8

ZTh =

P 9.49

VTh = 120/36.87◦ = 96 + j72 Ω. Isc

The Th´evenin voltage equals the open circuit voltage between the terminals a and b, which is the voltage across the capacitor. Identify and label the node voltages.

Write the two node voltage equations and the dependent source constraint equation: −2/45◦ +

V1 V1 − 10Ix + = 0; 20 j10

V2 V2 − 10Ix + = 0; 10 −j10 Ix =

V1 . 20

Solving these equations for V2 = VTh gives VTh = 10/45◦ V. To find the Th´evenin impedance, we remove the independent current source and apply a test voltage source at the terminals a, b. Thus

Problems It follows from the circuit that 10Ix = (20 + j10)Ix . Therefore Ix = 0 and IT =

ZTh =

VT , IT

VT VT + −j10 10.

therefore ZTh = (5 − j5) Ω.

P 9.50

V2 25 − 4V2 V2 − 4V2 + + = 0. 50 1000 −j250 Solving, V2 = −1 − j0.75 V = 1.25/216.87◦ V; Isc = −Iφ =

ZTh =

−25/0◦ = −25/0◦ mA; 1000

1.25/216.87◦ = 50/36.87◦ Ω = 40 + j30 Ω; −25 × 10−3 /0◦

IN = Isc = −25/0◦ mA. ZN = ZTh = 50/36.87◦ = 40 + j30 Ω.

9–35

9–36 P 9.51

CHAPTER 9. Sinusoidal Steady State Analysis ωL = j1.6 × 106 (25 × 10−6 ) = j40 Ω; 1 10−6 × 109 = = −j25 Ω. jωC j1.6(25)

VT = j40IT + 15I∆ + 25I∆ ; I∆ =

IT (−j25) −jIT = ; 25 − j25 1 − j1

VT = j40IT + 40

Zab =

P 9.52

(−jIT ) ; 1 − j1

VT = Zab = j40 + 20(−j)(1 + j) = 20 + j20 Ω = 28.28/45◦ Ω. IT

1 (10−3 )(109 ) = = 12 kΩ; ωC1 25(10/3) 1 (10−3 )(109 ) = = 24 kΩ. ωC2 25(5/3)

VT = (1 − j12)IT + 20IT (0.125); ZTh =

VT = 3.5 − j12 kΩ. IT

Problems P 9.53

9–37

[a]

IT =

VT VT − αVT /10 + ; 10 j10

1 IT (1 − α/10) (10 − α) + j10 = + = ; VT 10 j10 j100 VT 1000 + j100(10 − α) .·. ZTh = = . IT (10 − α)2 + 100 ZTh is real when α = 10. 1000 [b] ZTh = = 10 Ω. 100 [c] ZTh = 5 + j5; 1000 = 5; (10 − α)2 + 100 .·. 10 − α = ±10; α = 0;

(10 − α)2 = 100; α = 10 ∓ 10;

α = 20.

But the j term can only equal the real term with α = 0. Thus, α = 0. [d] ZTh will be inductive when α < 10. P 9.54

V1 − 100 V1 V1 + + = 0. j40 40 60 + j20 Solving for V1 yields V1 = 30 − j40 V; V1 Vo = (j20) = 60 + j20

!

j V1 ; 3+j

Vo = 15 + j5 V = 15.81/18.43◦ V.

9–38 P 9.55

CHAPTER 9. Sinusoidal Steady State Analysis The phasor domain circuit is as shown in the following diagram:

The node voltage equation is −10 +

V V V V − 100/−90◦ + + + = 0; 5 −j(20/9) j5 20

Solving, V = 10 − j30 = 31.62/−71.57◦ . Therefore v = 31.62 cos(50,000t − 71.57◦ ) V. P 9.56

jωL = j(5000)(0.4 × 10−3 ) = j2 Ω; 106 1 = −j = −j4 Ω; jωC (5000)(50) Vg1 = 10/53.13◦ = 6 + j8 V; Vg2 = 8/−90◦ = −j8 V.

Vo − 6 − j8 Vo Vo + (−j8) + + = 0. j2 6 −j4 Solving, Vo = 12/0◦ ; vo (t) = 12 cos 5000t V.

Problems

P 9.57

−15/0◦ +

I∆ =

Vo Vo − 2.5I∆ Vo + + = 0; 8 j5 −j10

Vo . −j10

Solving, Vo = 72 + j96 = 120/53.13◦ V. P 9.58

Vo Vo + + 32Io = 0; 25 −j50 (2 + j)Vo = −1600Io ; Vo = (−640 + j320)Io ; Io =

V1 − (Vo /4) ; j40

.·. V1 = (−160 + j120)Io . 17 =

V1 + Io = (−8 + j6)Io + Io = (−7 + j6)Io ; 20

.·. Io =

17 = −1.4 − j1.2 A = 1.84/ − 139.40◦ A. (−7 + j6)

Vo = (−640 + j320)Io = 1280 + j320 = 1319.39/14.04◦ V. P 9.59

The phasor domain circuit is as shown in the following diagram:

The mesh current equations are (5 − j2.22)I1 + (j2.22)I2 + (0)I3 = 50;

9–39

9–40

CHAPTER 9. Sinusoidal Steady State Analysis (j2.22)I1 + (j5 − j2.22)I2 + (−j5)I3 = 0; (0)I1 + (−j5)I2 + (20 + j5)I3 = j100. Solving, I1 = 8 + j6 A;

I2 = −5.5 + j1.5 A;

I3 = 0.5 + j3.5 A.

Therefore, V = 5(10 − I1 ) = 5(2 − j6) = 10 − j30 = 31.623/ − 71.565◦ V. v(t) = 31.623 cos(50,000t − 71.565◦ ) V. P 9.60

jωL = j(5000)(0.4 × 10−3 ) = j2 Ω; 106 1 = −j = −j4 Ω; jωC (5000)(50) Vg1 = 10/53.13◦ = 6 + j8 V; Vg2 = 8/−90◦ = −j8 V.

10/53.13◦ = (6 + j2)I1 − 6I2 ; 8/−90◦ = −6I1 + (6 − j4)I2 ; Vo = (I1 − I2 )6. Solving, Vo = 12/0◦ V; vo (t) = 12 cos 5000t V.

Problems P 9.61

Let Ia , Ib , and Ic be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 = (1 + j2)Ia + (3 − j5)(Ia − Ib ) and 0 = (3 − j5)(Ib − Ia ) + 2(Ib − Ic ). But Vx = −j5(Ia − Ib ), therefore Ic = −0.75[−j5(Ia − Ib )]. Solving for I = Ia = 29 + j2 = 29.07/3.95◦ A.

P 9.62

9–41

Va = 60/0◦ V;

Vb = 90/90◦ V;

jωL = j(4 × 104 )(125 × 10−6 ) = j5Ω; −j −j106 = = −j20 Ω. ωC 40,000(1.25)

60 = (20 + j5)Ia − j5Ib ; j90 = −j5Ia − j15Ib . Solving, Ia = 2.25 − j2.25 A;

Ib = −6.75 + j0.75 A;

Io = Ia − Ib = 9 − j3 = 9.49/ − 18.43◦ A; io (t) = 9.49 cos(40,000t − 18.43◦ ) A.

9–42 P 9.63

CHAPTER 9. Sinusoidal Steady State Analysis jωL = j104 (1.2 × 10−3 ) = j12 Ω; 1 −j106 = = −j20 Ω; jωC 5 × 104 Va = 100/ − 90◦ = −j100 V; Vb = 500/0◦ = 500 V.

The mesh current equations: (80 + j12)I1 + (−j12)I2 + (−80)I3 = −j100; (−j12)I1 + (20 − j8)I2 + (−20)I3 = 0; (−80)I1 + (−20)I2 + (100)I3 = −500; Solving, I1 = −14 − j17 A;

I2 = 5 − j25 A;

Thus, Ia = I1 = −14 − j17 = 22.02/ − 129.47◦ A; ia = 22.02 cos(10,000t − 129.47◦ ) A. Ib = −I3 = 15.2 + j18.6 = 24.02/50.74◦ A; ib = 24.02 cos(10,000t + 50.74◦ ) A.

I1 = −15.2 − j18.6 A.

Problems P 9.64

100/0◦ = (5 + j5)I1 − 5I2 − j5I3 ; 50/0◦ = −5I1 + (5 − j5)I2 + j5I3 ; −10/0◦ = −j5I1 + j5I2 + 5I3 . Solving, I1 = 58 − j20 A;

I2 = 58 + j10 A;

I3 = 28 + j0 A;

Ia = I3 + 2 = 30 + j0 A; Ib = I1 − I3 = 58 − j20 − 28 = 30 − j20 A; Ic = I2 − I3 = 58 + j10 − 28 = 30 + j10 A; Id = I1 − I2 = 58 − j20 − 58 − j10 = −j30 A. P 9.65

The impedances are ZL = j(5000)(0.4) = j2000 Ω;

Vo =

ZC =

−j = −j800 Ω. (5000)(250 × 10−9 )

600 + j2000 (75/0◦ ) = 98 + j36 = 104.4/20.17◦ V; 300 + j2000 + 600 − j800

vo = 104.4 cos(5000t + 20.17◦ ) V.

9–43

9–44

P 9.66

CHAPTER 9. Sinusoidal Steady State Analysis 1 106 = −j 4 = −j100 Ω; jωC 10 jωL = j(500)(1) = j500 Ω; Let Z1 = 50 − j100 Ω;

Z2 = 250 + j500 Ω

Ig = 125/0◦ mA; Io =

Ig Z1 125/0◦ (50 − j100) = Z1 + Z2 (300 + j400)

= −12.5 − j25 mA = 27.95/ − 116.57◦ mA; io = 27.95 cos(500t − 116.57◦ ) mA. P 9.67

[a] Superposition must be used because the frequencies of the two sources are different. [b] For ω = 200 rad/s:

Using voltage division, V1 =

j50k − j25 −j50 (100/135◦ ) = (100/135◦ ) = 70.71/90◦ V. (j50k − j25) + 50 50 − j50

Thus v1 = 70.71 cos(200t + 90◦ ) V. For ω = 100 rad/s:

Using voltage division, V2 =

50kj25 10 + j20 (50/45◦ ) = (50/45◦ ) = −35.355/0◦ V. (50kj25) − j50 10 − j30

Thus v2 = −35.355 cos 100t V. Therefore, vo (t) = v1 (t) + v2 (t) = [70.71 cos(200t + 90◦ ) − 35.355 cos 100t] V.

Problems P 9.68

9–45

[a] Superposition must be used because the frequencies of the two sources are different. [b] For ω = 500 rad/s:

Using voltage division, V1 =

100kj100 50 + j50 (−j40) = (−j40) = 40/0◦ V. (100kj100) − j100 50 − j50

Thus v1 = 40 cos 500t V. For ω = 250 rad/s:

Using voltage division, V2 =

100k − j200 80 − j40 (60/7.125◦ ) = (60/7.125◦ ) = 66.564/153.43◦ V. (100k − j200) + j50 80 + j10

Thus v2 = 66.564 cos(250t + 153.43◦ ) V. Therefore, vo (t) = v1 (t) + v2 (t) = [40 cos 500t + 66.564 cos(250t + 153.43◦ )] V. P 9.69

1 106 = = −j10 kΩ. jωC j100 Let Va = voltage across 1 µF capacitor, positive at upper terminal. Then: Vg = 1.2/0◦ V;

Va − 1.2/0◦ Va Va + + = 0; 10 −j10 10 0 − Va 0 − Vo + = 0; 10 200

.·. Va = (0.48 − j0.24) V;

Vo = −20Va ;

.·. Vo = −9.6 + j4.8 = 10.73/153.43◦ V. vo = 10.73 cos(100t + 153.43◦ ) V.

9–46 P 9.70

CHAPTER 9. Sinusoidal Steady State Analysis [a]

Va Va − 1.2/0◦ + jωCo Va + = 0; 10,000 10,000 Va =

1.2 ; 2 + j104 ωCo

Vo = −20Va Vo =

(see solution to Prob. 9.69);

−24 24/180◦ = ; 2 + j106 Co 2 + j106 Co

.·. denominator angle = 60◦ ; √ tan 60◦ = 3; 106 Co √ = 3, 2 √ √ 2 3 = 2 3 µF = 3.46 µF. or Co = 106 [b] Vo =

24/180◦ √ = 6/120◦ V. 2 + j2 3

vo = 6 cos(100t + 120◦ ) V. P 9.71

1 1012 = = −j10 kΩ; jωC1 j106 (100) 1012 1 = = −j20 kΩ. jωC2 j(106 )(50)

Problems

Va − 20 Va − Vo Va Va + + + = 0; −j20 20 40 10 .·. (−2 + j7)Va − jVo = j40. 0 − Va 0 − Vo + = 0; 10 −j10

.·. Va = −jVo ;

.·. (7 + j)Vo = j40. Vo =

j40 = 0.8 + j5.6 = 5.657/81.87◦ V; 7+j

vo (t) = 5.657 cos(106 t + 81.87◦ ) V. P 9.72

[a] Vg = 2/0◦ V; Vp =

80 Vg = 1.6/0◦ ; 100

Vn = Vp = 1.6/0◦ V;

1.6 1.6 − Vo + = 0; 160 Zp Zp =

(200)(1/jωC) ; 200 + (1/jωC)

1 109 = = −j105 = −j100 kΩ; jωC j105 (0.1) Zp =

200(−j100) = 40 − j80 kΩ; 200 − j100

Vo = 1.6 +

Zp = 2 − j0.8 = 2.15/ − 21.80◦ . 100

vo = 2.15 cos(105 t − 21.80◦ ) V.

9–47

9–48

CHAPTER 9. Sinusoidal Steady State Analysis [b] Vp = 0.8Vm /0◦ ;

Vn = Vp = 0.8Vm /0◦ ;

0.8Vm 0.8Vm − Vo + = 0; 160 40 − j80 40 − j80 .·. Vo = 0.8Vm + Vm (0.8) = 0.8Vm (1.25 − j0.5); 160 .·. |0.8Vm (1.25 − j0.5)| ≤ 5; .·. Vm ≤ 4.64 V. P 9.73

[a]

1 −j109 = = −j400 Ω; jωC (2 × 105 )(12.5) Vn Vn − Vo + = 0; 200 −j400 Vn Vn Vo = + ; −j400 200 −j400 Vo = Vn − j2Vn = (1 − j2)Vn ; Vp =

Vg (1/jωCo ) Vg = ; 500 + (1/jωCo ) 1 + j(500)(2 × 105 )Co

Vg = 10/0◦ V; Vp =

10/0◦ = Vn ; 1 + j108 Co

(1 − j2)10/0◦ .·. Vo = ; 1 + j108 Co √ 5(10) |Vo | = q = 10. 1 + 1016 Co2 Solving, Co = 20 nF. [b] Vo =

10(1 − j2) = 10/ − 126.87◦ ; 1 + j2

vo = 10 cos(2 × 105 t − 126.87◦ ) V. P 9.74

[a] jωL1 = j(50)(5) = j250 Ω; jωL2 = j(50)(20) = j1000 Ω; 1 109 = = −j500 Ω jωC j(50 × 103 )(40)

Problems

9–49

.·. Z22 = 75 + 300 + j1000 − j500 = 375 + j500 Ω; ∗ .·. Z22 = 375 − j500 Ω;

q

M = k L1 L2 = 10k × 10−3 ; ωM = (50)(10k) = 500k; "

500k Zr = 625

#2

(375 − j500) = k 2 (240 − j320) Ω;

Zin = 120 + j250 + 240k 2 − j320k 2 ; 1

|Zin | = [(120 + 240k 2 )2 + (250 − 320k 2 )2 ] 2 ; d|Zin | 1 1 = [(120 + 240k 2 )2 + (250 − 320k 2 )2 ]− 2 × dk 2 [2(120 + 240k 2 )480k + 2(250 − 320k 2 )(−640k)]; d|Zin | = 0 when dk 960k(120 + 240k 2 ) − 1280k(250 − 320k 2 ) = 0; √ .·. k 2 = 0.32; .·. k = 0.32 = 0.5657. [b] Zin (min) = 120 + 240(0.32) + j[250 − 0.32(320)] = 196.8 + j147.6 = 246/36.87◦ Ω; I1 (max) =

369/0◦ = 1.5/ − 36.87◦ A; 246/36.87◦

.·. i1 (peak) = 1.5 A. Note — You can test that the k value obtained from setting d|Zin |/dt = 0 leads to a minimum by noting 0 ≤ k ≤ 1. If k = 1, Zin = 360 − j70 = 366.74/ − 11◦ Ω. Thus, |Zin |k=1 > |Zin |k=√0.32 . If k = 0, Zin = 120 + j250 = 277.31/64.36◦ Ω. Thus, |Zin |k=0 > |Zin |k=√0.32 .

9–50 P 9.75

CHAPTER 9. Sinusoidal Steady State Analysis jωL1 = j(25 × 103 )(3.2 × 10−3 ) = j80 Ω; jωL2 = j(25 × 103 )(12.8 × 10−3 ) = j320 Ω; 109 1 = = −j160 Ω; jωC j(25 × 103 )(250) q

jωM = j(25 × 10 )k (3.2)(12.8) × 10−3 = j160k Ω; 3

Z22 = 40 + j320 − j160 = 40 + j160 Ω; ∗ Z22 = 40 − j160 Ω;

"

160k Zr = |40 + j160|

#2

(40 − j160) = 37.647k 2 − j150.588k 2 ;

Zab = 10 + j80 + 37.647k 2 − j150.588k 2 = (10 + 37.647k 2 ) + j(80 − 150.588k 2 ). Zab is resistive when 80 − 150.588k 2 = 0

or

k 2 = 0.53125.

.·. Zab = 10 + (37.647)(0.53125) = 30 Ω. P 9.76

√ [a] M = 0.4 0.0625 = 0.1 H,

ωM = 80 Ω;

Z22 = 40 + j800(0.125) + 360 + j800(0.25) = (400 + j300) Ω. Therefore |Z22 | = 500 Ω, 80 Zr = 500 

[b] I1 =

2

∗ Z22 = (400 − j300) Ω.

(400 − j300) = (10.24 − j7.68) Ω.

245.20 = 0.50/ − 53.13◦ A; 184 + 100 + j400 + Zτ

i1 = 0.5 cos(800t − 53.13◦ ) A. jωM j80 [c] I2 = I1 = (0.5/ − 53.13◦ ) = 0.08/0◦ A; Z22 500/36.87◦ 



i2 = 80 cos 800t mA.

Problems P 9.77

ZTh = 30 + j200 + (50/25)2 (15 − j20) = 90 + j120 Ω; VTh =

P 9.78

225/0◦ (j50) = 450/36.87◦ V. 15 + j20

[a] jωL2 = j(500)103 (500)10−6 = j250 Ω; 109 1 = = −j100 Ω; jωC j(500 × 103 )(20) Z22 = 150 + 50 + j250 − j100 = 200 + j150 Ω; ∗ Z22 = 200 − j150 Ω;

ωM = (500 × 103 )(100 × 10−6 ) = 50 Ω; 

Zr =

50 250

2

[200 − j150] = 8 − j6 Ω.

[b] Zab = R1 + jωL1 + 8 − j6; jωL1 = j(500 × 103 )(80 × 10−6 ) = j40 Ω; Zab = 20 + j34 Ω. P 9.79

In Eq. 9.34 replace ω 2 M 2 with k 2 ω 2 L1 L2 and then write Xab as Xab = ωL1 −

k 2 ω 2 L1 L2 (ωL2 + ωLL ) 2 R22 + (ωL2 + ωLL )2

(

= ωL1 1 −

k 2 ωL2 (ωL2 + ωLL ) } 2 R22 + (ωL2 + ωLL )2

For Xab to be negative requires 2 R22 + (ωL2 + ωLL )2 < k 2 ωL2 (ωL2 + ωLL ),

or 2 + (ωL2 + ωLL )2 − k 2 ωL2 (ωL2 + ωLL ) < 0, R22

which reduces to 2 R22 + ω 2 L22 (1 − k 2 ) + ωL2 ωLL (2 − k 2 ) + ω 2 L2L < 0.

But k ≤ 1, so it is impossible to satisfy the inequality. Therefore Xab can never be negative if XL is an inductive reactance.

9–51

9–52

CHAPTER 9. Sinusoidal Steady State Analysis

P 9.80

Zab =

V1 ; I1

V1 V2 =− ; 50 1

50I1 = −I2 ;

V2 −50V2 = 2500 . .·. Zab = −I2 /50 I2 V2 V3 = ; 1 25

I2 = 25I3

2500 V3 V3 /25 = .·. Zab = 2500 25I2 625 I3 = 4ZL = 4(200 + j150) = (800 + j600) Ω. P 9.81

[a]

N1 I1 = N2 I2 , Zab =

I2 =

N1 I1 ; N2

V2 V2 Vab = = ; I1 + I2 I1 + I2 (1 + N1 /N2 )I1

V1 N1 = , V2 N2

V1 =

N1 V2 ; N2

N1 V1 + V2 = ZL I1 = + 1 V2 ; N2 



Problems

Zab =

I1 ZL ; (N1 /N2 + 1)(1 + N1 /N2 )I1

.·. Zab =

ZL . [1 + (N1 /N2 )]2

[b] Assume dot on the N2 coil is moved to the lower terminal. Then N1 N1 V2 and I2 = − I1 . N2 N2 As before V2 Zab = and V1 + V2 = ZL I1 ; I1 + I2

V1 = −

.·. Zab =

P 9.82

ZL I1 V2 = . (1 − N1 /N2 )I1 [1 − (N1 /N2 )]2 I1

Zab =

ZL . [1 − (N1 /N2 )]2

Zab =

Vab V1 + V2 = ; I1 I1

[a]

V1 V2 = , N1 N2

V2 =

N1 I1 = N2 I2 ,

N2 V1 ; N1

I2 =

N1 I1 ; N2 

V2 = (I1 + I2 )ZL = I1

N1 ZL ; 1+ N2 

N1 N1 V1 + V2 = + 1 V2 = 1 + N2 N2 





(1 + N1 /N2 )2 ZL I1 .·. Zab = . I1 

Zab = 1 +

N1 N2

2

ZL .

2

ZL I1 ;

9–53

9–54

CHAPTER 9. Sinusoidal Steady State Analysis [b] Assume dot on N2 is moved to the lower terminal, then −V2 V1 = , N1 N2

V1 =

N1 I1 = −N2 I2 ,

−N1 V2 ; N2

I2 =

−N1 I1 . N2

As in part [a] V2 = (I2 + I1 )ZL Zab =

and Zab =

V1 + V2 ; I1

(1 − N1 /N2 )V2 (1 − N1 /N2 )(1 − N1 /N2 )ZL I1 = ; I1 I1

Zab = [1 − (N1 /N2 )]2 ZL . P 9.83

The phasor domain equivalent circuit is

Vo =

Vm − IRx ; 2

I=

Vm . Rx − jXC

As Rx varies from 0 to ∞, the amplitude of vo remains constant and its phase angle increases from 0◦ to −180◦ , as shown in the following phasor diagram:

P 9.84

[a] I =

200 200 + = (8 − j4) A; 25 j50

Vs = 200/0◦ + (1 + j2)(8 − j4) = 216 + j12 = 216.33/3.2◦ V.

Problems

9–55

[b] Use the capacitor to eliminate the j component of I, therefore Ic = j4 A,

Zc =

200 = −j50 Ω. j4

The capacitive reactance is −50 Ω. Vs = 200 + (1 + j2)8 = 208 + j16 = 208.6/4.4◦ V. [c] Let Ic denote the magnitude of the current in the capacitor branch. Then I = (8 − j4 + jIc ) = 8 + j(Ic − 4) A. Vs = 200/α = 200 + (1 + j2)[8 + j(Ic − 4)] = (216 − 2Ic ) + j(12 + Ic ). It follows that 200 cos α = (216 − 2Ic ) and 200 sin α = (12 + Ic ). Now square each term and then add to generate the quadratic equation 5Ic2 − 840Ic + 46,000 = 2002 ;

Ic = 84 ± 75.472 A.

Therefore Ic = 8.53 A (smallest value) and Zc = 200/j8.53 = −j23.45 Ω. Therefore, the capacitive reactance is −23.45 Ω. P 9.85

[a]

I` =

440 440 + = 20 − j20 A; 22 j22

V` = (0.2 + j1.6)(20 − j20) = 36 + j28 = 45.61/37.87◦ V; Vs = 440/0◦ + V` = 476 + j28 = 476.82/3.37◦ V. [b]

9–56

CHAPTER 9. Sinusoidal Steady State Analysis

[c] I` =

440 440 440 + + = 20 + j0 A; 22 j22 −j22

V` = (0.2 + j1.6)(20 + j0) = 4 + j32 = 32.25/82.87◦ ; Vs = 440 + V` = 444 + j32 = 445.15/4.12◦ .

P 9.86

[a]

120 = (R + 0.3 + j0.3)I1 − (0.2 + j0.2)I2 − RI3 ; 120 = −(0.2 + j0.2)I1 + (R + 0.3 + j0.3)I2 − RI3 . Subtracting the above two equations gives 0 = (R + 0.5 + j0.5)I1 − (R + 0.5 + j0.5)I2 ; .·. I1 = I2

so

In = I1 − I2 = 0 A.

[b] V1 = R(I1 − I3 );

V2 = R(I2 − I3 ).

Since I1 = I2 (from part [a]) V1 = V2 . [c]

240 = (275.2 + j0.2)Ia − 275Ib ;

Problems

9–57

0 = −275Ia + 285Ib . Solving, Ia = 24.3576 − j0.4946 A; Ib = 23.503 − j0.47726 A; I = Ia − Ib = 0.8546 − j0.01734 A; V1 = 250I = 213.65 − j4.335 = 213.694/ − 1.16◦ V; V2 = 25I = 21.365 − j0.4335 = 21.3694/ − 1.16◦ V. [d]

120 = (250.3 + j0.3)I1 − (0.2 + j0.2)I2 − 250I3 ; 120 = −(0.2 + j0.2)I1 + (25.3 + j0.3)I2 − 25I3 ; 0 = −250I1 − 25I2 + 285I3 . Solving, I1 = 23.94 − j0.515 A; I2 = 28.118 − j0.651 A; I3 = 23.468 − j0.509 A; V1 = 250(I1 − I3 ) = 118.4/ − 0.74◦ V; V2 = 25(I2 − I3 ) = 116.3/ − 1.75◦ V. [e] Because an open neutral can result in unbalanced voltages across the 120 V loads. P 9.87

[a] Let N1 be the number of turns on the primary winding; because the secondary winding is center-tapped, let 2N2 be the total turns on the secondary. Then 20,000 240 = ; N1 2N2

N2 3 .·. = = a. N1 500

9–58

CHAPTER 9. Sinusoidal Steady State Analysis In part c), IP = 2aIa ; 3 2N2 Ia = .·. IP = Ia N1 250 =

3 (24.3576 − j0.4946); 250

IP = 292.29 − j5.93 mA. In part d), IP N1 = I1 N2 + I2 N2 ; N2 (I1 + I2 ) .·. IP = N1 =

3 (23.94 − j0.515 + 28.12 − j0.651) 500

=

3 (52.06 − j1.166); 500

IP = 312.36 − j6.996 mA. [b] Yes, because the neutral conductor carries non-zero current whenever the load is not balanced. P 9.88

[a] The circuit is redrawn, with mesh currents identified:

The mesh current equations are: 125/0◦ = 86Ia − 4Ib − 80Ic ; 125/0◦ = −4Ia + 66Ib − 60Ic ; 0 = −80Ia − 60Ib + 240Ic .

Problems

9–59

Solving, Ia = 3.82653/0◦ A;

Ib = 4.2517/0◦ A;

Ic = 2.3384/0◦ A.

The branch currents are: I1 = Ia = 3.82653/0◦ A; I2 = Ib − Ia = 0.4252/180◦ A; I3 = −Ib = 4.2517/180◦ A; I4 = Ic = 2.3384/0◦ A; I5 = Ia − Ic = 1.4881/0◦ A; I6 = Ib − Ic = 1.9133/0◦ A. [b] Let N1 be the number of turns on the primary winding; because the secondary winding is center-tapped, let 2N2 be the total turns on the secondary. Therefore, 250 15,000 = N1 2N2

or

N2 1 = . N1 120

The ampere turn balance requires N1 IP = N2 I1 + N2 I3 . Therefore, IP = P 9.89

N2 1 (I1 + I3 ) = (3.8265 + 4.2517) = 67.32/0◦ mA. N1 120

[a]

The three mesh current equations are 125/0◦ = 86Ia − 4Ib − 80Ic ; 125/0◦ = −4Ia + 86Ib − 80Ic ; 0 = −80Ia − 80Ib + 260Ic .

9–60

CHAPTER 9. Sinusoidal Steady State Analysis Solving, Ia = 3.815/0◦ A;

Ib = 3.815/0◦ A;

Ic = 2.347/0◦ A;

.·. I2 = Ib − Ia = 0 A. [b] IP =

N2 N2 (I1 + I3 ) = (Ia + Ib ) N1 N1

1 (3.815 + 3.815) = 63.58/0◦ mA. 120 [c] Yes; when the two 125 V loads are equal, there is no current in the “neutral” line, so no power is lost to this line. Since you pay for power, the cost is lower when the loads are equal. =

P 9.90

No, the motor current drops to 5 A, well below its normal running value of 22.86 A.

P 9.91

After fuse A opens, the current in fuse B is only 15 A.

Sinusoidal Steady State Power Calculations

Assessment Problems 1 AP 10.1 [a] P = (100)(10) cos(50 − 15) = 500 cos 35◦ = 409.58 W (abs); 2 Q = 500 sin 35◦ = 286.79 VAR (abs). 1 [b] P = (40)(5) cos(−15 − 60) = 100 cos(−75◦ ) = 25.88 W (abs); 2 Q = 100 sin(−75◦ ) = −96.59 VAR (del). 1 [c] P = (400)(10) cos(30 − 150) = 2000 cos(−120◦ ) = −1000 W (del); 2 Q = 2000 sin(−120◦ ) = −1732.05 VAR (del). 1 [d] P = (200)(5) cos(160 − 40) = 500 cos(120◦ ) = −250 W (del); 2 Q = 500 sin(120◦ ) = 433.01 VAR (abs). AP 10.2 [a] pf = cos(θv − θi ) = cos(50 − 15) = cos(35◦ ) = 0.82 lagging; rf = sin(θv − θi ) = sin(35◦ ) = 0.57. [b] pf = cos(−15 − 60) = cos(−75◦ ) = 0.26 leading; rf = sin(−75◦ ) = −0.97. [c] pf = cos(30 − 150) = cos(−120◦ ) = −0.5 leading; rf = sin(−120◦ ) = −0.87.

10–1

10–2

CHAPTER 10. Sinusoidal Steady State Power Calculations [d] pf = cos(160 − 40) = cos(120◦ ) = −0.5 lagging; rf = sin(120◦ ) = 0.87.

0.24 Ip AP 10.3 From Ex. 9.4 Irms = √ = √ A; 3 3 P =

2 Irms R

0.0576 = (1000) = 19.2 W. 3 

AP 10.4 jωL = j25 Ω;

Io =



1 jωC

= −j50 Ω.

√ 150 = 2.4 + j1.2 = 7.2/26.565◦ A; 50 − j25

√ V50 = 50Io = 50 7.2/26.565◦ V;

√ Vj25 = (25/90◦ )Io = 25 7.2/116.565◦ V.

√ 1 1 √ [a] P = Vm Im cos(θv − θi ) = (50 7.2)( 7.2) cos(0) = 180 W. 2 2 √ 1 1 √ [b] Q = Vm Im sin(θv − θi ) = (25 7.2)( 7.2) sin(90) = 90 VAR. 2 2 [c] S = P + jQ = 180 + j90 VA; |S| = 201.25 VA. [d] pf = cos[tan−1 (Q/P )] = cos[tan−1 (0.5)] = cos 26.565◦ = 0.89 lagging. AP 10.5 [a] Z = (120 + j90)k(−j125) = 120 − j90 = 150/ − 36.87◦ Ω; Therefore, the line current is I` =

465/0◦ = 3.07/35.054◦ A (rms). 120 − j90 + 4 + j3

VL = ZI` = (150/ − 36.87◦ )(3.07/35.054◦ ) = 460.47/ − 1.82◦ V (rms); IL =

VL = 3.07/ − 38.69◦ A (rms). 120 + j90

Problems [b] SL = VL I∗L = (460.47/ − 1.82◦ )(3.07/38.69◦ ) = 1413.64/36.87◦ = (1130.91 + j848.19) VA; PL = 1130.91 W;

QL = 848.19 VAR;

[c] P` = |I` |2 (4) = (3.07)2 (4) = 37.7 W;

Q` = |I` |2 (3) = 28.27 VAR.

[d] Sg (delivering) = 465I∗` = (1168.61 − j819.91) VA. Therefore the source is delivering 1168.61 W and absorbing 819.91 magnetizing VAR. |VL |2 (460.47)2 [e] Qcap = = = −1696.26 VAR. −125 −125 Therefore the capacitor is delivering =1696.26 magnetizing VAR. Check:

28.27 + 848.19 + 819.91 = 1696.37 VAR (rounding) and 1130.91 + 37.7 = 1168.61 W.

AP 10.6 Series circuit derivation: S = 400I∗ = (6000 + j8000); Therefore I∗ = 15 + j20 = 25/53.13◦ A (rms). I = 25/ − 53.13◦ A (rms); Z=

V 400 = = 16/53.13◦ = (9.6 + j12.8) Ω; I 25/ − 53.13◦

Therefore R = 9.6 Ω,

XL = 12.8 Ω.

Parallel circuit derivation (400)2 = 26.67 Ω; 6000

P =

(400)2 ; R

therefore R =

Q=

(400)2 ; XL

therefore XL =

AP 10.7 [a] S1 = 16 + j18 kVA;

(400)2 = 20 Ω. 8000

S2 = 6 − j8 kVA;

S3 = 8 + j0 kVA;

ST = S1 + S2 + S3 = 30 + j10 kVA; 250I∗ = (30 + j10) × 103 ; Z=

.·. I = 120 − j40 A (rms);

250 = 1.875 + j0.625 Ω = 1.98/18.43◦ Ω. 120 − j40

10–3

10–4

CHAPTER 10. Sinusoidal Steady State Power Calculations [b] pf = cos(18.43◦ ) = 0.9487 lagging.

AP 10.8 [a] The phasor domain equivalent circuit and the Th´evenin equivalent are shown below: Phasor domain equivalent circuit after source transforming the voltage source and series-connected resistor:

Th´evenin equivalent: ZTh = 64 + (320kj160) − j640 = 128 − j512 = 527.76/ − 75.96◦ Ω. VTh = (320kj160)(0.25) = 16 + j32 = 35.777/63.435◦ V;

For maximum power transfer, ZL = (128 + j512) Ω. 35.777/63.435◦ [b] I = = 0.14/63.435◦ A; 256 Therefore P =

0.14 √ 2

!2

128 = 1.25 W.

[c] RL = |ZTh | = 527.76 Ω. 35.777/63.435◦ [d] I = = 0.043/101.417◦ A; 655.76 − j512 Therefore P =

0.043 √ 2

!2

(527.76) = 487.9 mW.

Problems AP 10.9

Mesh current equations: 180 = (3 + j4)I1 + j9(I1 − I2 ) − j3I1 + j3(I2 − I1 ); 0 = j9(I2 − I1 ) + j3I1 + 9I2 . In standard form: 180 = (3 + j7)I1 − j6I2 ; 0 = −j6I1 + (9 + j9)I2 . Solving, I2 = 12/0◦ A;

.·. P = (12)2 (9) = 1296 W.

AP 10.10 [a]

j80I1 − j100I2 + 75(I1 − I2 ) = 496; 0 = 75(I2 − I1 ) + j200I2 − j100I1 + 80I2 . In standard form: (75 + j80)I1 + (−75 − j100)I2 = 496; (−75 − j100)I1 + (155 + j200)I2 = 0. Solving, I1 = 8 − j6.2 A;

I2 = 4 − j3 = 5/ − 36.87◦ A;

1 .·. P = (25)(80) = 1000 W. 2

10–5

10–6

CHAPTER 10. Sinusoidal Steady State Power Calculations [b] I1 − I2 = 4 − j3.2 A; 1 P75 = |I1 − I2 |2 (75) = 984 W. 2 1 [c] Pg = (496)(8) = 1984 W; 2 X

Pabs = 1000 + 984 = 1984 W. (checks)

AP 10.11 [a] Open circuit voltage:

40/0◦ = 4(I1 + I3 ) + 12I3 + VTh ; I1 = −I3 ; 4

I1 = −4I3 .

Solving, VTh = 40/0◦ V. Short circuit current:

40/0◦ = 4I1 + 4I3 + I1 + V1 ; 4V1 = 16(I1 /4) = 4I1 ;

.·. V1 = I1 ;

.·. 40/0◦ = 6I1 + 4I3 . Also, 40/0◦ = 4(I1 + I3 ) + 12I3 . Solving, I1 = 6 A;

I3 = 1 A;

Isc = I1 /4 + I3 = 2.5 A;

Problems

RTh =

[b] I =

VTh 40 = = 16 Ω. Isc 2.5

40/0◦ = 1.25/0◦ A(rms); 32

P = (1.25)2 (16) = 25 W. [c]

40 = 4(I1 + I3 ) + 12I3 + 20; 4V1 = 4I1 + 16(I1 /4 + I3 );

.·. V1 = 2I1 + 4I3 ;

40 = 4I1 + 4I3 + I1 + V1 ; .·. I1 = 6 A;

I3 = −0.25 A;

I1 + I3 = 5.75/0◦ A.

P40V (developed) = 40(5.75) = 230 W; 25 .·. % delivered = (100) = 10.87%. 230

10–7

10–8

CHAPTER 10. Sinusoidal Steady State Power Calculations

Problems P 10.1

[a] V = 100/ − 45◦ V, I = 20/15◦ A. Therefore 1 A → B; P = (100)(20) cos[−45 − (15)] = 500 W, 2 Q = 1000 sin −60◦ = −866.03 VAR, B → A. [b] V = 100/ − 45◦ ,

I = 20/165◦ ;

P = 1000 cos(−210◦ ) = −866.03 W,

B → A;

Q = 1000 sin(−210◦ ) = 500 VAR, [c] V = 100/ − 45◦ ,

A → B.

I = 20/105◦ ;

P = 1000 cos(−150◦ ) = −866.03 W,

B → A;

Q = 1000 sin(−150◦ ) = −500 VAR,

B → A.

[d] V = 100/0◦ ,

I = 20/120;◦

P = 1000 cos(−120◦ ) = −500 W,

B → A;

Q = 1000 sin(−120◦ ) = −866.03 VAR, P 10.2

p = P + P cos 2ωt − Q sin 2ωt; dp = 0 when dt

cos 2ωt = √

B → A.

dp = −2ωP sin 2ωt − 2ωQ cos 2ωt; dt

− 2ωP sin 2ωt = 2ωQ cos 2ωt or

P ; + Q2

P2

sin 2ωt = − √

Q tan 2ωt = − . P

Q . + Q2

P2

Let θ = tan−1 (−Q/P ), then p is maximum when 2ωt = θ and p is minimum when 2ωt = (θ + π). Therefore pmax = P + P · √ and pmin = P − P · √

q P Q(−Q) √ − = P + P 2 + Q2 , P 2 + Q2 P 2 + Q2

q P Q √ − Q · = P − P 2 + Q2 . P 2 + Q2 P 2 + Q2

Problems P 10.3

10–9

[a] To find the power used by an appliance, divide the yearly kW-hours used by the product of the number of hours per month used and the number of months: 1080 k central AC = = 3000 W; (120)(3) dishwasher =

293 k = 813.89 W; (30)(12)

refrigerator =

533 k = 61.7 W; (720)(12)

microwave =

101 k = 935.2 W. (9)(12)

The total power consumed by these four appliances is P4 = 3000 + 813.89 + 61.7 + 035.2 = 4810.76 W. When 120 V(rms) is used, the amount of current required is 4810.8 = 40.1 A(rms). 120 Therefore, the 60 A current breaker will not interrupt the current. I4 =

[b] the power used by the dryer is dryer =

901 k = 3128.47 W. (24)(12)

With this fifth appliance added, the total power consumed is P5 = 4810.76 + 3128.47 = 7939.2 W. Now the current required is 7939.2 = 66.2 A(rms). 120 Now the 60 A current breaker will interrupt the current. So before turning on the dryer, the dishwasher must be turned off: I5 =

7939.2 − 813.89 = 7125.31 W;

the resulting current required is

7125.31 = 59.4 A(rms). 120 The 60 A circuit breaker will not interrupt the current if the dishwasher is off.

10–10 P 10.4

CHAPTER 10. Sinusoidal Steady State Power Calculations

[a] From the solution to Problem 9.57 we have:

Vo = 72 + j96 = 120/53.13◦ V; 1 1 Pg = − Vm Im cos(θv − θi ) = − (120)(15) cos(53.13◦ ) = −540 W; 2 2 1 1 Qg = − Vm Im sin(θv − θi ) = − (120)(15) sin(53.13◦ ) = −720 W; 2 2 Therefore, the independent current source is delivering 540 W and 720 magnetizing vars. I1 =

Vo = 15/53.13◦ A; 8

1 1 P8Ω = Vm Im cos(θv − θi ) = (120)(15) cos(0◦ ) = 900 W. 2 2 Therefore, the 8 Ω resistor is absorbing 900 W. I∆ =

Vo = −9.6 + j7.2 = 12/143.13◦ A; −j10

1 1 Qcap = Vm Im sin(θv − θi ) = (120)(12) sin(53.13◦ − 143.13◦ ) = −720 VAR. 2 2 Therefore, the −j10 Ω capacitor is delivering 720 magnetizing vars. 2.5I∆ = −24 + j18 V; I2 =

Vj5 = 72 + j96 + 24 − j18 = 123.69/39.1◦ V;

Vj5 123.69/39.1◦ = = 24.74/ − 50.91◦ A. j5 j5

1 1 Qj5 = Vm Im sin(θv − θi ) = (123.69)(24.74) sin(39.1◦ + 50.9◦ ) = 1530 VAR. 2 2 Therefore, the j5 Ω inductor is absorbing 1530 magnetizing vars. 1 1 P2.5I∆ = Vm Im cos(θv − θi ) = [(2.5)12](24.74) cos(143.13◦ + 50.9◦ ) = −360 W; 2 2 1 1 Q2.5I∆ = Vm Im sin(θv − θi ) = [(2.5)12](24.74) sin(143.13◦ + 50.9◦ ) = −90 W; 2 2 Thus the dependent source is delivering 360 W and 90 magnetizing vars. [b]

X

Pgen = 360 + 540 = 900 W =

X

Pabs .

Problems [c] P 10.5

X

Qgen = 720 + 90 + 720 = 1530 VAR =

X

10–11

Qabs .

[a] From the solution to Problem 9.62 we have

Ia = 2.25 − j2.25 = 3.182/ − 45◦ A;

Ib = −6.75 + j0.75 = 6.79152/173.66◦ A;

Io = 9 − j3 = 9.4868/ − 18.435◦ A; 1 1 P60V = − Vm Im cos(θv − θi ) = − (60)(3.182) cos(45◦ ) = −67.5 W; 2 2 1 1 Q60V = − Vm Im sin(θv − θi ) = − (60)(3.182) sin(45◦ ) = −67.5 VAR; 2 2 Thus, the 60 V source is developing 67.5 W and 67.5 magnetizing vars. 1 1 P90V = − Vm Im cos(θv − θi ) = − (90)(6.79152) cos(90◦ − 173.66◦ ) = −33.75 W; 2 2 1 1 Q90V = − Vm Im sin(θv − θi ) = − (90)(6.79152) sin(90◦ − 173.66◦ ) = 303.75 W; 2 2 Thus, the 90 V source is delivering 33.75 W and absorbing 303.75 magnetizing vars. V20Ω = 20Ia = 63.64/ − 45◦ V; 1 P20Ω = (63.64)(3.182) cos(0◦ ) = 101.25 W. 2 Thus the 20 Ω resistor is absorbing 101.25 W. V−j20Ω = −j20Ib = 135.83/83.66◦ V; 1 Q−j20Ω = (135.83)(6.79152) sin(−90◦ ) = −461.25 VAR. 2 Thus the −j20 Ω capacitor is developing 461.25 magnetizing vars. Vj5Ω = j5Io = 47.434/71.565◦ V; 1 Qj5Ω = (47.434)(9.4868) sin(90◦ ) = 225 VAR. 2 Thus the j5 Ω inductor is absorbing 225 magnetizing vars. [b]

X

Pdev = 67.5 + 33.75 = 101.25 W =

X

Pabs .

10–12 [c]

CHAPTER 10. Sinusoidal Steady State Power Calculations X

Qdev = 67.5 + 461.25 = 528.75 VAR;

X

P 10.6

Qabs = 225 + 303.75 = 528.75 VAR =

Ig = 30/0◦ mA;

X

Qdev .

1 106 = = −j1 Ω; jωC j(25 × 103 )(40)

jωL = j(25 × 103 )(40) × 10−6 = j1 Ω.

Z1 = −j1k(5 + j1) = 0.2 − j1 Ω;

Zeq = 2 + Z1 = 2.2 − j1 Ω;

Vg = Zeq Ig = 66 − j30 mA = 72.5/ − 24.444◦ mA; 1 1 Pg = Vg Ig cos(θv − θi ) = (0.0725)(0.03) cos(−24.444◦ ) = 990 µW. 2 2 P 10.7

[a] P =

1 (90)2 = 3 W; 2 1350

Q=

1 (90)2 = 4 VAR; 2 (1012.5)

From Problem 10.2,

pmax = P +

q

P 2 + Q2 = 3 +

q

(3)2 + (4)2 = 8 W.

[b] pmin = 3 − 5 = −2 W. [c] P = 3 W from (a). [d] Q = 4 VAR from (a). [e] absorb, because Q > 0. [f ] pf = cos(θv − θi ; ) I=

90 90 + = 0.0667 − j0.08889 = 111.11/ − 53.13◦ mA; 1350 j1012.5

.·. pf = cos(0 + 53.13◦ ) = 0.6 lagging. [g] rf = sin(53.13◦ ) = 0.8.

Problems

P 10.8

jωL = j10,000(10−3 ) = j10 Ω;

−15 + .·. Vo

1 106 = = −j40 Ω. jωC j10,000(2.5)

Vo Vo + 10(Vo / − j40) + = 0; −j40 20 + j10 "

#

1 1 + j0.25 + = 15; −j40 20 + j10

.·. Vo = 300 − j100 V; Vo = 2.5 + j7.5 A. .·. I∆ = −j40 Io = 15/0◦ − I∆ = 15 − 2.5 − j7.5 = 12.5 − j7.5 = 14.5774/ − 30.96◦ A; V20Ω = 20Io = 291.5476/ − 30.96◦ 1 1 P20Ω = Vm Im cos(θv − θi ) = (291.5476)(14.5774) cos(0◦ ) = 2125 W. 2 2 P 10.9

109 1 = = 2500 Ω; ωC (5000)(80) Zf =

−j2500(7500) = 750 − j2250 Ω; 7500 − j2500

Zi = 1500 Ω; Zf 750 − j2250 .·. = = 0.5 − j1.5. Zi 1500 V1kΩ = −

Zf Vg ; Zi

Vg = 4/0◦ V;

V1kΩ = (−0.5 + j1.5)(4) = −2 + j6 = 6.325/108.43◦ V; I1kΩ =

V1kΩ = 6.325/108.43◦ mA; 1000

1 1 P = Vm Im cos(θv − θi ) = (6.325)(6.325 × 10−3 ) cos(0◦ ) = 20 mW. 2 2

10–13

10–14

CHAPTER 10. Sinusoidal Steady State Power Calculations

P 10.10 [a] line loss = 50,000 − 40,000 = 10 kW;

Pline = V20Ω Iline cos(0◦ ) = (20|Ig |)(|Ig |) = 10,000; √ .·. |Ig | = 500 A (rms). Now find the load impedance: PL = (RL |Ig |)|Ig | cos(0◦ ) = 40,000 QL = (XL |Ig |)|Ig | sin(90◦ ) = 30,000 |Ig |2 XL = 30,000

40,000 .·. RL = = 80 Ω; 500 30,000 .·. XL = = 60 Ω; 500

.·. XL = 60 Ω.

Thus,

|Z| =

q

(100)2 + (60 − X` )2

2500

|Ig | = q

10,000 + (60 − X` )2

;

625 × 104 .·. 10,000 + (60 − X` )2 = = 12,500. 500 Solving,

(60 − X` ) = ±50.

Thus, X` = 10 Ω

or

X` = 110 Ω.

[b] If X` = 10 Ω: 2500 = 20 − j10 = 22.36/ − 26.565◦ A (rms); 100 + j50 √ Pg = −(2500)( 500) cos(26.565◦ ) = −50,000 W; √ Qg = −(2500)( 500) sin(26.565◦ ) = −25,000 VAR; Ig =

Thus, the voltage source is delivering 50 kW and 25 magnetizing kvars. √ √ Q−j10 = (10 500)( 500) sin(−90◦ ) = −5000 VAR.

Problems

10–15

Therefore the line reactance is generating 5 magnetizing kvars. √ √ Qj60 = (60 500)( 500) sin(90◦ ) = 30,000 VAR. Therefore the load reactance is absorbing 30 magnetizing kvars. X

Qdev = 25,000 kVAR =

X

Qabs.

If X` = 110 Ω: 2500 = 20 + j10 = 22.36/26.565◦ A (rms); Ig = 100 − j50 √ Pg = −(2500)( 500) cos(−26.565◦ ) = −50,000 W; √ Qg = −(2500)( 500) sin(−26.565◦ ) = 25,000 VAR; Thus, the voltage source is delivering 50 kW and absorbing 25 magnetizing kvars. √ √ Q−j110 = (110 500)( 500) sin(−90◦ ) = −55 kVAR. Therefore the line reactance is generating 55 magnetizing kvars. The load continues to absorb 30 magnetizing kvars. X

P 10.11 Wdc

Qdev = 55 kVAR =

Vdc2 T; = R

Ws =

X

Z to +T to

Qabs . vs2 dt; R

Z to +T 2 vs Vdc2 · .. T = dt. R R to

Vdc2

1 Z to +T 2 vs dt; = T to s

Vdc =

1 Z to +T 2 vs dt = Vrms . T to

P 10.12 [a] Ieff = 60/110 ∼ = 0.545 A; P 10.13 i(t) =

30 × 103 t = 750t 40

i(t) = M −

30 × 103 t 10

[b] Ieff = (60 + 80)/110 ∼ = 1.273 A. 0 ≤ t ≤ 40 ms;

40 ms ≤ t ≤ 50 ms;

i(t) = 0 when t = 50 ms; .·. M = 3000(50 × 10−3 ) = 150.

10–16

CHAPTER 10. Sinusoidal Steady State Power Calculations

i(t) = 150 − 3000t s

.·. Irms = Z 0.04

1 0.05

40 ms ≤ t ≤ 50 ms;

Z 0.04

(750)2 t2

dt +

0

(750) t dt = (750)

0

(150 −

3000t)2



dt .

0.04 3 2t

2 2

Z 0.05

3

0.04 =

12;

0

(150 − 3000t)2 = 22,500 − 9 × 105 t + 9 × 106 t2 ; Z 0.05

22,500 dt = 225;

0.04

Z 0.05

5

4 2

9 × 10 t dt = 45 × 10 t

0.04

0.05 =

405;

0.04

9 × 106

0.05

Z 0.05

t2 dt = 3 × 106 t3

0.04

.·. Irms =

q

= 183;

0.04

20{12 + (225 − 405 + 183)} =

√

300 = 17.32 A.

24 × 103 = 80 Ω. .·. R = 300

2 P 10.14 P = Irms R

P 10.15 [a] Area under one cycle of vg2 : A = (−25)2 (2) + (25)2 (2) = 2500; mean value = Vrms =

√

A = 312.5; 8

312.5 = 17.68 V(rms).

2 Vrms 312.5 = = 125 mW. R 2500 312.5 V2 = 500 Ω. [c] R = rms = P 0.625

[b] P =

P 10.16 [a] Area under one cycle of v 2 : A = 4(20)2 (20 × 10−6 ) + 2(100)2 (20 × 10−6 ) = 21,600(20 × 10−6 ). Mean value of vg2 : M.V. = .·. Vrms

A 21,600(20 × 10−6 ) = = 3600; 120 × 10−6 120 × 10−6 √ = 3600 = 60 V(rms).

Problems

[b] P =

10–17

2 Vrms 3600 = = 300 W. R 12

P 10.17 [a] ZL = j(2500)(0.405) = j1012.5 Ω. The phasor domain circuit is

P =

1 |V|2 1 902 = = 3 W. 2 R 2 1350

1 |V|2 1 902 = = 4 VAR. 2 X 2 1012.5 [c] S = P + jQ = 3 + j4 = 5/53.13◦ VA;

[b] Q =

so

|S| = 5 VA.

[d] pf = cos 53.13◦ = 0.6 lagging. P 10.18 [a] Let VL = VL /0◦ :

SL = 250(0.6 + j0.8) = 150 + j200 VA; I∗` =

200 150 +j ; VL VL

I` =

150 200 −j ; VL VL

200 150 240/θ = VL + −j (1 + j8); VL VL 



240VL /θ = VL2 + (150 − j200)(1 + j8) = VL2 + 1750 + j1000; 240VL cos θ = VL2 + 1750;

240VL sin θ = 1000;

(240)2 VL2 = (VL2 + 1750)2 + 10002 ; 57,600VL2 = VL4 + 3500VL2 + (3.0625 + 1) × 106 , or VL4 − 54,100VL2 + 4,062,500 = 0. Solving, VL2 = 27,050 ± 26,974.8;

VL = 232.43 V(rms) and VL = 8.67 V(rms).

10–18

CHAPTER 10. Sinusoidal Steady State Power Calculations If VL = 232.43 V(rms): 1000 sin θ = = 0.0179; (232.43)(240) If VL = 8.67 V(rms): 1000 = 0.4805; sin θ = (8.67)(240)

.·. θ = 1.03◦ .

.·. θ = 28.72◦ .

[b]

1 106 = −j10 Ω; P 10.19 [a] = jωC j105 jωL = j105 (50 × 10−6 ) = j5 Ω.

Z = −j10 + Ig =

(5)(j5) + 7.5 = 10 − j7.5 Ω; 5 + j5

50/0◦ = 3.2 + j2.4 A; 10 − j7.5

1 Sg = − Vg I∗g = −25(3.2 − j2.4) = −80 + j60 VA; 2 P = 80 W(dev); Q = 60 VAR(abs); |S| = |Sg | = 100 VA.

Problems Ig (j5) 1 = (3.2 + j2.4)(1 + j1) = 0.4 + j2.8 A; 5 + j5 2

[b] I1 =

1 P5Ω = |I1 |2 (5) = 20 W; 2 1 P7.5Ω = |Ig |2 (7.5) = 60 W; 2 X

Pabs = 20 + 60 = 80 W =

[c] Ij5 =

X

Pdev .

Ig 5 1 = (3.2 + j2.4)(1 − j1) = 2.8 − j0.4 A; 5 + j5 2

1 Qj5Ω = |Ij5 |2 (5) = 20 VAR(abs); 2 1 Q−j10Ω = |Ig |2 (−10) = −80 VAR(dev); 2 X

Qabs = 20 + 60 = 80 VAR =

X

Qdev .

P 10.20 [a]

Vo − 340 Vo Vo + + = 0; −j100 50 80 + j60 .·. Vo = 238 − j34 V(rms). Ig =

340 − 238 + j34 = 2.04 + j0.68 A(rms); 50

Sg = −Vg I∗g = −(340)(2.04 − j0.68) = −693.6 + j231.2 VA. [b] Source is delivering 693.6 W. [c] Source is absorbing 231.2 magnetizing VAR. Vo [d] I1 = = 0.34 + j2.38 A(rms); −j100 S1 = Vo I∗1 = (238 − j34)(0.34 − j2.38) = 0 − j578 VA;

10–19

10–20

CHAPTER 10. Sinusoidal Steady State Power Calculations

I2 =

Vo 238 − j34 = = 1.7 − j1.7 A(rms); 80 + j60 80 + j60

S2 = Vo I∗2 = (238 − j34)(1.7 + j1.7) = 462.4 + j346.8 VA; S50Ω = |Ig |2 (50) + j0 = (2.15)2 (50) = 231.2 W. [e]

X

Pdel = 693.6 W;

X

.·. [f ]

X

Pdiss = 462.4 + 231.2 = 693.6 W; X

Pdel =

X

Pdiss = 693.6 W.

Qabs = 231.2 + 346.8 = 578 VAR;

X

.·.

Qdel = 578 VAR; X

mag VAR del =

P 10.21 ST = 52,800 − j

X

mag VAR abs = 578.

52,800 (0.6) = 52,800 − j39,600 VA; 0.8

S1 = 40,000(0.96 + j0.28) = 38,400 + j11,200 VA; S2 = ST − S1 = 14,400 − j50,800 = 52,801.52/ − 74.17◦ VA; rf = sin(−74.17◦ ) = −0.9621; pf = cos(−74.17◦ ) = 0.2727 leading. P 10.22 [a] S1 = 24,960 + j47,040 VA; S2 =

|VL |2 (480)2 = = 23,040 − j23,040 VA; Z2∗ 5 + j5

S1 + S2 = 48,000 + j24,000 VA; 480I∗L = 48,000 + j24,000;

.·. IL = 100 − j50 A(rms);

Vg = VL + IL (0.02 + j0.20) = 480 + (100 − j50)(0.02 + j0.20) = 492 + j19 = 492.37/2.21◦ Vrms; |Vg | = 492.37 Vrms.

Problems

[b] T =

1 1 = = 16.67 ms; f 60

2.21◦ t = ; ◦ 360 16.67 ms

.·. t = 102.39 µs;

[c] VL lags Vg by 2.21◦ or 102.39 µs.

P 10.23 S1 = 15,000(0.6) + j15,000(0.8) = 9000 + j12,000 VA; S2 = 6000(0.8) − j6000(0.6) = 4800 − j3600 VA; ST = S1 + S2 = 13,800 + j8400 VA; ST = 200I∗ ;

therefore I∗ = 69 + j42

I = 69 − j42 A(rms);

Vs = 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.645/15.91◦ V (rms). P 10.24 [a] Z1 = 12 + j(2π)(60)(15 × 10−3 ) = 13.27/25.23◦ Ω; pf = cos(25.23◦ ) = 0.9 lagging; rf = sin(25.23◦ ) = 0.43; Z2 = 80 −

j = 184.08/ − 64.24◦ Ω; −6 2π(60)(16 × 10 )

pf = cos(−64.24◦ ) = 0.43 leading; rf = sin(−64.24◦ ) = −0.9; Z3 = 400 + Zp ; Zp =

jωL(1/jωC) jωL = jωL + 1/jωC 1 − ω 2 LC

=

j(120π)(20) = −j570.67 Ω; 1 − (120π)2 (20)(5 × 10−6 )

.·. Z3 = 400 − j570.67 = 696.90/ − 54.97◦ Ω. pf = cos(−54.97◦ ) = 0.57 leading; rf = sin(−54.97◦ ) = −0.82.

10–21

10–22

CHAPTER 10. Sinusoidal Steady State Power Calculations

[b] Y = Y1 + Y2 + Y3 ; Y1 =

1 ; 13.27/25.23◦

Y2 =

1 ; 184.08/ − 64.24◦

Y3 =

1 ; 696.90/ − 54.97◦

Y = 71.35 − j26.05 mS; Z=

1 = 13.16/20.06◦ Ω; Y

pf = cos(20.06◦ ) = 0.94 lagging; rf = sin(20.06◦ ) = 0.343. P 10.25

2400I∗1 = 24,000 + j18,000; I∗1 = 10 + j7.5;

.·. I1 = 10 − j7.5 A(rms);

2400I∗2 = 48,000 − j30,000; I∗2 = 20 − j12.5; I3 =

.·. I2 = 20 + j12.5 A(rms);

2400/0◦ = 40 + j0 A(rms); 60

I4 =

2400/0◦ = 0 − j5 A(rms); j480

Ig = I1 + I2 + I3 + I4 = 70 A(rms); Vg = 2400 + (70)(j10) = 2400 + j700 = 2500/16.26◦ V(rms). P 10.26 [a]

120I∗1 = 1800 + j600;

.·. I1 = 15 − j5 A(rms);

Problems .·. I2 = 10 + j7.5 A(rms);

120I∗2 = 1200 − j900; I3 =

10–23

240 240 + = 20 − j5 A(rms); 12 j48

Ig1 = I1 + I3 = 35 − j10 A(rms); Sg1 = −120(35 + j10) = −4200 − j1200 VA. Thus the Vg1 source is delivering 4200 W and 1200 magnetizing vars. Ig2 = I2 + I3 = 30 + j2.5 A(rms; ) Sg2 = −120(30 − j2.5) = −3600 + j300 VA. Thus the Vg2 source is delivering 3600 W and absorbing 300 magnetizing vars. [b]

X

P 10.27 S1 =

Pdev = 4200 + 3600 = 7800 W; X (240)2 = 7800 W = Pdev ; 12

X

Pabs = 1800 + 1200 +

X

Qdev = 1200 + 900 = 2100 VAR;

X

Qabs = 300 + 600 +

X (240)2 = 2100 VAR = Qdev . 48

336 k 101 k 284 k + + = 3603.7 + j0 VA; (720)(12) (9)(12) (9)(12)

3603.7 .·. I1 = = 30.03 A(rms). 120 S2 =

360 k 72 k 54 k + +4 = 505 + j0 VA; (240)(12) (150)(6) (60)(12)

505 .·. I2 = = 4.21 A(rms). 120 S3 =

108 k 901 k + = 3449.9 + j0 VA; (28)(12) (24)(12)

3449.9 .·. I3 = = 14.37 A(rms). 240 Ig1 = I1 + I3 = 44.4 A(rms); Ig2 = I2 + I3 = 18.58 A(rms). Neither current exceeds the 50 A limit, so no interruptions occur.

10–24

CHAPTER 10. Sinusoidal Steady State Power Calculations

P 10.28 [a] From the solution to Assessment Problem 10.7 we have IL = 120 − j40 A(rms); .·. Vs = 250/0◦ + (120 − j40)(0.01 + j0.08) = 254.4 + j9.2 = 254.57/2.07◦ V(rms). [b] |IL | =

q

16,000;

P` = (16,000)(0.01) = 160 W [c] Ps = 30,000 + 160 = 30.16 kW 30 [d] η = (100) = 99.47%. 30.16

Q` = (16,000)(0.08) = 1280 VAR. Qs = 10,000 + 1280 = 11.28 kVAR.

P 10.29 [a]

I1 =

5000 − j2000 = 40 − j16 A (rms); 125

I2 =

3750 − j1500 = 30 − j12 A (rms); 125

I3 =

8000 + j0 = 32 + j0 A (rms); 250

.·. Ig1 = 72 − j16 A (rms.) In = I1 − I2 = 10 − j4 A (rms); Ig2 = 62 − j12 A; Vg1 = 0.05Ig1 + 125 + j0 + 0.15In = 130.1 − j1.4 V(rms); Vg2 = −0.15In + 125 + j0 + 0.05Ig2 = 126.6 + j0 V(rms); Sg1 = [(130.1 − j1.4)(72 + j16)] = [9389.6 + j1980.8] VA; Sg2 = [(126.6)(62 + j12)] = [7849.2 + j1519.2] VA. Note: Both sources are delivering average power and magnetizing VAR to the circuit.

Problems [b] P0.05 = |Ig1 |2 (0.05) = 272 W; P0.15 = |In |2 (0.15) = 17.4 W; P0.05 = |Ig2 |2 (0.05) = 199.4 W; X

Pabs = 272 + 17.4 + 199.4 + 5000 + 3750 + 8000 = 17,238.8 W;

X

Pdel = 9389.6 + 7849.2 = 17,238.8 W =

X

Qabs = 2000 + 1500 = 3500 VAR;

X

Qdel = 1980.8 + 1519.2 = 3500 VAR =

P 10.30 [a] So = original load = 1800 + j Sf = final load = 2400 + j

X

X

Pabs ;

Qabs .

1800 (0.8) = 1800 + j2400 kVA; 0.6

2400 (0.28) = 2400 + j700 kVA; 0.96

.·. Qadded = 700 − 2400 = −1700 kVAR. [b] Deliver. [c] Sa = added load = 600 − j1700 = 1802.78/ − 70.56◦ kVA; pf = cos(−70.56) = 0.3328 leading. [d] I∗L =

(1800 + j2400) × 103 = 375 + j500 A(rms); 4800

IL = 375 − j500 = 625/ − 53.13◦ A(rms); |IL | = 625 A(rms). [e] I∗L =

(2400 + j700) × 103 = 500 + j145.83; 4800

IL = 500 − j145.83 = 520.83/ − 16.26◦ A(rms); |IL | = 520.83 A(rms). P 10.31 [a] Pbefore = (625)2 (0.02) = 7812.50 W; Pafter = (520.83)2 (0.02) = 5425.35 W.

10–25

10–26

CHAPTER 10. Sinusoidal Steady State Power Calculations

[b] Vs (before) = 4800 + (375 − j500)(0.02 + j0.16) = 4887.5 + j50 = 4887.5/0.59◦ V(rms); |Vs (before)| = 4887.76 V(rms); Vs (after) = 4800 + (500 − j145.83)(0.02 + j0.16) = 4833.33 + j77.08 = 4833.95/0.91◦ V(rms); |Vs (after)| = 4833.95 V(rms). P 10.32 [a] SL = 24 + j7 kVA; I∗L = 192 + j56 A(rms);

125I∗L = (24 + j7) × 103 ; .·. IL = 192 − j56 A(rms).

Vs = 125 + (192 − j56)(0.006 + j0.048) = 128.84 + j8.88 = 129.15/3.94◦ V(rms); |Vs | = 129.15 V(rms). [b] P` = |I` |2 (0.006) = |IL |2 (0.006) = (200)2 (0.006) = 240 W. [c]

(125)2 = −7000; XC −

1 = −2.23; ωC

XC = −2.23 Ω; C=

1 = 1188.36 µF. (2.23)(120π)

[d] I` = 192 + j0 A(rms); Vs = 125 + 192(0.006 + j0.048) = 126.152 + j9.216 = 126.49/4.18◦ V(rms); |Vs | = 126.49 V(rms). [e] P` = (192)2 (0.006) = 221.184 W. P 10.33 [a] I =

250/0◦ = 4 − j3 = 5/ − 36.87◦ A(rms); 40 + j30

P = (5)2 (1) = 25 W. [b] YL =

1 = 17.75 − j11.83 mS; 39 + j26

.·. XC =

1 = −84.5 Ω. −11.83 × 10−3

Problems

10–27

1 = 56.34 Ω. 17.75 × 10−3 250/0◦ [d] I = = 4.35/ − 3.99◦ A; 57.34 + j4

[c] ZL =

P = (4.35)2 (1) = 18.92 W. 18.92 (100) = 75.7%. 25 Thus the power loss after the capacitor is added is 75.7% of the power loss before the capacitor is added.

[e] % =

P 10.34

IL =

240,000 − j70,000 = 96 − j28 A(rms); 2500

IC =

2500 2500 =j = jIC ; −jXC XC

I` = 96 − j28 + jIC = 96 + j(IC − 28); Vs = 2500 + (1 + j8)[96 + j(IC − 28)] = (2820 − 8IC ) + j(740 + IC ); |Vs |2 = (2820 − 8IC )2 + (740 + IC )2 = (2500)2 ; .·. 65IC2 − 43,640IC + 2,250,000 = 0. IC =

43,640 ±

q

(43,640)2 − 4(65)(2,250,000) 2(65)

= 335.69 ± 279.42 = 56.27 A(rms)∗ . *Select the smaller value of IC to minimize the magnitude of I` . 2500 .·. XC = = 44.43; 56.27 .·. C =

1 = 59.7 µF. (44.43)(120π)

10–28

CHAPTER 10. Sinusoidal Steady State Power Calculations

P 10.35 [a] Using the results of Problem 9.78, I1 =

125/0◦ 125 = = 1 − j2 A(rms); 20 + j34 + 5 + j16 25 + j50

jωM j50 I1 = (1 − j2) Z22 200 + j150 = 0.44 − j0.08 = 0.45/ − 10.30◦ A(rms);

I2 =

VL = (150 − j100)(0.44 − j0.08) = 58 − j56 = 80.62/ − 43.99◦ V(rms); |VL | = 80.62 V(rms). [b] Pg (ideal) = 125(1) = 125 W; Pg (practical) = 125 − |I1 |2 (5) = 125 − 25 = 100 W; PL = |I2 |2 (150) = 30 W; % delivered =

30 (100) = 30%. 100

P 10.36 [a]

340/0◦ = 10Ig + j50Ig + j70(Ig − I2 ) − j30I2 +j70Ig − j40I2 + j100(Ig − I2 ); 0 = j100(I2 − Ig ) − j70Ig + j40I2 + j20I2 +j40(I2 − Ig ) − j30Ig + 40I2 ; Solving, Ig = 5 − j1 A(rms);

I2 = 6/0◦ A(rms);

P40Ω = (6)2 (40) = 1440 W. [b] Pg (developed) = (340)(5) = 1700 W. Vg 340 [c] Zab = − 10 = − 10 = 55.38 + j13.08 = 56.91/13.28◦ Ω. Ig 5−j

Problems [d] P10Ω = |Ig |2 (10) = 260 W; X

Pdiss = 1440 + 260 = 1700 W =

X

Pdev .

P 10.37 [a]

20 = j2(I1 − I2 ) + j1(I2 − I3 ) − j1(I1 − I3 ); 0 = 1I2 + j1(I2 − I3 ) + j1(I1 − I2 ) + j2(I2 − I1 ) − j1(I2 − I3 ); 0 = −j1(I3 − I1 ) + j1(I3 − I2 ) − j1(I1 − I2 ) + 1I3 . Solving, I1 = 20 − j20 A(rms);

I2 = 20 + j0 A(rms);

I3 = 0 A(rms).

Ia = I1 = 20 − j20 A

Ib = I1 − I2 = −j20 A;

Ic = I2 = 20 A

Id = I3 − I2 = −20 A;

Ie = I1 − I3 = 20 − j20 A

If = I3 = 0 A.

[b]

Va = 20 + j0 V

Vb = j2Ib − j1Id = 40 + j20 V;

Vc = 1Ic = 20 + j0 V

Vd = j1Id − j1Ib = −20 − j20 V;

Ve = −j1Ie = −20 − j20 V

Vf = 1If = 0 V;

Sa = −20I∗a = −400 − j400 VA; Sb = Vb I∗b = −400 + j800 VA;

10–29

10–30

CHAPTER 10. Sinusoidal Steady State Power Calculations Sc = Vc I∗c = 400 + j0 VA; Sd = Vd I∗d = 400 + j400 VA; Se = Ve I∗e = 0 − j800 VA; Sf = Vf I∗f = 0 + j0 VA.

[c]

X

Pdev = 400 W;

X

Pdiss = −400 + 400 + 400 = 400 W.

Note that the total power dissipated by the coupled coils is zero: −400 + 400 = 0 = Pb + Pd [d]

X

Qgen = 400 + 800 = 1200 VAR; Both the source and the capacitor are generating magnetizing vars. X

Qdiss = 400 + 800 = 1200 VAR;

X

Q dissipated by the coupled coils is Qb + Qd .

P 10.38 [a]

30 = 3000I1 + V1 + 1000(I1 − I2 ); 0 = 1000(I2 − I1 ) − V2 + 2000I2 ; 1 V2 = V1 ; 3

I2 = 3I1 .

Solving, V1 = 28.8 V(rms);

V2 = 9.6 V(rms);

I1 = 1.2 mA(rms);

I2 = 3.6 mA(rms);

V10mA = V1 + 1000(I1 − I2 ) = 26.4 V(rms); .·. P = −(26.4)(10 × 10−3 ) = −264 mW. Thus 264 mW is delivered by the current source to the circuit.

Problems

10–31

[b] I1kΩ = I1 − I2 = −2.4 mA(rms); .·. P1kΩ = (−0.0024)2 (1000) = 5.76 mW. P 10.39 [a]

a1 Ig = I1 ;

a2 Ig = I2 ;

P320 = |I1 |2 (320); .·.

a1 I1 = ; a2 I2

so

P80 = |I2 |2 (80);

|I2 |2 (80) = 16[|I1 |2 (320)]

P80 = 16P320 ;

thus

a1 I1 1 = = . a2 I2 8

The load impedances are matched to the source impedance: a21 (320) + a22 (80) = 136,000 .·.

a21 = 25

[b] Ig =

so

a21 (320) + (8a1 )2 (80) = 136,000;

so

a1 = 5

and

680/0◦ = 2.5/0◦ mA(rms); (136 + 136)103

I2 = 40Ig = 100 mA(rms); .·.

P80Ω = (0.1)2 (80) = 800 mW.

[c] I1 = 5Ig = 12.5/0◦ mA(rms); V320 = (12.5 × 10−3 )(320) = 4 V(rms). N1 = 1+ N2 

P 10.40 [a] Zab

.·. I1 =

2

(4 − j8) = 36 − j72 Ω;

250/0◦ = 5/36.87◦ A. 4 + j42 + 36 − j72

P4(left) = |I1 |2 (4) = (5)2 (4) = 100 W; I2 =

N1 I1 = 10/36.87◦ A; N2

.·. IL = 15/36.87◦ A(rms); P4(right) = (225)(4) = 900 W.

a2 = 8a1 = 40.

10–32

CHAPTER 10. Sinusoidal Steady State Power Calculations

[b] Pg = (250)(5) cos(36.87◦ ) = 1000 W(developed); X

P 10.41 [a]

Pabs = 100 + 900 = 1000 W =

X

Pdev .

240 − j80 − 480 240 − j80 + = 0; ZTh 100 −100(240 + j80) = 80 + j60 Ω; .·. ZTh = −(240 − j80) .·. ZL = 80 − j60 Ω.

[b] I =

480/0◦ = 3/0◦ A(rms); 160/0◦

P = (9)(80) = 720 W. [c] Construct the 80 Ω resistor by connecting a 33 Ω resistor and a 47 Ω resistor in series. 1 C= = 66.667 µF. (250)(60) Construct this capacitance (approximately) by combining three 22 µF and three 0.22 µF all in parallel to give 66.66 µF. P 10.42 [a] From Problem 9.77, ZTh = 90 + j120 Ω and VTh = 450/36.87◦ V. Thus, for ∗ maximum power transfer, ZL = ZTh = 90 − j120 Ω:

I2 =

450/36.87◦ = 2.5/36.87◦ = 2 + j1.5 A. 180

225/0◦ = (15 + j20)I1 − j50(2 + j1.5); 150 + j100 .·. I1 = = 6.8 − j2.4 A. 15 + j20 Sg (del) = 225(6.8 + j2.4) = 1530 + j540 VA; Pg = 1530 W.

Problems

10–33

[b] Ploss = |I1 |2 (15) + |I2 |2 (30) = 780 + 187.5 = 967.5 W; % loss =

967.50 (100) = 63.24%. 1530

P 10.43 ZL = |ZL |/θ◦ = |ZL | cos θ◦ + j|ZL | sin θ◦ . |VTh | Thus |I| = q . (RTh + |ZL | cos θ)2 + (XTh + |ZL | sin θ)2 Therefore P =

0.5|VTh |2 |ZL | cos θ . (RTh + |ZL | cos θ)2 + (XTh + |ZL | sin θ)2

Let D = demoninator in the expression for P, then dP (0.5|VTh |2 cos θ)(D · 1 − |ZL |dD/d|ZL |) ; = d|ZL | D2 dD = 2(RTh + |ZL | cos θ) cos θ + 2(XTh + |ZL | sin θ) sin θ; d|ZL | !

dP dD = 0 when D = |ZL | . d|ZL | d|ZL | Substituting the expressions for D and (dD/d|ZL |) into this equation gives us 2 2 the relationship RTh + XTh = |ZL |2 or |ZTh | = |ZL |. P 10.44 [a] ZTh =

(60,000)(j60,000) 1 + = −j40,000 + 30,000 + j30,000 = 30 − j10 kΩ; jωC 60,000 + j60,000

∗ .·. ZL = ZTh = 30 + j10 kΩ.

[b] VTh =

√ 90/0◦ (60) = 45(1 − j1) = 45 2/ − 45◦ V(rms). 60 + j60

√ √ 45 2/ − 45◦ I= = 0.75 2/ − 45◦ mA(rms); 60 × 103 |Irms | = 0.75 mA(rms); Pload = (0.75)2 × 10−6 (30 × 103 ) = 16.875 mW.

10–34

CHAPTER 10. Sinusoidal Steady State Power Calculations

[c] The closest resistor value from Appendix H is either 27 kΩ or 33 kΩ. Find the inductor value: 10,000L = 10,000

so

L = 1 H.

The closest inductor value is 10 mH, which has a reactance of 100 Ω. Choosing the 27 kΩ resistor, the current in the load is 45/ − 45◦ = 777.83/ − 35.15◦ µA(rms); 57,000 − j9900

Irms =

Pload = (777.83 × 10−6 )2 (27,000) = 16.3 mW

(instead of 16.875 mW).

Choosing the 33 kΩ resistor, the current in the load is 45/ − 45◦ = 705.63/ − 36.07◦ µA(rms); 63,000 − j9900

Irms =

Pload = (705.63 × 10−6 )2 (33,000) = 16.43 mW P 10.45 [a]

V1 V1 − 250 − 0.2Vσ + = 0; 5 − j10 10 + j5 Vσ =

−j5V1 −jV1 = ; 10 + j5 2 + j1

−0.2Vσ = .·. V1

"

j0.2V1 ; 2 + j1 #

1 j0.2 1 250 + + . = 5 − j10 2 + j1 10 + j5 5 − j10

Thus, V1 = 10(10 + j5). VTh =

j5 V1 = j50 = 50/90◦ V(rms). 10 + j5

(instead of 16.875 mW).

Problems Short circuit current:

Isc =

50 250/0◦ = A(rms); 15 − j10 3 − j2

ZTh =

VTh j50 = (3 − j2) = 2 + j3 Ω; Isc 50

.·. ZL = 2 − j3 Ω.

IL =

50/90◦ = 12.5/90◦ A(rms); 4

P = (12.5)2 (2) = 312.50 W. [b] VL = (2 − j3)(j12.5) = 37.5 + j25 V(rms).

I1 =

VL 37.5 + j25 = = 5 − j7.5 A(rms); j5 j5

I2 = I1 + IL = 5 − j7.5 + j12.5 = 5 + j5 A(rms); Vcs = VL + 10I2 = 37.5 + j25 + 50 + j50 = 87.5 + j75 V(rms); Vσ = −VL = −37.5 − j25; 0.2Vσ = −7.5 − j5; Scs = −Vcs I∗cs = −(87.5 + j75)(−7.5 + j5) = 1031.25 + j125 VA.

10–35

10–36

CHAPTER 10. Sinusoidal Steady State Power Calculations Therefore, the dependent source is absorbing 1031.25 W and 125 magnetizing vars. Only the independent voltage source is developing power. Ig = −0.2Vσ + I2 = 7.5 + j5 + 5 + j5 = 12.5 + j10 A; Sg = −250I∗g = −3125 + j2500 VA; .·. Pdev = 3125 W. % delivered =

312.5 (100) = 10%. 3125

Thus, 10% of the developed power is delivered to the load. Checks: √ P10Ω = (5 2)2 10 = 500 W; P2Ω = 312.5 W; √ P5Ω = ( 256.25)2 5 = 1281.25 W; .·.

X

Pdev =

X

Pabs = 500 + 312.5 + 1281.25 + 1031.25 = 3125 W.

VAR Check: The 250 V source is absorbing 2500 vars; the dependent current source is absorbing 125 vars; the j5 Ω inductor is absorbing |37.5 + j25|2 /5 = 406.25 vars. Thus, X

Qabs = 2625 + 406.25 = 3031.25 VAR;

X

Qdev = (12.5)2 (3) + 256.25(10) = 3031.25 VAR =

P 10.46 [a] ZTh = 8 + j15 +

X

Qabs .

(−j24)(18 + j6) = 24 + j7 = 25/16.26◦ Ω; 18 − j18

.·. R = |ZTh | = 25 Ω. [b] VTh =

I=

√ −j24 (630/0◦ ) = 420 − j420 = 420 2/ − 45◦ V(rms). 18 + j6 − j24

420 − j420 = 12/ − 53.13◦ ; 49 + j7

P = (12)2 (25) = 3600 W = 3.6 kW.

|I| = 12;

Problems [c] The resistor from Appendix H that is closest to 25 Ω is 22 Ω. P 10.47 [a]

Vφ − 600 Vφ + − 0.05Vφ = 0; 10 j10 .·. Vφ = 240 + j480 V(rms). VTh = Vφ + 0.05Vφ (−j20) = Vφ (1 − j1) = 720 + j240 V(rms). Short circuit current:

Isc = 0.05Vφ +

Vφ = (0.05 + j0.05)Vφ ; −j20

Vφ − 600 Vφ Vφ + + = 0; 10 j10 −j20 .·. Vφ = 480 + j240 V(rms). Isc = (0.05 + j0.05)(480 + j240) = 12 + j36 A(rms); ZTh =

720 + j240 VTh = = 12 − j16 = 20/ − 53.13◦ Ω; Isc 12 + j36

.·. Ro = 20 Ω. [b]

I=

√ 720 + j240 = 15 + j15 = 15 2/45◦ A(rms); 32 − j16

10–37

10–38

CHAPTER 10. Sinusoidal Steady State Power Calculations √ P = (15 2)2 (20) = 9000 W = 9 kW.

[c]

720 + j240 = 30 + j10 A(rms); 24 √ P = ( 1000)2 (12) = 12 kW. I=

[d]

Vφ − 200 − j600 Vφ − 600 Vφ + + = 0; 10 j10 −j20 .·. Vφ = 200 + j200 V(rms). 0.05Vφ = 10 + j10 A(rms); 10 + j10 + IC = 30 + j10; IL =

.·. IC = 20 + j0 A(rms);

Vφ = 20 − j20 A(rms); j10

IR = IC + IL = 40 − j20 A(rms); Ig = IR + 0.05Vφ = 50 − j10 A(rms); Sg = −600I∗g = −30,000 − j6000 VA; 600 = Vcs + 200 + j600;

Vcs = 400 − j600 V(rms);

Scs = (400 − j600)(10 − j10) = −2000 − j10,000 VA; X

Pdev = 30,000 + 2000 = 32,000 W = 32 kW;

% delivered to Zo =

12 (100) = 37.50%. 32

Problems Check: X

Pabs = 12,000 + |IR |2 (10) = 32 kW =

X

Qdev = 6000 + 10,000 + |IC |2 (20) = 24 kVAR;

X

Qabs = |IL |2 (10) + |Io |2 (16) = 24 kVAR =

X

Pdev ;

X

Qdev .

1 1 = 240 Ω; C= = 11.05 µF. ωC (240)(120π) 4800 4800 + = 30 − j20 A(rms); [b] Iwo = 160 j240

P 10.48 [a]

Vswo = 4800 + (30 − j20)(1 + j8) = 4990 + j220 = 4994.85/2.52◦ V(rms); Iw =

4800 4800 4800 + + = 30 + j0 A(rms); 160 j240 −j240

Vsw = 4800 + 30(1 + j8) = 4830 + j240 = 4835.96/2.84◦ V(rms); 

% increase =

4994.85 − 1 (100) = 3.29%. 4835.96 

[c] P`wo = |30 − j20|2 1 = 1300 W; P`w = 302 (1) = 900 W; 1300 − 1 (100) = 44.44%. % increase = 900 



P 10.49 [a] First find the Th´evenin equivalent: 106 1 = = −j100 Ω; jωC j104 ZTh =

300(−j100) = 30 − j90 Ω; 300 − j100

VTh =

150(−j100) = 15 − j45 V(rms); 300 − j100

jωL = j104 (6 × 10−3 ) = j60 Ω.

I=

15 − j45 1.5 = (13 − j9) A(rms); 40 − j30 25

10–39

10–40

CHAPTER 10. Sinusoidal Steady State Power Calculations |I| =

1.5 √ 250 A(rms); 25

P =

2.25 (250)(10) = 9 W. 625

[b] Since ZTh = 30 − j90 Ω and ω = 10,000 rad/s, set Lo as close as possible to 90/10,000 = 9 mH. Therefore, set Lo = 8 mH. Set Ro as close as possible to q √ Ro = (30)2 + (90 − 80)2 = 1000 = 31.62 Ω; .·. Ro = 20 Ω. [c] I =

15 − j45 3 − j9 = A(rms); 50 − j10 10 − j2 s

.·. |I| =

90 . 104

P = |I|2 (20) =

90 (20) = 17.31 W. 104

Yes; 17.31 W > 9 W. 15 − j45 1 − j3 [d] I = = A(rms); 60 4 √ !2 10 30 = 18.75 W. P = 4 [e] Ro = 30 Ω;

Lo = 9 mH.

[f ] Yes;

18.75 W > 17.31 W. √ P 10.50 [a] Lo = 8 mH; Ro = 302 + 102 = 31.62 Ω; √ 15(1 − j3) 15 10 / − 62.35◦ A(rms); = I= 61.62 − j10 62.43 √ !2 15 10 (31.62) = 18.26 W. P = 62.43 [b] Yes;

18.26 W > 17.31 W.

[c] Yes;

18.26 W < 18.75 W.

Problems

10–41

P 10.51 [a]

660 √ = 230I1 + j50I1 − j40(I2 − I1 ) + j100(I1 − I2 ) + j40I1 ; 2 0 = 100I2 + j100(I2 − I1 ) − j40I1 . Solving, I1 = 1.2245 − j0.4928 A(rms);

I2 = 1.2021 + j0.5122 A(rms);

.·. Vo = (1.30667)(100) = 130.667 V(rms). [b] P = (1.30667)2 (100) = 170.74 W. !

660 [c] Sg = − √ (1.2245 − j0.4928) = −571.46 + j229.985 VA 2 % delivered =

170.74 (100) = 29.88%. 571.46

P 10.52 [a] Open circuit voltage:

660 √ = 230I1 + j50I1 + j40I1 + j100I1 + j40I1 ; 2 √ 660/ 2 · . . I1 = = 1.0145 − j1.0145 A(rms). 230 + j230 VTh = j100I1 + j40I1 = j140I1 = 142.04 + j142.04 V(rms).

.·. Pg = −571.46 W;

10–42

CHAPTER 10. Sinusoidal Steady State Power Calculations Short circuit current:

660 √ = 230I1 + j50I1 − j40(Isc − I1 ) + j100(I1 − Isc ) + j40I1 ; 2 0 = j100(Isc − I1 ) − j40I1 . Solving, Isc = 2.78 − j0.411 A; ZTh =

IL =

VTh 142.04 + j142.04 = = 42.61 + j57.39 Ω. Isc 2.78 − j0.411

142.04 + j142.04 = 1.67 + j1.67 = 2.357/45◦ A(rms); 85.22

PL = (2.357)2 (42.61) = 236.72 W. [b] Using the equivalent circuit, S = −(142.04 + j142.04)(1.67 − j1.67) = −474.4 + j0 VA; Pdev = 474.4 W. [c] Begin by choosing the capacitor value from Appendix H that is closest to the required reactive impedance, assuming the frequency of the source is 60 Hz: 1 1 57.39 = so C= = 46.22 µF. 2π(60)C 2π(60)(57.39) Choose the capacitor value closest to this capacitance from Appendix H, which is 47 µF. Then, XL = −

1 = −56.44 Ω. 2π(60)(47 × 10−6 )

Problems Now set RL as close as possible to RL =

q

10–43

q

2 RTh + (XL + XTh )2 :

42.612 + (−56.44 + 57.39)2 = 42.62 Ω.

The closest single resistor value from Appendix H is 47 Ω. The resulting real power developed by the source is calculated below, using the Th´evenin equivalent circuit: 142.04 + j142.04 = 2.24/44.4◦ ; IL = 42.61 + j57.39 + 47 − j56.44 PL = (2.24)2 (47) = 236 W

(instead of 236.72 W).

P 10.53 [a] VTh = 210 V; V2 = − 41 V1 ; Short circuit equations:

I1 = 41 I2 .

840 = 80I1 − 20I2 + V1 ; 0 = 20(I2 − I1 ) + V2 ; .·. I2 = 14 A(rms); 210 = 30 

Pmax

RTh RL =

210 = 15 Ω. 14

2

15 = 735 W.

[b] VTh = −4(146/0◦ ) = −584/0◦ V (rms); I1 = −4I2 .

V2 = 4V1 ;

Short circuit equations: 146/0◦ = 80I1 − 20I2 + V1 ; 0 = 20(I2 − I1 ) + V2 ; .·. I2 = −146/365 = −0.40 A; −584 P = 2920 

2

1460 = 58.40 W.

P 10.54 [a] jωL1 = j(5000)(2 × 10−3 ) = j10 Ω; jωL2 = j(5000)(8 × 10−3 ) = j40 Ω; jωM = j10 Ω.

70 = (10 + j10)Ig + j10IL ;

RTh =

−584 = 1460 Ω. −0.4

10–44

CHAPTER 10. Sinusoidal Steady State Power Calculations 0 = j10Ig + (30 + j40)IL . Solving, Ig = 4 − j3 A;

IL = −1 A.

Thus, ig = 5 cos(5000t − 36.87◦ ) A; iL = 1 cos(5000t − 180◦ ) A. 2 M = √ = 0.5. L1 L2 16 [c] When t = 100π µs: [b] k = √

5000t = (5000)(100π) × 10−6 = 0.5π rad = 90◦ ; ig (100π µs) = 5 cos(53.13◦ ) = 3 A; iL (100π µs) = 1 cos(−90◦ ) = 0 A; 1 1 1 w = L1 i21 + L2 i22 + M i1 i2 = (2 × 10−3 )(9) + 0 + 0 = 9 mJ. 2 2 2 When t = 200π µs: 5000t = π rad = 180◦ ; ig (200π µs) = 5 cos(180 − 36.87◦ ) = −4 A; iL (200π µs) = 1 cos(180 − 180◦ ) = 1 A; 1 1 w = (2 × 10−3 )(16) + (8 × 10−3 )(1) + 2 × 10−3 (−4)(1) = 12 mJ. 2 2 [d] From (a), Im = |IL | = 1 A, 1 .·. P = (1)2 (30) = 15 W. 2 √ 70 [e] VTh = (j10) = 35 2/45◦ V; 10 + j10 10 √ ZTh = j40 + 10 2 √ .·. RL = 25 2 Ω.

!2

(10 − j10) = 5 + j35 =

√

1250/81.78◦ Ω;

Problems [f ]

√ 35 2/45◦ √ = 0.93/4.07◦ A; I= (5 + 25 2) + j35 √ 1 P = (0.93)2 (25 2) = 15.18 W. 2 ∗ [g] ZL = ZTh = 5 − j35 Ω. √ √ 35 2/45◦ [h] I = = 3.5 2/45◦ ; 10 √ 1 P = (3.5 2)2 (5) = 61.25 W. 2

P 10.55

VTh =

80 80 (j20) + (j20) = 128 + j64 V(rms). 10 + j20 10 + j20

80 = 10I1 + j20(I1 − Isc ) − j20Isc ; 0 = j20(Isc − I1 ) + j20Isc + j40Isc − j20(I1 − Isc ). Solving, Isc = 2.76 − j1.10 A;

ZTh =

128 + j64 = 32 + j36 Ω; 2.76 − j1.10

10–45

10–46

CHAPTER 10. Sinusoidal Steady State Power Calculations

128 + j64 .·. IL = = 2 + j1 A. 64

80 = 10I1 + j20(I1 − I2 ) − j20I2 ; I2 = 2 + j1 A. Solving, I1 = 4/0◦ A; Zg = 80/4 = 20 + j0 Ω. P 10.56 [a] VTh =

760/0◦ (j50) = 380/16.26◦ V(rms); 28 + j96 

ZTh = 31 + j100 +

50 100

2

(28 − j96) = 38 + j76 Ω;

.·. ZL = 38 − j76 Ω. IL =

380/16.26◦ = 4.8 + j1.4 = 5/16.26◦ A(rms); 76

PL = |IL |2 (38) = 950 W. [b]

760/0◦ = I1 (28 + j96) − j50(4.8 + j1.4); 690 + j240 .·. I1 = = 7.31/ − 54.56◦ = 4.24 − j5.95 A. 100/73.74◦ Sg (delivered) = 760(4.24 + j5.95) = 3219.36 + j4523.52 VA;

Problems Ploss = |I1 |2 (8) = 426.96 W; Pin (transformer) = 3219.36 − 426.96 = 2792.40 W; % delivered to ZL =

950 (100) = 34.02%. 2792.4

P 10.57 [a]

240 = 20I1 + j40(I1 − I2 ) − j60I2 ; 0 = j40(I2 − I1 ) + j60I2 + j160I2 + j60(I2 − I1 ) + 140I2 . Solving, I1 = 6.4 − j2.8 A(rms);

I2 = 2/0◦ A(rms);

Vo = 140I2 = 280/0◦ V(rms). [b] P = |I2 |2 (140) = 560 W. [c] Pg = (240)(6.4) = 1536 W; % delivered =

560 (100) = 36.46%. 1536

P 10.58 [a]

VTh =

240/0◦ 240/0◦ (j40) + (j60) = 480 + j240 V(rms). 20 + j40 20 + j40

240 = (20 + j40)I1 − j100Isc ;

10–47

10–48

CHAPTER 10. Sinusoidal Steady State Power Calculations 0 = −j100I1 + j320Isc . Solving, Isc = 3.15 − j1.377; ZTh =

480 + j240 VTh = = 100 + j120 = 156.20/50.19◦ Ω; Isc 3.15 − j1.377

.·. RL = 156.20 Ω. [b]

I=

536.66/26.57◦ = 1.90/1.47◦ ; 282.92/25.10◦

P = |I|2 (156.20) = 562.05 W. P 10.59 [a]

240 = 20I1 + j40(I1 − I2 ) + j80kI2 ; 0 = j40(I2 − I1 ) − j80kI2 + j160I2 + j80k(I1 − I2 ) + 140I2 , or 12 = (1 + j2)I1 + j(4k − 2)I2 ; 0 = j(4k − 2)I1 + [7 + j(10 − 8k)]I2 ; Solve the second equation for I1 , use it to eliminate I1 in the first equation, and solve for I2 to get I2 =

−j12(4k − 2) . (4k − 2)2 + (1 + j2)[7 + j(10 − 8k)]

I2 = 0 when 4k − 2 = 0; .·. k = 0.5.

Vo = 0 when I2 = 0;

Problems

10–49

[b] When I2 = 0, I1 =

12 = 2.4 − j4.8 A(rms); 1 + j2

Pg = (240)(2.4) = 576 W. Check: Ploss = |I1 |2 (20) = 576 W. P 10.60 [a] Replace the circuit to the left of the primary winding with a Th´evenin equivalent: VTh =

0.25/0◦ (j50) = 200 + j100 mV(rms); 25 + j50

ZTh = 20 +

(25)(j50) = 40 + j10 Ω. 25 + j50

Transfer the secondary impedance to the primary side: Zp =

1 XC (160 − jXC ) = 10 − j Ω. 16 16

Now match the load impedance to the source impedance by setting (XC /16) = 10 Ω: .·. C =

10−3 = 125 nF. (160)(50)

0.2 + j0.1 = 4 + j2 mA(rms); 50 √ |I| = 20 mA(rms);

[b] I =

P = (20 × 10−6 )(10) = 200 µW. Ro = 40 Ω; .·. Ro = 640 Ω. 16 0.2 + j0.1 [d] I = = 2.5 + j1.25 mA(rms); 80 [c]

P = |I|2 (40) = 312.50 µW.

10–50

CHAPTER 10. Sinusoidal Steady State Power Calculations

P 10.61

Vo = 4Va ;

4Io = Ia ;

Vb −Va = ; 1 2.5

therefore

Ib = −2.5Ia ;

Va = 250 Ω. Ia

therefore

Vb 250 = = 40 Ω. Ib 6.25

Therefore Ib = [100/(10 + 40)] = 2 A (rms); since the ideal transformers are lossless, P4kΩ = P40Ω , and the power delivered to the 4 kΩ resistor is 22 (40) or 160 W. P 10.62 [a]

Ia = −2.5Ib ;

10Ia + V1 = 100; .·.

10(−2.5Ib ) − V2 /2.5 = 100.

Ib = aIc = 0; .·.

10[−2.5(0)] − Voc /2.5a = 100;

Ia = −2.5Ib ;

V1 = −V2 /2.5;

10(−2.5Ib ) − V2 /2.5 = 100.

V2 = V3 /a = 0; .·.

V2 = Voc /a;

Voc = −250a.

10Ia + V1 = 100; .·.

V1 = −V2 /2.5;

Ib = aIsc ;

Isc = 100/(−2.5a) = −4/a.

10[−2.5(aIsc )] − 0 = 100;

Problems

10–51

Thus, ZTh =

−250a Voc = = 62.5a2 . Isc −4/a

For maximum power to the 4 kΩ load, 4000 = ZTh = 62.5a2 ;

so

a = 8.

[b] The circuit, with everything to the left of the 4 kΩ load resistor replaced by its Thevenin equivalent:

PL =

V2L (−1000)2 = = 250 W. 4000 4000

P 10.63 [a]

1 V2 = V1 ; 4

I2 = 4I1 ;

25 = 5I1 + V1 . 0 = −V2 + 1.25I2 ; .·. I1 = 1 A;

I2 = 4 A.

25 = (1)I1 + VTh + (1)I2 ;

.·. VTh = 20 V.

25 = (Isc + I1 )(1) + 4I1 + V1 ; 25 = (Isc + I1 )(1) + (Isc + 4I1 )(1);

10–52

CHAPTER 10. Sinusoidal Steady State Power Calculations V1 = 4I1 (0.25) + (Isc + 4I1 )(1). 4 Solving, Isc = 20 A; RTh =

20 VTh = = 1 Ω. Isc 20

P = (10)2 (1) = 100 W. [b]

25 = (10 + I1 )(1) + 4I1 + V1 ; V1 = 4I1 (0.25) + (4I1 + 10)(1). 4 Solving, I1 = −1 A; .·. Psource = (25)(10 − 1) = 225 W; % delivered =

100 (100) = 44.44%. 225

[c] Pdev = 25(10 − 1) = 225 W; P1Ω = (9)2 (1) = 81 W; P1Ω = (10)2 (1) = 100 W;

P4Ω = (−1)2 (4) = 4 W; P0.25Ω = (−4)2 (0.25) = 4 W;

P1Ω = (10 − 4)2 (1) = 36 W; X

Pdiss = 81 + 4 + 100 + 4 + 36 = 225 W =

X

Pdev .

Problems P 10.64 [a]

For maximum power transfer, Zab = 20 kΩ; N1 = 1− N2 

Zab .·. 1−



N1 1− N2

2

ZL ;

2

=

N1 = ±20; N2

N1 >0 N2 N2 =

20,000 = 400. 50 N1 = 1 ∓ 20; N2

N1 .·. = 21; N2

2520 N1 = = 120 turns. 21 21

[b]

P50Ω = P20kΩ =

(120)2 × 10−3 = 720 mW. 20

[c]

V1 + V2 = 120; V2 = − V1 −

V1 V2 =− ; N1 N2

N2 V1 V1 = − ; N1 21

V1 = 120; 21

.·. V1 = 126 V;

10–53

10–54

CHAPTER 10. Sinusoidal Steady State Power Calculations .·. V2 = −6 V. Check the power calculation: P50Ω =

36 = 0.72 W = 720 mW. 50

[d]

Vcs = 120 + (6)(5) = 150 V; Pcs (del) = (150)(16) = 2400 mW; % delivered =

720 (100) = 30%. 2400

20 = 32 + j124 + 5 

P 10.65 [a] ZTh

2

(3 − j4) = 80 + j60 = 100/36.87◦ Ω;

.·. Zab = 100 Ω. Zab =

ZL ; (1 + N1 /N2 )2

(1 + N1 /N2 )2 = 3600/100 = 36; .·. N1 /N2 = 5

or

N2 = N1 /5;

.·. N2 = 300 turns. [b] VTh =

240/0◦ (j20) = 960/36.87◦ V. 3 + j4

√ 960/36.87◦ = 1.6 10/18.43◦ A(rms); 180 + j60 √ |I| = 1.6 10 A(rms); I=

P = |I|2 (100) = 2560 W.

Problems [c]

240/0◦ = (3 + j4)I1 − j20(4.8 + j1.6); .·. I1 = 40.32 − j21.76 A(rms). Pgen = (240)(40.32) = 9676.80 W; Pdiss = 9676.80 − 2560 = 7116.80 W; % dissipated =

7116.80 (100) = 73.54%. 9676.80

N1 = 30,000 − j10,000 = 1 − N2 

P 10.66 [a] Zab

2

ZL ;

1 .·. ZL = (30,000 − j10,000) = 7500 − j2500 Ω. 4 [b]

I1 =

480 × 10−3 = 8/0◦ mA(rms). 60

N1 I1 = −N2 I2 ; I2 = −3I1 = −24/0◦ mA(rms); IL = I1 + I2 = −16/0◦ mA(rms); VL = (7500 − j2500)IL = −120 + j40 = 126.49/161.57◦ V(rms).

10–55

10–56

CHAPTER 10. Sinusoidal Steady State Power Calculations

30[6(44.28) + 18(15.77)] = 16.49 kWh. 1000 30[6(44.28) + 18(8.9)] [b] = 12.78 kWh. 1000 30[6(44.28) + 18(4.42)] [c] = 10.36 kWh. 1000 30[6(44.28) + 18(0)] [d] = 7.97 kWh. 1000 Note that this is about 48 % of the amount of total power consumed in part (a).

P 10.67 [a]

30[0.33(1433) + 23.67(3.08)] = 16.52 kWh. 1000 [b] The standby power consumed in one month by the microwave oven when in the ready state is

P 10.68 [a]

30[23.67(3.08)] = 2.187 kWh. 1000 This is (2.187/16.52) ∗ 100 = 13.2% of the total power consumed by the microwave in one month. Since it is not practical to unplug the microwave when you are not using it, this is the cost associated with having a microwave oven. P 10.69 jωL1 = j(2π)(60)(0.18) = j67.86 Ω; I=

120 = 1.75/ − 81.62◦ A(rms); 10 + j67.86

P = R1 |I|2 = 10(1.75)2 = 30.6 W. P 10.70 jωL1 = j(2π)(60)(0.18) = j67.86 Ω; I=

120 = 1.77/ − 89.93◦ A(rms); 0.08 + j67.86

P = R1 |I|2 = 0.08(1.77)2 = 250.6 mW. Note that while the current supplied by the voltage source is virtually identical to that calculated in Problem 10.69, the much smaller value of transformer resistance results in a much smaller value of real power consumed by the transformer. P 10.71 An ideal transformer has no resistance, so consumes no real power. This is one of the important characteristics of ideal transformers.

Balanced Three-Phase Circuits

Remember that all voltages and currents are rms values!

Assessment Problems AP 11.1 Make a sketch:

We know VAN and wish to find VCA . To do this, write a KVL equation to find VCA , and use the known phase angle relationships among the voltages. VCA = VCN − VAN . Since VAN , VBN , and VCN form a balanced set, and VAN = 80/50◦ V, and the phase sequence is positive, VCN = |VAN |//VAN + 120◦ = 80/50◦ + 120◦ = 80/170◦ V. Then, VCA = VCN − VAN = (80/170◦ ) − (80/50◦ ) = 138.56/ − 160◦ V.

11–1

11–2

CHAPTER 11. Balanced Three-Phase Circuits

AP 11.2 Make a sketch:

We know VBN and wish to find VBC . To do this, write a KVL equation to find VBC , and use the known phase angle relationships among the voltages. VBC = VBN + VNC = VBN − VCN . Since VAN , VBN , and VCN form a balanced set, and VBN = 150/ − 30◦ V, and the phase sequence is negative, VCN = |VBN |//VBN + 120◦ = 150/ − 30◦ + 120◦ = 150/90◦ V. Then, VBC = VBN − VCN = (150/ − 30◦ ) − (150/90◦ ) = 259.81/ − 60◦ V. AP 11.3 Make a sketch of the a-phase:

[a] Find the a-phase line current from the a-phase circuit: IaA =

125/0◦ 125/0◦ = 20 + j15 0.1 + j0.8 + 19.9 + j14.2

= 4 − j3 = 5/ − 36.87◦ A. Find the other line currents using the acb phase sequence: IbB = 5/ − 36.87◦ + 120◦ = 5/83.13◦ A; IcC = 5/ − 36.87◦ − 120◦ = 5/ − 156.87◦ A.

Problems

11–3

[b] The phase voltage at the source is Van = 125/0◦ V. Use Fig. 11.9(b) to find the line voltage, Van , from the phase voltage: √ Vab = Van ( 3/ − 30◦ ) = 216.51/ − 30◦ V. Find the other line voltages using the acb phase sequence: Vbc = 216.51/ − 30◦ + 120◦ = 216.51/90◦ V; Vca = 216.51/ − 30◦ − 120◦ = 216.51/ − 150◦ V. [c] The phase voltage at the load in the a-phase is VAN . Calculate its value using IaA and the load impedance: VAN = IaA ZL = (4 − j3)(19.9 + j14.2) = 122.2 − j2.9 = 122.23/ − 1.36◦ V. Find the phase voltage at the load for the b- and c-phases using the acb sequence: VBN = 122.23/ − 1.36◦ + 120◦ = 122.23/118.64◦ V; VCN = 122.23/ − 1.36◦ − 120◦ = 122.23/ − 121.36◦ V. [d] The line voltage at the load in the a-phase is VAB . Find this line voltage from the phase voltage at the load in the a-phase, VAN , using Fig. 11.9(b): √ VAB = VAN ( 3/ − 30◦ ) = 211.72/ − 31.36◦ V. Find the line voltage at the load for the b- and c-phases using the acb sequence: VBC = 211.72/ − 31.36◦ + 120◦ = 211.72/88.64◦ V;

AP 11.4 IaA

VCA = 211.72/ − 31.36◦ − 120◦ = 211.72/ − 151.36◦ V. √ √ = ( 3/ − 30◦ )IAB = ( 3/ − 30◦ ) · 25/40◦ = 43.3/10◦ A.

AP 11.5 IbB = 10/(−30◦ − 120◦ ) = 10/ − 150◦ ; " ! # ! / − 30◦ 1 ◦ √ / − 30 IbB = √ IBC = · 10/ − 150◦ 3 3 = 5.77/ − 180◦ A. "

AP 11.6 [a] ICA =

!

#

1 √ / − 90◦ [103.92/40◦ ] = 60/70◦ A. 3

Therefore Zφ =

2400/0◦ = 40/ − 70◦ Ω. 60/70◦

11–4

CHAPTER 11. Balanced Three-Phase Circuits "

[b] ICA =

!

#

1 √ / − 30◦ [103.92/40◦ ] = 60/ − 10◦ A. 3

Therefore Zφ = 40/ − 10◦ Ω. 240 240 + = 80 + j60 = 100/36.87◦ A. 3 −j4 √ √ Therefore |IaA | = 3Iφ = 3(100) = 173.21 A.

AP 11.7 Iφ =

AP 11.8 [a] S1 = 10,200(0.87) + j10,200(0.493) = 8874 + j5029.13 VA; S2 = 4200 + j1913.6 VA; √

3VL IL sin θ3 = 7250;

Therefore

sin θ3 = √

7250 = 0.517. 3(220)(36.8)

cos θ3 = 0.856.

Therefore 7250 P3 = × 0.856 = 12,003.9 W; 0.517 S3 = 12,003.9 + j7250 VA; ST = S1 + S2 + S3 = 25.078 + j14.192 kVA; Thus, PT = 25,078 W. 1 [b] ST /φ = ST = 8359.3 + j4730.7 VA; 3 220 ∗ √ IaA = (8359.3 + j4730.7); 3

I∗aA = 65.81 + j37.24 A;

IaA = 65.81 − j37.24 = 75.62/ − 29.51◦ A,

so

[c] pf = cos(0◦ − 29.51◦ ) = 0.87 lagging. AP 11.9 [a]

1 Ss/φ = (60)(0.96 − j0.28) × 103 = 19.2 − j5.6 kVA; 3

|IaA | = 75.62 A.

Problems 1 S1/φ = (45) = 15 + j0 kVA; 3 S2/φ = Ss/φ − S1/φ = 4.2 − j5.6 kVA; 4200 − j5600 √ .·. I∗2 = = 11.547 − j15.396 A. 630/ 3 I2 = 11.547 + j15.396 A; √ 630/ 3 VAN = = 11.34 − j15.12 Ω; Zy = I2 11.547 + j15.396 Z∆ = 3Zy = 34.02 − j45.36 Ω.

√ (630/ 3)2 [b] R = = 31.5 Ω; R∆ = 3R = 94.5 Ω; 4200 √ (630/ 3)2 = −23.625 Ω; X∆ = 3XL = −70.875 Ω. XL = −5600

11–5

11–6

CHAPTER 11. Balanced Three-Phase Circuits

Problems P 11.1

[a] First, convert the cosine waveforms to phasors: Va = 120/54◦ ;

Vb = 120/ − 66◦ ;

Vc = 120/174◦ .

Subtract the phase angle of the a-phase from all phase angles: /V0a = 54◦ − 54◦ = 0◦ ; /V0b = −66◦ − 54◦ = −120◦ ; /V0c = 174◦ − 54◦ = 120◦ . Compare the result to Eqs. 11.1 and 11.2: Therefore abc. [b] First, convert the cosine waveforms to phasors: Va = 3240/ − 26◦ ;

Vb = 3240/94◦ ;

Vc = 3240/ − 146◦ .

Subtract the phase angle of the a-phase from all phase angles: /V0a = −26◦ + 26◦ = 0◦ ; /V0b = 94◦ + 26◦ = 120◦ ; /V0c = −146◦ + 26◦ = −120◦ . Compare the result to Eqs. 11.1 and 11.2: Therefore acb. P 11.2

[a] Va = 339/0◦ V; Vb = 339/ − 120◦ V; Vc = 339/120◦ V. Balanced, positive phase sequence [b] Va = 622/0◦ V; Vb = 622/ − 240◦ V = 622/120◦ V; Vc = 622/240◦ V = 622/ − 120◦ V. Balanced, negative phase sequence [c] Va = 933/ − 90◦ V; Vb = 933/150◦ V; Vc = 933/30◦ V. Balanced, positive phase sequence

Problems

11–7

[d] Va = 170/ − 30◦ V; Vb = 170/90◦ V; Vc = 170/ − 150◦ V. Balanced, negative phase sequence [e] Unbalanced, due to unequal amplitudes. [f ] Unbalanced, due to unequal phase angle separation. P 11.3

Va = Vm /0◦ = Vm + j0; Vb = Vm / − 120◦ = −Vm (0.5 + j0.866); Vc = Vm /120◦ = Vm (−0.5 + j0.866); Va + Vb + Vc = (Vm )(1 + j0 − 0.5 − j0.866 − 0.5 + j0.866) = Vm (0) = 0.

P 11.4

I=

339/0◦ + 339/ − 120◦ + 339/120◦ = 0. 3(2 + j4)

P 11.5

I=

3394/70◦ + 3394/ − 140◦ + 3394/180◦ 4947.07/168.22◦ = = 322.746/156.91◦ A. 3(5 + j1) 15 + j3

P 11.6

[a] IaA =

240/0◦ = 2.4/ − 36.87◦ A; 80 + j60

IbB =

240/120◦ = 2.4/83.13◦ A; 80 + j60

IcC =

240/ − 120◦ = 2.4/ − 156.87◦ A; 80 + j60

Io = IaA + IbB + IcC = 0. [b] VAN = (79 + j55)IaA = (79 + j55)(2.4/ − 36.87◦ ) = 231.0/ − 2.02◦ V. [c] VBN = (79 + j52)IbB = 226.99/116.48◦ V; .·.

VAB = VAN − VBN = 393.6/ − 32.5◦ V.

[d] Unbalanced.

11–8 P 11.7

CHAPTER 11. Balanced Three-Phase Circuits Zga + Zla + ZLa = 80 + j60 Ω; Zgb + Zlb + ZLb = 40 + j30Ω; Zgc + Zlc + ZLc = 160 + j120Ω; VN − 480 VN − 480/ − 120◦ VN − 480/120◦ VN + + + = 0. 80 + j60 40 + j30 160 + j120 20 Solving for VN yields VN = 78.61/ − 122.69◦ V; Io =

P 11.8

P 11.9

VN = 3.93/ − 122.69◦ A. 20

[a] The circuit is unbalanced, because the impedance in each phase of the load is not the same. 240/0◦ = 2.4 − j7.2 A; [b] IaA = 10 + j30 IbB =

240/120◦ = 2.2 + j8.2 A; 20 + j20

IcC =

240/ − 120◦ = 2.96 − j4.48 A; 20 − j40

Io = IaA + IbB + IcC = 7.55 − j3.48 = 8.32/ − 24.75◦ A. √ [a] Van = 1/ 3/ − 30◦ Vab = 120/ − 30◦ V(rms). The a-phase circuit is

120/ − 30◦ = 2.4/ − 83.13◦ A(rms). 30 + j40 [c] VAN = (28 + j37)IaA = 111.36/ − 30.25◦ V(rms); √ VAB = 3/30◦ VAN = 192.88/ − 0.25◦ A(rms). [b] IaA =

Problems

11–9

P 11.10 Sketch the a-phase circuit:

[a] We can find the line current using Ohm’s law, since the a-phase line current is the current in the a-phase load. Then we can use the fact that IaA , IbB , and IcC form a balanced set to find the remaining line currents. Note that since we were not given any phase angles in the problem statement, we can assume that the phase voltage given, VAN , has a phase angle of 0◦ . 2400/0◦ = IaA (16 + j12) so IaA =

2400/0◦ = 96 − j72 = 120/ − 36.87◦ A. 16 + j12

With an acb phase sequence, /IbB = /IaA + 120◦

and /IcC = /IaA − 120◦

so IaA = 120/ − 36.87◦ A; IbB = 120/83.13◦ A; IcC = 120/ − 156.87◦ A. [b] The line voltages at the source are Vab Vbc , and Vca . They form a balanced set. To find Vab , use the a-phase circuit to find VAN , and use the relationship between phase voltages and line voltages for a Y-connection (see Fig. 11.9[b]). From the a-phase circuit, use KVL: Van = VaA + VAN = (0.1 + j0.8)IaA + 2400/0◦ = (0.1 + j0.8)(96 − j72) + 2400/0◦ = 2467.2 + j69.6 = 2468.18/1.62◦ V. From Fig. 11.9(b), √ Vab = Van ( 3/ − 30◦ ) = 4275.02/ − 28.38◦ V. With an acb phase sequence, /Vbc = /Vab + 120◦

and /Vca = /Vab − 120◦

11–10

CHAPTER 11. Balanced Three-Phase Circuits so Vab = 4275.02/ − 28.38◦ V; Vbc = 4275.02/91.62◦ V; Vca = 4275.02/ − 148.38◦ V.

[c] Using KVL on the a-phase circuit Va0 n = Va0 a + Van = (0.2 + j0.16)IaA + Van = (0.02 + j0.16)(96 − j72) + (2467.2 + j69.9) = 2480.64 + j83.52 = 2482.05/1.93◦ V. With an acb phase sequence, /Vb0 n = /Va0 n + 120◦

and /Vc0 n = /Va0 n − 120◦

so Va0 n = 2482.05/1.93◦ V; Vb0 n = 2482.05/121.93◦ V; Vc0 n = 2482.05/ − 118.07◦ V. P 11.11 VAN = 7967/0◦ V; VBN = 7967/ + 120◦ V; VCN = 7967/ − 120◦ V. VAB = VAN − VBN = 13,799.25/ − 30◦ V; VBC = VBN − VCN = 13,799.25/90◦ V; VCA = VCN − VAN = 13,799.25/ − 150◦ V. vAB = 13,799.25 cos(ωt − 30◦ ) V; vBC = 13,799.25 cos(ωt + 90◦ ) V; vCA = 13,799.25 cos(ωt − 150◦ ) V.

Problems P 11.12 [a]

IaA = √

12,800 = 32.84/ − 16.26◦ A(rms); 3(216 + j63)

|IaA | = |IL | = 32.84 A(rms). 12,800 √ + (32.84/ − 16.26◦ )(0.25 + j2) = 7416.61/0.47◦ ; 3 √ |Vab | = 3(7416.61) = 12,845.94 V(rms).

[b] Van =

P 11.13 [a] Van = Vbn (1/120◦ ) = 20/ − 210◦ = 20/150◦ V(rms); Zy = Z∆ /3 = 39 − j33 Ω. The a-phase circuit is

IaA =

20/150◦ = 0.4/ − 173.13◦ A(rms); 40 − j30

VAN = (39 − j33)IaA = 20.44/146.63◦ V(rms); √ VAB = 3/ − 30◦ VAN = 35.39/116.63◦ A(rms). 1 [b] IAB = √ / − 30◦ IaA = 0.23/156.87◦ A(rms). 3 [c] VAB = (117 − j99)IAB = 35.39/116.63◦ V(rms). P 11.14 [a] IAB =

33,000 = 88/ − 16.26◦ A; 360 + j105

IBC = 88/ − 136.26◦ A; ICA = 88/103.74◦ A.

11–11

11–12

CHAPTER 11. Balanced Three-Phase Circuits

[b] IaA =

√

3/ − 30◦ IAB = 152.42/ − 46.26◦ A;

IbB = 152.42/ − 166.26◦ A; IcC = 152.42/73.74◦ A. [c]

Van = 19,052.56/ − 30◦ + (0.1 + j1.0)(152.42/ − 46.26◦ ) = 19,110.40/ − 29.57◦ V. Vab =

√

3/30◦ Van = 33,100.18/0.43◦ V;

Vbc = 33,100.18/ − 119.57◦ V; Vca = 33,100.18/120.43◦ V. P 11.15 [a]

IAN =

7500/0◦ = 75/16.26◦ A; 96 − j28

|IAN | = 75 A. √ 7500 3/30◦ [b] IAB = = 86.60/13.74◦ A; 144 + j42 |IAB | = 86.60 A. [c] IaA =

7500 7500 + = 217.02/ − 5.55◦ A; 96 − j28 48 + j14

|IaA | = 217.02 A.

Problems [d] Van = (216 − j21)(j1.5) + 7500/0◦ = 7538.47/2.46◦ V; √ |Vab | = 3(7538.47) = 13,057.01 V. P 11.16 Zy = Z∆ /3 = 4 + j3 Ω. The a-phase circuit is

IaA =

120/80◦ = 18.74/41.34◦ A(rms); (1 + j1) + (4 + j3)

1 IAB = √ /30◦ IaA = 10.82/71.34◦ A(rms). 3 √ 110 P 11.17 Van = 1/ 3/ − 30◦ Vab = √ /90◦ V; 3 Zy = Z∆ /3 = 4 − j5 Ω. The a-phase circuit is

Zeq = (10 + j8)k(4 − j5) = 5.2 − j2.4 Ω; VAN

5.2 − j2.4 = (1.8 + j0.4) + (5.2 − j2.4)

VAB =

√

!

110 √ /90◦ = 49.96/ − 98.83◦ V; 3

3/30◦ VAN = 86.53/ − 68.83◦ V.

P 11.18 [a] IAB =

720/0◦ = 144/ − 16.26◦ A; 4.8 + j1.4

IBC =

720/ − 120◦ = 36/ − 83.13◦ A; 16 − j12

ICA =

720/120◦ = 20.36/75◦ A. 25 + j25

11–13

11–14

CHAPTER 11. Balanced Three-Phase Circuits

[b] IaA = IAB − ICA = 138.24 − j40.32 − 5.27 − j19.67 = 132.97 − j59.99 = 145.88/ − 24.28◦ A. IbB = IBC − IAB = 4.31 − j35.74 − 138.24 + j40.32 = −133.93 + j4.58 = 134.01/178.04◦ A. IcC = ICA − IBC = 5.27 + j19.67 − 4.31 + j35.74 = 0.96 + j55.41 = 55.42/89.01◦ A. P 11.19 [a]

[b] IaA = √

34,500 = 19.92/16.26◦ A; 3(960 − j280)

|IaA | = 19.92 A. [c] VAN = (959 − j288)(19.92/16.26◦ ) = 19,944.71/ − 0.46◦ V; √ |VAB | = 3|VAN | = 34,545.25 V. [d] Van = (959.8 − j281.6)(19.92/16.26◦ ) = 19,923.71/ − 0.09◦ V; √ |Vab | = 3|Van | = 34,508.88 V. |IaA | [e] |IAB | = √ = 11.50 A. 3 [f ] |Iba | = |IAB | = 11.50 A. P 11.20 [a] Since the phase sequence is acb (negative) we have: Van = 7200/30◦ V; Vbn = 7200/150◦ V; Vcn = 7200/ − 90◦ V;

Problems 1 ZY = Z∆ = 1.8 + j9.0 Ω/φ. 3

√ [b] Vab = 7200/30◦ − 7200/150◦ = 7200 3/0◦ V. Since the phase sequence is negative, it follows that √ Vbc = 7200 3/120◦ V; √ Vca = 7200 3/ − 120◦ V. [c]

√ 7200 3 Iba = = 452.91/ − 78.69◦ A; 5.4 + j27 √ 7200 3/ − 120◦ Iac = = 452.91/ − 198.69◦ A; 5.4 + j27 IaA = Iba − Iac = 784.46/ − 48.69◦ A.

11–15

11–16

CHAPTER 11. Balanced Three-Phase Circuits Since we have a balanced three-phase circuit and a negative phase sequence we have: IbB = 784.46/71.31◦ A; IcC = 784.46/ − 168.69◦ A.

[d]

IaA =

7200/30◦ = 784.46/ − 48.69◦ A. 1.8 + j9

Since we have a balanced three-phase circuit and a negative phase sequence we have: IbB = 784.46/71.31◦ A; IcC = 784.46/ − 168.69◦ A. P 11.21 [a]

[b] IaA =

7200/30◦ = 7.2/13.74◦ A; 960 + j280

VAN = (957 + j259)(7.2/13.74◦ ) = 7138.28/28.88◦ V; √ |VAB | = 3(7138.28) = 12,363.87 V.

Problems

11–17

7.2 [c] |Iba | = √ = 4.16 A. 3 [d] Van = (958.2 + j271)(7.20/13.74◦ ) = 7169.65/29.54◦ V; √ |Vab | = 3(7169.65) = 12,418.20 V. P 11.22 [a] IAB =

69,000/0◦ = 92/ − 36.87◦ A; 600 + j450

IBC = 92/ − 156.87◦ A; ICA = 92/83.13◦ A. √ [b] IaA = 3/ − 30◦ IAB = 159.35/ − 66.87◦ A; IbB = 159.35/ − 186.87◦ A; IcC = 159.35/53.13◦ A. [c] Iba = IAB = 92/ − 36.87◦ A; Icb = IBC = 92/ − 156.87◦ A; Iac = ICA = 92/83.13◦ A. P 11.23 Let pa , pb , and pc represent the instantaneous power of phases a, b, and c, respectively. Then assuming a positive phase sequence, we have pa = van iaA = [Vm cos ωt][Im cos(ωt − θφ )]; pb = vbn ibB = [Vm cos(ωt − 120◦ )][Im cos(ωt − θφ − 120◦ )]; pc = vcn icC = [Vm cos(ωt + 120◦ )][Im cos(ωt − θφ + 120◦ )]. The total instantaneous power is pT = pa + pb + pc , so pT = Vm Im [cos ωt cos(ωt − θφ ) + cos(ωt − 120◦ ) cos(ωt − θφ − 120◦ ) + cos(ωt + 120◦ ) cos(ωt − θφ + 120◦ )]. Now simplify using trigonometric identities. In simplifying, collect the coefficients of cos(ωt − θφ ) and sin(ωt − θφ ). We get pT = Vm Im [cos ωt(1 + 2 cos2 120◦ ) cos(ωt − θφ ) +2 sin ωt sin2 120◦ sin(ωt − θφ )] = 1.5Vm Im [cos ωt cos(ωt − θφ ) + sin ωt sin(ωt − θφ )] = 1.5Vm Im cos θφ .

11–18

CHAPTER 11. Balanced Three-Phase Circuits

P 11.24 The a-phase of the circuit is shown below:

I1 = I∗2

250/ − 60◦ = 10/ − 113.13◦ A (rms); 15 + j20

500/45◦ = = 2/105◦ A (rms); ◦ / 250 − 60

I = I1 + I2 = 10/ − 113.13◦ + 2/ − 105◦ = 11.98/ − 111.78◦ A (rms); Sa = VI∗ = (250/ − 60◦ )(11.98/111.78◦ ) = 2995/51.78◦ VA; ST = 3Sa = 8985/51.78◦ VA. P 11.25 [a] S1 = 72 − j21 kVA; S2 = 120 + j90 kVA; S3 = 168 + j36 kVA; ST = S1 + S2 + S3 = 360 + j105 kVA; ST /φ = 120 + j35 kVA. Single phase equivalent circuit:

120,000 + j35,000 .·. I∗aA = = 48 + j14; 2500 .·. IaA = 48 − j14 A = 50/ − 16.26◦ A; Van = 2500 + (1 + j5)(48 − j14) = 2618 + j226 = 2627.74/4.93◦ V; √ .·. |Vab | = 3(2627.74) = 4551.4 V.

Problems [b] PL /φ = 120 kW; PS /φ = 120,000 + |IaA |2 (1) = 122,500 W = 122.5 kW; 120 η= 100 = 97.96%. 122.5 



P 11.26 [a] ST ∆ = 24,000/53.13◦ − 12,000/36.87◦ = 12,924.37/68.2◦ VA; S∆ = ST ∆ /3 = 4308.12/68.2◦ VA. [b] |Van | =

4000/36.87◦ 40/60◦

|Vline | = |Vab | = P 11.27

I∗aA

√

= 100 V (rms);

√ 3|Van | = 100 3 = 173.205 V (rms).

(48 + j36)103 = = 100 + j75; 480

IaA = 100 − j75 A; Van = 480 + j0 + (100 − j75)(0.2 + j0.4) = 530 + j25 V.

Ic =

530 + j25 = −5 + j106 A; −j5

Ina = IaA + Ic = 95 + j31 = 99.93/18.07◦ A. [b] Sg/φ = (530 + j25)(95 − j31) = 51,125 − j14,055 VA; SgT = 3Sg/φ = 153,375 − j42,165 VA. Therefore, the source is delivering 153,375 W and absorbing 42,165 vars. [c] Pdel = 153,375 W. Pabs = 3(48,000) + 3|IaA |2 (0.2) = 144,000 + 9375 = 153,375 W = Pdel .

11–19

11–20

CHAPTER 11. Balanced Three-Phase Circuits

[d] Qdel = 3|Ic |2 (5) = 168,915 VAR; Qabs = 3(36,000) + 42,165 + 3|IaA |2 (0.4) = 168,915 VAR = Qdel . P 11.28 From the solution to Problem 11.18 we have: SAB = (720/0◦ )(144/16.26◦ ) = 99,532.9 + j29,030.04 VA; SBC = (720/ − 120◦ )(36/83.13◦ ) = 20,735.97 − j15,552.04 VA; SCA = (720/120◦ )(20.36/ − 75◦ ) = 10,365.62 + j10,365.62 VA. P 11.29 [a]

IaA =

14,000/0◦ = 8/ − 16.26◦ A; 1680 + j490

IaA ICA = √ /150◦ = 4.62/133.74◦ A. 3 [b] Sg/φ = −14,000I∗aA = −107,520 − j31,360 VA; .·. Pdeveloped/phase = 107.52 kW; Pabsorbed/phase = |IaA |2 1677 = 107.328 kW; % delivered = P 11.30 [a] |S| = Q= [b] pf =

√

107.328 (100) = 99.82%. 107.52

3(480)(400) = 332,553.755 VA;

q

(332,553.755)2 − (300,000)2 = 143,499.13 VAR.

300,000 = 0.9021 lagging. 332,553.755

Problems

11–21

P 11.31 The complex power of the source per phase is Ss = 30,000/( cos−1 0.8) = 30,000/36.87◦ = 24,000 + j18,000 kVA. This complex power per phase must equal the sum of the per-phase power of the two loads: Ss = S1 + S2

so

24,000 + j18,000 = 20,000 + S2 ;

.·. S2 = 4000 + j18,000 VA. |Vrms |2 . Z2∗

Also,

S2 =

|Vrms | =

415.69 |Vload | √ = √ = 240 V(rms). 3 3

Thus,

Z2∗ =

(240)2 |Vrms |2 = = 0.68 − j3.05 Ω; S2 4000 + j18,000

.·. Z2 = 0.68 + j3.05 Ω. P 11.32 [a]

7200/0◦ I1 = = 21.6 − j7.2 A; 300 + j100 I2 =

7200/0◦ = 3.2 + j1.6 A; 1800 − j900

I∗3 =

37,440 + j31,680 = 5.2 + j4.4; 7200

I3 = 5.2 − j4.4 A. IaA = I1 + I2 + I3 = 30 − j10 A =

√

1000/ − 18.43◦ A.

Van = 7200 + j0 + (30 − j10)(1 + j10) = 7330 + j290 V; Sφ = Van I∗aA = (7330 + j290)(30 + j10) = 217,000 + j82,000 VA; ST = 3Sφ = 651 + j246 kVA.

11–22

CHAPTER 11. Balanced Three-Phase Circuits

[b] S1/φ = 7200(21.6 + j7.2) = 155.52 + j51.84 kVA; S2φ = 7200(3.2 − j1.6) = 23.04 − j11.52 kVA; S3φ = 37.44 + j31.68 kVA; Sφ (load) = 216 + j72 kVA. 216 % delivered = (100) = 99.54%. 217 



P 11.33 [a] POUT = 746 × 200 = 149,200 W; PIN = 149,200/(0.96) = 155,416.67 W; √ 3VL IL cos θ = 155,416.67; 155,416.67 IL = √ = 468.91 A. 3(208)(0.92) √ √ [b] Q = 3VL IL sin φ = 3(208)(468.91)(0.39) = 66,207.79 VAR. !

2450 /0◦ V; P 11.34 [a] VAN = √ VAN I∗aA = Sφ = 144 + j192 kVA. 3 Therefore (144 + j192)1000 √ I∗aA = = (101.8 + j135.7) A; 2450/ 3 IaA = 101.8 − j135.7 = 169.67/ − 53.13◦ A; |IaA | = 169.67 A. [b] P =

(2450)2 ; R

Q=

therefore R =

(2450)2 ; X

(2450)2 = 41.68 Ω. 144,000

therefore X =

(2450)2 = 31.26 Ω. 192,000

√ VAN 2450/ 3 [c] Zφ = = = 8.34/53.13◦ = (5 + j6.67) Ω; IaA 169.67/ − 53.13◦ .·. R = 5 Ω, X = 6.67 Ω. P 11.35 [a] From Problem 11.34, IaA = (101.8 − j135.7) A. Therefore Icap = j135.7 A Thus CY =

and

ZCY

1 = 254.5 µF. (10.42)(2π)(60)

ZC∆ = (−j10.42)(3) = −j31.26 Ω; Therefore C∆ =

254.5 = 84.84 µF. 3

√ 2450/ 3 = = −j10.42 Ω. j135.7

Problems [b] CY = 254.5 µF. [c] |IaA | = 101.8 A. 1 P 11.36 Sg/φ = (78)(0.8 − j0.6) × 103 = 20,800 − j15,600 VA; 3 I∗aA =

20,800 − j15,600 = 100 − j75 A; 208

IaA = 100 + j75 A.

VAN = 208 − (100 + j75)(0.04 + j0.20) = 219 − j23 = 220.20/ − 6◦ V; |VAB | =

√

3(220.20) = 381.41 V.

[b] SL/φ = (219 − j23)(100 − j75) = 20,175 − j18,725 VA; SL = 3SL/φ = 60,525 − j56,175 VA. Check: Sg = 3(20,800 − j15,600) = 62,400 − j46,800 VA; P` = 3|IaA |2 (0.04) = 1875 W; Pg = PL + P` = 60,525 + 1875 = 62,400 W (checks); Q` = 3|IaA |2 (0.20) = 9375 VAR; Qg = QL + Q` = −56,175 + 9375 = −46,800 VAR (checks).

11–23

11–24

CHAPTER 11. Balanced Three-Phase Circuits

P 11.37 |Iline | =

1200 √ = 10 A(rms); 208/ 3

√ |V | 208/ 3 |Zy | = = = 12; |I| 10 Zy = 12/25◦ Ω; Z∆ = 3Zy = 36/25◦ = 32.63 + j15.21 Ω/φ. P 11.38

7200I∗1 = (230.4 − j67.2)103 ; I∗1 = 32 − j9.33 A; I1 = 32 + j9.33 A. 1 Zy = Z∆ = 207.36 + j60.48 Ω; 3 I2 =

7200/0◦ = 32 − j9.33 A; 207.36 + j60.48

.·. IaA = I1 + I2 = 64 + j0 A. Van = 7200 + j0 + 64(0.5 + j4) = 7236.53/2.03◦ V; |Vab | =

√

3|Van | = 12,534.04 V.

Problems P 11.39 Assume a ∆-connect load (series): 1 Sφ = (190.44 × 103 )(0.8 − j0.6) = 50,784 − j38,088 VA; 3 ∗ Z∆φ =

|13,800|2 = 3000/36.87◦ Ω; 50,784 − j38,088

Z∆φ = 3000/ − 36.87◦ = 2400 − j1800 Ω.

Now assume a Y-connected load (series): 1 ZY φ = Z∆φ = 800 − j600 Ω. 3

Now assume a ∆-connected load (parallel): Pφ =

|13,800|2 ; R∆

R∆φ =

Qφ =

|13,800|2 = 3750 Ω; 50,784

|13,800|2 ; X∆

|13,800|2 X∆ φ = = −5000 Ω. −38,088

11–25

11–26

CHAPTER 11. Balanced Three-Phase Circuits

Now assume a Y-connected load (parallel): 1 RY φ = R∆φ = 1250 Ω; 3 1 XY φ = X∆φ = −1666.67 Ω. 3

P 11.40 [a]

1 1188 (0.8) 103 = 396,000 + j528,000 VA; 1188 + j 3 0.6 

SL/φ = I∗aA =



396,000 + j528,000 = 60 + j80 A; 6600

IaA = 60 − j80 A; Van = 6600 + (60 − j80)(0.5 + j4) = 6950 + j200 = 6952.88/1.65◦ V. |Vab | =

√

3(6952.88) = 12,042.74 V.

Problems [b]

I1 = 60 − j80 A

(from part [a]);

1 S2 = 0 − j (1920) × 103 = −j640,000 VAR; 3 I∗2 =

−j640,000 = −j96.97 A; 6600

.·. I2 = j96.97 A. IaA = 60 − j80 + j96.97 = 60 + j16.97 A; Van = 6600 + (60 + j16.97)(0.5 + j4) = 6562.12 + j248.485 = 6566.82/2.17◦ V. |Vab | =

√

3(6566.82) = 11,374.07 V.

[c] |IaA | = 100 A; Ploss/φ = (100)2 (0.5) = 5000 W; Pg/φ = 396,000 + 5000 = 401 kW; %η =

396 (100) = 98.75%. 401

[d] |IaA | = 62.354 A; P`/φ = (3887.98)(0.5) = 1943.99 W; %η =

396,000 (100) = 99.51%. 397,944

[e] Zcap/Y = −j

6600 = −j68.062 Ω; 96.97

Zcap/∆ = 3Zcap/Y = −j204.187 Ω; .·.

1 = 204.187; ωC

C=

1 = 12.99 µF. (204.187)(120π)

11–27

11–28

CHAPTER 11. Balanced Three-Phase Circuits

P 11.41 Zφ = |Z|/θ =

VAN ; IaA

θ = /VAN − /IaA ; θ1 = /VAB − /IaA . For a positive phase sequence, /VAB = /VAN + 30◦ . Thus, θ1 = /VAN + 30◦ − /IaA = θ + 30◦ . Similarly, Zφ = |Z|/θ =

VCN ; IcC

θ = /VCN − /IcC ; θ2 = /VCB − /IcC . For a positive phase sequence, /VCB = /VBA − 120◦ = /VAB + 60◦ ; /IcC = /IaA + 120◦ . Thus, θ2 = /VAB + 60◦ − (/IaA + 120◦ ) = θ1 − 60◦ = θ + 30◦ − 60◦ = θ − 30◦ . P 11.42 Wm1 = |VAB ||IaA | cos(/VAB − /IaA ) = (199.58)(2.4) cos(65.68◦ ) = 197.26 W; Wm2 = |VCB ||IcC | cos(/VCB − /IcC ) = (199.58)(2.4) cos(5.68◦ ) = 476.64 W; CHECK: W1 + W2 = 673.9 = (2.4)2 (39)(3) = 673.9 W.

Problems P 11.43 [a] W2 − W1 = VL IL [cos(θ − 30◦ ) − cos(θ + 30◦ )] = VL IL [cos θ cos 30◦ + sin θ sin 30◦ − cos θ cos 30◦ + sin θ sin 30◦ ] = 2VL IL sin θ sin 30◦ = VL IL sin θ. Therefore

√

3(W2 − W1 ) =

√

3VL IL sin θ = QT .

[b] Zφ = (8 + j6) Ω; √ QT = 3[2476.25 − 979.75] = 2592 VAR; QT = 3(12)2 (6) = 2592 VAR;

(checks).

Zφ = (8 − j6) Ω; √ QT = 3[979.75 − 2476.25] = −2592 VAR; QT = 3(12)2 (−6) = −2592 VAR; (checks). √ Zφ = 5(1 + j 3) Ω; √ QT = 3[2160 − 0] = 3741.23 VAR; √ QT = 3(12)2 (5 3) = 3741.23 VAR; (checks). Zφ = 10/75◦ Ω; √ QT = 3[−645.53 − 1763.63] = −4172.80 VAR; QT = 3(12)2 [−10 sin 75◦ ] = −4172.80 VAR; √ 3(W2 − W1 ) 873,290.66 P 11.44 tan φ = = = 1.1918; W1 + W2 732,777.88

(checks).

.·. φ = 50◦ ; √ .·. 7600 3|IL | cos 80◦ = 114,291.64. |IL | = 50 A; |Z| =

7600 = 152 Ω 50

.·. Z = 152/50◦ Ω.

P 11.45 From the solution to Prob. 11.18 we have IaA = 145.88/ − 24.28◦ A

and

IbB = 134.01/178.04◦ A.

11–29

11–30

CHAPTER 11. Balanced Three-Phase Circuits

[a] W1 = |Vac | |IaA | cos(θac − θaA ) = 720(145.88) cos(−60◦ + 24.28◦ ) = 85,274.70 W. [b] W2 = |Vbc | |IbB | cos(θbc − θbB ) = 720(134.01) cos(−120◦ − 178.04◦ ) = 45,357.50 W. [c] W1 + W2 = 130,632 W; PAB = (144)2 (4.8) = 99,532.8 W; PBC = (36)2 (16) = 20,736 W; PCA = (20.36)2 (25) = 10,363.2 W; PAB + PBC + PCA = 130,632; therefore W1 + W2 = Ptotal . P 11.46 [a] Zφ = 600 + j450 = 750/36.87◦ Ω; Sφ =

(69 × 103 )2 = 5,078,400 + j3,808,800 VA; 750/ − 36.87◦

ST = 3Sφ = 15,235,200 + j11,426,400 VA. √ [b] Wm1 = (69,000) 3(92) cos(0 + 66.87◦ ) = 4,318,082.44 W; √ Wm2 = (69,000) 3(92) cos(60 − 53.13◦ ) = 10,916,117.56 W. Check:

PT = 15,235,200 W = Wm1 + Wm2 .

P 11.47 [a] W1 = |VBA ||IbB | cos θ. Positive phase sequence, using the equivalent Y-connected load impedances: √ VBA = 480 3/ − 150◦ V; IaA =

480/0◦ = 24/ − 30◦ A; 20/30◦

IbB = 24/ − 150◦ A; √ W1 = (24)(480) 3 cos 0◦ = 19,953.23 W; W2 = |VCA ||IcC | cos θ. √ VCA = 480 3/150◦ V; IcC = 24/90◦ A; √ W2 = (24)(480) 3 cos 60◦ = 9976.61 W.

Problems √ [b] Pφ = (24)2 (20) cos 30◦ = 5760 3 W; √ PT = 3Pφ = 17,280 3 W; √ √ √ W1 + W2 = 11,520 3 + 5760 3 = 17,280 3 W; .·. W1 + W2 = PT P 11.48 IaA =

(checks).

VAN = |IL |/−θφ A; Zφ

Zφ = |Z|/θφ ,

VBC = |VL |/ − 90◦ V.

Wm = |VL | |IL | cos[−90◦ − (−θφ )] = |VL | |IL | cos(θφ − 90◦ ) = |VL | |IL | sin θφ . Therefore

√

3Wm =

√

3|VL | |IL | sin θφ = Qtotal .

P 11.49 [a] Z = 96 + j72 = 120/36.87◦ Ω; VAN = 720/0◦ V;

.·. IaA = 6/ − 36.87◦ A; √ = 720 3/ − 90◦ V.

VBC = VBN − VCN √ Wm = (720 3)(6) cos(−90 + 36.87◦ ) = 4489.48 W; √ 3Wm = 7776 VAR. [b] Qφ = (36)(72) = 2592 VAR; √ QT = 3Qφ = 7776 VAR = 3Wm . P 11.50 [a] Negative phase sequence: √ VAB = 480 3/ − 30◦ V; √ VBC = 480 3/90◦ V; √ VCA = 480 3/ − 150◦ V; √ √ 480 3/ − 30◦ IAB = = 8 3/0◦ A; ◦ 60/ − 30 √ √ 480 3/90◦ IBC = = 20 3/60◦ A; ◦ / 24 30

11–31

11–32

CHAPTER 11. Balanced Three-Phase Circuits

ICA

√ √ 480 3/ − 150◦ = = 6 3/ − 150◦ A. 80/0◦

IaA = IAB + IAC √ √ = 8 3/0◦ + 6 3/30◦ = 23.44/12.81◦ A; IcC = ICB + ICA √ √ = 20 3/ − 120◦ + 6 3/ − 150◦ = 43.95/ − 126.79◦ A. √ Wm1 = 480 3(23.44) cos(−30 − 12.81◦ ) = 14,296.61 W; √ Wm2 = 480 3(43.95) cos(−90 + 126.79◦ ) = 29,261.53 W. [b] Wm1 + Wm2 = 43,558.14 W; √ PA = (8 3)2 (60 cos 30◦ ) = 9976.61 W; √ PB = (20 3)2 (24 cos 30◦ ) = 24,941.53 W; √ PC = (6 3)2 (80) = 8640 W; PA + PB + PC = 43,558.14 = Wm1 + Wm2 . P 11.51 [a] I∗aA =

(192 + j56)103 = 41.67/16.26◦ A; 4800

IaA = 41.67/ − 16.26◦ A; IbB = 41.67/ − 136.26◦ A. √ VAB = 4800 3/30◦ V; √ VBC = 4800 3/ − 90◦ V. √ W1 = (4800 3)(41.67) cos 46.26◦ = 239,502.58 W. [b] Current coil in line aA, measure IaA . Voltage coil across AC, measure VAC . [c] IaA = 41.67/ − 16.76◦ A; √ VCA = 4800 3/150◦ V; √ .·. VAC = 4800 3/ − 30◦ V; √ W2 = (4800 3)(41.67) cos 13.74◦ = 336,497.42 W. [d] W1 + W2 = 576,000 = 576kW; PT = 600(0.96) = 576 kW = W1 + W2 .

Problems

P 11.52 [a] ZY =

Z∆ = 276 − j207 = 345/ − 36.87◦ Ω; 3

IaA =

6900/0◦ = 20/36.87◦ A; 345/ − 36.87◦

IbB = 20/ − 83.13◦ A; √ VAC = 6900 3/ − 30◦ V; √ VBC = 6900 3/ − 90◦ V; √ W1 = (6900 3)(20) cos(−30 − 36.87◦ ) = 93,893.10 W; √ W2 = (6900 3)(20) cos(−90 + 83.13◦ ) = 237,306.90 W. [b] W1 + W2 = 331,200 W; PT = 3(20)2 (276) = 331,200 W. √ [c] 3(W1 − W2 ) = −248,400 VAR; QT = 3(20)2 (−207) = −248,400 VAR. P 11.53 [a]

[b]

[c]

11–33

11–34

CHAPTER 11. Balanced Three-Phase Circuits

[d]

P 11.54 [a] Q =

|V|2 ; XC

(13,800)2 = 158.70 Ω; .·. |XC | = 1.2 × 106 1 1 = 158.70; C= = 20.06 µF. ωC 2π(50)(158.70) √ (13,800/ 3)2 1 [b] |XC | = = (158.70); 6 1.2 × 10 3 .·.

.·. C = 3(20.06) = 60.2 µF. P 11.55 Before the capacitors are added the total line loss is PL = 3|150.61 + j150.61|2 (0.6) = 81.66 kW. After the capacitors are added the total line loss is PL = 3|150.61|2 (0.6) = 40.83 kW. Note that adding the capacitors to control the voltage level also reduces the amount of power loss in the lines, which in this example is cut in half. P 11.56 If the capacitors remain connected when the substation drops its load, the expression for the line current becomes 13,800 ∗ √ IaA = −j1.2 × 106 3 or Hence

I∗aA = −j150.61 A; IaA = j150.61 A.

Problems

11–35

Now, Van =

13,800 ◦ √ /0 + (0.6 + j4.8)(j150.61) = 7244.49 + j90.37 = 7245.05/0.71◦ V. 3

The magnitude of the line-to-line voltage at the generating plant is |Vab | =

√

3(7245.05) = 12,548.80 V.

This is a problem because the voltage is below the acceptable minimum of 13 kV. Thus when the load at the substation drops off, the capacitors must be switched off. P 11.57 [a]

13,800 ∗ √ IaA = 75 × 103 + j180 × 103 − j1200 × 103 ; 3 √ √ 75 3 − j1020 3 = 9.41 − j128.02 A; I∗aA = 13.8 .·. IaA = 9.41 + j128.02 A. 13,800 ◦ √ /0 + (0.6 + j4.8)(9.41 + j128.02) 3 = 7358.58 + j121.98 = 7359.59/0.95◦ V;

Van =

.·. |Vab | =

√

3(7359.59) = 12,747.2 V.

[b] Yes, the magnitude of the line-to-line voltage at the power plant is less than the allowable minimum of 13 kV. P 11.58 [a]

13,800 ∗ √ IaA = (75 + j180) × 103 ; 3 √ √ 75 3 + j180 3 ∗ = 9.41 + j22.59 A; IaA = 13.8 .·. IaA = 9.41 − j22.59 A. 13,800 ◦ √ /0 + (0.6 + j4.8)(9.41 − j22.59) 3 = 8081.51 + j31.61 = 8081.57/0.22◦ V;

Van =

.·. |Vab | = [b] Yes:

√

3(8081.57) = 13,997.7 V.

13 kV < 13,997.7 < 14.6 kV.

[c] Ploss = 3|9.41 + j128.02|2 (0.6) = 29.66 kW. [d] Ploss = 3|9.41 − j22.59|2 (0.6) = 1.078 kW.

11–36

CHAPTER 11. Balanced Three-Phase Circuits

[e] Yes, the voltage at the generating plant is at an acceptable level and the line loss is greatly reduced. P 11.59 [a] The capacitor from Appendix H whose value is closest to 20.06 µF is 22 µF. |XC | = Q=

1 1 = = 144.686 Ω; ωC 2π(50)(22 × 10−6 )

|V |2 (13,800)2 = = −1,316,226.8 VAR/φ. XC −144.686

QT = 3Q = −3,948,680.4 VAR. [b] I∗aA =

1,200,000 + j1,200,000 − j1,316,226.8 √ = 150.6 − j14.588 A; 13,800/ 3

13,800 ◦ √ /0 + (0.6 + j4.8)(150.6 + j14.588) = 8021.2/5.23◦ ; 3 √ |Vab | = 3(8021.2) = 13,893 V.

Van =

This voltage falls within the allowable range of 13 kV to 14.6 kV. P 11.60 [a] The capacitor from Appendix H whose value is closest to 60.17 µF is 47 µF. |XC | =

1 1 = = 67.7255 Ω; ωC 2π(50)(47 × 10−6 )

|V |2 (13,800)2 Q= = = −937,313.02 VAR. 3XC 3(−67.7255) [b] I∗aA =

1,200,000 + j262,687 √ = 150.6 + j32.97 A; 13,800/ 3

13,800 ◦ √ /0 + (0.6 + j4.8)(150.6 − j32.97) = 8246.1/4.89◦ ; 3 √ |Vab | = 3(8246.1) = 14,282.7 V. Van =

This voltage falls within the allowable range of 13 kV to 14.6 kV.

Introduction to the Laplace Transform

Assessment Problems AP 12.1 [a] sin(ωt + θ) = (sin ωt cos θ + cos ωt sin θ). Therefore L{sin(ωt + θ)} = cos θL{sin ωt} + sin θL{cos ωt} ω cos θ + s sin θ . = s2 + ω 2 [b] cos(ωt + θ) = (cos ωt cos θ − sin ωt sin θ). Therefore L{cos(ωt + θ)} = cos θL{cos ωt} − sin θL{sin ωt} s cos θ − ω sin θ = . s2 + ω 2 AP 12.2 [a] Let f (t) = te−at : F (s) = L{te−at } = Now,

1 . (s + a)2

L{t2 f (t)} = (−1)2

d2 F (s) . ds2

d2 1 6 }= 2 = . 2 ds (s + a) (s + a)4 "

So,

2

−at

L{t · te

[b] f1 (t) = e−at cos ωt;

#

F1 (s) =

F (s) = sF1 (s) − f1 (0− ) =

s+a ; (s + a)2 + ω 2

−a2 − sa − ω 2 s(s + a) − 1 = . (s + a)2 + ω 2 (s + a)2 + ω 2

12–1

12–2

CHAPTER 12. Introduction to the Laplace Transform [c] f1 (t) = e−at sin ωt; F (s) =

AP 12.3 F (s) =

F1 (s) =

ω ; (s + a)2 + ω 2

F1 (s) ω = . s s[(s + a)2 + ω 2 ]

K1 K2 K3 + + ; s+1 s+2 s+4

8s2 + 37s + 32 K1 = = 1; (s + 2)(s + 4) s=−1

8s2 + 37s + 32 K2 = = 5; (s + 1)(s + 4) s=−2

8s2 + 37s + 32 = 2; K3 = (s + 1)(s + 2) s=−4

f (t) = [e−t + 5e−2t + 2e−4t ]u(t). AP 12.4 F (s) =

10s2 + 76s + 160 K1 K2 K3 = + + ; (s + 2)(s + 4)(s + 6) s+2 s+4 s+6

K1 =

40 − 152 + 160 = 6; 2(4)

K3 =

360 − 456 + 160 = 8; (−4)(−2)

K2 =

160 − 304 + 160 = −4; (−2)(2)

f (t) = [6e−2t − 4e−4t + 8e−6t ]u(t). AP 12.5 [a] From Example 12.2, V (s) = [b] V (s) = K1 =

s2

Idc /C 15,000 = 2 . + (1/RC)s + (1/LC) s + 1250s + 250,000

15,000 K1 K2 = + . (s + 250)(s + 1000) s + 250 s + 1000 15,000 = 20; 750

V (s) =

K2 =

15,000 = −20. −750

20 20 − . s + 250 s + 1000

Therefore,

v(t) = (20e−250t − 20e−1000t )u(t) V.

Problems

AP 12.6 F (s) =

12–3

K1 K2 K2∗ + + ; s+1 s+2−j s+2+j

22s2 + 60s + 58 K1 = = 10; s2 + 4s + 5 s=−1

22s2 + 60s + 58 = 6 + j8 = 10/53.13◦ ; (s + 1)(s + 2 + j) s=−2+j

K2 =

f (t) = [10e−t + 20e−2t cos(t + 53.13◦ )]u(t). AP 12.7 [a] From Example 12.2, V (s) = [b] V (s) = K=

s2

15,000 Idc /C = 2 . + (1/RC)s + (1/LC) s + 800s + 250,000

K K∗ 15,000 = + . (s + 400 − j300)(s + 400 + j300) s + 400 − j300 s + 400 + j300

15,000 = −j25 = 25/ − 90◦ . j600

V (s) =

25/ − 90◦ 25/90◦ + . s + 400 − j300 s + 400 + j300 v(t) = 2(25)e−400t cos(300t − 90◦ ) = (50e−400t sin 300t)u(t) V.

Therefore, AP 12.8 F (s) =

K1 K3 K2 + + ; 2 s (s + 2) s+2

80(s + 3) K1 = = 60; (s + 2)2 s=0

K2 =

80(s + 3) = −40; s s=−2 "

#

"

80 80(s + 3) d 80(s + 3) K3 = = − ds s s s2

#

= −60; s=−2

f (t) = [60 − 40te−2t − 60e−2t ]u(t). AP 12.9 [a] From Example 12.2, V (s) =

s2

15,000 Idc /C = 2 . + (1/RC)s + (1/LC) s + 1000s + 250,000

12–4

CHAPTER 12. Introduction to the Laplace Transform

[b] V (s) =

15,000 K1 K2 = + . 2 2 (s + 500) (s + 500) s + 500

K2 = 0; V (s) =

K1 = 15,000. 15,000 . (s + 500)2

Therefore, AP 12.10 F (s) =

(s2

14,400 14,400 = 2 + 60s + 2500) (s + 30 − j40)2 (s + 30 + j40)2

K1 K2 K1∗ + + (s + 30 − j40)2 (s + 30 − j40) (s + 30 + j40)2

=

+ K1 =

v(t) = 15,000te−500t u(t) V.

K2∗ ; (s + 30 + j40)

14,400 = −2.25 and K1∗ = −2.25; (j80)2 "

14,400 d K2 = ds (s + 30 + j40)2

#

= s=−30+j40

−2(14,400) = −j0.05625 = 0.05625/ − 90◦ ; (j80)3

K2∗ = j0.05625; Therefore f (t) = [−4.5te−30t cos 40t + 0.1125e−30t cos(40 − 90◦ )] u(t) = e−30t [0.1125 sin 40t − 4.5t cos 40t] u(t). AP 12.11

25 F (s) = s2 + 15s + 54

25s2 + 395s + 1494 25s2 + 375s + 1350 20s + 144

F (s) = 25 +

20s + 144 K1 K2 = 25 + + ; + 15s + 54 s+6 s+9

s2

20s + 144 K1 = = 8; s + 9 s=−6

K2 =

20s + 144 = 12; s + 6 s=−9

f (t) = 25δ(t) + [8e−6t + 12e−9t ]u(t).

Problems

12–5

5s − 15

AP 12.12 F (s) = s2 + 7s + 10

5s3 + 20s2 − 49s − 108 5s3 + 35s2 + 50s −15s2 − 99s − 108 −15s2 − 105s − 150 6s + 42

F (s) = 5s − 15 +

K1 K2 6s + 42 = 5s − 15 + + ; (s + 2)(s + 5) s+2 s+5

K1 =

6s + 42 = 10; s + 5 s=−2

K2 =

6s + 42 = −4; s + 2 s=−5





f (t) = 5δ 0 (t) − 15δ(t) + [10e−2t − 4e−5t ]u(t). AP 12.13 [a] Factoring the numerator, 10s2 + 210s + 980 = 10(s + 7)(s + 14). There are zeros at −7 and −14. Factoring the denominator, s3 + 14s2 + 50s = s(s + 7 + j)(s + 7 − j). There are poles at 0, −7 − j and −7 + j. [b] Factoring the numerator, 5s2 + 50s + 445 = 5(s + 5 + j8)(s + 5 − j8). There are zeros at −5 − j8 and −5 + j8. Factoring the denominator, 4s3 + 40s2 + 176s + 480 = 4(s + 6)(s + 2 + j4)(s + 2 − j4). There are poles at −6, −2 + j4 and −2 − j4. [c] Factoring the numerator, 25s2 + 200s + 400 = 25(s + 4)2 . There are two zeros at −4. Factoring the denominator, s4 + 8s3 + 38s2 + 56s + 25 = (s + 3 + j4)(s + 3 − j4)(s + 1)2 . There are poles at −3 + j4, −3 − j4, and two poles at −1.

12–6

CHAPTER 12. Introduction to the Laplace Transform

AP 12.14 lim f (t) = lim sF (s) t→0

s→∞

"

= lim

1/s→0

10[1 + (7.6/s) + (16/(s2 ))] = 10; [1 + (2/s)][1 + (4/s)][1 + (6/s)] #

.·. f (0+ ) = 10. 10s3 + 76s2 + 160s lim f (t) = lim sF (s) = lim = 0; t→∞ s→0 s→0 (s + 2)(s + 4)(s + 6) "

#

.·. f (∞) = 0. "

lim f (t) = lim sF (s) = lim t→0

s→∞

1/s→0

(80/s) + (240/s2 ) = 0; [1 + (2/s)]2 #

.·. f (0+ ) = 0. "

#

80s + 240 lim f (t) = lim sF (s) = lim = 60; t→∞ s→0 s→0 (s + 2)2 .·. f (∞) = 60. "

lim f (t) = lim sF (s) = lim s→∞ t→0

1/s→0

(14, 400/s3 ) = 0; [1 + (60/s) + (2500/s2 )]2 #

.·. f (0+ ) = 0. "

#

14,400s lim f (t) = lim sF (s) = lim = 0; t→∞ s→0 s→0 (s2 + 60s + 2500)2 .·. f (∞) = 0.

Problems

12–7

Problems P 12.1

[a] f (t) = 120 + 30t f (t) = 120 − 30t f (t) = −360 + 30t f (t) = 0

− 4 s ≤ t ≤ 0; 0 ≤ t ≤ 8 s; 8 s ≤ t ≤ 12 s;

elsewhere.

f (t) = (120 + 30t)[(u(t + 4) − u(t)] + (120 − 30t)[u(t) − u(t − 8)] +(−360 + 30t)[u(t − 8) − u(t − 12)]. [b] f (t) = 50 sin π2 t[u(t) − u(t − 4)] = (50 sin π2 t)u(t) − (50 sin π2 t)u(t − 4). [c] f (t) = (30 − 3t)[u(t) − u(t − 10)]. P 12.2

[a] (50 + 2.5t)[u(t + 20) − u(t)] + (50 − 5t)[u(t) − u(t − 10)] = (2.5t + 50)u(t + 20) − 7.5tu(t) + (5t − 50)u(t − 10). [b] (5t + 45)[u(t + 9) − u(t + 6)] + 15[u(t + 6) − u(t + 3)] − 5t[u(t + 3) − u(t − 3)] −15[u(t − 3) − u(t − 6)] + (5t − 45)[u(t − 6) − u(t − 9)] = 5(t + 9)u(t + 9) − 5(t + 6)u(t + 6) − 5(t + 3)u(t + 3) + 5(t − 3)u(t − 3) +5(t − 6)u(t − 6) − 5(t − 9)u(t − 9).

P 12.3

12–8 P 12.4

CHAPTER 12. Introduction to the Laplace Transform [a]

[b] f (t) = 30t[u(t) − u(t − 2)] + 60[u(t − 2) − u(t − 4)] +60 cos( π4 t − π)[u(t − 4) − u(t − 8)] +(30t − 300)[u(t − 8) − u(t − 10)]. P 12.5

F (s) =

Z −ε/2 −ε

Z ε/2 Z ε 4 −st −4 −st 4 −st e dt + e dt + e dt. 3 3 ε ε −ε/2 ε/2 ε3

Therefore F (s) =





4 sε [e − 2esε/2 + 2e−sε/2 − e−sε ]; 3 sε

L{δ 00 (t)} = lim F (s). ε→0

After applying L’Hopital’s rule three times, we have 2s 3s 2s s s sesε − esε/2 − e−sε/2 + se−sε = . ε→0 3 4 4 3 2 

lim

Therefore L{δ 00 (t)} = s2 .







Problems P 12.6

As ε → 0 the amplitude → ∞; the duration → 0; and the area is independent of ε, i.e., A=

Z ∞

ε 1 dt = 1. π ε2 + t2

−∞

1 1 1 [a] A = bh = (2ε) = 1. 2 2 ε [b] 0.  

P 12.7

 

 

[b] ∞. P 12.8

esε − e−sε 1 −st e dt = 2εs −ε 2ε The limit as ε → 0 is indeterminate so use l’Hopital’s rule:

F (s) =

Z ε

1 sesε + se−sε 1 2s F (s) = lim = · = 1. ε→0 2s 1 2s 1 "

P 12.9

12–9

[a] I =

Z 4

#

3

(t + 4)δ(t) dt +

−2

Z 4

4(t3 + 4)δ(t − 2) dt

−2

= 4 + 4(8 + 4) = 52. [b] I =

Z 4

2

t δ(t) dt +

−3

Z 4

t2 δ(t + 2.5) dt + 0

−3

= 02 + (−2.5)2 + 0 = 6.25. !   3 + j0 −jt0 3 1 1 Z ∞ (3 + jω) jtω P 12.10 f (t) = · πδ(ω) · e dω = πe = . 2π −∞ (4 + jω) 2π 4 + j0 8

dn f (t) P 12.11 L dtn (

)

= sn F (s) − sn−1 f (0− ) − sn−2 f 0 (0− ) − · · · .

Therefore L{δ n (t)} = sn (1) − sn−1 δ(0− ) − sn−2 δ 0 (0− ) − sn−3 δ 00 (0− ) − · · · = sn . P 12.12 [a] Let dv = δ 0 (t − a) dt,

v = δ(t − a),

du = f 0 (t) dt.

u = f (t), Therefore Z ∞

0

f (t)δ (t − a) dt = f (t)δ(t − a)

−∞

∞

−∞

= 0 − f 0 (a).

−

Z ∞ −∞

δ(t − a)f 0 (t) dt

12–10

CHAPTER 12. Introduction to the Laplace Transform 0

[b] L{δ (t)} = P 12.13 [a] L{t} =

Z ∞

Z ∞ 0−

d(e−st ) dt = − dt "

0

−st

δ (t)e

#

h

= − −se−st t=0

i t=0

∞ 1 e−st 1 =0− (−st − 1) (0 − 1) = . s2 s2 s2 0 1 therefore L{te−at } = . (s + a)2

te−st dt =

0

1 [b] L{t} = 2 ; s

ejωt − e−jωt . j2 Therefore

[c] sin ωt =

1 j2

L{sin ωt} = =

!

1 1 − s − jω s + jω

!

=

1 j2

!

ω . + ω2

s2

eβt + e−βt . 2 Therefore,

[d] cosh βt =

1Z L{cosh βt} = 2

∞ 0−

[e−(s−β)t + e−(s+β)t ]dt

1 e−(s−β)t ∞ e−(s+β)t = + 2 −(s − β) 0− −(s + β) "

=

1 1 + s−β s+β

1 2

!

=

s2

∞ # − 0

s . − β2

eβt − e−βt . 2 Therefore,

[e] sinh βt =

L{sinh βt} =

1Z 2

∞ 0−

h

1 e−(s−β)t = 2 −(s − β) "

= (

1 2

i

e−(s−β)t − e−(s+β)t dt #∞ 0−

1 e−(s+β)t − 2 −(s + β)

1 1 − s−β s+β

)

"

!

d −at s P 12.14 [a] L (te )u(t) = − 0. dt (s + a)2 (

)

d −at s L (te )u(t) = . dt (s + a)2

=

(s2

#∞ 0−

β . − β 2)

2jω 2 s + ω2



= s.

Problems

[b]

d −at (te ) = −ate−at + e−at . dt L{−ate−at + e−at } =

−a 1 −a s+a + = + . 2 2 (s + a) (s + a) (s + a) (s + a)2

)

(

s d −at (te ) = . .·. L dt (s + a)2 P 12.15 L{e−at f (t)} =

Z ∞ 0−

Z ∞

P 12.16 L{f (at)} =

0−

CHECKS

[e−at f (t)]e−st dt =

Z ∞ 0−

f (t)e−(s+a)t dt = F (s + a).

f (at)e−st dt

Let u = at,

u = 0−

du = a dt,

when t = 0− ,

and u = ∞ when t = ∞. Therefore L{f (at)} = Z t

Z ∞ 0−

f (u)e−(u/a)s

1 du = F (s/a). a a

F (s) 1 = . s s(s + a) 0− Z t 1 e−at −ax . [b] e dx = − a a 0−

P 12.17 [a] L

−ax

e



dx =

1 e−at L − a a (

)

(

1 1 1 1 = − . = a s s+a s(s + a) 



)

sω sω d sin ωt P 12.18 [a] L u(t) = 2 − sin(0) = 2 . 2 dt s +ω s + ω2 s2 s2 −ω 2 d cos ωt u(t) = 2 − cos(0) = 2 −1= 2 . [b] L dt s + ω2 s + ω2 s + ω2 )

(

d3 (t2 ) 2 3 u(t) = s [c] L − s2 (0) − s(0) − 2(0) = 2. 3 3 dt s ωs d sin ωt [d] = (cos ωt) · ω, L{ω cos ωt} = 2 ; dt s + ω2 (

)





d cos ωt = −ω sin ωt dt L{−ω sin ωt} = −

ω2 s2 + ω 2

d3 (t2 u(t)) = 2δ(t); dt3

L{2δ(t)} = 2.

12–11

12–12

CHAPTER 12. Introduction to the Laplace Transform

P 12.19 [a] f (t) = 4t[u(t) − u(t − 4)] +(32 − 4t)[u(t − 4) − u(t − 12)] +(4t − 64)[u(t − 12) − u(t − 16)] = 4tu(t) − 8(t − 4)u(t − 4) +8(t − 12)u(t − 12) − 4(t − 16)u(t − 16); 4[1 − 2e−4s + 2e−12s − e−16s ] .·. F (s) = . s2 [b]

f 0 (t) = 4[u(t) − u(t − 4)] − 4[u(t − 4) − u(t − 12)] +4[u(t − 12) − u(t − 16)] = 4u(t) − 8u(t − 4) + 8u(t − 12) − 4u(t − 16); L{f 0 (t)} =

4[1 − 2e−4s + 2e−12s − e−16s ] . s

[c]

f 00 (t) = 4δ(t) − 8δ(t − 4) + 8δ(t − 12) − 4δ(t − 16); L{f 00 (t)} = 4[1 − 2e−4s + 2e−12s − e−16s ]. P 12.20 [a] L{−20e−5(t−2) u(t − 2)} =

−20e−2s . (s + 5)

Problems [b] First rewrite f (t) as f (t) = (8t − 8)u(t − 1) + (24 − 8t − 8t + 8)u(t − 2) +(8t − 40 − 24 + 8t)u(t − 4) − (8t − 40)u(t − 5) = 8(t − 1)u(t − 1) − 16(t − 2)u(t − 2) +16(t − 4)u(t − 4) − 8(t − 5)u(t − 5); 8[e−s − 2e−2s + 2e−4s − e−5s ] .·. F (s) = . s2 P 12.21 [a]

Z t 0−

x dx =

t2 L 2 (

t2 . 2

)

1Z 2

=

∞ 0−

t2 e−st dt ∞ #

1 e−st 2 2 = (s t + 2st + 2) 3 − 2 −s 0 "

1 1 (2) = 3 . 3 2s s Z t  1 .·. L x dx = 3 . s 0− =

[b] L

x dx =

L{t} 1/s2 1 = = 3. s s s

Z t



Z t 0−

.·. L



0−

0

P 12.22 [a] L{f (t)} =

x dx =

1 . s3

CHECKS

Z e−st dt + −ε 2ε

Z ε

∞

− ae−a(t−ε) e−st dt

ε

1 sε a = (e − e−sε ) − e−sε = F (s). 2sε s+a a s lim F (s) = 1 − = . ε→0 s+a s+a 

[b] L{e−at } =



1 . s+a

Therefore L{f 0 (t)} = sF (s) − f (0− ) = P 12.23 [a] f1 (t) = e−at sinh βt;

F1 (s) =

F (s) = sF1 (s) − f1 (0− ) =

s s −0= . s+a s+a

β ; (s + a)2 − β 2

sβ sβ −0= . 2 2 (s + a) − β (s + a)2 − β 2

12–13

12–14

CHAPTER 12. Introduction to the Laplace Transform

[b] f1 (t) = e−at cos ωt; F (s) = [c]

F1 (s) =

s+a ; (s + a)2 + ω 2

s+a F1 (s) = . s s[(s + a)2 + ω 2 ]

d −at [e sinh βt] = βe−at cosh βt − ae−at sinh βt; dt Therefore F (s) = Z t 0−

e−ax cos ωx dx =

β(s + a) aβ sβ − = . 2 2 2 2 (s + a) − β (s + a) − β (s + a)2 − β 2 −ae−at cos ωt + ωe−at sin ωt + a . a2 + ω 2

Therefore −a(s + a) ω2 a 1 + + F (s) = 2 a + ω 2 (s + a)2 + ω 2 (s + a)2 + ω 2 s "

= P 12.24 [a]

#

s+a . s[(s + a)2 + ω 2 ]

dF (s) d = ds ds

Z ∞ 0−



f (t)e−st dt = −

Therefore L{tf (t)} = −

0−

tf (t)e−st dt

dF (s) . ds

d2 F (s) Z ∞ 2 [b] = t f (t)e−st dt; 2 − ds 0 Therefore

Z ∞

d3 F (s) Z ∞ 3 = −t f (t)e−st dt. 3 − ds 0

Z ∞ dn F (s) n = (−1) tn f (t)e−st dt = (−1)n L{tn f (t)}. dsn 0−

[c] L{t5 } = L{t4 t} = (−1)4 d L{t sin βt} = (−1) ds 1

d4 ds4



1 s2



β 2 s + β2

=

120 ; s6

!

=

(s2

L{te−t cosh t}: . From Problem 12.13(d), s ; F (s) = L{cosh t} = 2 s −1 dF (s2 − 1)1 − s(2s) s2 + 1 = = − ; ds (s2 − 1)2 (s2 − 1)2 Therefore

−

dF s2 + 1 = 2 . ds (s − 1)2

2βs + β 2 )2

Problems

12–15

Thus L{t cosh t} =

s2 + 1 (s2 − 1)2

L{e−t t cosh t} = P 12.25 [a]

Z ∞

F (u)du =

(s + 1)2 + 1 s2 + 2s + 2 = . [(s + 1)2 − 1]2 s2 (s + 2)2

Z ∞ Z ∞

s

0−

s

= = [b] L{t sin βt} =

Z ∞



f (t)e−ut dt du =

Z ∞

0−

f (t)

−ut

e

du dt =

Z ∞ Z ∞ 0−



f (t)e−ut du dt

s

e−tu ∞ dt f (t) −t s 0− "

Z ∞

s

#

f (t) −e−st dt = L . f (t) − −t t 0 #

"

Z ∞

(

)

2βs ; (s2 + β 2 )2 (

t sin βt therefore L t

)

=

Z ∞" s

#

2βu du. (u2 + β 2 )2

Let ω = u2 + β 2 , then ω = s2 + β 2 when u = s, and ω = ∞ when u = ∞; also dω = 2u du. Thus (

t sin βt L t

)

=β

"

Z ∞ s2 +β 2

#

dω −1 ∞ β = β . 2 2= 2 2 ω ω s + β2 s +β 



P 12.26 [a] For t ≥ 0+ : Rio + L io = C

dio + vo = 0; dt

dvo dt

d2 vo dio =C 2 ; dt dt

dvo d2 vo · . . RC + LC 2 + vo = 0 dt dt or 1 d2 vo R dvo + + vo = 0. 2 dt L dt LC R 1 [b] s2 Vo (s) − sVdc − 0 + [sVo (s) − Vdc ] + Vo (s) = 0; L LC Vo (s) s2 +

R 1 s+ = Vdc (s + R/L); L LC

Vo (s) =

Vdc [s + (R/L)] . + (R/L)s + (1/LC)]



[s2



12–16

CHAPTER 12. Introduction to the Laplace Transform

P 12.27 [a] For t ≥ 0+ : vo dvo +C + io = 0; R dt vo = L

dvo d2 io =L 2; dt dt

dio ; dt

.·.

d2 io L dio + LC 2 + io = 0, R dt dt

or

d2 io 1 dio 1 + + io = 0. 2 dt RC dt LC

[b] s2 Io (s) − sIdc − 0 +

1 1 [sIo (s) − Idc ] + Io (s) = 0; RC LC

Io (s) s2 +

1 1 s+ = Idc (s + 1/RC); RC LC

Io (s) =

Idc [s + (1/RC)] . + (1/RC)s + (1/LC)]



P 12.28 io = C

[s2



dvo ; dt

.·. Io (s) = sCVo (s) =

s2

sIdc . + (1/RC)s + (1/LC)

vo − Vdc 1Zt dvo P 12.29 [a] + vo dx + C = 0; R L 0 dt .·. [b] Vo +

dvo RZ t vo dx + RC = Vdc . vo + L 0 dt Vdc R Vo + RCsVo = ; L s s

.·.

sLVo + RVo + RCLs2 Vo = LVdc ;

.·.

Vo (s) =

[c] io =

s2

(1/RC)Vdc . + (1/RC)s + (1/LC)

1Zt vo dx; L 0

Io (s) =

Vo Vdc /RLC = . 2 sL s[s + (1/RC)s + (1/LC)]

Problems

P 12.30 [a] C

dv1 v1 − v2 + = ig ; dt R

v2 − v1 1Zt =0 v2 dτ + L 0 R or dv1 v1 v2 + − = ig ; C dt R R v1 v2 1Zt − + + v2 dτ = 0. R R L 0 [b] CsV1 (s) + −

V1 (s) V2 (s) − = Ig (s); R R

V1 (s) V2 (s) V2 (s) + + =0 R R sL

or (RCs + 1)V1 (s) − V2 (s) = RIg (s); −sLV1 (s) + (R + sL)V2 (s) = 0. Solving, V2 (s) =

C[s2

sIg (s) . + (R/L)s + (1/LC)]

P 12.31 [a] 180 = 100i1 + 15 0 = 20 [b]

di1 di2 + 10 ; dt dt

di2 di1 + 10 + 200i2 . dt dt

180 = (15s + 100)I1 (s) + 10sI2 (s); s 0 = 10sI1 (s) + (20s + 200)I2 (s).

[c] Solving the second equation in (b), I2 (s) =

−10s I1 (s); 20s + 200

Substituting into the first equation in (b), 180 (−100s2 ) = (15s + 100) I1 (s); s (20s + 200)

12–17

12–18

CHAPTER 12. Introduction to the Laplace Transform Therefore, I1 (s) = = I2 (s) = =

180(20s + 200) 3600(s + 10) = 2 s[(15s + 100)(20s + 200) − 100s ] 200s(s2 + 25s + 100) 18(s + 10) ; s(s + 5)(s + 20) −10s −180s(s + 10) I1 (s) = 20s + 200 20s(s + 5)(s + 10)(s + 20) −9 . (s + 5)(s + 20)

P 12.32 ig (t) = 5 cos 10tu(t); 1 = 64; LC

1 = 40; RC

Therefore V =

P 12.33

R = 5000; L Vo (s) =

s2

15(s + 5000) + 5000s + 4 × 106 √

6.25 × 106 − 4 × 106 s2 = −4000 rad/s

15(s + 5000) K1 K2 = + (s + 1000)(s + 4000) s + 1000 s + 4000

15(4000) = 20 V; 3000

Vo (s) =

1 = 40. C

(40)(5)s2 200s2 = . (s2 + 40s + 64)(s2 + 100) (s2 + 40s + 64)(s2 + 100)

s1 = −1000 rad/s;

K1 =

5s2 . s2 + 100

1 = 4 × 106 LC

s1,2 = −2500 ±

Vo (s) =

so Ig (s) =

K2 =

20 5 − s + 1000 s + 4000

vo (t) = [20e−1000t − 5e−4000t ]u(t) V

15(1000) = −5 V −3000

Problems

P 12.34

1 = 10,000; RC Io (s) =

s2

0.1(s + 10,000) ; + 10,000s + 16 × 106

s1 = −2000; Io (s) =

1 = 16 × 106 ; LC

s2 = −8000;

0.1(s + 10,000) K1 K2 = + ; (s + 2000)(s + 8000) s + 2000 s + 8000

K1 =

0.1(8000) = 0.133; 6000

K2 =

0.1(2000) = −0.033; −6000

Io (s) =

0.133 0.033 − ; s + 2000 s + 8000

io (t) = [133.33e−2000t − 33.33e−8000t ]u(t) mA. P 12.35 [a]

109 1 = = 104 ; RC (4 × 103 )(25) 1 109 = = 16 × 106 ; LC (2.5)(25) Vo (s) =

40 × 106 Idc s + 10,000s + 16 × 106

40 × 106 Idc = (s + 2000)(s + 8000)

K1 =

=

120,000 (s + 2000)(s + 8000)

=

K1 K2 + . s + 2000 s + 8000

120,000 = 20; 6000

Vo (s) =

K2 =

120,000 = −20; −6000

20 20 − ; s + 2000 s + 8000

vo (t) = [20e−2000t − 20e−8000t ]u(t) V.

12–19

12–20

CHAPTER 12. Introduction to the Laplace Transform

[b] Io (s) =

3 × 10−3 s (s + 2000)(s + 8000)

=

K1 K2 + . s + 2000 s + 8000

K1 =

−(3 × 10−3 )(2000) = −10−3 ; 6000

K2 =

(3 × 10−3 )(−8000) = 4 × 10−3 ; −6000

Io (s) =

−10−3 4 × 10−3 + ; s + 2000 s + 8000

io (t) = (4e−8000t − e−2000t )u(t) mA. [c] io (0) = 4 − 1 = 3 mA. Yes. The initial inductor current is zero because there is no initial energy in the circuit, the initial resistor current is zero because the initial capacitor voltage is zero (no initial energy). Thus at t = 0 the source current appears in the capacitor. P 12.36 [a]

109 1 = = 1250 × 104 ; LC (0.8)(100) 1 106 = = 1000; RC (10)(100) 70,000 . + 1000s + 1250 × 104 ) √ = −500 ± 25 × 104 − 1250 × 104 = −500 ± j3500 rad/s;

Vo (s) = s1,2

Vo (s) =

(s2

70,000 (s + 500 − j3500)(s + 500 + j3500)

K K∗ = + ; s + 500 − j3500 s + 500 + j3500 K=

70,000 = 10/ − 90◦ ; (j7000)

Vo (s) =

10/ − 90◦ 10/90◦ + ; s + 500 − j3500 s + 500 + j3500

vo (t) = [20e−500t cos(3500t − 90◦ )]u(t) V = [20e−500t sin 3500t]u(t) V.

Problems 87,500 s(s + 500 − j3500)(s + 500 + j3500)

[b] Io (s) =

K1 K2 K2∗ + + . s s + 500 − j3500 s + 500 + j3500

=

87,500 = 7 mA; 1250 × 104 87,500 K2 = = 3.5/171.87◦ mA; (−500 + j3500)(j7000)

K1 =

io (t) = [7 + 7e−500t cos(3500t + 171.87◦ )]u(t) mA. P 12.37

1 = 2 × 106 ; C V2 (s) =

s2

1 = 4 × 106 ; LC

R = 5000; L

Ig =

30,000 . + 5000s + 4 × 106

s1 = −1000;

s2 = −4000;

V2 (s) =

30,000 (s + 1000)(s + 4000)

=

10 10 − ; s + 1000 s + 4000

v2 (t) = [10e−1000t − 10e−4000t ]u(t) V. P 12.38 [a] I1 (s) =

K2 K3 K1 + + ; s s + 5 s + 20

K1 =

(18)(10) = 1.8; (5)(20)

K3 =

(10)(−10) = −0.6; (−20)(−15)

K2 =

(18)(5) = −1.2; (−5)(15)

1.8 1.2 0.6 I1 (s) = − − ; s s + 5 s + 20 



i1 (t) = (1.8 − 1.2e−5t − 0.6e−20t )u(t) A. K2 K1 + ; s + 5 s + 20 −9 −9 K1 = = −0.6; K2 = = 0.6; 15 −15   −0.6 0.6 I2 (s) = + ; s + 5 s + 20 I2 (s) =

i2 (t) = (0.6e−20t − 0.6e−5t )u(t) A.

0.015 ; s

12–21

12–22

CHAPTER 12. Introduction to the Laplace Transform

[b] i1 (∞) = 1.8 A;

i2 (∞) = 0 A.

[c] Yes, at t = ∞ 180 = 1.8 A. 100 Since i1 is a dc current at t = ∞ there is no voltage induced in the 15 H inductor; hence, i2 = 0. Also note that i1 (0) = 0 and i2 (0) = 0. Thus our solutions satisfy the condition of no initial energy stored in the circuit. i1 =

P 12.39 From Problem 12.32: V (s) =

200s2 ; (s2 + 40s + 64)(s2 + 100)

s2 + 40s + 64 = (s + 38.33)(s + 1.67);

s2 + 100 = (s − j10)(s + j10).

Therefore V (s) =

200s2 (s + 38.33)(s + 1.67)(s − j10)(s + j10)

K2 K3 K3∗ K1 + + + . = s + 1.67 s + 38.33 s − j10 s + j10 200s2 K1 = = 0.15; 2 (s + 38.33)(s + 100) s=−1.67

K2 =

200s2 = −5.11; (s + 1.67)(s2 + 100) s=−38.33

200s2 K3 = = 2.49/ − 5.14◦ . (s + 1.67)(s + 38.33)(s + j10) s=j10

Therefore v(t) = [4.98 cos(10t − 5.14◦ ) + 0.15e−1.67t − 5.11e−38.33t ]u(t) V. ( −1

P 12.40 f (t) = L

K K∗ + s + α − jβ s + α + jβ

= Ke−αt ejβt + K ∗ e−αt e−jβt = |K|e−αt [ejθ ejβt + e−jθ e−jβt ] = |K|e−αt [ej(βt+θ) + e−j(βt+θ) ] = 2|K|e−αt cos(βt + θ).

)

Problems " n

n

P 12.41 [a] L{t f (t)} = (−1)

dn F (s) . dsn #

1 then F (s) = , s

Let f (t) = 1,

dn F (s) (−1)n n! = . dsn s(n+1)

thus

Therefore L{t } = (−1)

(−1)n n! n! = (n+1) . (n+1) s s

It follows that L{t(r−1) } =

(r − 1)! sr

" n

and L{t(r−1) e−at } =

n

#

(r − 1)! . (s + a)r

K K Ktr−1 e−at . L{tr−1 e−at } = = L (r − 1)! (s + a)r (r − 1)! (

Therefore ( −1

[b] f (t) = L

K K∗ + . (s + α − jβ)r (s + α + jβ)r )

Therefore f (t) =

Ktr−1 −(α−jβ)t K ∗ tr−1 −(α+jβ)t e + e (r − 1)! (r − 1)!

i |K|tr−1 e−αt h jθ jβt = e e + e−jθ e−jβt (r − 1)!

2|K|tr−1 e−αt = cos(βt + θ). (r − 1)! "

P 12.42 [a] F (s) =

#

6s2 + 26s + 26 K1 K2 K3 = + + ; (s + 1)(s + 2)(s + 3) s+1 s+2 s+3

K1 =

6 − 26 + 26 = 3; (1)(2)

K3 =

54 − 78 + 26 = 1; (−2)(−1)

K2 =

24 − 52 + 26 = 2; (−1)(1)

Therefore f (t) = [3e−t + 2e−2t + e−3t ] u(t). [b] F (s) =

K1 K2 K3 K4 + + + s s+2 s+4 s+6

13s3 + 134s2 + 392s + 288 K1 = = 6; (s + 2)(s + 4)(s + 6) s=0 13s3 + 134s2 + 392s + 288 K2 = = 4; s(s + 4)(s + 6) s=−2

)

12–23

12–24

CHAPTER 12. Introduction to the Laplace Transform 13s3 + 134s2 + 392s + 288 K3 = = 2; s(s + 2)(s + 6) s=−4

13s3 + 134s2 + 392s + 288 = 1; K4 = s(s + 2)(s + 4) s=−6

f (t) = [6 + 4e−2t + 2e−4t + e−6t ]u(t). 10(s2 + 119) ; (s + 5)(s2 + 10s + 169) √ = −5 ± 25 − 169 = −5 ± j12;

[c] F (s) = s1,2

F (s) =

K1 K2 K2∗ + + ; s + 5 s + 5 − j12 s + 5 + j12

K1 =

10(25 + 119) = 10; 25 − 50 + 169

K2 =

10[(−5 + j12)2 + 119] = j4.17 = 4.17/90◦ ; (j12)(j24)

Therefore f (t) = [10e−5t + 8.33e−5t cos(12t + 90◦ )] u(t)

[d] s1,2

= [10e−5t − 8.33e−5t sin 12t] u(t). √ = −7 ± 49 − 625 = −7 ± j24;

F (s) = =

56s2 + 112s + 5000 s(s + 7 − j24)(s + 7 + j24) K2 K2∗ K1 + + ; s s + 7 − j24 s + 7 + j24

K1 =

5000 = 8; 625

K2 =

56(−7 + j24)2 + 112(−7 + j24) + 5000 (−7 + j24)j48

= 24 + j7 = 25/16.26◦ ; .·. f (t) = [8 + 50e−7t cos(24t + 16.26◦ )]u(t). P 12.43 [a] F (s) = K1 =

K1 K1∗ + ; s + 6 − j8 s + 6 + j8 480 = −j30 = 30/ − 90◦ ; s + 6 + j8 s=−6+j8

f (t) = [60e−6t cos(8t − 90◦ )]u(t) = [60e−6t sin 8t]u(t).

Problems

[b] F (s) =

K1 K2 K2∗ + + ; s s + 5 − j5 s + 5 + j5

s2 + 15s + 30 K1 = 2 = 0.6; s + 10s + 50 s=0

s2 + 15s + 30 = 0.2 − j0.7 = 0.728/ − 74.0546◦ ; s(s + 5 + j5) s=−5+j5

K2 =

f (t) = [0.6 + 1.456e−5t cos(5t − 74.0546◦ )]u(t). [c] F (s) =

K2 K2∗ K1 + + ; s + 5 s + 10 − j15 s + 10 + j15

10s2 + 30s + 400 K1 = 2 = 2; s + 20s + 325 s=−5

10s2 + 30s + 400 K2 = = 4 + j4.33 = 5.9/47.3◦ ; (s + 5)(s + 10 + j15) s=−10+j15

f (t) = [2e−5t + 11.8e−10t cos(15t + 47.3◦ )]u(t). [d] F (s) =

K1∗ K2 K2∗ K1 + + + ; s + 3 − j5 s + 3 + j5 s + 6 − j4 s + 6 + j4

10(s + 3)2 K1 = = 0.833; (s + 3 + j5)(s2 + 12s + 52) s=−3+j5 10(s + 3)2 = 1.09/ − 143.13◦ ; K2 = 2 (s + 6 + j4)(s + 6s + 34) s=−6+j4

f (t) = [1.67e−3t cos 5t + 2.18e−6t cos(4t − 143.13◦ )]u(t). P 12.44 [a] F (s) =

K1 K2 K3 + + . 2 s s s + 10

8(s2 − 5s + 50) 400 K1 = = = 40; s + 10 10 s=0

d K2 = ds

(

8(s2 − 5s + 50) s + 10

)

s=0

8(s + 10)(2s − 5) − 8(s2 − 5s + 50)(1) = (s + 10)2 s=0

=

10(−40) − 8(50) = −8; 100

8(s2 − 5s + 50) 8(100 + 50 + 50) K3 = = = 16; 2 s 100 s=−10

12–25

12–26

CHAPTER 12. Introduction to the Laplace Transform

F (s) =

40 8 16 − + . 2 s s s + 10

f (t) = [40t − 8 + 16e−10t ]u(t). [b] F (s) = K0 =

K0 K1 K2 4s2 + 7s + 1 = + + ; s(s + 1)2 s (s + 1)2 s + 1 1 = 1; (1)2

K1 =

d 4s2 + 7s + 1 K2 = ds s "

=

4−7+1 = 2; −1

# s=−1

s(8s + 7) − (4s2 + 7s + 1) = s2 s=−1

1+2 = 3; 1

Therefore f (t) = [1 + 2te−t + 3e−t ] u(t). √ [c] s1,2 = −2 ± 4 − 5 = −2 ± j1 F (s) = K1 =

K1 K2 K3 K3∗ + + + . s2 s s + 2 − j1 s + 2 + j1

50 = 10; 5

d K2 = ds

(

s3 − 6s2 + 15s + 50 s2 + 4s + 5

)

s=0

(s2 + 4s + 5)(3s2 − 12s + 15) − (s3 − 6s2 + 15s + 50)(2s + 4) = (s2 + 4s + 5)2 s=0

=

5(15) − 50(4) = −5; 25

s3 − 6s2 + 15s + 50 K3 = s2 (s + 2 + j1) s=−2+j1

(−2 + j1)3 = −2 + j11; K3 =

(−2 + j1)2 = 3 − j4;

−2 + j11 − 6(3 − j4) + 15(−2 + j1) + 50 (3 − j4)(j2)

= 3 + j4 = 5/53.13◦ ; F (s) =

10 5 5/53.13◦ 5/ − 53.13◦ − + + . s2 s s + 2 − j1 s + 2 + j1

f (t) = [10t − 5 + 10e−2t cos(t + 53.13◦ )]u(t).

Problems [d] s1,2 = −1 ± F (s) =

√

1 − 5 = −1 ± j2

K1 K1∗ K2 K2∗ + + + . (s + 1 − j2)2 (s + 1 + j2)2 s + 1 − j2 s + 1 + j2

16s3 + 72s2 + 216s − 128 ; (s + 1 + j2)2 s=−1+j2

K1 =

(−1 + j2)3 = 11 − j2; K1 =

(−1 + j2)2 = −3 − j4;

176 − j32 − 216 − j288 − 216 + j432 − 128 −16

= 24 − j7 = 25/ − 16.26◦ ; d K2 = ds

(

16s3 + 72s2 + 216s − 128 (s + 1 + j2)2 s=−1+j2

)

(s + 1 + j2)2 (48s2 + 144s + 216) = (s + 1 + j2)4 s=−1+j2

−

(16s3 + 72s2 + 216s − 128)2(s + 1 + j2) (s + 1 + j2)4 s=−1+j2

=

(j4)2 (−144 − j192 − 144 + j288 + 216) − (−384 + j112)(j8) (j4)4

=

2048 + j1536 = 8 + j6 = 10/36.87◦ ; 256

25/16.26◦ 10/36.87◦ 25/ − 16.26◦ 10/ − 36.87◦ + + F (s) = + . (s + 1 − j2)2 (s + 1 + j2)2 s + 1 − j2 s + 1 + j2 f (t) = [50te−t cos(2t − 16.26◦ ) + 20e−t cos(2t + 36.87)]u(t). P 12.45 [a] F (s) =

K1 K2 K3 K3∗ + + + . s2 s s + 1 − j2 s + 1 + j2

100(s + 1) K1 = 2 = 20; s + 2s + 5 s=0 "

#

"

d 100(s + 1) 100 100(s + 1)(2s + 2) K2 = = 2 − 2 ds s + 2s + 5 s + 2s + 5 (s2 + 2s + 5)2 = 20 − 8 = 12; 100(s + 1) K3 = 2 = −6 + j8 = 10/126.87◦ ; s (s + 1 + j2) s=−1+j2

f (t) = [20t + 12 + 20e−t cos(2t + 126.87◦ )]u(t).

# s=0

12–27

12–28

CHAPTER 12. Introduction to the Laplace Transform

[b] F (s) =

K1 K2 K3 K4 + + + . 3 2 s (s + 1) (s + 1) s+1

40(s + 2) K1 = = 80; (s + 1)3 s=0

40(s + 2) K2 = = −40; s s=−1

"

#

"

40 40(s + 2) d 40(s + 2) = − K3 = ds s s s2 "

1 d 40 40(s + 2) K4 = − 2 ds s s2

#

= −40 − 40 = −80; s=−1

#

"

1 −40 40 80(s + 2) = − 2 + 2 s2 s s3

# s=−1

1 = (−40 − 40 − 80) = −80; 2

f (t) = [80 − 20t2 e−t − 80te−t − 80e−t ]u(t). [c] F (s) =

s+8 5s2 + 29s + 32 5s2 + 29s + 32 = ; =5− 2 (s + 2)(s + 4) s + 6s + 8 (s + 2)(s + 4)

s+8 K1 K2 = + (s + 2)(s + 4) s + 2 s + 4; K1 =

−2 + 8 = 3; 2

K2 =

−4 + 8 = −2. −2

Therefore, F (s) = 5 −

2 3 + ; s+2 s+4

f (t) = 5δ(t) + [−3e−2t + 2e−4t ]u(t). [d] F (s) =

4(s + 1) 4 2s3 + 8s2 + 2s − 4 = 2s − 2 + = 2s − 2 + ; 2 s + 5s + 4 (s + 1)(s + 4) s+4

f (t) = 2

dδ(t) − 2δ(t) + 4e−4t u(t). dt

480 480 = ; + 12s + 100 (s + 6 + j8)(s + 6 − j8) This function has no zeros and complex conjugate poles at −6 ± j8.

P 12.46 F (s) =

s2

Problems

F (s) =

=

12–29

10(s + 3)2 (s2 + 6s + 34)(s2 + 12s + 52)

10(s + 3)2 ; (s + 3 + j5)(s + 3 − j5)(s + 6 + j4)(s + 6 − j4)

This function has two zeros at −3, and complex conjugate poles at −3 ± j5 and −6 ± j4. 100(s + 1) 100(s + 1) = ; s2 (s2 + 2s + 5) s2 (s + 1 + j2)(s + 1 − j2) This function has a zero at −1, two poles at 0, and complex conjugate poles at −1 ± j2.

P 12.47 F (s) =

F (s) =

40(s + 2) ; s(s + 1)3

This function has a zero at −2, a pole at 0 and three poles at −1. P 12.48 sIo (s) =

s2

Idc s[s + (1/RC)] ; + (1/RC)s + (1/LC)

lim sIo (s) = 0,

s→0

lim sIo (s) = Idc ,

s→∞

P 12.49 sIo (s) =

s2

lim sIo (s) = 0, lim sIo (s) = Idc ,

s→∞

lim sVo (s) = 0, lim sVo (s) = 0,

s→∞

s2

.·. vo (∞) = 0; .·. vo (0+ ) = 0;

Vdc /RLC Vdc = , 1/LC R

lim sIo (s) = 0,

s→∞

.·. io (0+ ) = Idc .

Vdc /RLC ; + (1/RC)s + (1/LC)

lim sIo (s) =

s→0

.·. io (∞) = 0;

sVdc /RC ; + (1/RC)s + (1/LC)

s2

s→0

sIo (s) =

.·. io (0+ ) = Idc .

Idc s[s + (1/RC)] ; + (1/RC)s + (1/LC)

s→0

P 12.50 sVo (s) =

.·. io (∞) = 0;

Vdc .·. io (∞) = ; R

.·. io (0+ ) = 0.

12–30

CHAPTER 12. Introduction to the Laplace Transform

200s3 = 0; P 12.51 [a] lim sV (s) = lim 4 s→∞ s→∞ s [1 + (40/s) + (64/s2 )][1 + (100/s2 )] #

"

Therefore v(0+ ) = 0. [b] Yes, all of the poles of V are in the left-half of the complex plane. Therefore, 200s3 lim sV (s) = lim = 0; s→0 s→0 (s2 + 40s + 64)(s2 + 100) "

#

Therefore v(∞) = 0. P 12.52 [a] sF (s) =

6s3 + 26s2 + 26s ; (s + 1)(s + 2)(s + 3)

lim sF (s) = 0,

s→0

lim sF (s) = 6,

s→∞

.·. f (∞) = 0; .·. f (0+ ) = 6.

13s3 + 134s2 + 392s + 288 [b] sF (s) = ; (s + 2)(s2 + 10s + 24) lim sF (s) = 6,

s→0

lim sF (s) = 13,

s→∞

[c] sF (s) =

lim sF (s) = 0, lim sF (s) = 10,

s→∞

s→0

.·. f (0+ ) = 10.

5000 = 8, 625

lim sF (s) = 56,

s→∞

s2

.·. f (∞) = 8;

.·. f (0+ ) = 56.

480s ; + 12s + 100

lim sF (s) = 0,

s→0

lim sF (s) = 0,

s→∞

.·. f (∞) = 0;

56s2 + 112s + 5000 ; s2 + 14s + 625

lim sF (s) =

P 12.53 [a] sF (s) =

.·. f (0+ ) = 13.

10s(s2 + 119) ; (s + 5)(s2 + 10s + 169)

s→0

[d] sF (s) =

.·. f (∞) = 6;

.·. f (∞) = 0; .·. f (0+ ) = 0.

Problems

12–31

s2 + 15s + 30 ; s2 + 10s + 50 lim sF (s) = 0.6, .·. f (∞) = 0.6;

[b] sF (s) = s→0

lim sF (s) = 1,

s→∞

[c] sF (s) =

.·. f (0+ ) = 1.

10s3 + 30s2 + 400s ; (s + 5)(s2 + 20s + 325)

lim sF (s) = 0,

s→0

.·. f (∞) = 0; .·. f (0+ ) = 10.

lim sF (s) = 10,

s→∞

[d] sF (s) =

10s(s + 3)2 ; (s2 + 6s + 34)(s2 + 12s + 52)

lim sF (s) = 0,

s→0

lim sF (s) = 0,

s→∞

.·. f (∞) = 0; .·. f (0+ ) = 0.

8(s2 − 5s + 50) . s(s + 10) F (s) has a second-order pole at the origin so we cannot use the final value theorem here.

P 12.54 [a] sF (s) =

lim sF (s) = 8,

s→∞

[b] sF (s) =

.·. f (0+ ) = 8.

4s2 + 7s + 1 ; (s + 1)2

lim sF (s) = 1,

s→0

lim sF (s) = 4,

s→∞

.·. f (∞) = 1; .·. f (0+ ) = 4.

s3 − 6s2 + 15s + 50 ; s(s2 + 4s + 5) F (s) has a second-order pole at the origin so we cannot use the final value theorem here.

[c] sF (s) =

lim sF (s) = 1

s→∞

[d] sF (s) =

16s4 + 72s3 + 216s2 − 128s . (s2 + 2s + 5)2

lim sF (s) = 0,

s→0

lim sF (s) = 16,

s→∞

.·. f (0+ ) = 1.

.·. f (∞) = 0; .·. f (0+ ) = 16.

12–32

CHAPTER 12. Introduction to the Laplace Transform 100(s + 1) ; s(s2 + 2s + 5) F (s) has a second-order pole at the origin so we cannot use the final value theorem here.

P 12.55 [a] sF (s) =

lim sF (s) = 0,

s→∞

[b] sF (s) =

.·. f (0+ ) = 0.

40(s + 2) ; (s + 1)3

lim sF (s) = 80,

.·. f (∞) = 80;

lim sF (s) = 0,

.·. f (0+ ) = 0.

s→0

s→∞

[c] This F (s) function is an improper rational function, and thus the corresponding f (t) function contains impulses (δ(t)). Neither the initial value theorem nor the final value theorem may be applied to this F (s) function! [d] This F (s) function is an improper rational function, and thus the corresponding f (t) function contains impulses (δ(t)). Neither the initial value theorem nor the final value theorem may be applied to this F (s) function! P 12.56 [a] ZL = j120π(0.01) = j3.77 Ω;

ZC =

−j = −j26.526 Ω. 120π(100 × 10−6 )

The phasor-transformed circuit is

IL =

1 = 36.69/56.61◦ mA; 15 + j3.77 − j26.526

.·. iL(ss) (t) = 36.69 cos(120πt + 56.61◦ ) mA. [b] The steady-state response is the second term in the expression for iL (t) derived in the Practical Perspective. This matches the steady-state response just derived in part (a). P 12.57 The transient and steady-state components are both proportional to the magnitude of the input voltage. Therefore, K=

40 = 0.947. 42.26

Problems

12–33

So if we make the amplitude of the sinusoidal source 0.947 instead of 1, the current will not exceed the 40 mA limit. A plot of the current through the inductor is shown below with the amplitude of the sinusoidal source set at 0.947.

P 12.58 We begin by using the Laplace transform of the describing differential equation for the circuit in Fig. 12.19. Change the right-hand side so it is the Laplace transform of Kte−500t to give: 15IL (s) + 0.01sIL (s) + 104

K IL (s) = . s (s + 500)2

Solving for IL (s), 100Ks K1 K1∗ IL (s) = 2 = + (s + 1500s + 106 )(s + 500)2 s + 750 − j661.44 s + 750 + j661.44 +

K3 K2 + . 2 (s + 500) s + 500

100Ks = 0.1512K / − 172.82◦ mA; K1 = 2 (s + 750 + j661.44)(s + 500) s=−750+j661.44

K2 =

100Ks = −100K mA; 2 6 (s + 1500s + 10 ) s=−500 "

100Ks d K3 = 2 ds (s + 1500s + 106 )

#

= 0.3K mA. s=−500

Therefore, iL (t) = K[0.3024e−750t cos(661.44t − 172.82◦ ) − 100te−500t + 0.3e−500t ]u(t) mA.

12–34

CHAPTER 12. Introduction to the Laplace Transform

Plot the expression above with K = 1:

The maximum value of the inductor current is 0.0242K mA. Therefore, K=

40 = 1652.9. 0.0242

So the inductor current rating will not be exceeded if the input to the RLC circuit is 1652te−500t V.

The Laplace Transform in Circuit Analysis

Assessment Problems AP 13.1 [a] Y = Z=

1 1 C[s2 + (1/RC)s + (1/LC)] + + sC = ; R sL s s/C 4 × 106 s 1 = 2 = 2 . Y s + (1/RC)s + (1/LC) s + 2000s + 64 × 104

[b] zero at z1 = 0, poles at −p1 = −400 rad/s and −p2 = −1600 rad/s. AP 13.2 [a]

2.5s −150 Vo = 5 (16 × 10 )/s + 5000 + 2.5s s 

= = [b] Vo =

s2

−150s + 2000s + 64 × 104

−150s . (s + 400)(s + 1600)

K1 K2 + . s + 400 s + 1600

−150s K1 = = 50; s + 1600 s=−400

13–1



13–2

CHAPTER 13. The Laplace Transform in Circuit Analysis −150s K2 = = −200; s + 400 s=−1600

Vo =

200 50 − . s + 400 s + 1600

vo (t) = (50e−400t − 200e−1600t )u(t) V. AP 13.3 [a]

V V 0.0384 + = ; 6 1250 (s/10) + (2.5 × 10 /s) s Therefore, V = [b] I =

48(s2 + 25 × 106 ) . s(s + 2500)(s + 10,000)

480 V = . 6 (s/10) + (2.5 × 10 /s) (s + 2500)(s + 10,000)

[c] v(t) = (48 − 80e−2500t + 80e−10,000t )u(t) V. [d] i(t) = (64e−2500t − 64e−10,000t )u(t) mA. AP 13.4 Begin by transforming the circuit from the time domain to the s domain:

Io =

=

5 s2 + 502 400 4 + 0.24s + s

=

(s2

+

502 )(s2

20.833s + 16.67s + 1666.67)

K1 K1∗ K2 K2∗ + + + . s − j50 s + j50 s + 8.33 − j39.965 s + 8.33 + j39.965

Problems

20.833s = 0.00884/ − 135◦ ; K1 = (s + j50)(s2 + 16.67s + 1666.67) s=j50

20.833s K2 = 2 = 0.00903/46.194◦ . 2 (s + 50 )(s + 8.33 + j39.965) s=−8.33+j39.965

Therefore, io (t) = [17.68 cos(50t − 135◦ ) + 18.06e−8.33t cos(39.965t + 46.194◦ )] mA. AP 13.5 [a]

0 = 0.5s(I1 − 30/s) +

2500 (I1 − I2 ) + 100I1 ; s

2500 −375 = (I2 − I1 ) + 50(I2 − 30/s) s s or (s2 + 200s + 5000)I1 − 5000I2 = 30s; −50I1 + (s + 50)I2 = 22.5. ∆=

2 (s

N1 =

I1 =

+ 200s + 5000)

30s 22.5

−50 −5000 (s + 50)

−5000 (s + 50)

= s(s + 100)(s + 150);

= 30(s2 + 50s + 3750);

N1 30(s2 + 50s + 3750) = . ∆ s(s + 100)(s + 150)

13–3

13–4

CHAPTER 13. The Laplace Transform in Circuit Analysis

N2 =

2 (s

+ 200s + 5000) −50

30s 22.5

= 22.5s2 + 6000s + 112,500;

N2 22.5s2 + 6000s + 112,500 I2 = = . ∆ s(s + 100)(s + 150) [b] I1 =

30(s2 + 50s + 3750) K1 K2 K3 = + + . s(s + 100)(s + 150) s s + 100 s + 150

K1 = 7.5;

K2 = −52.5;

K3 = 75;

i1 (t) = [7.5 − 52.5e−100t + 75e−150t ]u(t) A. I2 =

22.5s2 + 6000s + 112,500 K1 K2 K3 = + + . s(s + 100)(s + 150) s s + 100 s + 150

K1 = 7.5;

K2 = 52.5;

K3 = −37.5;

i2 (t) = [7.5 + 52.5e−100t − 37.5e−150t ]u(t) A. AP 13.6

VTh =

10s 40 400 40 · = = ; 10s + 1000 s 10s + 1000 s + 100

ZTh = 1000 + 1000k10s = 1000 +

I=

10,000s 2000(s + 50) = . 10s + 1000 s + 100

40/(s + 100) 40s = (5 × 105 )/s + 2000(s + 50)/(s + 100) 2000s2 + 600,000s + 5 × 107 0.02s K1 K1∗ = 2 = + . s + 300s + 25,000 s + 150 − j50 s + 150 + j50

Problems 0.02s K1 = = 31.62 × 10−3 /71.57◦ ; s + 150 + j50 s=−150+j50

i(t) = 63.25e−150t cos(50t + 71.57◦ )u(t) mA. AP 13.7 [a] s-domain equivalent circuit is

Note:

i2 (0+ ) = −

20 = −2 A. 10

24 = (120 + 3s)I1 + 3sI2 + 6 s 0 = −6 + 3sI1 + (360 + 15s)I2 + 36. In standard form, (s + 40)I1 + sI2 = (8/s) − 2; sI1 + (5s + 120)I2 = −10. ∆=

s + 40 s

(8/s) − 2 −10

N1 =

I1 = [b] I1 =

s 5s + 120

= 4(s + 20)(s + 60);

s 5s + 120

−200(s − 4.8) ; s

=

N1 −50(s − 4.8) = . ∆ s(s + 20)(s + 60)

K1 K2 K3 + + . s s + 20 s + 60 −50(−20) + 240 = −1.55; (−20)(40)

K1 =

240 = 0.2; 1200

K3 =

−50(−60) + 240 = 1.35; (−60)(−40)

K2 =

i1 (t) = [0.2 − 1.55e−20t + 1.35e−60t ]u(t) A.

13–5

13–6

CHAPTER 13. The Laplace Transform in Circuit Analysis

AP 13.8 [a] The s-domain circuit with the voltage source acting alone is

V0 V 0 − 60/s V 0 s + + = 0; 10 80 20 + 10s .·. V 0 =

480(s + 2) K1 K2 K3 = + + . s(s + 4)(s + 6) s s+4 s+6

K1 =

480(2) = 40; (4)(6)

V0 =

120 −160 40 + + ; s s+4 s+6

K2 =

480(−2) = 120; (−4)(2)

K3 =

480(−4) = −160; (−6)(−2)

.·. v 0 (t) = [40 + 120e−4t − 160e−6t ]u(t) V. [b] The s-domain circuit with the current source acting alone is

V 00 V 00 s V 00 − 30/s + + = 0; 10 80 10(s + 2) .·. V 00 =

240 K1 K2 K3 = + + . s(s + 4)(s + 6) s s+4 s+6

K1 =

240 = 10; (4)(6)

V 00 =

10 −30 20 + + ; s s+4 s+6

K2 =

240 = −30; (−4)(2)

.·. v 00 (t) = [10 − 30e−4t + 20e−6t ]u(t) V.

K3 =

240 = 20; (−6)(−2)

Problems

13–7

[c] v(t) = v 0 (t) + v 00 (t) = [50 + 90e−4t − 140e−6t ]u(t) V. AP 13.9 [a]

Vo s Vo + = Ig . 0.05s + 500 12.5 × 106 Therefore,

Vo 12.5 × 106 (s + 10,000) = H(s) = 2 . Ig s + 10,000s + 250 × 106

[b] −z1 = −10,000 rad/s; −p1 = −5000 + j15,000 rad/s; AP 13.10 [a] Vo = =

−p2 = −5000 − j15,000 rad/s.

12.5 × 106 (s + 10,000) 1 · s2 + 10,000s + 250 × 106 s

Ko K1 K1∗ + + ; s s + 5000 − j15,000 s + 5000 + j15,000

Ko = 500;

K1 = 1250/3/ − 126.87◦ ;

K1∗ = 1250/3/126.87◦ ;

vo = [500 + (2500/3)e−5000t cos(15,000t − 126.87◦ )]u(t) V. 12.5 × 106 (s + 10,000) K2 K2∗ [b] Vo = 2 ·1= + ; s + 10,000s + 250 × 106 s + 5000 − j15,000 s + 5000 + j15,000 K2 = 6.59 × 106 / − 18.43◦ ;

K2∗ = 6.59 × 106 /18.43◦ ;

vo = [13.2e−5000t cos(15,000t − 18.43◦ )]u(t) MV. AP 13.11 [a] H(s) = L{h(t)} = L{vo (t)}. √ √ vo (t) = ( 1000/3) cos θe−t cos 3t − ( 1000/3) sin θe−t sin 3t = 10e−t cos 3t + (10/3)e−t sin 3t. Therefore H(s) = = [b] Vo =

10(s + 1) (10/3)(3) + (s + 1)2 + (3)2 (s + 1)2 + (3)2 10(s + 2) . + 2s + 10

s2

10(s + 2) 1 Ko K1 K1∗ · = + + ; s2 + 2s + 10 s s s + 1 − j3 s + 1 + j3

Ko = 2;

K1 = 5/3/ − 126.87◦ ;

K1∗ = 5/3/126.87◦ ;

vo = [2 + (10/3)e−t cos(3t − 126.87◦ )]u(t) V.

13–8

CHAPTER 13. The Laplace Transform in Circuit Analysis

AP 13.12 From Assessment Problem 13.9: H(s) =

12.5 × 106 (s + 10,000) . s2 + 10,000s + 250 × 106

Therefore H(j20,000) =

12.5 × 106 (10,000 + j20,000) = 1118.034/ − 63.43◦ . 250 × 106 − 400 × 106 + j200 × 106

Thus, vo = (0.01)(1118.034) cos(20,000t − 63.43◦ ) = 11.8 cos(20,000t − 63.43◦ ) V. AP 13.13 [a]

Vp =

0.01s s Vg = Vg ; 80 + 0.01s s + 8000

Vn = Vp ; Vn Vn − Vo + + (Vn − Vo )8 × 10−9 s = 0; 5000 25,000 5Vn + Vn − Vo + (Vn − Vo )2 × 10−4 s = 0; 6Vn + 2 × 10−4 sVn = Vo + 2 × 10−4 sVo ; 2 × 10−4 Vn (s + 30,000) = 2 × 10−4 Vo (s + 5000); s + 30,000 s + 30,000 Vo = Vn = s + 5000 s + 5000 

H(s) =



Vo s(s + 30,000) = . Vg (s + 5000)(s + 8000)

sVg ; s + 8000 

Problems

[b] vg = 0.6u(t);

Vg =

0.6 ; s

Vo =

0.6(s + 30,000) K1 K2 = + ; (s + 5000)(s + 8000) s + 5000 s + 8000

K1 =

0.6(25,000) = 5; 3000

K2 =

0.6(22,000) = −4.4; −3000

.·. vo (t) = (5e−5000t − 4.4e−8000t )u(t) V. [c] Vg = 2 cos 10,000t V; H(jω) = .·.

j10,000(30,000 + j10,000) = 2.21/ − 6.34◦ ; (5000 + j10,000)(8000 + j10,000)

vo = 4.42 cos(10,000t − 6.34◦ ) V.

13–9

13–10

CHAPTER 13. The Laplace Transform in Circuit Analysis

Problems P 13.1 P 13.2

1Zt i= vdτ + I0 ; L 0−

1 therefore I = L 



V s



+

I0 V I0 = + . s sL s

−LI0 −I0 = ; ZN = sL. sL s Therefore, the Norton equivalent is the same as the circuit in Fig. 13.4. IN =



P 13.3

VTh = Vab = CV0

P 13.4

[a]

Z=

1 sC



=

V0 ; s

ZTh =

1 . sC

(R + sL)(1/sC) (1/C)(s + R/L) = 2 ; R + sL + (1/sC) s + (R/L)s + (1/LC)

1000 R = = 2000; L 0.5

1 106 = = 5 × 106 ; LC 0.2

2.5 × 106 (s + 2000) Z= 2 . s + 2000s + 5 × 106 √ [b] s1,2 = −1000 ± 106 − 5 × 106 = −1000 ± j2000; Z=

2.5 × 106 (s + 2000) ; (s + 1000 − j2000)(s + 1000 + j2000)

−z1 = −2000 rad/s;

−p1 = −1000 + j2000 rad/s;

−p2 = −1000 − j2000 rad/s. P 13.5

[a] Z = R + sL +

1 L[s2 + (R/L)s + (1/LC)] = sC s

5[s2 + 2000s + 107 ] . s √ [b] s1,2 = −1000 ± 106 − 107 = −1000 ± j3000 rad/s. Zeros at −1000 + j3000 rad/s and −1000 − j3000 rad/s; Pole at 0. =

Problems

P 13.6

[a] Z = 2000 +

13–11

1 4 × 107 s = 2000 + 2 Y s + 80,000s + 25 × 108

2000(s2 + 105 s + 25 × 108 ) 2000(s + 50,000)2 = . s2 + 80,000s + 25 × 108 s2 + 80,000s + 25 × 108 [b] −z1 = −z2 = −50,000 rad/s; =

−p1 = −40,000 − j30,000 rad/s; −p2 = −40,000 + j30,000 rad/s. P 13.7

Z1 = 0.5s +

2s + 50 s2 + 75s + 1350 1 + 2 = ; 25 s + 25s + 100 25(s2 + 25s + 100

Yab =

Zab

s2 + 25s + 100 2(50/s) = ; (2 + 50/s) 2s + 50

25(s2 + 25s + 100) 25(s + 5)(s + 20) = 2 = . s + 75s + 1350 (s + 30)(s + 45)

Zeros at −5 rad/s and −20 rad/s; poles at −30 rad/s and −45 rad/s. P 13.8

Transform the Y-connection of the two resistors and the inductor into the equivalent delta-connection:

where Za =

(s)(1) + (1)(s) + (1)(1) 2s + 1 = ; s s

Zb = Zc =

(s)(1) + (1)(s) + (1)(1) = 2s + 1. 1

Then Zab = Za k[(1/skZc ) + (1/skZb )] = Za k2(1/skZb ); 1/skZb =

1 (2s + 1) s 1 + 2s + 1 s

=

2s + 1 ; +s+1

2s2

13–12

CHAPTER 13. The Laplace Transform in Circuit Analysis 

Zab =

2s + 1 2(2s + 1) k 2 s 2s + s + 1 

2(2s + 1)2 2 = = . 2 (2s + 1)(2s + s + 1) + 2s(2s + 1) s+1 No zeros; one pole at −1 rad/s. P 13.9

[a]

Vo =

(Ig /s2 C)(sL) Ig /C = 2 ; R + sL + (1/sC) s + (R/L)s + (1/LC)

9.6 × 10−3 Ig = = 3000 × 103 = 3 × 106 ; −9 C 3.2 × 10 R 7000 = = 14,000; L 0.5

1 2 = × 109 = 625 × 106 ; LC 3.2

Vo =

3 × 106 . s2 + 14,000s + 625 × 106

[b] sVo =

3 × 106 s ; s2 + 14,000s + 625 × 106 .·. vo (∞) = 0;

lim sVo = 0;

s→0

.·. vo (0+ ) = 0.

lim sVo = 0;

s→∞

[c] s1,2 = −7000 ±

√

49 × 106 − 625 × 106 = −7000 ± j24,000 rad/s;

Vo =

3,000,000 ; (s + 7000 − j24,000)(s + 7000 + j24,000)

K1 =

3 × 106 = −j62.5 = 62.5/ − 90◦ ; j48,000

vo = 125e−7000t cos(24,000t − 90◦ ) = [125e−7000t sin 24,000t]u(t) V.

Problems

P 13.10 IC =

Ig Vo − s sL

9.6 × 10−3 2 3 × 106 = − s s (s + 7000 − j24,000)(s + 7000 + j24,000) "

#

=

6 × 106 9.6 × 10−3 − s s(s + 7000 − j24,000)(s + 7000 + j24,000)

=

9.6 × 10−3 K1 K2 K2∗ − − − . s s s + 7000 − j24,000 s + 7000 + j24,000

K1 =

6 × 106 = 9.6 × 10−3 ; 6 625 × 10

K2 =

6 × 106 (−7000 + j24,000)(j48,000)

=

6 = 5 × 10−3 /163.74◦ ; (−7 + j24)(j48)

9.6 × 10−3 9.6 × 10−3 .·. IC = − − [conjugate terms] s s −5/163.74◦ = + conjugate s + 7000 − j24,000

#

5/ − 16.26◦ = + conjugate s + 7000 − j24,000

#

"

× 10−3

"

× 10−3 .

iC = 10e−7000t cos(24,000t − 16.26◦ )u(t) mA. Check: iC (0+ ) = 10 cos(−16.26◦ ) = 9.6 mA (ok); iC (∞) = 0 (ok). P 13.11 [a] At t = 0− ,

0.2v1 = (0.8)v2 ;

v1 = 4v2 ;

Therefore v1 (0− ) = 80 V = v1 (0+ );

v1 + v2 = 100 V;

v2 (0− ) = 20 V = v2 (0+ ).

13–13

13–14

CHAPTER 13. The Laplace Transform in Circuit Analysis

[b] I =

(80/s) + (20/s) 20 × 10−3 ; = 5000 + [(5 × 106 )/s] + (1.25 × 106 /s) s + 1250

80 5 × 106 − V1 = s s

20 × 10−3 s + 1250

20 1.25 × 106 V2 = − s s [c] i = 20e−1250t u(t) mA;

!

=

20 × 10−3 s + 1250

80 ; s + 1250

!

=

20 . s + 1250

v1 = 80e−1250t u(t) V;

v2 = 20e−1250t u(t) V. P 13.12 [a] For t < 0:

vC (0− ) vC (0− ) − 120 + = 0; 30 50 8vC (0− ) = 2160; .·. vC (0− ) = 270 V;

−12 +

io (0− ) =

270 − 120 = 3 A. 50

Problems

[b] Io =

(270/s) + 0.06 − (120/s) 40 + 0.02s + (106 /5s)

=

3(s + 2500) s2 + 2000s + 107

=

3(s + 2500) ; (s + 1000 − j3000)(s + 1000 + j3000)

K1 =

√ 3(1500 + j3000) = 0.75 5/ − 26.57◦ . j6000

[c] io (t) = 3.35e−1000t cos(3000t − 26.57◦ )u(t) A. P 13.13

Vo =

(30/s)(5 × 106 /s) 1500 + 0.1s + (5 × 106 /s)

=

1500 × 106 s(s2 + 15,000s + 50 × 106 )

=

1500 × 106 s(s + 5000)(s + 10,000)

=

K1 K2 K3 + + . s s + 5000 s + 10,000

K1 =

1500 × 106 = 30; (5000)(10,000)

K2 =

1500 × 106 = −60; (−5000)(5000)

K3 =

1500 × 106 = 30; (−5000)(−10,000)

Vo =

30 60 30 − + ; s s + 5000 s + 10,000

vo (t) = [30 − 60e−5000t + 30e−10,000t ]u(t) V.

13–15

13–16

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.14 Since we already have the solution for vo (t) when the initial voltage is zero, we will use superposition to determine the contribution of the initial voltage of −20 V. Vo1 = output when vC (0) = 0; Vo2 = output when vC (0) = −20 V.

4 × 10−6 +

.·. Vo2 = =

Vo2 s Vo2 + = 0; 6 5 × 10 1500 + 0.1s

s2

−20(s + 15,000) + 15,000s + 50 × 106

K2 K1 + . s + 5000 s + 10,000

K1 =

−20(10,000) = −40; 5000

K2 =

−20(5000) = 20; −5000

Vo2 =

−40 20 + . s + 5000 s + 10,000

From the solution to Problem 13.13 we have Vo1 =

30 60 30 − + ; s s + 5000 s + 10,000

Vo = Vo1 + Vo2 ; 30 100 50 .·. Vo = − + ; s s + 5000 s + 10,000 vo (t) = [30 − 100e−5000t + 50e−10,000t ]u(t) V.

Problems P 13.15 [a] io (0− ) =

13–17

48 × 10−3 = 12 mA = ρ. 4

Vo − 48/s Vo + ρL + = 0; (1/sC) R + sL .·. Vo =

48(s + R/L) − ρ/C . + (R/L)s + (1/LC)

s2

When the numerical values are substituted we get Vo =

48(s + 4875) ; (s + 4000 − j3000)(s + 4000 + j3000)

K1 =

48(875 + j3000) = 25/ − 16.26◦ ; j6000

vo (t) = 50e−4000t cos(3000t − 16.26◦ )u(t) V. Check: vo (0+ ) = 50 cos(−16.26◦ ) = 48 V, which agrees with the fact that the initial capacitor voltage is zero. ρ[s + (48/ρL)] 48/s + ρL [b] Io = = 2 ; R + sL + (1/sC) s + (R/L)s + (1/LC) 12 × 10−3 (s + 8000) Io = ; (s + 4000 − j3000)(s + 4000 + j3000) 12 × 10−3 (4000 + j3000) = 10 × 10−3 / − 53.13◦ ; K1 = j6000 io (t) = 20e−4000t cos(3000t − 53.13◦ )u(t) mA.

13–18

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.16 [a]

The two node voltage equations are 5 V2 V2 − V1 V2 − (15/s) V1 − V2 + V1 s = and + + = 0. s s 3 s 15 Solving for V1 and V2 yields V1 =

5(s + 3) , 2 s(s + 2.5s + 1)

2.5(s2 + 6) . s(s2 + 2.5s + 1)

V2 =

[b] The partial fraction expansions of V1 and V2 are 15 50/3 5/3 15 125/6 25/3 − + and V2 = − + . s s + 0.5 s + 2 s s + 0.5 s + 2 It follows that   50 −0.5t 5 −2t + e u(t) V and v1 (t) = 15 − e 3 3

V1 =

125 −0.5t 25 −2t e + e u(t) V. v2 (t) = 15 − 6 3 



[c] v1 (0+ ) = 15 −

50 5 + = 0; 3 3

v2 (0+ ) = 15 − [d] v1 (∞) = 15 V;

125 25 + = 2.5 V. 6 3 v2 (∞) = 15 V.

P 13.17 [a]

I=

Vdc /s Vdc /L = 2 ; R + sL + (1/sC) s + (R/L)s + (1/LC)

Vdc = 40; L

R = 1.2; L

1 = 1.0; LC

Problems

I=

13–19

40 K1 K1∗ = + ; (s + 0.6 − j0.8)(s + 0.6 + j0.8) s + 0.6 − j0.8 s + 0.6 + j0.8

K1 =

40 = −j25 = 25/ − 90◦ ; j1.6

K1∗ = 25/90◦ .

[b] i = 50e−0.6t cos(0.8t − 90◦ ) = [50e−0.6t sin 0.8t]u(t) A. 160s [c] V = sLI = (s + 0.6 − j0.8)(s + 0.6 + j0.8) K1 K1∗ = + ; s + 0.6 − j0.8 s + 0.6 + j0.8 K1 =

160(−0.6 + j0.8) = 100/36.87◦ . j1.6

[d] v(t) = [200e−0.6t cos(0.8t + 36.87◦ )]u(t) V. P 13.18 [a]

−3 V1 − Vo V1 − (15/s) + + = 0; s s 4 Vo Vo − V1 Vo − (15/s) + + = 0. 2 s 2/s Simplfying, (s + 4)V1 − 4Vo = 27; (s2 + s + 2)Vo − 2V1 = 15s; ∆=

s + 4 −2 s2

−4 + s + 2

= s(s + 2)(s + 3; )

N2 =

s + 4 −2



Vo =

N2 15s2 + 60s + 54 K1 K2 K3 = = + + . ∆ s(s + 2)(s + 3) s s+2 s+3

27

15s

= 15s2 + 60s + 54;

13–20

CHAPTER 13. The Laplace Transform in Circuit Analysis 60 − 120 + 54 = 3; (−2)(1)

K1 =

54 = 9; (2)(3)

K3 =

135 − 180 + 54 = 3; (−3)(−1)

K2 =

9 3 3 .·. Vo = + + . s s+2 s+3 [b] vo (t) = (9 + 3e−2t + 3e−3t )u(t) V. [c] At t = 0+ :

vo (0+ ) = 15 V(checks). At t = ∞:

vo (∞) vo (∞) − 15 −3+ = 0; 2 4 .·. 3vo (∞) = 27; .·. vo (∞) = 9 V(checks). P 13.19 For t < 0:

vo (0− ) + 150 vo (0− ) vo (0− ) + + = 0; 4000 15,000 10,000 .·. vo (0− ) = −90 V;

.·. iL (0− ) = −6 mA.

Problems For t > 0 :

Vo (Vo + 90/s)s Vo − 30 × 10−3 + 4+ = 0; 5s + 15,000 10 50 × 106 Vo =

s2

30(1000 − 3s) + 8000s + 25 × 106

30(1000 − 3s) ; (s + 4000 − j3000)(s + 4000 + j3000)

=

K1 =

30(1000 + 12,000 − j9000) = 79.06/ − 124.70◦ V; j6000

vo (t) = 158.11e−4000t cos(3000t − 124.70◦ )u(t) V. Check: vo (0) = 158.11 cos(−124.70◦ ) = −90 V. P 13.20 [a]

Vo − 300/s 12.5Vo Vo s + + = 0; 4000 s 21,000s + 2 × 108 .·. Vo =

12(21s + 20 × 104 ) K1 K2 = + . (s + 10,000)(s + 40,000) s + 10,000 s + 40,000

K1 = −4; Vo =

K2 = 256;

−4 256 + ; s + 10,000 s + 40,000

vo (t) = (256e−40,000t − 4e−10,000t )u(t) V.

13–21

13–22

CHAPTER 13. The Laplace Transform in Circuit Analysis

[b] Io =

Vo 12.5Vo = ; 0.08s s

Io =

150(21s + 20 × 104 ) K1 K2 K3 = + + . s(s + 10,000)(s + 40,000) s s + 10,000 s + 40,000

K1 = 75 × 10−3 ;

K2 = 5 × 10−3 ;

K3 = −80 × 10−3 ;

io (t) = (75 + 5e−10,000t − 80e−40,000t )u(t) mA. [c] At t = 0+ the circuit is

300 (21) = 252 V; .·. vo (0+ ) = 25

io (0+ ) = 0.

Both values agree with our solutions for vo and io . At t = ∞ the circuit is

.·. vo (∞) = 0;

io (∞) = 75 mA.

Both values agree with our solutions for vo and io . P 13.21 [a] For t < 0:

iL (0− ) = i1 =

500 = 5 A; 30 + 100k(50 + 100) + 10

5(100) = 2 A; 250

Problems vC (0− ) = 500 − 5(30) − 2(100) = 500 − 350 = 150 V. For t > 0:

[b]

Vo + 0.01 Vo − 150/s + = 0; 10 + 0.002s 40 + 2 × 105 /s 

Vo

150 0.01 1 s − + ; = 5 5 10 + 0.002s 40s + 2 × 10 40s + 2 × 10 10 + 0.002s

Vo = Vo =



s2

−50(s + 5000) −50(s + 5000) ; = 8 + 25,000s + 10 (s + 5000)(s + 20,000)

−50 . s + 20,000

[c] vo (t) = −50e−20,000t u(t) V. P 13.22 [a] For t < 0:

1 1 1 1 1 = + + = ; Re 12.5 50 10 5

Re = 5 kΩ;

v1 = −20(5) = −100 V; iL (0− ) =

−100 × 10−3 = −10 mA; 10

vC (0− ) = −v1 = 100 V.

13–23

13–24

CHAPTER 13. The Laplace Transform in Circuit Analysis For t = 0+ :

s-domain circuit:

where R = 10 kΩ; L = 4 H; [b]

C = 10 nF; and

γ = 100 V;

ρ = 10 mA.

Vo Vo ρ + Vo sC − γC + − = 0; R sL s .·. Vo =

s2

γ[s + (ρ/γC)] . + (1/RC)s + (1/LC)

ρ 10 × 10−3 = = 104 ; γC (100)(10)10−9 109 1 = 5 = 104 ; RC 10 1 109 = = 25 × 106 ; LC 40 Vo = [c] IL =

Vo ρ Vo 10 × 10−3 − = − ; sL s 4s s

IL = [d] Vo = =

100(s + 104 ) . s2 + 104 s + 25 × 106

25(s + 104 ) 10−2 −0.01(s + 7500) − = . 2 4 6 s(s + 10 s + 25 × 10 ) s (s + 5000)2 100(s + 104 ) s2 + 104 s + 25 × 106 K1 K2 100(s + 104 ) = + ; 2 2 (s + 5000) (s + 5000) s + 5000

K1 = 100(5000) = 5 × 105 ;

Problems

K2 =

d [100(s + 10,000)]s=−5000 = 100; ds

Vo =

5 × 105 100 + ; 2 (s + 5000) s + 5000

vo = [5 × 105 te−5000t + 100e−5000t ]u(t) V. −0.01(s + 7500) (s + 5000)2

[e] IL =

K1 K2 + ; (s + 5000)2 (s + 5000)

=

K1 = −0.01(2500) = −25; K2 =

d [−0.01(s + 7500)]s=−5000 = −0.01; ds

"

#

−25,000 10 IL = − × 10−3 ; 2 (s + 5000) s + 5000 iL = −[25,000t + 10]e−5000t u(t) mA. 50,000 . (s + 30)2

P 13.23 [a] Vg =

Io =

50,000 10,000 = ; (s + 30)2 (5s + 400) (s + 30)2 (s + 80)

Vo = 5sIo = [b] Io =

50,000s . (s + 30)2 (s + 80)

K1 K2 K3 + + . 2 (s + 30) s + 30 s + 80

K1 =

10,000 = 200; 50

K2 =

d 10,000 ds s + 80 

K3 =

 s=−30

10,000 = 4; (−50)2

= −4;

13–25

13–26

CHAPTER 13. The Laplace Transform in Circuit Analysis

Io =

200 4 4 − + ; 2 (s + 30) s + 30 s + 80

io (t) = [200te−30t − 4e−30t + 4e−80t ]u(t) A. Vo =

K1 K2 K3 + + . (s + 30)2 s + 30 s + 80

50,000(−30) = −30,000; 50   d 50,000s K2 = = 1600; ds s + 80 s=−30

K1 =

K3 =

50,000(−80) = −1600; (−50)2

vo (t) = [−30,000te−30t + 1600e−30t − 1600e−80t ]u(t) V. P 13.24 Transforming the circuit into the s domain for t > 0:

1 40 s = + + 40 × 106 625 s 

Zeq

=

−1

40 × 106 s ; s2 + 64,000s + 16 × 108

Zeq I= s/40

0.024s s2 + 40,0002

!

=

384 × 105 s (s2 + 40,0002 )(s + 32,000 − j24,000)(s + 32,000 + j24,000)

=

K1∗ K2 K2∗ K1 + + + ; s − j40,000 s + j40,000 s + 32,000 − j24,000 s + 32,000 + j24,000

K1 =

384 × 105 s = −j0.0075 = 0.0075/ − 90◦ ; (s + j40,000)(s2 + 64,000s + 16 × 108 ) s=j40,000

384 × 105 s = j0.0125 = 0.0125/90◦ . K2 = 2 2 (s + j40,000 )(s + 32,000 + j24,000) s=−32,000+j24,000

Therefore, i(t) = (15 sin 40,000t − 25e−32,000t sin 24,000t)u(t) mA.

Problems P 13.25 [a] iL (0− ) = iL (0+ ) = 5 A, down; vC (0− ) = vC (0+ ) = 0.

Vo Vo −5 + + Ia = ; 20 0.0075s s V1 (5s) V1 = Ia = + 106 50 Vo + 20Iφ = V1 ;

250s + 106 V1 ; 50 × 106 !

Vo + 20

V1 = V1 ; 50

.·. 0.6V1 = Vo ;

Vo Vo 250s + 106 −5 .·. + + V = ; o 20 0.0075s 30 × 106 s .·. (s2 + 10,000s + 16 × 106 )Vo = −6 × 105 ; Vo = [b] Vo =

−6 × 105 . s2 + 10,000s + 16 × 106

−6 × 105 K1 K2 = + . (s + 2000)(s + 8000) (s + 8000) (s + 2000)

K1 =

−6 × 105 = 100; −6000

K2 =

−6 × 105 = −100; 6000

vo (t) = [100e−8000t − 100e−2000t ]u(t) V. P 13.26 [a]

Vo − 5/s Vo − 10Iφ + 0.02V∆ + = 0; 40 s + (200/s)

13–27

13–28

CHAPTER 13. The Laplace Transform in Circuit Analysis "

#

Vo − 10Iφ s; V∆ = s + (200/s)

Iφ =

(5/s) − Vo . 40

Solving for Vo yields: Vo =

3s2 + 25s + 500 3s2 + 25s + 500 = . s(s2 + 25s + 100) s(s + 5)(s + 20)

Vo =

K1 K2 K3 + + ; s s + 5 s + 20

K1 =

3s2 + 25s + 500 = 5; (s + 5)(s + 20) s=0

3s2 + 25s + 500 = −6; K2 = s(s + 20) s=−5

3s2 + 25s + 500 K3 = = 4; s(s + 5) s=−20

5 −6 4 .·. Vo = + + ; s s + 5 s + 20 .·. vo (t) = [5 − 6e−5t + 4e−20t ]u(t) V. [b] From the equation at t = 0+ , From the circuit,

vo = 5 − 6 + 4 = 3 V.

vo = v∆ + 10iφ ; iφ =

5 − vo ; 40

(5 − vo ) .·. vo = v∆ + 10 = v∆ + 1.25 − 0.25vo ; 40 .·. 1.25vo − 1.25 = v∆ ; vo − 5 + 0.02v∆ = 0; 40 vo − 5 + 0.8v∆ = 0; vo − 5 + vo − 1 = 0

so

vo = 3 V(checks.)

Problems At t = ∞, the circuit is

From the equation for vo (t), vo (∞) = 5 V. From the circuit, v∆ = 0,

.·. vo = 5 V(checks).

iφ = 0

P 13.27 [a]

Vφ Vφ − (600/s) Vφ − Vo + + = 0; 10/s 10 20s Vo Vo − Vφ Vφ + + = 0. 140 20s 4 Simplfying, (2s2 + 2s + 1)Vφ − Vo = 1200; (35s − 7)Vφ + (s + 7)Vo = 0; ∆=

2 2s + 2s + 1 35s − 7



−1

s + 7 1200 0

= 2s(s2 + 8s + 25);

N2 =

2 2s + 2s + 1 35s − 7

Vo =

N2 −21,000s + 4200 −4200(5s − 1) = = . 2 ∆ s(s + 8s + 25) s(s2 + 8s + 25)

= −42,000s + 8400;

[b] vo (0+ ) = s→∞ lim sVo = 0; vo (∞) = lim sVo = s→0

4200 = 168 V. 25

13–29

13–30

CHAPTER 13. The Laplace Transform in Circuit Analysis

[c] At t = 0+ the circuit is

At t = ∞ the circuit is

Vφ − 600 Vφ Vφ + + = 0; 10 140 4 .·. Vφ = 168 V = Vo (∞) [d] Vo =

(checks).

−21,000s + 4200 K1 K2 K2∗ N2 = = + + . ∆ s(s2 + 8s + 25) s s + 4 − j3 s + 4 + j3

K1 =

4200 = 168; 25

K2 =

−21,000(−4 + j3) + 4200 = −84 + j3612 = 3612.98/91.33◦ ; (−4 + j3)(j6)

vo (t) = [168 + 7225.95e−4t cos(3t + 91.33◦ )]u(t) V. Check:

vo (0+ ) = 0 V;

vo (∞) = 168 V.

P 13.28 [a]

20Iφ + 25s(Io − Iφ ) + 25(Io − I1 ) = 0;

Problems 25s(Iφ − Io ) + Iφ − I1 =

50 Iφ + 5I1 + 25(I1 − Io ) = 0; s

100 ; s

100 .·. I1 = Iφ − . s

Simplifying, (−5s − 1)Iφ + (5s + 5)Io = −500/s; (5s2 + 6s + 10)Iφ + (−5s2 − 5s)Io = 600; ∆=

−5s − 1 2 5s + 6s + 10

N2 =

Io =

5s + 5 −5s2 − 5s

−5s − 1 2 5s + 6s + 10

= −25(s2 + 3s + 2);



−500/s 600



=−

500 2 (s − 4.8s − 10; ) s

N2 20s2 − 96s − 200 = . ∆ s(s + 1)(s + 2)

[b] io (0+ ) = s→∞ lim sIo = 20 A; io (∞) = lim sIo = s→0

−200 = −100 A. 2

[c] At t = 0+ the circuit is

20iφ + 5i1 = 0;

iφ − i1 = 100;

.·. 20iφ + 5(iφ − 100) = 0; .·. iφ = io (0+ ) = 20 A(checks).

25iφ = 500;

13–31

13–32

CHAPTER 13. The Laplace Transform in Circuit Analysis As t → ∞ the circuit is

io (∞) = −100 A(checks). [d] Io =

K1 K2 K3 20s2 − 96s − 200 = + + . s(s + 1)(s + 2) s s+1 s+2

K1 =

−200 = −100; (1)(2)

K3 =

80 + 192 − 200 = 36; (−2)(−1)

Io =

K2 =

20 + 96 − 200 = 84; (−1)(1)

−100 84 36 + + ; s s+1 s+2

io (t) = (−100 + 84e−t + 36e−2t )u(t) A. io (∞) = −100 A(checks). io (0+ ) = −100 + 84 + 36 = 20 A(checks). P 13.29 vC (0− ) = vC (0+ ) = 0. Find the Th´evenin equivalent with respect to the capacitor:

vTh vTh 9vTh + + = −0.005; 20,000 200,000 200,000 .·. vTh = −50 V; isc = −5 mA;

.·. RTh = 10 kΩ.

Problems

Vo = =

−50/s 107 · 10,000 + (107 /s) s −50 × 103 −50 50 = + ; s(s + 1000) s s + 1000

vo (t) = [−50 + 50e−1000t ]u(t) V. P 13.30

Vo Vo 15 = + 0.4Vφ + ; s 1.6 + 5/s 0.2s Vφ =

5/s 5Vo Vo = ; 1.6 + 5/s 1.6s + 5

15 Vo s 2Vo 5Vo .·. = + + s 1.6s + 5 1.6s + 5 s "

= Vo

s(s + 2) + 5(1.6s + 5) ; s(1.6s + 5)

#

15(1.6s + 5) = Vo (s2 + 10s + 25); 15(1.6s + 5) K1 K2 .·. Vo = = + . 2 2 (s + 5) (s + 5) s+5 K1 = 15(−8 + 5) = −45; Vo =

K2 = 24;

−45 24 + ; 2 (s + 5) s+5

vo (t) = [−45te−5t + 24e−5t ]u(t) V.

13–33

13–34

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.31 [a]

100 100 I1 + (I1 − I2 ) + 50(I1 − 4/s) = 0; s s 100 100 (I2 − 4/s) + (I2 − I1 ) + 50I2 = 0. s s Simplifying, (s + 4)I1 − 2I2 = 4; 8 −2I1 + (s + 4)I2 = ; s ∆=

(s + 4) −2

N1 =

I1 =

N2 =

I2 =

4 8/s



−2

(s + 4)

=

= s2 + 8s + 12 = (s + 2)(s + 6; )

4s2 + 16s + 16 4(s + 2)2 = ; s s

N1 4(s + 2)2 4(s + 2) 4/3 8/3 = = = + . ∆ s(s + 2)(s + 6) s(s + 6) s s+6 (s + 4) −2



4

8/s

=

16s + 32 16(s + 2) = ; s s

N2 16(s + 2) 16 8/3 8/3 = = = − . ∆ s(s + 2)(s + 6) s(s + 6) s s+6

Ia = I1 = Ib =

−2 (s + 4)

4/3 8/3 + ; s s+6

4 8/3 8/3 − I1 = − . s s s+6

Problems

13–35

[b] ia (t) = (4/3)(1 + 2e−6t )u(t) A; ib (t) = (8/3)(1 − e−6t )u(t) A. 100 100 [c] Va = Ia = s s =

4/3 8/3 + s s+6

!

400/3 800/3 400/3 400/9 400/9 + = + − . s2 s(s + 6) s2 s s+6

100 100 Vb = (I2 − I1 ) = s s =

!

400/3 1600/3 400/3 800/9 800/9 − = − + . 2 s s(s + 6) s2 s s+6

100 100 Vc = (4/s − I2 ) = s s =

4/3 16/3 − s s+6

4/3 8/3 + s s+6

!

400/3 800/3 400/3 400/9 400/9 + = + − . 2 s s(s + 6) s2 s s+6

[d] va (t) = (400/9)(3t + 1 − e−6t )u(t) V; vb (t) = (400/9)(3t − 2 + 2e−6t )u(t) V; vc (t) = (400/9)(3t + 1 − e−6t )u(t) V. [e] Calculating the time when a capacitor’s voltage drop first reaches 1000 V: For va (t) or vc (t) : 

1000

9 = 3t + 1 − e−6t = 22.5; 400 

3t − e−6t = 21.5; t = 7.17 s. For vb (t) : 3t − 2 + 2e−6t = 22.5; 3t + 2e−6t = 24.5; t = 8.17 s. Thus, the capacitors whose voltage drops are designated va and vc will break down first, at a time of 7.17 s.

13–36

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.32 [a]

Ye =

12.5s 10s 2.5s + 9 = ; 9 10 10 109

Ze =

109 80 × 106 = . 12.5s s

[b] I1 =

4 × 10−3 80/s = ; 20,000 + (100 × 106 /s) s + 5000

V1 =

4 × 10−3 20 × 106 80,000 · = ; s + 5000 s s(s + 5000)

V2 =

320,000 4 × 10−3 80 × 106 · = . s + 5000 s s(s + 5000)

[c] i1 (t) = 4e−5000t u(t) mA; V1 =

16 16 − ; s s + 5000

v1 (t) = (16 − 16e−5000t )u(t) V;

V2 =

64 64 − ; s s + 5000

v2 (t) = (64 − 64e−5000t )u(t) V.

[d] i1 (0+ ) = 4 mA; i1 (0+ ) =

80 × 10−3 = 4 mA(checks); 20

v1 (0+ ) = 0; v1 (∞) =

v2 (0+ ) = 0(checks);

12.5 (80) = 16 V; 50 + 12.5

v2 (∞) =

50 (80) = 64 V(checks); 50 + 12.5

Problems v1 (∞) + v2 (∞) = 80 V(checks); (50 × 10−9 )v1 (∞) = 800 nC; (12.5 × 10−9 )v2 (∞) = 800 nC(checks). P 13.33 [a] For t < 0:

v1 = 75 − vo ; .·. vo = 60 V;

50vo = 200(75 − v0 ); v1 = 15 V.

For t > 0:

[b] Io =

75/s 300 = (25 × 106 /s) + 6250 + 0.25s s2 + 25,000s + 108

=

300 20 × 10−3 20 × 10−3 = − ; (s + 5000)(s + 20,000) s + 5000 s + 20,000

io (t) = (20e−5000t − 20e−20,000t )u(t) mA. [c] Vo =

60 20 × 106 300 − · s s (s + 5000)(s + 20,000) "

60 60 80 20 = − − + s s s + 5000 s + 20,000 =

−20 80 + ; s + 5000 s + 20,000

vo (t) = (80e−5000t − 20e−20,000t )u(t) V.

#

13–37

13–38

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.34 [a]

With no load across terminals a − b Vx = 20/s: 1 20 20 − VTh s + 1.2 − VTh = 0; 2 s s 





Therefore VTh =

Vx = 5IT







20(s + 2.4) . s(s + 2)

and ZTh =

VT . IT

Solving for IT gives (VT − 5IT )s + VT − 6IT . 2 Therefore IT =

14IT = VT s − 5sIT + 2VT ;

therefore ZTh =

5(s + 2.8) . s+2

Problems

13–39

[b]

I=

20(s + 2.4) VTh = . ZTh + 2 + s s(s + 3)(s + 6)

P 13.35 Begin by transforming the circuit from the time domain to the s domain:

30k0.0012s =

30s . s + 25,000

Use voltage division to find Vo (s): 30s s + 25,000 Vo = 6 10 30s + s s + 25,000

2 × 106 s2 + 50,0002

!

s2 = 2 (s + 33,333.33s + 0.833 × 109 ) =

2 × 106 s2 + 50,0002

!

K1 K1∗ + s + 16,666.67 − j23,570.226 s + 16,666.67 + j23,570.226 +

K2 K2∗ + . s − j50,000 s + j50,000

2 × 106 s2 K1 = (s + 16,666.67 + j23,570.226)(s2 + 50,0002 ) s=−16,666.67+j23,570.226

= 15/180◦ ;

13–40

CHAPTER 13. The Laplace Transform in Circuit Analysis

2 × 106 s2 K2 = 2 (s + 33,333.33s + 0.833 × 109 )(s + j50,000) s=j50,000

= 21.21/ − 45◦ . Therefore, vo (t) = [−30e−16,666.67t cos(23,570.226t) + 42.43 cos(50,000t − 45◦ )] V. P 13.36 vC = 12 × 105 te−5000t V, dvC iC = C dt

C = 5 µF;

therefore

!

= 6e−5000t (1 − 5000t) A.

iC > 0 when 1 > 5000t or iC > 0 when 0 < t < 200 µs and iC < 0 when t > 200 µs. iC = 0 when 1 − 5000t = 0,

or t = 200 µs.

dvC = 12 × 105 e−5000t [1 − 5000t]; dt .·. iC = 0 when

dvC = 0. dt

P 13.37 [a] The s-domain equivalent circuit is

I=

Vg /L Vg = , R + sL s + (R/L)

Vg =

Vm (ω cos φ + s sin φ) ; s2 + ω 2

K0 K1 K1∗ + + . I= s + R/L s − jω s + jω Vm (ωL cos φ − R sin φ) K0 = , R 2 + ω 2 L2

Vm /φ − 90◦ − θ(ω) √ K1 = 2 R 2 + ω 2 L2

where tan θ(ω) = ωL/R. Therefore, we have i(t) =

Vm (ωL cos φ − R sin φ) −(R/L)t Vm sin[ωt + φ − θ(ω)] √ . e + R 2 + ω 2 L2 R 2 + ω 2 L2

Problems Vm sin[ωt + φ − θ(ω)]. + ω 2 L2 Vm (ωL cos φ − R sin φ) −(R/L)t e . [c] itr = R 2 + ω 2 L2 Vg [d] I = , Vg = Vm /φ − 90◦ . R + jωL [b] iss (t) = √

R2

Therefore I = √

Vm /φ − 90◦ Vm /φ − θ(ω) − 90◦ . √ = 2 2 2 2 2 2 / R + ω L θ(ω) R +ω L

Therefore iss = √

Vm sin[ωt + φ − θ(ω)]. R 2 + ω 2 L2

[e] The transient component vanishes when ωL cos φ = R sin φ or

tan φ =

ωL R

or φ = θ(ω).

P 13.38 The s-domain equivalent circuit is

V1 − 12/s V1 + 2.4 V1 + + = 0; 10 + (250/s) 2s 2s + 50 V1 =

−300(s + 25) −300 = 2 ; 2 (s + 25)(s + 10s + 125) s + 10s + 125

Io =

−300 (2s + 50)(s2 + 10s + 125)

=

−150 (s + 25)(s + 5 − j10)(s + 5 + j10)

K1 K2 K2∗ = + + . s + 25 s + 5 − j10 s + 5 + j10 K1 =

−150 = −300 × 10−3 ; 625 − 250 + 125

K2 =

√ −150 = 150 5 × 10−3 /63.43◦ ; (−5 + j10 + 25)(j20)

√ io (t) = [−300e−25t + 300 5e−5t cos(10t + 63.43◦ )]u(t) mA.

13–41

13–42

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.39 [a] i2 = 1.25e−t − 1.25e−3t ;

so

di2 = −1.25e−t + 3.75e−3t . dt

di2 = 0 when dt

Therefore

1.25e−t = 3.75e−3t

or e2t = 3,

t = 0.5(ln 3) = 549.31 ms.

i2 (max) = 1.25[e−0.549 − e−3(0.549) ] = 481.13 mA. [b] Solving the mesh current equations from Example 13.7 are (3 + 2s)I1 + 2sI2 = 10;

and

2sI1 + (12 + 8s)I2 = 10.

Solving for I1 , I1 =

5(s + 2) . (s + 1)(s + 3)

A partial fraction expansion leads to the expression 2.5 2.5 + . s+1 s+3 Therefore we get I1 =

i1 = 2.5[e−t + e−3t ]u(t) A. di1 di1 (0.54931) = −2.5[e−t + 3e−3t ]; = −2.89 A/s. dt dt [d] When i2 is at its peak value, [c]

di2 = 0. dt Therefore L2 [e] i2 (max) =

di2 dt

!

M = 0 and i2 = − 12 



−2(−2.89) = 481.13 mA. (checks) 12

P 13.40 [a]

88.4 = 136I1 − sI1 + 1.5s(I1 − I2 ); s 0 = 1.5s(I2 − I1 ) + 11sI2 + 5000I2 . Simplifying, 88.4 = (0.5s + 136)I1 − 1.5sI2 ; s

!

di1 . dt

Problems

13–43

0 = −1.5sI1 + (12.5s + 5000)I2 ; ∆=

0.5s + 136 −1.5s

N1 =

I1 = [b] sI1 =

88.4/s 0

−1.5s 12.5s + 5000

12.5s + 5000

−1.5s

=

= 4(s + 200)(s + 850);

1105(s + 400) ; s

N1 276.25(s + 400) = . ∆ s(s + 200)(s + 850) 276.25(s + 400) ; (s + 200)(s + 850)

lim sI1 = i1 (∞) = 650 mA;

s→0

lim sI1 = i1 (0) = 0.

s→∞

[c] I1 =

K1 K2 K3 + + . s s + 200 s + 850

K1 = 650 × 10−3 ;

K2 = −425 × 10−3 ;

K3 = −225 × 10−3 ;

i1 (t) = (650 − 425e−200t − 225e−850t )u(t) mA. P 13.41 [a] From the solution to Problem 13.40 we have N2 =

0.5s + 136 −1.5s

.·. I2 = =

88.4/s 0

= 132.6;

132.6 33.15 = 4(s + 200)(s + 850) (s + 200)(s + 850) 51 × 10−3 51 × 10−3 − ; s + 200 s + 850

i2 (t) = (51e−200t − 51e−850t )u(t) mA. [b] Reversing the dot on the 12.5 H coil will reverse the sign of M , thus the circuit becomes

The two simultaneous equations are 88.4 = (136 + 0.5s)I1 + 1.5sI2 ; s

13–44

CHAPTER 13. The Laplace Transform in Circuit Analysis 0 = 1.5sI1 + (12.5s + 5000)I2 . When these equations are compared to those derived in Problem 13.40 we see the only difference is the algebraic sign of the 1.5s term. Thus reversing the dot will have no effect on I1 and will reverse the sign of I2 . Hence, i2 (t) = (−51e−200t + 51e−850t )u(t) mA.

1 1 P 13.42 [a] w = L1 i21 + L2 i22 + M i1 i2 ; 2 2 1 1 w = (40)(9) + (90)(4) + 30(6) × 10−3 = 540 mJ. 2 2 



[b] The s-domain circuit:

(600 + 0.04s)I1 − 0.03sI2 = 0.18; −0.03sI1 + (0.09s + 1350)I2 = −0.27; ∆=

0.04(s + 15,000) −0.03s

0.09(s + 15,000)

−0.03s

= 27 × 10−4 (s + 10,000)(s + 30,000; )

N1 =

0.18 −0.27

−0.03s 0.09(s + 15,000)

I1 =

N1 3 = . ∆ s + 10,000

N2 =

0.04(s + 15,000) −0.03s

I2 =

N2 −2 = . ∆ s + 10,000

[c] i1 (t) = 3e−10,000t u(t) A;

0.18 −0.27

= 81 × 10−4 (s + 30,000; )

= −54 × 10−4 (s + 30,000);

i2 (t) = −2e−10,000t u(t) A.

Problems [d] p600Ω = (600)(9e−20,000t ) = 5400e−20,000t W; p1350Ω = (1350)(4e−20,000t ) = 5400e−20,000t W; w600 = w1350 =

5400 × 10−3 = 270 mJ; 20 5400 × 10−3 = 270 mJ; 20

wT = 540 mJ. [e] With the dot reversed, 1 1 w = L1 i21 + L2 i22 − M i1 i2 = 180 + 180 − 180 = 180 mJ. 2 2 The s-domain equivalent circuit is

Solving for I1 and I2 yields I1 =

3 ; s + 30,000

I2 =

−2 ; s + 30,000

.·. i1 (t) = 3e−30,000t u(t) A; w600 = 5400

Z ∞

w1350 = 5400

e−60,000t dt = 90 mJ;

0

Z ∞

wT = 180 mJ. P 13.43 For t < 0:

i2 (t) = −2e−30,000t u(t) A.

0

e−60,000t dt = 90 mJ;

13–45

13–46

CHAPTER 13. The Laplace Transform in Circuit Analysis

For t > 0+ :

L1 + M = 3.6 + 1.2 = 4.8 H; 15 × 4.8 = 72;

L2 + M = 2.4 + 1.2 = 3.6 H;

15 × 3.6 = 54.

12Io + 4.8sIo − 72 + (Io − I2 )(6 − 1.2s) = 0; (6 − 1.2s)(I2 − Io ) + 3.6sI2 − 54 = 0;

.·. ∆ =

60 45

No =

Io =

3(s + 5) −(5 − s)

−(5 − s) 2(s + 2.5)

−(5 − s)

= 75(s + 7);

2(s + 2.5)

No 75(s + 7) = ∆ 5(s + 1)(s + 10) =

K1 =

= 5(s + 1)(s + 10);

K1 K2 + . (s + 1) (s + 10)

(15)(6) = 10; 9

Problems

K2 =

(15)(−3) = 5; −9

10 5 .·. Io = + ; s + 1 s + 10 .·. io (t) = [10e−t + 5e−10t ]u(t) A. P 13.44 [a] The s-domain circuit with the voltage source acting alone is

V 0 − (20/s) V0 V 0s + + = 0; 2 1.25s 20 V0 =

200 100/3 100/3 = − . (s + 2)(s + 8) s+2 s+8

With the current source acting alone,

V 00 V 00 V 00 s 5 + + = ; 2 1.25s 20 s V 00 =

50/3 50/3 100 = − ; (s + 2)(s + 8) s+2 s+8

Vo = V 0 + V 00 =

50 50 − . s+2 s+8

[b] vo = [50e−2t − 50e−8t ]u(t) V.

13–47

13–48

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.45

Va − 0.4/s Va s (Va − Vo )s + + = 0; 6 5000 250 × 10 250 × 106 (0 − Va )s (0 − Vo ) + = 0; 250 × 106 25,000 −104 Vo ; s

Va =

.·. Vo (s2 + 20,000s + 500 × 106 ) = −20,000; Vo =

−20,000 . (s + 10,000 − j20,000)(s + 10,000 + j20,000)

K1 =

−20,000 = j0.5 = 0.5/90;◦ j40,000

vo (t) = e−10,000t cos(20,000t + 90◦ ) = −e−10,000t sin(20,000t)u(t) V. P 13.46 [a] Let Va be the s-domain voltage across the 0.2 µF capacitor, positive at the upper terminal and let Vb be the s-domain voltage across the 200 kΩ resistor, positive at the upper terminal. With Vg = 8/s V, Va s Va − Vg Va + + = 0; 6 5 × 10 400,000 400,000 −Va sVb − 7 = 0; 400,000 10

12.5 .·. Va = Vg . s + 25

−25 −312.5 .·. Vb = Va = Vg ; s s(s + 25)

Vb sVb (Vb − Vo )s + 7 + = 0; 200,000 10 107 "

2(s + 25) 2(s + 25) .·. Vo = Vb = s s

#"

−312.5 s(s + 25)

# 

8 −5000 = . s s3

Problems [b] vo (t) = −2500t2 u(t) V. [c] The op amp will saturate when vo = −12.5 V. −12.5 = −2500t2 ; P 13.47 [a] Vo = − Zf =

t2 = 0.005;

Zf Vg ; Zi 108

s+

h

109 (10)(2)×104

i

=

108 ; s + 5000

8000 109 Zi = s+ s (50)(8000) Vg =

!

8000 (s + 2500); s

=

20,000 ; s2

.·. Vo = [b] Vo =

.·. t = 0.071 = 71 ms.

−250 × 106 . s(s + 2500)(s + 5000)

K1 K2 K3 + + . s s + 2500 s + 5000

K1 =

−250 × 106 = −20; (5000)(2500)

K2 =

−250 × 106 = 40; (−2500)(2500)

K3 =

−250 × 106 = −20; (−5000)(−2500)

.·. vo (t) = (−20 + 40e−2500t − 20e−5000t )u(t) V. [c] −20 + 40e−2500ts − 20e−5000ts = −5; .·. 40e−2500ts − 20e−5000ts = 15. Let x = e−2500ts . Then 40x − 20x2 = 15;

or x2 − 2x + 0.75 = 0.

Solving, x = 1 ± 0.5

so

.·. e−2500ts = 0.5;

x = 0.5; .·. ts =

ln 2 × 10−6 = 277.26 µs. 0.0025

13–49

13–50

CHAPTER 13. The Laplace Transform in Circuit Analysis

[d] vg = m tu(t); Vo = = K1 =

Vg =

m ; s2

−108 s m · 2 8000(s + 2500)(s + 5000) s −12,500m . s(s + 2500)(s + 5000) −12,500m = −m × 10−3 ; (2500)(5000)

.·. −5 = −m × 10−3

.·. m = 5000 V/s.

P 13.48 [a]

Vp − Vg2 Vp s + = 0; 50 × 106 20,000

Vp =

2500 Vg2 ; s + 2500

Vp − Vg1 Vp − Vo (Vp − Vo )s = 0; + + 80,000 20,000 108 (s + 6250)Vp − (s + 5000)Vo = 1250Vg1 ; (s + 6250)(2500) .·. (s + 5000)Vo = Vg2 − 1250Vg1. (s + 2500) Vg1 =

16 ; s

.·. Vo = = K1 =

8 Vg2 = ; s 7500 × 104 s(s + 2500)(s + 5000) K2 K3 K1 + + . s s + 2500 s + 5000

7500 × 104 750 = = 6; (2500)(5000) 125

Problems

K2 =

7500 × 104 = −12; (−2500)(2500)

K3 =

7500 × 104 = 6; (−5000)(−2500)

13–51

vo = [6 − 12e−2500t + 6e−5000t ]u(t) V. [b] 6 − 12e−2500ts + 6e−5000ts = 5;

let x = e−2500ts ;

6 − 12x + 6x2 = 5. 1 = 0. 6

x2 − 2x + x=1−

q

5/6 = 0.0871.

.·. e−2500ts = 0.0871;

ts = 976.15 µs.

1 1/RC 1/sC = = . R + 1/sC RsC + 1 s + 1/RC There are no zeros, and a single pole at −1/RC rad/s. R R/L [b] = . R + sL s + R/L There are no zeros, and a single pole at −R/L rad/s.

P 13.49 [a]

[c] There are several possible solutions. For the RL circuit choose L = 10 mH. Then R = (5000)(0.01) = 50 Ω. Use 2 100 Ω resistors in parallel to get the equivalent 50 Ω resistance needed. For the RC circuit choose C = 1 µF. Then R = 1/(5000 × 10−6 ) = 200 Ω. Use 2 100 Ω resistors in series to get the 200 Ω resistance required. R RsC s = = . R + 1/sC RsC + 1 s + 1/RC There is a single zero at 0 rad/s, and a single pole at −1/RC rad/s. s sL = . [b] R + sL s + R/L There is a single zero at 0 rad/s, and a single pole at −R/L rad/s.

P 13.50 [a]

[c] There are several possible solutions. For the RL circuit choose L = 10 mH. Then R = (250)(0.01) = 2.5 Ω. Use 4 10 Ω resistors in parallel to get the equivalent 2.5 Ω resistance needed. For the RC circuit choose C = 1 µF. Then R = 1/(250 × 10−6 ) = 4 kΩ. Use a 1.8 kΩ and a 2.2 kΩ resistor in series to get the 4 kΩ resistance required.

13–52

P 13.51 [a]

CHAPTER 13. The Laplace Transform in Circuit Analysis 1/sC + sL s2 + 1/LC = 2 . 1/sC + sL + R s + (R/L)s + 1/LC q

There are two zeros at j 1/LC rad/s, and two poles: −p1 = −(R/2L) +

q

−p2 = −(R/2L) −

q

(R/2L)2 − (1/LC) (R/2L)2 − (1/LC).

[b] There are several possible solutions. One is R = 100 Ω + 150 Ω = 250 Ω;

L = 10 mH;

C = 1 µF.

These component values yield the following poles: −p1 = −5000 rad/s

− p2 = −20,000 rad/s.

and

[c] There are several possible solutions. One is R = 100 Ω + 100 Ω = 200 Ω;

L = 10 mH;

C = 1 µF.

These component values yield the following poles: −p1 = −10,000 rad/s

and

− p2 = −10,000 rad/s.

[d] There are several possible solutions. One is R = 120 Ω;

L = 10 mH;

C = 1 µF.

These component values yield the following poles: −p1 = −6000 + j8000 rad/s P 13.52 [a]

and

− p2 = −6000 − j8000 rad/s.

Vo Rk1/sC 1/LC = = 2 . Vi (Rk1/sC) + sL s + (1/RC)s + 1/LC There are no zeros and two poles: p1 = −(1/2RC) +

q

p2 = −(1/2RC) −

q

(1/2RC)2 − (1/LC); (1/2RC)2 − (1/LC).

[b] There are several qpossible solutions. Any selection of R, L, and C such that 1/2RC > 1/LC will yield two real, distinct poles. When L = 1 mH and C = 0.1 µF, R must be less than 50 Ω. For example, let R = 10 Ω;

L = 1 mH;

C = 0.1 µF.

These component values yield the following poles: −p1 = −10,102.05 rad/s

and

− p2 = −989,897.95 rad/s.

Problems [c] To get repeated real poles, 1/2RC = R=

q

13–53

1/LC, so choose

1 = 50 Ω. 2(100,000)(0.1 × 10−6 )

We can create a resistance of 50 Ω using two parallel-connected 100 Ω resistors. Then R = 50 Ω;

L = 1 mH;

C = 0.1 µF,

These component values yield the following poles: −p1 = −100,000 rad/s

and

− p2 = −100,000 rad/s,

[d] There are several possible solutions. Any resistor with a value greater than 50 Ω will yield two complex conjugate poles. For example, let R = 100 Ω;

L = 1 mH;

C = 0.1 µF,

These component values yield the following poles: −p1 = −50,000 + j86,602.54 rad/s P 13.53 [a]

− p2 = −50,000 − j86,602.54 rad/s,

1/sC 1 Vo = = ; Vi R + 1/sC RCs + 1 H(s) =

[b]

and

(1/RC) 250 = ; s + (1/RC) s + 250

−p1 = −250 rad/s.

R RCs s Vo = = = Vi R + 1/sC RCs + 1 s + (1/RC)

s ; z1 = 0, −p1 = −250 rad/s. s + 250 Vo sL s s ; [c] = = = Vi R + sL s + R/L s + 3 × 106 =

z1 = 0; [d]

−p1 = −3 × 106 rad/s.

Vo R R/L 3 × 106 = = = ; Vi R + sL s + (R/L) s + 3 × 106 −p1 = −3 × 106 rad/s.

[e]

Vo s Vo Vo − Vi + 5+ = 0; 6 2.5 × 10 10 25 × 103

13–54

CHAPTER 13. The Laplace Transform in Circuit Analysis sVo + 25Vo + 100Vo = 100Vi H(s) =

Vo 100 = Vi s + 125

−p1 = −125 rad/s. P 13.54 [a]

4 800,000Vi Vi = Vo + ; 8 800,000 + (16 × 106 /s) 0.5Vi −

sVi = Vo ; s + 20

Vo −0.5(s − 20) .·. = H(s) = . Vi (s + 20) [b] −z1 = 20 rad/s; −p1 = −20 rad/s. P 13.55 [a] Zi = 10,000 + Zf =

109 104 (s + 5000) = ; 20s s

25,000 250 × 106 = ; (25,000)(4 × 10−9 )s + 1 s + 10,000

H(s) = −

Zf −25,000s = . Zi (s + 5000)(s + 10,000)

[b] Zero at s = 0; Poles at −p1 = −5000 rad/s and −p2 = −10,000 rad/s. [c] If vg (t) = u(t), Vg (s) = 1/s. Then, Vo =

−25,000 (s + 5000)(s + 10,000)

=

5 −5 + . s + 5000 s + 10,000

Therefore, vo (t) = (−5e−5000t + 5e−10,000t )u(t) V. Note that the op amp will never saturate because the output voltage will always stay within the two power supply voltages.

Problems P 13.56 [a]

Va − Vg Va s (Va − Vo )s + + = 0; 8 8000 5 × 10 1.25 × 108 Vo −Va s − = 0; 8 5 × 10 62,500

Va =

−8000Vo ; s

−8000Vo .·. (5s + 62,500) − 4sVo = 62,500Vg ; s −15,625s Vo .·. H(s) = = 2 . Vg s + 10,000s + 125 × 106 √ s1,2 = −5000 ± 25 × 106 − 125 × 106 = −5000 ± j10,000; H(s) =

−15,625s . (s + 5000 − j10,000)(s + 5000 + j10,000)

[b] −p1 = −5000 + j10,000 rad/s; −p2 = −5000 − j10,000 rad/s; z = 0. [c] If vi (t) = u(t), Vi (s) = 1/s. Then, Vo =

−15,625 (s + 5000 − j10,000)(s + 5000 + j10,000)

=

K1 K1∗ + . s + 5000 − j10,000 s + 5000 + j10,000

−15,625 = 0.78125/90◦ . K1 = s + 5000 + j10,000 s=−5000+j10,000

Therefore, vo (t) = 1.5625e−5000t cos(10,000t + 90◦ ) V.

13–55

13–56

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.57 [a] Let R1 = 40 kΩ; R2 = 10 kΩ; C2 = 500 nF; and Cf = 250 nF. Then (R2 + 1/sC2 )1/sCf (s + 1/R2 C2 )  =  ; Zf =  C +C R2 + sC1 2 + sC1 f Cf s s + C22Cf Rf2 1 = 4 × 106 ; Cf 1 = 200 rad/s; R2 C2 750 × 10−9 C2 + Cf = = 600 rad/s; C2 Cf R2 1.25 × 10−9 4 × 106 (s + 200) .·. Zf = Ω; s(s + 600) Zi = R1 = 40 × 103 Ω; H(s) =

−Zf −100(s + 200) Vo = = . Vg Zi s(s + 600)

[b] −z1 = −200 rad/s; −p1 = 0;

−p2 = −600 rad/s.

[c] If vg (t) = u(t) then Vg (s) = 1/s. Then, Vo =

−100(s + 200) −33.33 −0.11 0.11 = + + . 2 2 s (s + 600) s s s + 600

Thus, vo = (−33.33t − 0.11 + 0.11e−600t )u(t) V. Note that the op amp will saturate once the output voltage exceeds the power supply voltages. P 13.58 [a]

Vo Vo + + Vo (50 × 10−9 )s = Ig ; 10,000 2s .·. Vo =

20 × 106 s · Ig s2 + 2000s + 10 × 106

60 × 10−3 s Vo ; Io = 4 ; 2 6 s + 16 × 10 10 2000s .·. H(s) = 2 . s + 2000s + 107

Ig =

Problems

13–57

(2000s)(60 × 10−3 s) ; (s + 1000 − j3000)(s + 1000 + j3000)(s2 + 16 × 106 )

[b] Io =

Io =

120s2 . (s + 1000 − j3000)(s + 1000 + j3000)(s + j4000)(s − j4000)

[c] Damped sinusoid of the form M e−1000t cos(3000t + θ1 ). [d] Steady-state sinusoid of the form N cos(4000t + θ2 ). [e] Io =

K1 K1∗ K2 K2∗ + + + . s + 1000 − j3000 s + 1000 + j3000 s − j4000 s + j4000

K1 =

120(−1000 + j3000)2 = 20 × 10−3 /163.74◦ ; (j6000)(−1000 − j1000)(−j1000 + j7000)

K2 =

120(−16 × 106 ) = 24 × 10−3 / − 36.87◦ ; (j8000)(1000 + j1000)(j1000 + j7000)

io (t) = [40e−1000t cos(3000t + 163.74◦ ) + 48 cos(4000t − 36.87◦ )] mA. Test: io (0) = 0.040 cos(163.74◦ ) + 0.048 cos(−36.87◦ ) = −0.0384 + 0.0384 = 0; Z=

1 ; Y

Y =

1 1 1 1 + j0.75 + + = ; 10,000 j8000 −j5000 10,000

10,000 .·. Z = = 8000/ − 36.87◦ Ω. 1 + j0.75 Vo = Ig Z = (60 × 10−3 /0◦ )(8000/ − 36.87◦ ) = 480/ − 36.87◦ V; Io =

Vo = 48/ − 36.87◦ mA; 104

ioss = 48 cos(4000t − 36.87◦ ) mA(checks). P 13.59 [a]

1000(Io − Ig ) + 500Io + µ(Ig − Io )(1000) + 10sIo = 0;

13–58

CHAPTER 13. The Laplace Transform in Circuit Analysis .·. Io =

100(1 − µ) Ig ; s + 100(1.5 − µ)

.·. H(s) =

100(1 − µ) . s + 100(1.5 − µ)

[b] µ < 1.5. [c] µ

H(s)

Io

−0.5 150/(s + 200)

1500/s(s + 200)

0 100/(s + 150)

1000/s(s + 150)

1.0 0

0

1.5 −50/s

−500/s2

2.0 −100/(s − 50) −1000/s(s − 50) µ = −0.5: Io =

7.5 7.5 − ; s (s + 200)

io = [7.5 − 7.5e−200t ]u(t), A.

µ = 0: Io =

20/3 20/3 − ; s s + 150

µ=1:

io =

20 [1 − e−150t ]u(t), A. 3

io = 0 A.

µ = 1.5: −500 ; s2 µ = 2: Io =

Io =

io = −500t u(t) A.

20 20 − ; s s − 50

P 13.60

Vg = 12sI1 − 8sIo ;

io = 20[1 − e50t ]u(t) A.

Problems 0 = −8sI1 + (12s + 8000 + 4 × 106 /3s)Io ; ∆=

12s −8s

No =

Io =

−8s 12s + 8000 + 4 × 106 /3s

12s −8s

Vg 0

= 80(s + 200)(s + 1000);

= 8sVg ;

No 8sVg = ; ∆ 80(s + 200)(s + 1000)

H(s) =

Io 0.1s = . Vg (s + 200)(s + 1000)

z1 = 0;

−p1 = −200 rad/s;

−p2 = −1000 rad/s.

P 13.61 [a] 0 ≤ t ≤ 20:

y(t) =

Z t

(5)(1)(dλ) = 5λ

0

t =

5t.

0

20 ≤ t ≤ 40:

y(t) =

Z 20 t−20

(5)(1)(dλ) = 5λ

20

t−20

= 5(40 − t).

13–59

13–60

CHAPTER 13. The Laplace Transform in Circuit Analysis t ≥ 40 :

y(t) = 0.

[b] 0 ≤ t ≤ 5:

y(t) =

Z t

20 dλ = 20λ

0

t =

20t.

0

5 ≤ t ≤ 20:

y(t) =

Z t t−5

t

20 dλ = 20λ

= 100.

t−5

20 ≤ t ≤ 25:

y(t) =

Z 20

20 dλ = 20λ

t−5

t ≥ 25 :

20

t−5

y(t) = 0.

= 20(25 − t).

Problems

13–61

[c] The expressions are 0≤t≤1:

y(t) =

Z t 0

1 ≤ t ≤ 20 :

y(t) =

t

100 dλ = 100λ = 100t; 0

Z t t−1

20 ≤ t ≤ 21 :

y(t) =

t

100 dλ = 100λ

Z 20

100 dλ = 100λ

t−1

21 ≤ t < ∞ :

= 100;

t−1

20

= 100(21 − t);

t−1

y(t) = 0.

[d]

[e] Yes, note that h(t) is approaching 20δ(t), therefore y(t) must approach 20x(t), i.e. y(t) =

Z t

h(t − λ)x(λ) dλ →

0

Z t

20δ(t − λ)x(λ) dλ

0

→ 20x(t). This can be seen in the plot, e.g., in part (c), y(t) ∼ = 20x(t). P 13.62 [a]

y(t) = 0

t < 0;

13–62

CHAPTER 13. The Laplace Transform in Circuit Analysis 0≤t≤8:

y(t) =

Z t

400 dλ = 400t;

0

8 ≤ t ≤ 16 :

y(t) =

Z 8

400 dλ = 400(8 − t + 8) = 400(16 − t);

t−8

16 ≤ t < ∞ :

y(t) = 0.

[b]

y(t) = 0

t < 0;

0≤t≤8:

y(t) =

Z t

200 dλ = 200t;

0

8 ≤ t ≤ 10 :

y(t) =

Z t

200 dλ = 200(t − t + 8) = 1600;

t−8

10 ≤ t ≤ 18 :

y(t) =

Z 10 t−8

18 ≤ t < ∞ :

y(t) = 0.

200 dλ = 200(18 − t);

Problems [c]

y(t) = 0

t < 0;

0 ≤ t ≤ 0.5 :

y(t) =

Z t

400 dλ = 400t;

0

0.5 ≤ t ≤ 8 :

y(t) =

Z t

400 dλ = 400(t − t + 0.5) = 200;

t−0.5

8 ≤ t ≤ 8.5 :

y(t) =

Z 8

400 dλ = 400(8.5 − t);

t−0.5

8.5 ≤ t < ∞ :

y(t) = 0.

P 13.63 [a] −1 ≤ t ≤ 4: vo = 20

Z t+1 0

t+1

3λ dλ = 30λ2

= 30t2 + 60t + 30 V.

0

4 ≤ t ≤ 7: Z 5

vo = 20

= 30λ

Z t+1

3λ dλ + 20

0 2

(20 − λ) dλ

5

5 0

2

+400λ

t+1

−10λ

5

= −10t + 380t − 610 V.

2

t+1 5

13–63

13–64

CHAPTER 13. The Laplace Transform in Circuit Analysis 7 ≤ t ≤ 12: Z 5

vo = 20

= 30λ

Z t+1

3λ dλ + 20

t−7

5

5

t+1

2

+400λ

t−7

(20 − λ) dλ

−10λ

2

5

t+1 5

2

= −40t + 800t − 2080 V. 12 ≤ t ≤ 19: Z t+1

vo = 20

t−7

t+1

(20 − λ) dλ = 400λ

t−7

t+1

−10λ2

t−7

= −160t + 3680 V. 19 ≤ t ≤ 27: Z 20

vo = 20

(20 − λ) dλ = 400λ

t−7

t−7

= 10t2 − 540t + 7290 V. [b]

P 13.64 H(s) =

Vo 1 = ; Vi s+1

20

h(t) = e−t .

−10λ

2

20

t−7

Problems For 0 ≤ t ≤ 1:

vo =

Z t

e−λ dλ = (1 − e−t ) V.

0

For 1 ≤ t ≤ ∞: vo =

Z t

e−λ dλ = (e − 1)e−t V.

t−1

P 13.65 H(s) =

s 1 Vo = =1− ; Vi s+1 s+1

h(t) = δ(t) − e−t ;

h(λ) = δ(λ) − e−λ . For 0 ≤ t ≤ 1: vo =

Z t 0

[δ(λ) − e−λ ] dλ = [1 + e−λ ] |t0 = e−t V.

For 1 ≤ t ≤ ∞: vo =

Z t t−1

t

(−e−λ ) dλ = e−λ

P 13.66 [a] h(λ) =

5 λ 10

= (1 − e)e−t V.

t−1

0 ≤ λ ≤ 10 s;

h(λ) = 10 −

5 λ 10

h(λ) = 0

20 ≤ λ ≤ ∞.

10 ≤ λ ≤ 20 s;

0 ≤ t ≤ 10 s: vo =

Z t 0

λ2 t (0.5λ)(4) dλ = 2 = t2 V. 2 0

13–65

13–66

CHAPTER 13. The Laplace Transform in Circuit Analysis 10 ≤ t ≤ 20 s: vo =

Z 10

2λ dλ +

0

Z t

4(10 − 0.5λ) dλ;

10

vo = 100 + 40t − 400 − t2 + 100 = 40t − 200 − t2 V. 20 ≤ t ≤ ∞: vo =

Z 10

2λ dλ +

0

Z 20

4(10 − 0.5λ) dλ;

10

vo = 100 + 400 − (400 − 100) = 200 V. [b]

0 ≤ λ ≤ 2.5 s;

[c] h(λ) = 8λ

h(λ) = 40 − 8λ

2.5 ≤ λ ≤ 5 s;

5 ≤ λ ≤ ∞.

h(λ) = 0

0 ≤ t ≤ 2.5 s: vo =

Z t

32λ dλ = 16t2 V.

0

2.5 ≤ t ≤ 5 s: vo =

Z 2.5

32λ dλ +

0

Z t

4(40 − 8λ) dλ = 160t − 200 − 16t2 V.

2.5

5 ≤ t ≤ ∞: vo =

Z 2.5 0

32λ dλ +

Z 5 2.5

4(40 − 8λ) dλ = 200 V.

Problems

13–67

[d] The waveform in part (c) is closer to replicating the input waveform because in part (c) h(λ) is closer to being an ideal impulse response. That is, the area was preserved as the base was shortened. P 13.67 [a] From Problem 13.53(a) 250 H(s) = ; s + 250 h(λ) = 250e−250λ . 0 ≤ t ≤ 4 ms: vo =

Z t

16(250)e−250λ dλ = 16(1 − e−250t ) V.

0

4 ms ≤ t ≤ ∞: vo =

Z t

16(250)e−250λ dλ = 16(e − 1)e−250t V.

t−0.004

[b]

2500 s + 2500 0 ≤ t ≤ 4 ms:

P 13.68 [a] H(s) =

vo =

Z t

.·. h(λ) = 2500e−2500λ .

16(2500)e−2500λ dλ = 16(1 − e−2500t ) V.

0

4 ms ≤ t ≤ ∞: vo =

Z t t−0.004

16(2500)e−2500λ dλ = 16(e10 − 1)e−2500t V.

13–68

CHAPTER 13. The Laplace Transform in Circuit Analysis

[b] Decrease. [c] The circuit with R = 10 kΩ. P 13.69 [a]

vo =

Z t

20(20e−2λ ) dλ

0

e−2λ t = 400 = −200[e−2t − 1] −2 0

= 200(1 − e−2t ) V, [b]

0 ≤ t ≤ ∞.

Problems 0 ≤ t ≤ 1: vo =

Z t 0

t

400(1 − λ) dλ = 400(λ − λ2 /2) = 200t(2 − t) V; 0

1 ≤ t ≤ ∞: vo =

Z 1

2

400(1 − λ) dλ = 400(λ − λ /2)

0

1 =

200 V.

0

[c]

P 13.70 H(s) =

Vo 8s 0.8s = = Vi 50 + 10s s+5

4 5 = 0.8 − ; = 0.8 1 − s+5 s+5 



h(t) = 0.8δ(t) − 4e−5t . vo = =

Z t

75[0.8δ(λ) − 4e−5λ ] dλ

0

Z t

Z t

60δ(λ) dλ − 300

0

e−5λ dλ

0

e−5λ t = 60 − 300 −5 0

= 60 + 60[e−5t − 1] = 60e−5t V

P 13.71 [a] H(s) = =

0 ≤ t ≤ ∞.

Vo 1/LC = 2 Vi s + (R/L)s + (1/LC) 25 25 = . s2 + 10s + 25 (s + 5)2

h(λ) = 25λe−5λ u(λ).

13–69

13–70

CHAPTER 13. The Laplace Transform in Circuit Analysis

0 ≤ t ≤ 0.10s: Z t

vo = 250

λe−5λ dλ

0

t e−5λ = 250 (−5λ − 1) 25 0 (

)

= 10[1 − e−5t (5t + 1)] V. 0.1 ≤ t ≤ ∞: Z t

vo = 250

λe−5λ dλ

t−0.1

t e−5λ = 250 (−5λ − 1) 25 t−0.1 (



)

= −10e−5t [(5t + 1) − e0.5 (5t + 0.5)] V. [b]

P 13.72 [a] Io =

800Ig 20Ig = ; 25 + 0.025s s + 1000

Io 800 = H(s) = ; Ig s + 1000

Problems h(λ) = 800e−1000λ u(λ). 0 ≤ t ≤ 1 ms:

io =

Z t

−6

−1000λ

(10 × 10 )(800)e

0

e−1000λ t dλ = 0.008 −1000 0

= 8(1 − e−1000t ) µA. 1 ms ≤ t ≤ 4 ms:

io =

Z t−1×10−3

(−5 × 10−6 )(800e−1000λ dλ)

0

Z t

+

t−1×10−3

(10 × 10−6 )(800e−1000λ dλ)

−3 e−1000λ t−1×10 e−1000λ = −0.004 +0.008 −1000 0 −1000

t

t−1×10−3

13–71

13–72

CHAPTER 13. The Laplace Transform in Circuit Analysis h

i

i

h

= 4 e−1000(t−0.001) − 1 − 8 e−1000t − e−1000(t−0.001) ; = [12e−1000(t−0.001) − 8e−1000t − 4] µA. 4 ms ≤ t ≤ ∞:

io =

Z t−0.001

−1000λ

− 0.004e

dλ +

Z t

t−0.004

" −1000λ

= 4e

0.008e−1000λ dλ

t−0.001

t−0.001

−1000λ

−8e

t−0.004

−1000(t−0.001)

= [12e

[b] Vo = 0.025sIo =

t

#

× 10−6 ;

t−0.001

−1000(t−0.004)

− 4e

− 8e−1000t ] µA.

20sIg ; s + 1000

Vo 20s 20,000 = H(s) = = 20 − ; Ig s + 1000 s + 1000 h(λ) = 20δ(λ) − 20,000e−1000λ .

Problems 0 < t < 0.001 s:

vo =

Z t

(10 × 10−6 )[20δ(λ) − 20,000e−1000λ ] dλ

0 −6

= 200 × 10

e−1000λ t − 0.2 −1000 0

= 200 × 10−6 + 200 × 10−6 [e−1000t − 1] = 200e−1000t µV. 0.001 s < t < 0.004 s:

vo =

Z t−0.001

(−5 × 10−6 )[20δ(λ) − 20,000e−1000λ ] dλ

0

Z t

+

(10 × 10−6 )(−20,000e−1000λ ) dλ

t−0.001 −6

= −100 × 10

e−1000λ t−0.001 e−1000λ + 0.1 −0.2 −1000 0 −1000

t

t−0.001

13–73

13–74

CHAPTER 13. The Laplace Transform in Circuit Analysis = −100 × 10−6 − 0.1 × 10−3 e−1000(t−0.001) + 0.1 × 10−3 +0.2 × 10−3 e−1000t − 0.2 × 10−3 e−1000(t−0.001) = 200e−1000t − 300e−1000(t−0.001) µV. 0.004 s < t < ∞:

vo =

Z t−0.001

(−5 × 10−6 )(−20,000e−1000λ ) dλ

t−0.004

Z t

+

(10 × 10−6 )(−20,000e−1000λ ) dλ

t−0.001

= 200e−1000t − 300e−1000(t−0.001) + 100e−1000(t−0.004) µV. [c] At t = 0.001− : io = 8(1 − e−1 ) = 5.06 µA;

i20Ω = (10 − 5.06) = 4.94 µA;

.·. vo = 20(4.94 × 10−6 ) − 5(5.06 × 10−6 ) = 73.58 µV. From the solution for vo we have vo (0.001− ) = 200e−1 = 73.58 µV (checks). At t = 0.001+ : io (0.001+ ) = io (0.001− ) = 5.06 µA; i20Ω = (−5 − 5.06) µA = −10.06 µA; .·. vo (0.001+ ) = 20(−10.06 × 10−6 ) − 5(5.06 × 10−6 ) = −226.42 µV. From the solution for vo we have vo (0.001+ ) = 200e−1 − 300 = −226.42 µV (checks).

Problems At t = 0.004− : io = 12e−3 − 8e−4 − 4 = −3.55 µA; i20Ω = (−5 + 3.55) = −1.45 µA; vo = 20(−1.45 × 10−6 ) − 5(−3.55 × 10−6 ) = −11.27 µV. From the solution for vo , vo (0.004− ) = 200e−4 − 300e−3 = −11.27 µV (checks). At t = 0.004+ : io (0.004+ ) = io (0.004− ) = −3.55 µA;

i20Ω = 3.55 µA;

vo (0.004+ ) = 20(3.55 × 10−6 ) + 5(3.55 × 10−6 ) = 88.73 µV. From the solution for vo , vo (0.004+ ) = 200e−4 − 300e−3 + 100 = 88.73 µV(checks). 60 Ig ; 100 Vg .·. Io = ; 50

P 13.73 [a] Io =

Ig =

Vg ; 30

H(s) =

Io 1 = ; Vg 50

h(λ) = 0.02δ(λ). [b]

0 < t < 0.5 s :

io =

Z t

100[0.02δ(λ)] dλ = 2 A.

0

0.5 s ≤ t ≤ 1.0 s:

io =

Z t−0.5 0

−100[0.02δ(λ)] dλ = −2 A.

13–75

13–76

CHAPTER 13. The Laplace Transform in Circuit Analysis 1s < t < ∞ :

vo = 0.

[c]

Yes, because the circuit has no memory. P 13.74 [a]

Vo − Vg Vo s Vo + + = 0; 40 8 10 (5s + 5)Vo = Vg ; H(s) =

Vo 0.2 = ; Vg s+1

h(λ) = 0.2e−λ u(λ).

[b] 0 ≤ t ≤ 0.5 s:

vo =

Z t

100(0.2e−λ ) dλ = 20

0

vo = 20 − 20e−t V,

e−λ t ; −1 0

0 ≤ t ≤ 0.5 s.

Problems

13–77

0.5 s ≤ t ≤ 1 s:

vo =

Z t−0.5

−λ

(−100)(0.2e

) dλ +

Z t

0

100(0.2e−λ ) dλ

t−0.5

e−λ t e−λ t−0.5 +20 −1 0 −1 t−0.5

= − 20



= 40e−(t−0.5) − 20e−t − 20 V,

0.5 s ≤ t ≤ 1 s.

1 s ≤ t ≤ ∞: vo =

Z t−0.5

−λ

(−100)(0.2e

) dλ +

Z t

t−1

100(0.2e−λ ) dλ

t−0.5

e−λ e−λ t−0.5 = −20 +20 −1 t−1 −1

t

t−0.5

= 40e−(t−0.5) − 20e−(t−1) − 20e−t V,

1 s ≤ t ≤ ∞.

[c]

[d] No, the circuit has memory because of the capacitive storage element. P 13.75 [a] Y (s) =

Z ∞

y(t)e−st dt.

0

Y (s) =

Z ∞

−st

Z ∞

e

0

= =

Z ∞ 0

h(λ)x(t − λ) dλ dt

0

Z ∞Z ∞ 0



e−st h(λ)x(t − λ) dλ dt

0

Z ∞

h(λ)

0

e−st x(t − λ) dt dλ.

13–78

CHAPTER 13. The Laplace Transform in Circuit Analysis But x(t − λ) = 0 when t < λ. Therefore Y (s) =

Z ∞

h(λ)

0

Let u = t − λ; Y (s) =

Z ∞

Z ∞

0

=

Z ∞

Z ∞

u = 0 when t = λ;

u = ∞ when t = ∞.

e−s(u+λ) x(u) du dλ

0

h(λ)e−sλ

0

=

e−st x(t − λ) dt dλ.

λ

du = dt;

h(λ)

Z ∞

Z ∞

e−su x(u) du dλ

0

h(λ)e−sλ X(s) dλ = H(s) X(s).

0

Note on x(t − λ) = 0,

t < λ.

We are using one-sided Laplace transforms; therefore h(t) and x(t) are assumed zero for t < 0.

[b] F (s) =

a 1 a = · = H(s)X(s); 2 s(s + a) s (s + a)2

.·. h(t) = u(t),

x(t) = at e−at u(t).

Problems

.·. f (t) =

Z t

(1)aλe

0

=

t

e−aλ (−aλ − 1) dλ = a a2 0 "

−aλ

1 1 −at [e (−at − 1) − 1(−1)] = [1 − e−at − ate−at ] a a

1 1 −at = − e − te−at u(t). a a 



Check: F (s) =

K0 K1 K2 a = + + ; 2 2 s(s + a) s (s + a) s+a

1 K0 = ; a 

f (t) =

 

K1 = −1;

1 1 − te−at − e−at u(t). a a

P 13.76 vi = 10 sin λ [u(λ) − u(λ − π)]; H(s) =

d a K2 = ds s

1 ; s+1

h(λ) = e−λ ; h(t − λ) = e−(t − λ) = e−t eλ .



s=−a

1 =− ; a

13–79

13–80

CHAPTER 13. The Laplace Transform in Circuit Analysis

−t

vo = 10e

Z t

" −t

= 10e

eλ sin λ dλ

0

t eλ (sin λ − cos λ) 2 0

#

= 5e−t [et (sin t − cos t) + 1] = 5(sin t − cos t + e−t ) V; vo (2.2) = 7.539 V. P 13.77

Vo =

40 × 103 Ig (10 × 103 ); 50 × 103 + 12.5 × 106 /s

Vo 8000s = H(s) = ; Ig s + 250 250 2 × 106 H(s) = 8000 1 − = 8000 − ; s + 250 s + 250 



h(t) = 8000δ(t) − 2 × 106 e−250t .

Problems

vo =

Z 5×10−3

(−10 × 10−3 )[8000δ(λ) − 2 × 106 e−250λ ] dλ

0

Z 7×10−3

+

5×10−3

(5 × 10−3 )[−2 × 106 e−250λ ] dλ Z 5×10−3

= −80 + 20,000

Z 7×10−3

e−250λ dλ − 10,000

0

5×10−3

e−250λ dλ

= −80 − 80(e−1.25 − 1) + 40(e−1.75 − e−1.25 ) = −120e−1.25 + 40e−1.75 = −27.43 V. Alternate: Ig =

Z 2×10−3

(5 × 10−3 )e−st dt +

Z 8×10−3

0

2×10−3

(−10 × 10−3 )e−st dt

5 15 −2×10−3 s 10 −8×10−3 s = + e × 10−3 ; − e s s s 



Vo = Ig H(s) =

8 −3 −3 [5 − 15e−2×10 s + 10e−8×10 s ] s + 250 −3 s

40 120e−2×10 = − s + 250 s + 250

vo (t) = 40e−250t − 120e−250(t−2×10 +80e−250(t−8×10

−3 )

−3

80e−8×10 s + ; s + 250 −3 )

u(t − 2 × 10−3 )

u(t − 8 × 10−3 ) V;

vo (7 × 10−3 ) = 40e−1.75 − 120e−1.25 + 0 = −27.43 V (checks). P 13.78 Vo =

75 100 25 192 × 106 − + = ; s s + 800 s + 3200 s(s + 800)(s + 3200)

240 Vo = H(s)Vg = H(s) ; s 

.·. H(s) =

H(j1600) =



800,000 . (s + 800)(s + 3200) 8 × 105 = 0.125/ − 90◦ ; (800 + j1600)(3200 + j1600)

.·. vo (t) = (40)(0.125) cos(1600t − 90◦ ) V = 5 sin 1600t V.

13–81

13–82

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.79 H(j8000) = =

104 (6000 + j8000) −64 × 106 + j7 × 106 + 88 × 106 107 (6 + j8) = 4/36.87◦ ; 106 (24 + j7)

.·. vo (t) = 50 cos(8000t + 36.87◦ ) V. P 13.80 H(j20) =

25(8 + j20) = 0.44/ − 33.57◦ ; −400 + j1200 + 150

.·. io (t) = 4.4 cos(20t − 33.57◦ ) A. P 13.81 [a] H(s) =

−Zf ; Zi

Zf =

(1/Cf ) 4 × 109 = ; s + (1/Rf Cf ) s + 16,000

Zi =

Ri [s + (1/Ri Ci )] 25,000(s + 8000) = ; s s

H(s) =

−16 × 104 s . (s + 8000)(s + 16,000)

[b] H(j8000) =

√ −16 × 104 (j8000) = 40/ − 161.57◦ ; (8000 + j8000)(16,000 + j8000)

√ √ vo (t) = (200 10) × 10−3 ( 40) cos(8000t − 161.57◦ ) = 4 cos(8000t − 161.57◦ ) V. P 13.82 [a] Let R1 = 10 kΩ, Then Vn = Vp = Also

R2 = 50 kΩ,

C = 400 pF,

R2 C = 2 × 10−5 .

Vg R2 . R2 + (1/sC)

Vn − Vg Vn − Vo + = 0, R1 R1

therefore Vo = 2Vn − Vg . Now solving for Vo /Vg , we get H(s) = It follows that H(j50,000) =

R2 Cs − 1 . R2 Cs + 1

j−1 = j1 = 1/90.◦ j+1

Therefore vo = 10 cos(50,000t + 90◦ ) V.

Problems

[b] Replacing R2 by Rx gives us H(s) =

Rx Cs − 1 . Rx Cs + 1

Therefore H(j50,000) =

j20 × 10−6 Rx − 1 Rx + j50,000 = . −6 j20 × 10 Rx + 1 Rx − j50,000

Thus, 50,000 = tan 60◦ = 1.7321, Rx

Rx = 28,867.51 Ω.

P 13.83 [a] The s-domain circuit is

The node-voltage equation is So V =

ρR s + (R/Le )

V V V ρ + + = sL1 R sL2 s

where Le =

L1 L2 ; L1 + L2

Therefore v = ρRe−(R/Le )t u(t) V. [b] I1 =

V V ρ[s + (R/L2 )] K0 K1 + = = + . R sL2 s[s + (R/Le )] s s + (R/Le )

K0 =

ρL1 ; L1 + L2

K1 =

Thus we have i1 = [c] I2 =

ρL2 ; L1 + L2

ρ [L1 + L2 e−(R/Le )t ]u(t) A. L1 + L2

V (ρR/L2 ) K2 K3 = + . = sL2 s[s + (R/Le )] s s + (R/Le )

K2 =

ρL1 ; L1 + L2

Therefore i2 =

K3 =

−ρL1 ; L1 + L2

ρL1 [1 − e−(R/Le )t ]u(t). L1 + L2

[d] λ(t) = L1 i1 + L2 i2 = ρL1 .

13–83

13–84

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.84 [a] As R → ∞, v(t) → ρLe δ(t) since the area under the impulse generating function is ρLe . ρL1 i1 (t) → u(t) A as R → ∞; L1 + L2 i2 (t) →

ρL1 u(t) A as R → ∞. L1 + L2

[b] The s-domain circuit is

V V ρ + = ; sL1 sL2 s

so V =

ρL1 L2 = ρLe ; L1 + L2

Therefore v(t) = ρLe δ(t) V ρL1 = sL2 L1 + L2 

I1 = I2 =

Thus i1 = i2 =

1 ; s

 

ρL1 u(t) A. L1 + L2

P 13.85 Original charge on C1 ; q1 = V0 C1 . The charge transferred to C2 ; q2 = V0 Ce = The charge remaining on C1 ;

q10

V0 C1 q2 = Therefore V2 = C2 C1 + C2

V0 C12 = q1 − q2 = . C1 + C2 q10 V0 C1 and V1 = = . C1 C1 + C2

P 13.86 [a]

Vo = =

V0 C1 C2 . C1 + C2

0.75 · 20 × 10−3 s 100 + 25 × 10−3 s 0.6s 2400 = 0.6 − ; s + 4000 s + 4000

Problems

13–85

vo (t) = 0.6δ(t) − 2400e−4000t u(t) V. [b] At t = 0 the voltage impulse establishes a current in the inductors; thus iL (0) =

103 Z 0+ 750 × 10−3 δ(t) dt = 30 A. 25 0−

It follows that since iL (0− ) = 0 that diL (0) = 30δ(t); dt .·. vo (0) = (20 × 10−3 )(30δ(t)) = 0.6δ(t). This agrees with our solution. At t = 0+ our circuit is

.·. iL (t) = 30e−t/τ A,

t ≥ 0+ ;

τ = L/R = 0.25 ms; .·. iL (t) = 30e−4000t A, vo (t) = 20 × 10−3

t ≥ 0+ ;

diL = −2400e−4000t V, dt

t ≥ 0+ ;

which agrees with our solution. P 13.87 [a] After making a source transformation, the circuit is as shown. The impulse current will pass through the capacitive branch since it appears as a short circuit to the impulsive current,

So vo (0+ ) = 106

Z 0+ " 0−

#

δ(t) dt = 1000 V; 1000

Therefore wC = (0.5)Cv 2 = 0.5 J.

13–86

CHAPTER 13. The Laplace Transform in Circuit Analysis

[b] iL (0+ ) = 0;

therefore wL = 0 J. Vo Vo [c] Vo (10−6 )s + + = 10−3 ; 250 + 0.05s 1000 Therefore 1000(s + 5000) Vo = 2 s + 6000s + 25 × 106 =

K1∗ K1 + . s + 3000 − j4000 s + 3000 + j4000

K1 = 559.02/ − 26.57◦ ;

K1∗ = 559.02/26.57◦ ;

vo = [1118.03e−3000t cos(4000t − 26.57◦ )]u(t) V. [d] The s-domain circuit is

Vo s Vo Vo + + = 10−3 . 6 10 250 + 0.05s 1000 Note that this equation is identical to that derived in part [c], therefore the solution for Vo will be the same. P 13.88 [a]

Vo = =

2.7 2 × 106 · 54 × 103 + 25 × 106 /s + 2 × 106 /s s 5.4 × 106 100 ; = 3 6 54 × 10 s + 27 × 10 s + 500

vo (t) = 100e−500t u(t) V.

Problems

13–87

[b] At t = 0 the impulsive current passes through the two capacitors. The voltage on the 0.04 µF capacitor at t = 0+ is v0.04 = 25 × 106

Z 0+ 0−

50 × 10−6 δ(t) dt = 1250 V.

The voltage on the 0.5 µF capacitor at t = 0+ is v0.5 = 2 × 106

Z 0+ 0−

50 × 10−6 δ(t) dt = 100 V.

Note this agrees with our solution. At t = 0+ the circuit is

The equivalent capacitance is (0.04)(0.5) × 10−12 1 Ce = = µF. −6 0.54 × 10 27 Thus, the time constant is τ = 54 × 103 Ce = 2 ms. Therefore, 1/τ = 500, which agrees with our solution. It follows that vR (t) = 1350e−500t V,

t ≥ 0+ .

Therefore 0.04 vo (t) = vR = 100e−500t V, t ≥ 0+ , 0.54 which also agrees with our solution. P 13.89 [a] For t < 0:

Req = 4 kΩk20 kΩk80 kΩ = 3.2 kΩ; i1 (0− ) =

64,000 = 3.2 A; 20,000

i2 (0− ) =

v = 20(3200) = 64 kV; 64,000 = 0.8 A. 80,000

13–88

CHAPTER 13. The Laplace Transform in Circuit Analysis

[b] For t > 0: i1 + i2 = 0; 400(∆i1 ) = 100(∆i2 ); i1 (0− ) + ∆i1 + i2 (0− ) + ∆i2 = 0;

therefore ∆i1 = −0.8 A;

i1 (0+ ) = 3.2 − 0.8 = 2.4 A.

∆i2 = −3.2 A; [c] i2 (0− ) = 0.8 A.

[d] i2 (0+ ) = 0.8 − 3.2 = −2.4 A. [e] The s-domain equivalent circuit for t > 0 is

I1 =

2.4 1.2 = ; 0.5s + 100,000 s + 2 × 105 5

i1 (t) = 2.4e−2×10 t u(t) A. 5

[f ] i2 (t) = −i1 (t) = −2.4e−2×10 t u(t) A. −0.32(s + 650 × 103 ) [g] V = −1.28 + (0.4s + 20,000)I1 = s + 2 × 105 = −0.32 −

144,000 ; s + 2 × 105 5

v(t) = [−0.32δ(t)] − [144,000e−2×10 t u(t)] V. P 13.90 [a]

50 = 2sI1 − sI2 ;

Problems

13–89

1.5 0 = −sI1 + 2s + I2 ; s 

∆=



2s −s

−s (2s + 1.5/s)

= 4s2 + 3 − s2 = 3(s2 + 1);

50 0

(2s + 1.5/s)

= 100s +

N1 =

−s

75 100s2 + 75 100(s2 + 0.75) = = ; s s s

N1 100(s2 + 0.75) 100 s2 + 0.75 I1 = = · = ∆ s(3)(s2 + 1) 3 s(s2 + 1) K1 K1∗ K0 + + . s s−j s+j

=

"



#

100 −1 + 0.75 12.5 ◦ /0 ; K1 = = 3 (j1)(j2) 3

100 0.75 = 25; K0 = 3 1 

25 .·. i1 = 25 + cos t u(t) A. 3 

[b] N2 =

2s −s



50 0

= 50s;

N2 50s 50 K1 K1∗ s = = = + . ∆ 3(s2 + 1) 3 s2 + 1 s−j s+j 

I2 =

50 K1 = 3 

i2 =

j1 j2

!

=



25 ◦ /0 ; 3

50 cos t u(t) A. 3 

1.5 50 K1 K1∗ 1.5 s 25 I2 = = + . [c] Vo = = s s 3 s2 + 1 s2 + 1 s−j s+j 

K1 =







25 = −j12.5 = 12.5/ − 90◦ ; j2

vo = 25 cos(t − 90◦ ) = 25 sin t; vo = [25 sin t]u(t) V. [d] Let us begin by noting i1 jumps from 0 to (100/3) A between 0− and 0+ and in this same interval i2 jumps from 0 to (50/3) A. Therefore in the

13–90

CHAPTER 13. The Laplace Transform in Circuit Analysis derivatives of i1 and i2 there will be impulses of (100/3)δ(t) and (50/3)δ(t), respectively. Thus di1 100 50 = δ(t) − sin t A/s; dt 3 3 di2 50 100 = δ(t) − sin t A/s. dt 3 3 From the circuit diagram we have 50δ(t) = 2 =

di1 di2 − dt dt

200 100 50δ(t) 100 δ(t) − sin t − + sin t 3 3 3 3

= 50δ(t). Thus our solutions for i1 and i2 are in agreement with known circuit behavior. Let us also note the impulsive voltage will impart energy into the circuit. Since there is no resistance in the circuit, the energy will not dissipate. Thus the fact that i1 , i2 , and vo exist for all time is consistent with known circuit behavior. Also note that although i1 has a dc component, i2 does not. This follows from known transformer behavior. Finally we note the flux linkage prior to the appearance of the impulsive voltage is zero. Now since v = dλ/dt, the impulsive voltage source must be matched to an instantaneous change in flux linkage at t = 0+ of 50. For the given polarity dots and reference directions of i1 and i2 we have λ(0+ ) = L1 i1 (0+ ) + M i1 (0+ ) − L2 i2 (0+ ) − M i2 (0+ ); λ(0+ ) = 2 =

P 13.91 [a] Z1 =



100 50 50 100 + −2 − 3 3 3 3 







300 150 − = 50. (checks) 3 3

1/C1 20 × 1010 = Ω; s + 1/R1 C1 s + 20 × 104

Z2 =



1/C2 5 × 1010 = Ω; s + 1/R2 C2 s + 12,500





Problems Vo − 10/s Vo + = 0; Z2 Z1 Vo (s + 12,500) Vo (s + 20 × 104 ) 10 (s + 20 × 104 ) + = ; 5 × 1010 20 × 1010 s 20 × 1010 Vo =

2(s + 200,000) K1 K2 = + . s(s + 50,000) s s + 50,000

K1 =

2(200,000) = 8; 50,000

K2 =

2(150,000) = −6; −50,000

.·. vo = [8 − 6e−50,000t ]u(t) V. [b] I0 =

V0 2(s + 200,000)(s + 12,500) = Z2 s(s + 50,000)5 × 1010 "

162,500s + 25 × 108 1+ s(s + 50,000)

"

K1 K2 1+ + . s s + 50,000

−12

= 40 × 10

−12

= 40 × 10

K1 = 50,000;

#

#

K2 = 112,500;

io = 40δ(t) + [2 × 106 + 4.5 × 106 e−50,000t ]u(t) pA. [c] When Z1 =

C1 = 80 pF, 125 × 108 Ω. s + 12,500

Vo (s + 12,500) Vo (s + 12,500) 10 (s + 12,500) + = ; 500 × 108 125 × 108 s 125 × 108 40 .·. Vo + 4V0 = ; s 8 Vo = ; s vo = 8u(t) V. V0 8 (s + 12,500) 12,500 −12 I0 = = = 160 × 10 1 + ; Z2 s 5 × 1010 s 

io (t) = 160δ(t) + 2 × 106 u(t) pA.



13–91

13–92

CHAPTER 13. The Laplace Transform in Circuit Analysis

P 13.92 Let a =

1 1 = . R1 C1 R2 C2

Then Z1 =

1 C1 (s + a)

and

Z2 =

1 . C2 (s + a)

Vo Vo 10/s + = ; Z2 Z1 Z1 Vo C2 (s + a) + V0 C1 (s + a) = (10/s)C1 (s + a); 10 C1 Vo = ; s C1 + C2 



Thus, vo is the input scaled by the factor P 13.93 [a] For t < 0,

125v1 = 500v2 ;

v1 + v2 = 25;

C1 . C1 + C2

therefore v1 = 4v2 ;

therefore v1 (0− ) = 20 V.

[b] v2 (0− ) = 5 V. [c] v3 (0− ) = 0 V. [d] For t > 0:

I=

25/s × 10−6 = 2 × 10−6 ; 12.5/s

i(t) = 2δ(t) µA. 109 Z 0+ 2 × 10−6 δ(t) dt + 20 = −16 + 20 = 4 V. 125 0− 109 Z 0+ [f ] v2 (0+ ) = − 2 × 10−6 δ(t) dt + 5 = −4 + 5 = 1 V. 500 0−

[e] v1 (0+ ) = −

Problems

[g] V3 =

13–93

2.5 × 106 5 · 2 × 10−6 = s s

v3 (t) = 5u(t) V;

v3 (0+ ) = 5 V.

Check: v1 (0+ ) + v2 (0+ ) = v3 (0+ ). P 13.94 [a] The circuit parameters are 1202 1202 1202 Ra = = 20 Ω Rb = = 32 Ω Xa = = 40 Ω. 720 450 360 The branch currents are 120/0◦ 120/0◦ = −j3 = 3/ − 90◦ A(rms); = 6/0◦ A(rms) I2 = I1 = j40 20 120/0◦ I3 = = 3.75/0◦ A(rms); 32 .·. Io = I1 + I2 + I3 = 9.75 − j3 = 10.2/ − 17.1◦ A(rms). Therefore, √ i2 = 3 2 cos(ωt − 90◦ ) A Thus, i2 (0− ) = i2 (0+ ) = 0 A

√ io = 10.2 2 cos(ωt − 17.1◦ ) A.

and

and

√ io (0− ) = io (0+ ) = 9.75 2 A.

[b] Begin by using the s-domain circuit in Fig. 13.60 to solve for Vo symbolically. Write a single node voltage equation: Vo − (Vg + L` Io ) Vo Vo + + = 0; sL` Ra sLa .·. Vo =

(Ra /L` )Vg + Io Ra , s + [Ra (La + L` )]/La L`

√ where L` = 1/60π H, La = 1/3π H, Ra = 20 Ω, and I0 Ra = 195 2 V. Also, Vg = 120/0◦ + (9.75 − j3)(j2) = 126 + j19.5V(rms). Inverse phasor-transforming gives us √ vg (t) = 2[126 cos(120πt) − 19.5 sin(120πt)]V(rms). The Laplace transform of vg (t) is then √ 2(126s − 2340π) Vg = . s2 + 14,400π 2 Thus,

√ √ √ 1200π(126 2s − 2340π 2) 195 2 Vo = + (s + 1260π)(s2 + 14,400π 2 ) s + 1260π

13–94

CHAPTER 13. The Laplace Transform in Circuit Analysis √ K1 K2 K2∗ 195 2 = + + + . s + 1260π s − j120π s + j120π s + 1260π The coefficients are √ K1 = −120.67 2 V

√ √ K2 = 60.44 2/3.357◦ V K2∗ = 60.44 2/ − 3.357◦ . √ √ Note that K1 + 195 2 = 74.33 2 V. Thus, the inverse transform of Vo is √ √ vo = 74.33 2e−1260πt + 120.88 2 cos(120πt + 3.357◦ ) V. Initially, √ √ √ vo (0+ ) = 74.33 2 + 120.88 2 cos 3.357◦ = 195 2 V.

√ Note that at t = 0+ the initial value of i0 , which is 9.75 in the √ 2 A, exists √ 20 Ω resistor Ra . Thus, the initial value of Vo is (9.75 2)(20) = 195 2 V. [c] The phasor domain equivalent circuit has a j2 Ω inductive impedance in series with the parallel combination of a 20 Ω resistive impedance and a j40 Ω inductive impedance (remember that ω = 120π rad/s). From part (b) Vg = 126 + j19.5 = 127.5/8.8◦ V(rms). The node voltage equation in the phasor domain circuit is V0 V0 − 127.5/8.8◦ V0 + + = 0; j2 20 j40 .·. V0 = 120.88/3.357◦ V(rms). √ Therefore, v0 = 120.88 2 cos(120πt + 3.357◦ ) V, agreeing with the steady-state component of the result in part (b). [d] A plot of v0 , generated in Excel, is shown below.

Problems P 13.95 [a] At t = 0− the phasor domain equivalent circuit is

I1 =

−j120 = −j6 = 6/ − 90◦ A (rms); 20

I2 =

−j120 = −3 = 3/180◦ A (rms); j40

I3 =

−j120 = −j3.75 = 3.75/ − 90◦ A (rms); 32

IL = I1 + I2 + I3 = −3 − j9.75 = 10.2/ − 107.103◦ A (rms); √ iL = 10.2 2 cos(120πt − 107.103◦ )A. √ iL (0− ) = iL (0+ ) = −3 2A; √ i2 = 3 2 cos(120πt + 180◦ )A. √ i2 (0− ) = i2 (0+ ) = −3 2A; Vg = Vo + j2IL ; Vg = −j120 + 19.5 − j6 = 19.5 − j126 = 127.5/ − 81.2◦ V (rms); √ vg = 127.5 2 cos(120πt − 81.2◦ ) √ = 127.5 2[cos 120πt cos 81.2◦ + sin 120πt sin 81.2◦ ] √ √ = 19.5 2 cos 120πt + 126 2 sin 120πt V; √ √ 19.5 2s + 126 2(120π) · . . Vg = . s2 + (120π)2 s-domain circuit:

13–95

13–96

CHAPTER 13. The Laplace Transform in Circuit Analysis where 1 1 H; La = H; Ra = 20 Ω; 60π 3π √ √ iL (0) = −3 2 A; i2 (0) = −3 2 A. Ll =

The node voltage equation is 0=

Vo − (Vg + iL (0)Ll ) Vo Vo + i2 (0)La + + . sLl Ra sLa

Solving for Vo yields Vo =

Vg Ra /Ll Ra [iL (0) − i2 (0)] + . [s + Ra (Ll + La )/La Ll ] [s + Ra (Ll + La )/Ll La ]

Ra = 1200π; Ll 1 1 10( 60π + 3π ) Ra (Ll + La ) = = 1260π; 1 1 Ll La ( 3π )( 60π ) √ √ iL (0) − i2 (0) = −3 2 + 3 2 = 0; √ √ 1200π[19.5 2s + 126 2(120π)] .·. Vo = 2 (s + 1260π)[s + (120π)2 ] K2 K2∗ K1 = + + ; s + 1260π s − j120π s + j120π √ √ K1 = −7.08 2 K2 = 60.44 2/ − 86.643◦ ; √ √ .·. vo (t) = −7.08 2e−1260πt + 120.88 2 cos(120πt − 86.643◦ )V.

Check: √ vo (0) = (−7.08 + 7.08) 2 = 0. [b]

Problems

13–97

√ [c] In Problem 13.94, the line-to-neutral voltage spikes at 195 2 V. Here the line-to-neutral voltage has no spike. Thus the amount of voltage disturbance depends on what part of the cycle the sinusoidal steady-state voltage is switched. P 13.96 [a] First find Vg before Rb is disconnected. The phasor domain circuit is

120/θ◦ 120/θ◦ 120/θ◦ + + Ra Rb jXa 120/θ◦ = [(Ra + Rb )Xa − jRa Rb ]. Ra Rb Xa

IL =

Since Xl = 2 Ω we have Vg = 120/θ◦ +

120/θ◦ (2)[Ra Rb + j(Ra + Rb )Xa ]; Ra Rb Xa

Ra = 20 Ω;

Rb = 32 Ω;

Xa = 40 Ω;

Vg = 120/θ◦ + 120/θ◦ (0.05 + j0.1625) = 120/θ◦ (1.05 + j0.1625) = 127.5/(θ + 8.8)◦ ; √ vg = 127.5 2 cos(120πt + θ + 8.8◦ )V. Let β = θ + 8.8◦ . Then √ vg = 127.5 2(cos 120πt cos β − sin 120πt sin β)V. Therefore

√ 127.5 2(s cos β − 120π sin β) Vg = . s2 + (120π)2 The s-domain circuit becomes

where ρ1 = iL (0+ ) and ρ2 = i2 (0+ ).

13–98

CHAPTER 13. The Laplace Transform in Circuit Analysis The s-domain node voltage equation is Vo Vo + ρ2 La Vo − (Vg + ρ1 Ll ) + + = 0. sLl Ra sLa Solving for Vo yields Vo =

Vg Ra /Ll + (ρ1 − ρ2 )Ra [s +

(La +Ll )Ra ] La Ll

.

Substituting the numerical values Ll =

1 H; 60π

La =

1 H; 3π

Ra = 20 Ω;

gives Vo =

1200πVg + 20(ρ1 − ρ2 ) . (s + 1260π)

Now determine the values of ρ1 and ρ2 . ρ1 = iL (0+ ) IL = =

and

ρ2 = i2 (0+ );

120/θ◦ [(Ra + Rb )Xa − jRa Rb ] Ra Rb Xa 120/θ◦ (52(40) − j640) 640(40)

= 10.2/(θ − 17.1)◦ A(rms); √ .·. iL = 10.2 2 cos(120πt + θ − 17.1◦ )A. √ iL (0+ ) = ρ1 = 10.2 2 cos(θ − 17.2◦ )A; √ √ .·. ρ1 = 9.75 2 cos θ + 3 2 sin θA. 120/θ◦ = 3/(θ − 90)◦ ; j40 √ i2 = 3 2 cos(120πt + θ − 90◦ )A; √ ρ2 = i2 (0+ ) = 3 2 sin θA; √ .·. ρ1 − ρ2 = 9.75 2 cos θ. √ (ρ1 − ρ2 )Ra = 195 2 cos θ.

I2 =

Rb = 32 Ω;

Problems

13–99

√ 195 1200π 2 cos θ · Vg + .·. Vo = s + 1260π s + 1260π √ √ " # 1200π 127.5 2(s cos β − 120π sin β) 195 2 cos θ = + s + 1260π s2 + 14,400π 2 s + 1260π √ K1 + 195 2 cos θ K2 K2∗ = + + . s + 1260π s − j120π s + j120π Now

√ (1200π)(127.5 2)[−1260π cos β − 120π sin β] K1 = 12602 π 2 + 14,400π 2 √ −1200(127.5 2)[1260 cos β + 120 sin β] . = 12602 + 14,400

Since β = θ + 8.8◦ , K1 reduces to √ √ K1 = −120.67 2 cos θ + 7.08 2 sin θ. From the partial fraction expansion√for Vo we see vo (t) will go directly into steady state when K1 = −195 2 cos θ. It follows that √ √ 7.08 2 sin θ = −74.33 2 cos θ or

tan θ = −10.5;

Therefore,

θ = −84.5597◦ .

[b] When θ = −84.5597◦ , β = −75.7623◦ . √ 2)[−120π sin(−75.7623◦ ) + j120π cos(−75.7623◦ ) 1200π(127.5 · . . K2 = (1260π + j120π)(j240π) √ 637.5 2(116.314 + j29.5135) = −120 + j1260 √ = 60.44 2/ − 81.20◦ ; √ .·. vo = 120.88 2 cos(120πt − 81.20◦ ) V t > 0 = 170.95 cos(120πt − 81.20◦ ) V t > 0. [c] vo = 169.71 cos(120πt − 84.5597◦ )V vo = 170.95 cos(120πt − 81.20◦ )V

t < 0; t > 0.

13–100

CHAPTER 13. The Laplace Transform in Circuit Analysis

Introduction to Frequency-Selective Circuits

Assessment Problems AP 14.1 fc = 4 kHz, ωc =

1 ; RC

ωc = 2πfc = 8π krad/s; C = 10 nF;

1 1 = = 3978.9 Ω. .·. R = 3 ωc C (8π × 10 )(10 × 10−9 ) AP 14.2 [a] ωc = 2πfc = 2π(10,000) = 20π krad/s; R = Lωc = (10−3 )(20,000π) = 62.8 Ω. [b] H(jω) =

ωc 20,000π = . ωc + jω 20,000π + jω

When ω = 2πf = 2π(100,000) = 200,000π rad/s, H(j200,000π) =

20,000π 1 = = 0.0995/ − 84.3◦ ; 20,000π + j200,000π 1 + j10

.·. |H(j200,000π)| = 0.0995. [c] .·. θ(200,000π) = −84.3◦ . AP 14.3 ωc =

15,000 R = = 1.5 Mrad/s. L 10 × 10−3

AP 14.4 [a] ωc =

1 1 = = 425.5 rad/s. RC (50)(47 × 10−6 )

14–1

14–2

CHAPTER 14. Introduction to Frequency-Selective Circuits 1 = 21.3 rad/s. × 10−6 ) 1 [c] ωc = = 4.26 rad/s. (5000)(47 × 10−6 ) [b] ωc =

(103 )(47

AP 14.5 Let Z represent the parallel combination of (1/sC) and RL . Then RL . (RL Cs + 1)

Z=

Thus H(s) = =

where K =

AP 14.6 ωo2 =

1 LC

(1/RC) s+

R+RL RL



1 RC



=

(1/RC) s+

1 K



1 RC

,

RL . R + RL

so C =

ωo ωo = β R/L

Q=

Z RL = R+Z R(RL Cs + 1) + RL

1 1 = = 40.53 nF; 2 3 ωo L (50π × 10 )2 (1 × 10−3 )

so R =

ωo L (50π × 103 )(1 × 10−3 ) = = 19.63 Ω. Q 8

AP 14.7 ωo = 2π(8000) = 16,000π rad/s; β = 2π(500) = 1000π rad/s; β=

1 RC

so R =

1 1 = = 12.73 Ω; βC (1000π)(25 × 10−6 )

ωo2 =

1 LC

so L =

AP 14.8 ωo2 =

1 LC

so C =

β=

1 RC

so R =

C = 25 × 10−6 F;

1 ωo2 C

=

1 (16,000π)2 (25

× 10−6 )

= 15.83 µH.

1 1 = = 1.0132 µF; 2 4 ωo L (10 π)2 (10−3 )

1 1 = = 785.4 Ω. βC 400π(1.0132 × 10−6 )

Problems AP 14.9 ωo2 =

1 LC

1

so L =

ωo2 C

=

1 (16,000π)2 (25

14–3

= 15.83 mH.

× 10−9 )

In Example 14.6(d), Q = fo /β = 5000/200 = 25. In this problem, fo = 8 kHz but Q is unchanged, so calculate the new bandwidth: β=

fo 8000 = = 320 Hz. Q 25

1 1 .·. R = = = 19.89 kΩ. βC 2π(320)(25 × 10−9 ) AP 14.10 ωo = 30,000π rad/s; L = 10 µH; ωo2 =

1 LC

Q=

ωo = RCωo ; β

so C =

1 = 11.26 µF; ωo2 L

Q 6 .·. R = = = 5.65 Ω. ωo C (30,000π)(11.26 × 10−6 ) AP 14.11 ωo = 2πfo = 2π(4000) = 8π krad/s; Q=

ωo ωo = β (1/RC)

ωo2 =

1 LC

so L =

so C = 1 ωo2 C

=

R = 30 Ω;

Q = 6;

Q 6 = = 7.96 µF; Rωo (30)(8000π) 1

(8000π)2 (7.96

× 10−6 )

= 198.9 µH.

14–4

CHAPTER 14. Introduction to Frequency-Selective Circuits

Problems P 14.1

R 1.5 × 103 = = 6000 rad/s; L 0.25 6000 .·. fc = = 954.93 Hz. 2π R/L 6000 [b] H(s) = = ; s + R/L s + 6000

[a] ωc =

H(jω) = H(jωc ) =

6000 ; 6000 + jω 6000 = 0.7071/ − 45◦ ; 6000 + j6000

H(j0.3ωc ) = H(j3ωc ) =

6000 = 0.9578/ − 16.70◦ ; 6000 + j1800

6000 = 0.3162/ − 71.57◦ . 6000 + j18,000

[c] vo (ωc ) = 35.36 cos(6000t − 45◦ ) V; vo (0.3ωc ) = 47.89 cos(1800t − 16.70◦ ) V; vo (3ωc ) = 15.81 cos(18,000t − 71.57◦ ) V.

P 14.2

Vo R [a] H(U L) (s) = = = Vi R + sL

R L

. R L   R RL Vo RkRL L R + RL [b] H(L) (s) = = =  . R RL Vi RkRL + sL s+ L R + RL   R RL R ωc(L) = [c] ωc(U L) = ; so the cutoff frequencies are different. L L R + RL H(0)(U L) = 1;

s+

H(0)(L) = 1

so the passband gains are the same.

1000 = 50,000 rad/s. 0.02 [e] From part (c), ωc(L) < ωc(U L) . Therefore,

[d] ωc(U L) =

ωc(L) = 50,000 − 0.1(50,000) = 45,000 rad/s; 1000 RL 0.02 1000 + RL 

45,000 =



so

.·. 0.1RL = 900 so RL ≥ 9 kΩ.

RL = 0.9; 1000 + RL

Problems

P 14.3

R (R/L) Vo = = . Vi sL + R + Rl s + (R + Rl )/L (R/L) ; [b] H(jω) =  R+R  l + jω L

[a] H(s) =

|H(jω)| = r

(R/L)  R+Rl 2 L

;

+ ω2

|H(jω)|max occurs when ω = 0. R . R + Rl R R/L [d] |H(jωc )| = √ = r  2(R + Rl ) R+Rl 2

[c] |H(jω)|max =

L

R + Rl 2 .·. ωc2 = ; L 1575 = 6300 rad/s; [e] ωc = 0.25 6000 H(jω) = . 6300 + jω 



; + ωc2

.·. ωc = (R + Rl )/L.

H(j0) = 0.9524; 0.9524 √ / − 45◦ = 0.6734/ − 45◦ ; 2 6000 H(j1890) = = 0.9122/ − 16.70◦ ; 6300 + j1890 H(j6300) =

H(j18,900) = P 14.4

R so R = Lωc = (0.01)(2500) = 25 Ω. L = 100k25 = 20 Ω;

[a] ωc = [b] Req

ωc = P 14.5

6000 = 0.3012/ − 71.57◦ . 6300 + j18,900

Req 20 = = 2000 rad/s. L 0.01

[a] ωc = 0.95(2500) = 2375 =

Req 0.01

Then, 23.75 = 25kRL = Solving, RL = 475 Ω.

25RL . 25 + RL

so Req = 2375(0.01) = 23.75 Ω.

14–5

14–6

CHAPTER 14. Introduction to Frequency-Selective Circuits

[b] H(s) =

ωc 2375 = ; s + ωc s + 2375

|H(j0)| = P 14.6

2375 = 1. 2375

R = 5000π rad/s; L R = Lωc = (0.025)(5000π) = 392.7 Ω.

[a] ωo =

[b] Re = 392.7k750 = 257.745 Ω; Re = 10,309.8 rad/s; L ωloaded .·. floaded = = 1640.85 Hz. 2π [c] Use a 390 Ω resistor, which gives ωc = R/L = 390/0.025 = 15,600 rad/s = 2482.8 Hz. ωloaded =

P 14.7

[a] ωc =

109 = 12,500 rad/s; 80 × 103

fc = 1989.44 Hz. [b] H(jω) =

12,500 ; 12,500 + jω

.·. H(jωc ) = 0.7071/ − 45◦ . H(j0.2ωc ) = H(j8ωc ) =

12,500 = 0.9806/ − 11.31◦ ; 12,500 + j2500

12,500 = 0.1240/ − 82.87◦ . 12,500 + j100,000

[c] vo (ωc ) = 339.41 cos(12,500t − 45◦ ) mV; vo (0.2ωc ) = 470.68 cos(2500t − 11.31◦ ) mV; vo (8ωc ) = 59.54 cos(100,000t − 82.87◦ ) mV. P 14.8

[a] Let Z =

RL (1/SC) RL = . RL + 1/SC RL Cs + 1

Then H(s) = = =

Z Z +R RL RRL Cs + R + RL (1/RC)  R + RL s+ . RRL C 

Problems (1/RC)

[b] |H(jω)| = q

ω 2 + [(R + RL )/RRL C]2

14–7

.

|H(jω)| is maximum at ω = 0. RL . R + RL RL (1/RC) [d] |H(jωc )| = √ =q ; 2 2(R + RL ) ωC + [(R + RL )/RRL C]2 [c] |H(jω)|max =

1 R + RL = (1 + (R/RL )) . .·. ωc = RRL C RC 20 [e] ωc = 12,500 1 + = 13,333.33 rad/s. 300 

H(j0) =

300 = 0.9375; 320

H(jωc ) =

12,500 = 0.6629/ − 45◦ ; 13,333.33 + j13,333.33

H(j0.2ωc ) = H(j8ωc ) = P 14.9



12,500 = 0.9193/ − 11.31◦ ; 13,333.33 + j2666.67

12,500 = 0.1163/ − 82.87◦ . 13,333.33 + j106,666.67

[a] ωc = 2π(500) = 3141.59 rad/s. 1 1 1 so R = = = 6366 Ω. [b] ωc = RC ωc C (3141.59)(50 × 10−9 ) [c]

1/sC 1/RC 3141.59 Vo = = = . Vi R + 1/sC s + 1/RC s + 3141.59 Vo (1/sC)kRL 1/RC 3141.59 [e] H(s) = = = . =   R + RL Vi R + (1/sC)kRL s + 2(3141.59) s+ 1/RC RL [f ] ωc = 2(3141.59) = 6283.19 rad/s. [d] H(s) =

[g] H(0) = 1/2. P 14.10 [a] fc =

160 × 103 = 25.46 kHz. 2π

14–8

CHAPTER 14. Introduction to Frequency-Selective Circuits

[b]

1 = 160 × 103 ; RC R=

1 = 250 Ω. (160 × 103 )(25 × 10−9 )

1 R [c] ωc = 1+ ; RC RL 

.·.



R = 0.08 RL

[d] H(j0) =

.·. RL = 12.5R = 3125 Ω.

RL 3125 = = 0.9259; R + RL 3375

H(j0) = 0.9259. P 14.11 [a] R = ωc L = (160 × 103 )(25 × 10−3 ) = 4000 Ω = 4 kΩ. R RL · = 150,000; [b] L R + RL .·.

RL 150,000 = = 0.9375; R + RL 160,000

.·. 0.0625RL = (0.9375)(4000); P 14.12 [a] H(s) =

.·. RL = 60 kΩ.

s s sL = = . R + sL s + R/L s + 20,000

R = 20,000 rad/s. L j jR/L 1 [c] |H(jR/L)| = = = √ . j + 1 jR/L + R/L 2 [b] ωc =

RL Vo RL ksL R + RL P 14.13 [a] H(s) = = =   R RL Vi R + RL ksL s+ L R + RL 



s

=

1 s 2

1 s + (20,000) 2

.

R RL 1 = (20,000) = 10,000 rad/s. L R + RL 2 1 [c] ωc(L) = ωc(U L) . 2 [d] The gain in the passband is also reduced by a factor of 1/2 for the loaded filter. 

[b] ωc =



Problems

P 14.14 [a]

1 109 = = 10 krad/s; RC (40 × 103 )(2.5) fc =

5000 = 1591.55 Hz. π jω ; 10,000 + jω

[b] H(jω) =

j10,000 = 0.7071/45◦ ; 10,000 + j10,000

H(jωc ) =

H(j0.1ωc ) =

j1000 = 0.0995/84.29◦ ; 10,000 + j1000

H(j10ωc ) =

j100,000 = 0.9950/5.71◦ . 10,000 + j100,000

[c] vo (ωc ) = 565.69 cos(10,000t + 45◦ ) mV. vo (0.1ωc ) = 79.60 cos(1000t + 84.29◦ ) mV; vo (10ωc ) = 796.03 cos(100,000t + 5.71◦ ) mV. P 14.15 [a] H(s) =

R Vo = Vi R + Rc + (1/sC)

R s · . R + Rc [s + (1/(R + Rc )C)] R jω [b] H(jω) = · ; R + Rc jω + (1/(R + Rc )C) =

|H(jω)| =

ω R . ·q 2 R + Rc ω + (R+R1c )2 C 2

The magnitude will be maximum when ω = ∞. R [c] |H(jω)|max = . R + Rc Rωc q [d] |H(jωc )| = . (R + Rc ) ωc2 + [1/(R + Rc )C]2 .·. |H(jω)| = √ .·. ωc2 = or ωc =

R 2(R + Rc )

1 (R + Rc )2 C 2

1 . (R + Rc )C

when ω = ωc .

14–9

14–10

CHAPTER 14. Introduction to Frequency-Selective Circuits

[e] ωc =

1 109 = = 8000 rad/s; (R + RC )C (50 × 103 )(2.5)

40 j8000 H(jωc ) = = 0.5657/45◦ . 50 8000 + j8000 

P 14.16 [a]



H(j0.1ωc ) =

(0.8)j800 = 0.0796/84.29◦ ; 8000 + j800

H(j10ωc ) =

(0.8)j80,000 = 0.7960/5.71◦ . 8000 + j80,000

1 = 2π(800) = 1600π rad/s; RC .·. R =

109 = 9.95 kΩ. (1600π)(20)

[b] Re = 9.95k68 = 8.68 kΩ; ωc =

109 = 5761.84 rad/s; (8.68)(103 )(20)

fc =

5761.84 = 917.03 Hz. 2π

P 14.17 [a] ωc = 2π(300) = 1884.96 rad/s. 1 1 1 [b] ωc = so R = = = 5305 Ω. RC ωc C (1884.96)(100 × 10−9 ) [c]

Vo s s R = = = . Vi R + 1/sC s + 1/RC s + 1884.96 s Vo RkRL s [e] H(s) = = = = .   R + RL Vi RkRL + (1/sC) s + 2(1884.96) s+ 1/RC RL [f ] ωc = 2(1884.96) = 3769.91 rad/s. [d] H(s) =

[g] H(∞) = 1. P 14.18 ωo = fo =

√

ωc1 ωc2 =

q

(180)(200) = 189.74 krad/s;

ωo = 30.20 kHz. 2π

Problems

14–11

β = 200 − 180 = 20 krad/s = 3.18 kHz. Q=

P 14.19 β =

ωo 189.74 30.20 = = 9.49 = . β 20 3.18 ωo 80 = = 10 krad/s; Q 8 

1 ωc2 = 80  + 16 fc2 =

s

2

10,000 = 1.59 kHz; 2π

 

= 85.16 krad/s;

2



85.16 = 13.55 kHz; 2π 

s

1 ωc1 = 80 − + 16 fc1 =

1 1+ 16 

β=

1 1+ 16 



= 75.16 krad/s;

75.16 = 11.96 kHz. 2π q

P 14.20 [a] ωo =

1/LC

Q=

ωo β

β=

R L

[b] ωc1,2

so L =

so β =

=

1 (20,000)2 (50

× 10−9 )

= 50 mH.

ωo 20,000 = = 4000 rad/s. Q 5

so R = Lβ = (50 × 10−3 )(4000) = 200 Ω.

β =± + 2

s

β2 4000 + ωo2 = ± + 2 2

ωc1 = 18,099.75 rad/s P 14.21 H(jω) =

1 ωo2 C

20,0002



4000 2

2

+ 20,0002 = ±2000 + 20,099.75.

ωc2 = 22,099.75 rad/s.

jω(4000) . − ω 2 + jω(4000)

[a] H(j20,000) = Vo = (1)Vi

s

20,0002

j20,000(4000) = 1; − 20,0002 + j(20,000)(4000)

.·. vo (t) = 200 cos 20,000t mV.

14–12

CHAPTER 14. Introduction to Frequency-Selective Circuits

[b] H(j18,099.75) =

20,0002

1 Vo = √ /45◦ Vi 2 [c] H(j22,099.75) =

.·. vo (t) = 141.42 cos(18,099.75t + 45◦ ) mV. 1 j22,099.75(4000) / − 45◦ ; √ = 20,0002 − 22,099.752 + j(22,099.75)(4000) 2

1 Vo = √ / − 45◦ Vi 2 [d] H(j2000) =

.·. vo (t) = 141.42 cos(22,099.75t − 45◦ ) mV.

20,0002

j2000(4000) = 0.02/88.8◦ ; − 20002 + j(2000)(4000) .·. vo (t) = 4 cos(2000t + 88.8◦ ) mV.

Vo = 0.02/88.8◦ Vi [e] H(j200,000) =

20,0002

j200,000(4000) = 0.02/ − 88.8◦ ; 2 − 200,000 + j(200,000)(4000) .·. vo (t) = 4 cos(200,000t − 88.8◦ ) mV.

Vo = 0.02/ − 88.8◦ Vi P 14.22 H(s) = 1 −

H(jω) =

j18,099.75(4000) 1 = √ /45◦ ; 2 − 18,099.75 + j(18,099.75)(4000) 2

(R/L)s s2 + (1/LC) = ; s2 + (R/L)s + (1/LC) s2 + (R/L)s + (1/LC)

20,0002 − ω 2 . 20,0002 − ω 2 + jω(4000)

[a] H(j20,000) = Vo = (0)Vi

20,0002 − 20,0002 = 0; 20,0002 − 20,0002 + j(20,000)(4000) .·. vo (t) = 0 mV.

[b] H(j18,099.75) =

20,0002 − 18,099.752 1 = √ / − 45◦ ; 2 2 20,000 − 18,099.75 + j(18,099.75)(4000) 2

1 Vo = √ / − 45◦ Vi 2 [c] H(j22,099.75) = 1 Vo = √ /45◦ Vi 2 [d] H(j2000) =

.·. vo (t) = 141.42 cos(18,099.75t − 45◦ ) mV.

20,0002 − 22,099.752 1 /45◦ ; √ = 20,0002 − 22,099.752 + j(22,099.75)(4000) 2 .·. vo (t) = 141.42 cos(22,099.75t + 45◦ ) mV.

20,0002 − 20002 = 0.9998/ − 1.16◦ ; 20,0002 − 20002 + j(2000)(4000)

Vo = 0.9998/ − 1.16◦ Vi

.·. vo (t) = 199.96 cos(2000t − 1.16◦ ) mV.

Problems

[e] H(j200,000) =

20,0002 − 200,0002 = 0.9998/1.16◦ ; 20,0002 − 200,0002 + j(200,000)(4000) .·. vo (t) = 199.96 cos(200,000t + 1.16◦ ) mV.

Vo = 0.9998/1.16◦ Vi P 14.23 [a] β = ωc2 − ωc1 Then, β − + 2

v u u t

β 2

ωo2 = ωc1 ωc2 .

and

!2

ωc2 − ωc1 + ωo2 = − + 2

s 

ωc2 − ωc1 2

2

ωc1 − ωc2 = + 2

s

2 2 ωc2 − 2ωc1 ωc2 + ωc1 + 4ωc1 ωc2 4

ωc1 − ωc2 = + 2

s

2 2 ωc2 + ωc1 + 2ωc1 ωc2 4

ωc1 − ωc2 + = 2

s

(ωc1 + ωc2 )2 4

=

+ ωc1 ωc2

ωc1 − ωc2 ωc1 + ωc2 + = ωc1 . 2 2

Similarly, v u

β u β +t 2 2

!2

s



ωc2 − ωc1 2

2

+ ωc1 ωc2

s

2 2 ωc2 − 2ωc1 ωc2 + ωc1 + 4ωc1 ωc2 4

ωc2 − ωc1 = + 2

s

2 2 ωc2 + ωc1 + 2ωc1 ωc2 4

ωc2 − ωc1 = + 2

s

(ωc1 + ωc2 )2 4

ωc2 − ωc1 ωc1 + ωc2 + = ωc2 . 2 2



[b]

ωc2 − ωc1 + ωo2 = + 2

ωc2 − ωc1 = + 2

=

ωo  −

1 + 2Q

v u u t

1 1+ 2Q

!2

  

=

v u −ωo u + tω 2

2Q

o

ωo + 2Q

!2

.

But Q = ωo /β. Thus v u −ωo u + tω 2

2Q

o

14–13

ωo + 2Q

!2

−ωo = 2(ωo /β)

v u u + tω 2 o

ωo + 2(ωo /β)

!2

14–14

CHAPTER 14. Introduction to Frequency-Selective Circuits β =− + 2 = ωc1

v u u t

β 2

!2

+ ωo2

from part (a).

Similarly,   ωo 

1 + 2Q

v u u t

1 1+ 2Q

!2

  

=

v u ωo u + tω 2 o

2Q

ωo + 2Q

!2

.

But Q = ωo /β. Thus v u

ωo ωo u + tωo2 + 2Q 2Q β = + 2 = ωc2 P 14.24 [a] ωo2 =

v u u t

β 2

!2

v u

u ωo ωo = + tωo2 + 2(ωo /β) 2(ωo /β)

!2

+ ωo2

from part (a).

1 109 = = 25 × 108 ; LC L(25)

Q2 =

R2 C ; L 

!2

1 [b] ωc2 = 50  + 20

.·. L =

109 = 16 mH; 625 × 108

LQ2 (0.016)102 = 64 × 106 ; .·. R2 = = −9 C 25 × 10

.·. R = 8 kΩ.



s

1  1+ = 52.56 krad/s; 400

.·. fc2 = 8.37 kHz. 

1 ωc1 = 50 − + 20

s



1  1+ = 47.56 krad/s; 400

.·. fc1 = 7.57 kHz. ωo [c] β = = 5000 rad/s = 795.77 Hz. Q Check: β = fc2 − fc1 = 795.77 Hz. P 14.25 [a] We need ωc = 50,000 rad/s. There are several possible approaches – this one starts by choosing L = 1 mH. Then, 1 C= = 0.4 µF. 2 50,000 (0.001) Use the closest value from Appendix H, which is 0.47 µF to give s

ωo =

1 = 46,126.56 rad/s. (0.001)(0.47 × 10−6 )

Problems

Then, R =

14–15

Q 10 = = 461.3 Ω. ωo C (46,126.56)(0.47 × 10−6 )

Use the closest value from Appendix H, which is 470 Ω to give Q = 470(46,126.56)(0.47 × 10−6 ) = 10.2. [b] % error in ωo =

46,126.56 − 50,000 (100) = −7.75%; 50,000

% error in Q = P 14.26 [a] L =

10.2 − 10 (100) = 2%. 10

1 = 3.17 mH; [2π(20,000)]2 (20 × 10−9 )

R=

ωo L 40π × 103 (3.17 × 10−3 ) = = 79.58 Ω. Q 5 

s

1 [b] fc1 = 20 − + 10 

1 [c] fc2 = 20  + 10

s



1  1+ = 18.10 kHz. 100 

1  1+ = 22.10 kHz. 100

[d] β = fc2 − fc1 = 4 kHz or fo 20 = = 4 kHz. β= Q 5 P 14.27 [a] We need ωc close to 2π(20,000) = 125,663.706 rad/s. There are several possible approaches – this one starts by choosing L = 10 mH. Then, 1 C= = 6.33 nF. [2π(20,000)]2 (0.01) Use the closest value from Appendix H, which is 0.0047 µF to give s

ωo =

1 = 145,864.99 rad/s or fo = 23,215.13 Hz. (0.01)(4.7 × 10−9 )

Then, R =

ωo L (145,864.99)(0.01) = = 291.73 Ω. Q 5

Use the closest value from Appendix H, which is 270 Ω to give (145,864.99)(0.01) = 5.4. 270 23,215.13 − 20,000 [b] % error in fc = (100) = 16.1%; 20,000 Q=

% error in Q =

5.4 − 5 (100) = 8%. 5

14–16

CHAPTER 14. Introduction to Frequency-Selective Circuits 1 (103 )(1012 ) = = 2.56 × 1012 ; LC (312.5)(1.25)

P 14.28 [a] ωo2 =

ωo = 1.6 × 106 rad/s; fo = [b] Q =

800 = 254.65 kHz. π

ωo L (1.6 × 106 )(312.5 × 10−3 ) = = 8. R + Ri (50 + 12.5)103 

800  1 [c] fc1 = − + π 16 

800  1 [d] fc2 = + π 16

s

s



1  1+ = 239.23 kHz. 256 

1  = 271.06 kHz. 1+ 256

[e] β = fc2 − fc1 = 31.83 kHz or 100 ωo β= = 200 krad/s = kHz = 31.83 kHz. Q π (R/L)s

. 1 + + LC For the numerical values in Problem 14.28 we have

P 14.29 [a] H(s) =

(R+Ri ) s L

s2

16 × 104 s H(s) = 2 ; s + 2 × 105 s + 2.56 × 1012 .·. H(jω) = H(jωo ) =

j16 × 104 ω . (2.56 × 1012 − ω 2 ) + j2 × 105 ω

j16 × 104 (1.6 × 106 ) = 0.8/0◦ ; j2 × 105 (1.6 × 106 )

.·. vo (t) = 640 cos ωt mV. 

1 [b] ωc1 = 1.6 × 106 − + 16 H(jωc1 ) =

s



1  = 1.5 × 106 rad/s. 1+ 256

j16 × 104 (1.5 × 106 ) 2.56 × 1012 − 1.52 × 1012 + j2 × 105 (1.5 × 106 )

= 0.57/45◦ ; .·. vo (t) = 452.55 cos(1.5 × 106 t + 45◦ ) mV.

Problems 

1 [c] ωc2 = 1.6 × 106  + 16



s

1  1+ = 1.7 × 106 rad/s. 256

j16 × 104 (1.7 × 106 ) 2.56 × 1012 − 1.72 × 1012 + j2 × 105 (1.7 × 106 )

H(jωc2 ) =

= 0.57/ − 45◦ ; .·. vo (t) = 452.55 cos(1.7 × 106 t − 45◦ ) mV. P 14.30 [a]

ωo2

1 (106 )(109 ) = = = 1012 ; LC (40)(25)

ωo = 106 rad/s = 1 Mrad/s. 500 kHz = 159.15 kHz. π [c] Q = ωo RC = (106 )(300)(25 × 10−9 ) = 7.5.

[b] fo =





s

1  1+ = 935.55 krad/s. 225

1 [d] ωc1 = 106 − + 15

[e] .·. fc1 = 148.90 kHz. 

1 [f ] ωc2 = 106  + 15



s

1  = 1068.89 krad/s. 1+ 225

[g] .·. fc2 = 170.12 kHz. ωo [h] β = = 133.33 krad/s or 21.22 kHz. Q P 14.31 [a]

Z 1 Vo = where Z = Vi Z +R Y and Y = sC + H(s) = =

RL

RLCs2

=

RL Ls + (R + RL )Ls + RRL

(1/RC)s s2 + 

=

1 1 LCRL s2 + sL + RL + = . sL RL RL Ls

h

R+RL RL





1 RC

i



s+

1 LC



RL R+RL 1 s R+RL RL RC h  i 1 1 L s2 + R+R s + LC RL RC

s2

Kβs , + βs + ωo2

K=

RL , R + RL

β=

1 . (RkRL )C

14–17

14–18

CHAPTER 14. Introduction to Frequency-Selective Circuits

R + RL [b] β = RL 1 [c] βU = ; RC 

1 . RC



R R + RL βU = 1 + βU . .·. βL = RL RL 

[d] Q =







ωo ωo RC =  R+R  . L β RL

[e] QU = ωo RC; .·. QL = [f ] H(jω) =



RL 1 QU . QU = R + RL [1 + (R/RL )] 

Kjωβ ; ωo2 − ω 2 + jωβ

H(jωo ) = K. Let ωc represent a cutoff frequency. Then K Kωc β |H(jωc )| = √ = q ; 2 (ωo2 − ωc2 )2 + ωc2 β 2 1 ωc β .·. √ = q . 2 (ωo2 − ωc2 )2 + ωc2 β 2 Squaring both sides leads to (ωo2 − ωc2 )2 = ωc2 β 2 or (ωo2 − ωc2 ) = ±ωc β; .·. ωc2 ± ωc β − ωo2 = 0 or β ωc = ∓ ± 2

s

β2 + ωo2 . 4

The two positive roots are β ωc1 = − + 2

s

β2 + ωo2 4

β and ωc2 = + 2

where R β = 1+ RL 



1 1 and ωo2 = . RC LC

s

β2 + ωo2 4

Problems 1 (106 )(1012 ) = = 625 × 1012 ; LC (400)(4)

P 14.32 ωo2 =

ωo = 25 Mrad/s. Qu = ωo RC = (25 × 106 )(100 × 103 )(4 × 10−12 ) = 10; .·.



RL 10 = 9; R + RL 

P 14.33 [a] ωo2 =

.·. RL = 9R = 900 kΩ.

(103 )(1012 ) 1 = = 25 × 1012 ; LC (10)(4)

ωo = 5 Mrad/s. R + RL 1 6.25 [b] β = · = RL RC 5.0 5 ωo [c] Q = = = 20. β 0.25 RL [d] H(jωo ) = = 0.8/0◦ ; R + RL 



1012 (4)(1.25 × 106 )

!

= 250 krad/s.

.·. vo (t) = 600 cos(5 × 106 t) mV. 

[e] β = 1 +

R RL



1 1.25 = 1+ (200 × 103 ) rad/s; RC RL 



ωo = 5 × 106 rad/s. Q= [f ]

ωo 25 = β 1 + (1.25/RL )

where RL is in megohms.

14–19

14–20

CHAPTER 14. Introduction to Frequency-Selective Circuits

P 14.34 [a]

[b] L =

1 109 = = 3.2 × 10−3 = 3.2 mH; ωo2 C (625 × 108 )5

R=

800 ωo L = = 80 Ω. Q 10

[c] Re = 80k320 = 64 Ω; Re + Ri = 64 + 36 = 100 Ω; Qsystem = [d] βsystem =

ωo L 800 = = 8. Re + Ri 100 ωo

Qsystem

βsystem (kHz) =

250 × 103 = = 31.25 krad/s; 8

31.25 = 4.97 kHz = 4973.59 Hz. 2π

P 14.35 [a] In analyzing the circuit qualitatively we visualize vi as a sinusoidal voltage and we seek the steady-state nature of the output voltage vo . At zero frequency the inductor provides a direct connection between the input and the output, hence vo = vi when ω = 0. At infinite frequency the capacitor provides the direct connection, hence vo = vi when ω = ∞. At the resonant frequency of the parallel combination of L and C the impedance of the combination is infinite and hence the output voltage will be zero when ω = ωo . At frequencies on either side of ωo the amplitude of the output voltage will be nonzero but less than the amplitude of the input voltage. Thus the circuit behaves like a band-reject filter. [b] Let Z represent the impedance of the parallel branches L and C, thus Z=

sL sL(1/sC) = 2 . sL + 1/sC s LC + 1

Then Vo R R(s2 LC + 1) H(s) = = = Vi Z +R sL + R(s2 LC + 1) =

[s2 + (1/LC)] s2 +



1 RC



s+



1 LC

;

Problems

H(s) =

s2 + ωo2 . s2 + βs + ωo2

[c] From part (b) we have H(jω) =

ωo2 − ω 2 . ωo2 − ω 2 + jωβ

It follows that H(jω) = 0 when ω = ωo . 1 .·. ωo = √ . LC ωo2 − ω 2

[d] |H(jω)| = q

(ωo2 − ω 2 )2 + ω 2 β 2

;

1 |H(jω)| = √ when ω 2 β 2 = (ωo2 − ω 2 )2 2 or ± ωβ = ωo2 − ω 2 , thus ω 2 ± βω − ωo2 = 0. The two positive roots of this quadratic are −β + ωc1 = 2

v u u t

v u

β u β ωc2 = + t 2 2

β 2

!2

+ ωo2 ;

!2

+ ωo2 .

Also note that since β = ωo /Q 

v u

u 1  −1 ωc1 = ωo  + t1 + 2Q 2Q 

ωc2 = ωo  

1 + 2Q

v u u t

1+

1 2Q

!2

  ;

!2

  .

[e] It follows from the equations derived in part (d) that β = ωc2 − ωc1 = 1/RC. [f ] By definition Q = ωo /β = ωo RC = P 14.36 [a] ωo2 =

q

1 (106 )(1012 ) = = 64 × 1012 ; LC (625)(25)

.·. ωo = 8 Mrad/s.

R2 C/L.

14–21

14–22

CHAPTER 14. Introduction to Frequency-Selective Circuits

ωo = 1.27 MHz. 2π [c] Q = ωo RC = (8 × 106 )(80 × 103 )(25 × 10−12 ) = 16. [b] fo =



[d] ωc1 = 8 × 106 − [e] fc1 =

1 + 32



s

1+

1  = 7.75 Mrad/s. 1024

ωc1 = 1.234 MHz. 2π 



s

1 1  [f ] ωc2 = 8 × 106  + 1 + = 8.25 Mrad/s. 32 1024 ωc1 [g] fc2 = = 1.31 MHz. 2π [h] β = fc2 − fc1 = 79.58 kHz or β=

ωo 500 × 103 = = 79.58 kHz. 2πQ 2π

P 14.37 [a] ωo = 2πfo = 100π krad/s; L= R=

1 ωo2 C

=

1 (100π ×

103 )2 (100

× 10−9 )

= 101.32 µH;

Q 8 = = 254.65 Ω. 3 ωo C (100π × 10 )(100 × 10−9 ) 

1 [b] fc2 = 50k  + 16 

fc1 = 50k −



s

1  1+ = 53.22 kHz; 256

1 + 16



s

1+

1  = 46.97 kHz. 256

[c] β = fc2 − fc1 = 6.25 kHz or fo 50k β= = = 6.25 kHz. Q 8 P 14.38 [a] Re = 254.65k932 = 200 Ω; Q = ωo Re C = 100π × 103 (200)(0.1)10−6 = 2π = 6.28. [b] β =

fo 50,000 = = 7.96 kHz. Q 2π 

1 [c] fc2 = 50,000  + 4π

s



1  1+ = 54.14 kHz. 16π 2

Problems 

1 [d] fc1 = 50,000 − + 4π

14–23



s

1  1+ = 46.18 kHz. 16π 2

P 14.39 [a] We need ωc = 2π(50,000) = 314,159.265 rad/s. There are several possible approaches – this one starts by choosing L = 10 µH. Then, 1 = 1.01 µF. [2π(50,000)]2 (10 × 10−6 )

C=

Use the closest value from Appendix H, which is 1 µF, to give s

ωc =

1 10−6 )(1

(10 ×

Then, R =

× 10−6 )

= 316,227.766 rad/s so fc = 50,329.2 Hz.

Q 8 = = 25.3 Ω. ωo C (316,227.766)(1 × 10−6 )

Use the closest value from Appendix H, which is 22 Ω, to give Q = 22(316,227.766)(1 × 10−6 ) = 6.957. [b] % error in fc = % error in Q = P 14.40 H(s) =

50,329.2 − 50,000 (100) = 0.66%; 50,000 6.957 − 8 (100) = −13.04%. 8

1 s2 + LC 1 . s2 + R s + LC L

s

[a] ωo =

1 = LC

s

1 = 5000 rad/s. (0.005)(8 × 10−6 )

ωo = 795.8 Hz. 2π s s L 0.005 [c] Q = = = 2.5. 2 2 R C (10) (8 × 10−6 ) R 10 [d] β = = = 2000 rad/s. L 0.005 [b] fo =

β (Hertz) =

2000 = 318.31 Hz. 2π

v u

−β u β [e] ωc1 = +t 2 2 −2000 = + 2 [f ] fc1 =

!2

+ ωo2 s



2000 2

2

4099.02 = 652.38 Hz. 2π

+ 50002 = 4099.02 rad/s.

14–24

CHAPTER 14. Introduction to Frequency-Selective Circuits

β [g] ωc2 = + 2

v u u t

2000 = + 2 [h] fc2 =

!2

β 2

+ ωo2 s 

2000 2

2

+ 50002 = 6099.02 rad/s.

6099.02 = 970.69 Hz. 2π

P 14.41 [a] H(jω) =

1 − ω2 LC 1 R − ω2 + j ω LC L

=

50002 − ω 2 . 50002 − ω 2 + j2000ω

ωo = 5000 rad/s : H(jωo ) =

50002 − 50002 = 0. 50002 − 50002 + j2000(5000)

ωc1 = 4099.02 rad/s : H(jωc1 ) =

50002 − 4099.022 = 0.7071/ − 45◦ . 50002 − 4099.022 + j2000(4099.02)

ωc2 = 6099.02 rad/s : H(jωc2 ) =

50002 − 6099.022 = 0.7071/45◦ . 50002 − 6099.022 + j2000(6099.02)

0.1ωo = 500 rad/s : H(j0.1ωo ) =

50002 − 5002 = 0.9992/ − 2.314◦ . 50002 − 5002 + j2000(500)

10ωo = 50,000 rad/s : H(j10ωo ) =

50002 − 50,0002 = 0.9992/2.314◦ . 50002 − 50,0002 + j2000(50,000)

[b] ω = ωo = 5000 rad/s :

Vo = H(j5000)Vi = 0Vi ;

vo (t) = 0. ω = ωc1 = 4099.02 rad/s : Vo = H(j4099.02)Vi = (0.7071/ − 45◦ )(2) = 1.414/ − 45◦ ; vo (t) = 1.414 cos(4099.02t − 45◦ ) V. ω = ωc2 = 6099.02 rad/s :

Problems

14–25

Vo = H(j6099.02)Vi = (0.7071/45◦ )(2) = 1.414/45◦ ; vo (t) = 1.414 cos(6099.02t + 45◦ ) V. ω = 0.1ωo = 500 rad/s : Vo = H(j500)Vi = (0.9992/ − 2.314◦ )(2) = 1.998/ − 2.314◦ ; vo (t) = 1.998 cos(500t − 2.314◦ ) V. ω = 10ωo = 50,000 rad/s : Vo = H(j50,000)Vi = (0.9992/2.314◦ )(2) = 1.998/2.314◦ ; vo (t) = 1.998 cos(50,000t + 2.314◦ ) V. q

P 14.42 [a] ωo =

1/LC

Q=

ωo β

β=

R L

[b] ωc1,2

so L =

so β =

ωo 4000 = = 6000 rad/s; Q 2/3

so R = Lβ = (781.25 × 10−3 )(6000) = 4687.5 Ω.

β =± + 2

s

6000 β2 + ωo2 = ± + 2 2

ωc1 = 2000 rad/s P 14.43 H(jω) =

1 1 = = 781.25 mH; ωo2 C (4000)2 (80 × 10−9 )

s



6000 2

2

+ 40002 = ±3000 + 5000;

ωc2 = 8000 rad/s.

ωo2 − ω 2 40002 − ω 2 = ωo2 − ω 2 + jωβ 40002 − ω 2 + jω(6000)

[a] H(j4000) = Vo = (0)Vi [b] H(j2000) =

40002 − 40002 =0 40002 − 40002 + j(4000)(6000) .·. vo (t) = 0 V 40002 − 20002 1 = √ / − 45◦ 2 2 4000 − 2000 + j(2000)(6000) 2

1 Vo = √ / − 45◦ Vi 2

.·. vo (t) = 88.39 cos(2000t − 45◦ ) V

14–26

CHAPTER 14. Introduction to Frequency-Selective Circuits 40002 − 80002 1 = √ /45◦ 2 2 4000 − 8000 + j(8000)(6000) 2

[c] H(j8000) =

1 Vo = √ /45◦ Vi 2 [d] H(j500) =

.·. vo (t) = 88.39 cos(8000t + 45◦ ) V

40002 − 5002 = 0.982/ − 10.78◦ 2 2 4000 − 500 + j(500)(6000)

Vo = 0.982/ − 10.78◦ Vi [e] H(j32,000) =

40002 − 32,0002 = 0.982/10.78◦ 40002 − 32,0002 + j(32,000)(6000) .·. vo (t) = 122.8 cos(32,000t + 10.78◦ ) V

Vo = 0.982/10.78◦ Vi P 14.44 H(jω) =

.·. vo (t) = 122.8 cos(500t − 10.78◦ ) V

jω(6000) jωβ = ωo2 − ω 2 + jωβ 40002 − ω 2 + jω(6000)

[a] H(j4000) =

40002

j(4000)(6000) =1 − 40002 + j(4000)(6000)

.·. vo (t) = 125 cos 4000t V

Vo = (1)Vi [b] H(j2000) =

j(2000)(6000) 1 = √ /45◦ 2 − 2000 + j(2000)(6000) 2

40002

1 Vo = √ /45◦ Vi 2 [c] H(j8000) =

.·. vo (t) = 88.39 cos(2000t + 45◦ ) V

40002

j(8000)(6000) 1 = √ / − 45◦ 2 − 8000 + j(8000)(6000) 2

1 Vo = √ / − 45◦ Vi 2 [d] H(j500) =

40002

.·. vo (t) = 88.39 cos(8000t − 45◦ ) V

j(500)(6000) = 0.187/79.22◦ 2 − 500 + j(500)(6000)

Vo = 0.187/79.22◦ Vi [e] H(j32,000) =

40002

.·. vo (t) = 23.375 cos(500t + 79.22◦ ) V

j(32,000)(6000) = 0.187/ − 79.22◦ 2 − 32,000 + j(32,000)(6000)

Vo = 0.187/ − 79.22◦ Vi

.·. vo (t) = 23.375 cos(32,000t − 79.22◦ ) V

Problems

P 14.45 [a] Let Z = Z=

RL (sL + (1/sC)) ; RL + sL + (1/sC)

RL (s2 LC + 1) . s2 LC + RL Cs + 1

Vo s2 RL CL + RL Then H(s) = = . Vi (R + RL )LCs2 + RRL Cs + R + RL Therefore 

H(s) = =

RL [s2 + (1/LC)]  h  i · RRL s 1 R + RL s2 + R+R + L LC L 

K(s2 + ωo2 ) , s2 + βs + ωo2

RL where K = ; R + RL

ωo2

1 ; = LC

RRL β= R + RL

1 . LC   RRL 1 [c] β = . R + RL L ωo ωo L [d] Q = = . β [RRL /(R + RL )] [b] ωo = √

[e] H(jω) =

K(ωo2 − ω 2 ) ; (ωo2 − ω 2 ) + jβω

H(jωo ) = 0. [f ] H(j0) =

Kωo2 RL = K = . ωo2 R + RL h

i

K (ωo /ω)2 − 1

i o; [g] H(jω) = nh (ωo /ω)2 − 1 + jβ/ω

H(j∞) =

−K RL =K= . −1 R + RL

K(ωo2 − ω 2 ) ; (ωo2 − ω 2 ) + jβω Let ωc represent a cutoff frequency. Then

[h] H(jω) =

K |H(jωc )| = √ ; 2 K K(ωo2 − ωc2 ) .·. √ = q . 2 (ωo2 − ωc2 )2 + ωc2 β 2





1 . L

14–27

14–28

CHAPTER 14. Introduction to Frequency-Selective Circuits Squaring both sides leads to (ωo2 − ωc2 )2 = ωc2 β 2 or (ωo2 − ωc2 ) = ±ωc β; .·. ωc2 ± ωc β − ωo2 = 0 or β ωc = ∓ ± 2

s

β2 + ωo2 . 4

The two positive roots are β ωc1 = − + 2

s

β2 + ωo2 4

β and ωc2 = + 2

s

β2 + ωo2 , 4

where β= P 14.46 [a] ωo2 =

RRL 1 1 · and ωo2 = . R + RL L LC 1 1 = = 1010 ; −3 LC (400 × 10 )(250 × 10−12 )

ωo = 105 = 100 krad/s = 15.9 kHz. β=

1 (5000)(20,000) 1 RRL · = · = 104 rad/s = 1.59 kHz; R + RL L 25,000 0.4

Q=

105 ωo = 4 = 10. β 10

[b] H(j0) =

RL 20,000 = = 0.8; R + RL 25,000 RL = 0.8. R + RL

H(j∞) = 

105  1 [c] fc2 = + 2π 20 

s

105  1 fc1 = − + 2π 20 Check: ωo [d] Q = = β



1  = 16.73 kHz; 1+ 400 s



1  1+ = 15.14 kHz. 400

β = fc2 − fc1 = 1.59 kHz. 105 RRL ·1 R+RL L

40(R + RL ) 5 =8 1+ RRL RL where RL is in kilohms. 

=



Problems

14–29

[e]

P 14.47 [a] ωo2 =

1 = 1012 ; LC

.·. L =

1 × 10−12 )

(1012 )(400

RL = 0.96; R + RL .·.

RL = 24R 

[b] β = Q=

= 2.5 mH.

.·. 0.04RL = 0.96R; 36,000 = 1.5 kΩ. .·. R = 24

1 RL R · = 576 × 103 ; R + RL L 

ωo 106 = = 1.74. β 576 × 103 4 × 106

P 14.48 [a] |H(jω)| = q

(4 × 106 − ω 2 )2 + (500ω)2

.·.

= 1;

16 × 1012 = (4 × 106 − ω 2 )2 + (500ω)2 = −8 × 106 ω 2 + ω 4 + 25 × 104 ω 2 = 0;

.·.

ω 2 = 8 × 106 − 25 × 104

so

ω = 2783.88 rad/s.

[b] From the equation for |H(jω)| in part (a), the frequency for which the magnitude is maximum is the frequency for which the denominator is minimum. This is the frequency at which √ (4 × 106 − ω 2 )2 = 0 so ω = 4 × 106 = 2000 rad/s. [c] |H(j2000)| = q

P 14.49 [a] H(s) =

4 × 106 (4 × 106 − 20002 )2 + [500(2000)]2

sL R + sL +

1 sC

=

= 4.

s2 LC s2 = RsC + s2 LC + 1 s2 + R s+ L

1 LC

.

14–30

CHAPTER 14. Introduction to Frequency-Selective Circuits

[b] When s = jω is very small (think of ω approaching 0), H(s) ≈

s2 1 LC

= 0.

[c] When s = jω is very large (think of ω approaching ∞), s2 = 1. s2 [d] The magnitude of H(s) approaches 0 as the frequency approaches 0, and approaches 1 as the frequency approaches ∞. Therefore, this circuit is behaving like a high pass filter when the output is the voltage across the inductor. ωc2 1 [e] |H(jωc )| = q =√ ; 2 (250,000 − ωc2 )2 + (8000ωc )2 H(s) ≈

2ωc4 = (250,000 − ωc2 )2 + (8000ωc )2 = 625 × 108 − 500,000ωc2 + ωc4 + 64 × 106 ωc2 . ωc4 − 635 × 105 ωc2 − 625 × 108 = 0.

Simplifying,

Solve for ωc2 and then ωc : ωc2 = 63,500,984.2367

so

ωc = 7968.75 rad/s.

fc = 1268.3 Hz P 14.50 [a] H(s) =

1 sC

R + sL +

1 sC

=

1 1 LC = RsC + s2 LC + 1 s2 + R s+ L

1 LC

.

[b] When s = jω is very small (think of ω approaching 0), H(s) ≈

1 LC 1 LC

= 1.

[c] When s = jω is very large (think of ω approaching ∞), H(s) ≈

1 LC s2

= 0.

[d] The magnitude of H(s) approaches 1 as the frequency approaches 0, and approaches 0 as the frequency approaches ∞. Therefore, this circuit is behaving like a low pass filter when the output is the voltage across the capacitor. 250,000 1 [e] |H(jωc )| = q =√ ; 2 (250,000 − ωc2 )2 + (8000ωc )2 1250 × 108 = (250,000 − ωc2 )2 + (8000ωc )2 = 625 × 108 − 500,000ωc2 + ωc4 + 64 × 108 ωc2 .

Problems

14–31

ωc4 + 635 × 105 ωc2 − 625 × 108 = 0.

Simplifying,

Solve for ωc2 and then ωc : ωc2 = 984.2367

so

ωc = 31.37 rad/s.

f = 4.99 Hz. P 14.51 [a] Use the cutoff frequencies to calculate the bandwidth: ωc1 = 2π(697) = 4379.38 rad/s

ωc2 = 2π(941) = 5912.48 rad/s;

β = ωc2 − ωc1 = 1533.10 rad/s.

Thus

Calculate inductance and capacitance: C= L=

1 1 = = 1.087 µF; Rβ (600)(1533.1) 1 Cωc1 ωc2

=

1 = 35.5 mH. (1.087 × 10−6 )(4379.38)(5912.48)

[b] At the outermost two frequencies in the low-frequency group (687 Hz and 941 Hz) the amplitudes are |Vpeak | √ = 0.707|Vpeak | 2

|V697Hz | = |V941Hz | =

because these are cutoff frequencies. We calculate the amplitudes at the other two low frequencies: ωβ |V | = (|Vpeak |)(|H(jω)|) = |Vpeak | q . (ωo2 − ω 2 )2 + (ωβ)2 Therefore (4838.05)(1533.10)

|V770Hz | = (|Vpeak |) q

(5088.522 − 4838.052 )2 + [(4838.05)(1533.10)]2

= 0.948|Vpeak |, and (5353.27)(1533.10)

|V852Hz | = (|Vpeak |) q

(5088.522 − 5353.272 )2 + [(5353.27)(1533.10)]2

= 0.948|Vpeak |. It is not a coincidence that these two magnitudes are the same. The frequencies in both bands of the DTMF system were carefully chosen to produce this type of predictable behavior with linear filters. In other words, the frequencies were chosen to be equally far apart with respect to the response produced by a linear filter. Most musical scales consist of

14–32

CHAPTER 14. Introduction to Frequency-Selective Circuits tones designed with this same property – note intervals are selected to place the notes equally far apart. That is why the DTMF tones remind us of musical notes! Unlike musical scales, DTMF frequencies were selected to be harmonically unrelated, to lower the risk of misidentifying a tone’s frequency if the circuit elements are not perfectly linear.

[c] The high-band frequency closest to the low-frequency band is 1209 Hz. The amplitude of a tone with this frequency is |V1209Hz | = (|Vpeak |) q

(7596.37)(1533.10) (5088.522 − 7596.372 )2 + [(7596.37)(1533.10)]2

= 0.344|Vpeak |. This is less than one half the amplitude of the signals with the low-band cutoff frequencies, ensuring adequate separation of the bands. P 14.52 The cutoff frequencies and bandwidth are ωc1 = 2π(1209) = 7596 rad/s;

ωc2 = 2π(1633) = 10.26 krad/s;

β = ωc2 − ωc1 = 2664 rad/s. Telephone circuits always have R = 600 Ω. Therefore, the filters inductance and capacitance values are C=

1 1 = = 625.6 nF; Rβ (600)(2664)

L=

1 1 = = 20.51 mH. 2 ωc1 ωc2 C (2π) (1209)(1633)(625.6 × 10−9 )

At the highest of the low-band frequencies, 941 Hz, the amplitude is |Vω | = |Vpeak | q where |Vω | = q

ωβ (ωo2 − ω 2 )2 + ω 2 β 2

ωo =

√

ωc1 ωc2 . Thus,

|Vpeak |(5912)(2664) [(8828)2 − (5912)2 ]2 + [(5912)(2664)]2

= 0.344 |Vpeak |. Again it is not coincidental that this result is the same as the response of the low-band filter to the lowest of the high-band frequencies.

Problems

14–33

P 14.53 From Problem 14.51 the response to the largest of the DTMF low-band tones is 0.948|Vpeak |. The response to the 20 Hz tone is |V20Hz | = q

|Vring |(125.6)(1533) [(50892 − 125.62 )2 + [(125.6)(1533)]2 ]

= 0.00744|Vring |; .·.

|V20Hz | |V20Hz | 0.00744|Vring | 1 = = = . |V770Hz | |V852Hz | 0.948|Vpeak | 3

.·. |V20Hz | = 42.5|V770Hz |. Thus, the 20Hz signal can be 42.5 times as large as the DTMF tones.

Active Filter Circuits

Assessment Problems AP 15.1 H(s) =

−8000 −(1/R1 C) = . s + (1/R2 C) s + 2000

1 = 8000; R1 C .·. R1 =

C = 40 nF;

1 = 3125 Ω. (8000)(40 × 10−9 )

1 = 2000; R2 C .·. R2 =

AP 15.2 H(s) =

1 = 12.5 kΩ. (2000)(40 × 10−9 )

−(R2 /R1 )s ; s + (1/R1 C)

1 = 1 rad/s; R1 C R2 = 1, R1 .·.

R1 = 1 Ω,

.·. C = 1 F.

.·. R2 = R1 = 1 Ω;

Hprototype (s) =

−s . s+1

15–1

15–2

CHAPTER 15. Active Filter Circuits

AP 15.3 ωc = 2πfc = 2π(4000) = 8000π rad/s; .·. kf = 8000π = 25,132.74. C kf km

C0 =

.·. km =

.·.

10 × 10−6 =

(10 ×

10−6 )(25,132.74)

1 ; kf km

1

= 3.98.

AP 15.4 [a] For a 2nd order Butterworth low pass filter H(s) =

s2

1 √ . + 2s + 1

For the circuit in Fig. 15.21 H(s) =

1 R 2 C1 C2

s2 +



2 RC1



s+



1 R2 C

.

1 C2

Equate the transfer functions. For R = 1 Ω, √ 2 2 = 2, .·. C1 = √ = 1.414 F; RC1 2 1 R2 C1 C2

= 1,

1 .·. C2 = = 0.707 F. C1

[b] For a 2nd order Butterworth high pass filter s2 √ H(s) = 2 . s + 2s + 1 For the circuit in Fig. 15.25 H(s) =

s2 s2 +



2 R2 C



s+



1 R1 R2 C 2

.

Equate the transfer functions. For C = 1F, √ √ 2 = 2, .·. R2 = 2 = 1.414 Ω; R2 C 1 = 1, R1 R2 C 2

1 .·. R1 = √ = 0.707 Ω. 2

AP 15.5 From the statement of the problem, K = 10 ( = 20 dB). Therefore for the prototype bandpass circuit R1 =

Q 16 = = 1.6 Ω; K 10

Problems

R2 =

Q 16 = Ω; −K 502

2Q2

R3 = 2Q = 32 Ω. The scaling factors are kf =

ωo0 = 2π(6400) = 12,800π; ωo

km =

1 C = = 1243.40. C 0 kf (20 × 10−9 )(12,800π)

Therefore, R10 = km R1 = (1.6)(1243.30) = 1.99 kΩ; R20 = km R2 = (16/502)(1243.40) = 39.63 Ω; R30 = km R3 = 32(1243.40) = 39.79 kΩ. AP 15.6 ωo = 2000π rad/s; ωo0 = 2000π; ωo

.·.

kf =

km =

1 105 C = = ; C 0 kf (15 × 10−9 )(2000π) 3π

R 0 = km R =

σ =1−

105 (1) = 10,610 Ω; 3π

1 1 =1− = 0.9875; 4Q 4(20)

σR0 = 10,478 Ω; C 0 = 15 nF; 2C 0 = 30 nF.

(1 − σ)R0 = 133 Ω;

15–3

15–4

CHAPTER 15. Active Filter Circuits

Problems P 15.1

[a] ωc =

1 R2 C

K=

R2 R1

so R2 = so R1 =

1 1 = = 6366 Ω; ωc C 2π(500)(50 × 10−9 ) R2 6366 = = 637 Ω. K 10

[b] Both the cutoff frequency and the passband gain are changed. P 15.2

[a] 10(0.2) = 2 V so Vcc ≥ 2 V. −10(2π)(500) [b] H(jω) = ; jω + 2π(500) H(j1000π) =

−10(1000π) 10 = −5 + j5 = √ /135◦ ; 1000π + j1000π 2

10 Vo = √ /135◦ Vi 2 [c] H(j100π) =

−10(1000π) = 9.95/174.3◦ ; 1000π + j100π

Vo = 9.95/174.3◦ Vi [d] H(j10,000π) =

so vo (t) = 1.99 cos(100πt + 174.3◦ ) V.

−10(1000π) = 0.995/95.7◦ ; 1000π + j10,000π

Vo = 0.995/95.7◦ Vi P 15.3

so vo (t) = 1.414 cos(1000πt + 135◦ ) V.

so vo (t) = 199 cos(10,000πt + 95.7◦ ) mV.

[a] K = 10(10/20) = 3.1623 =

R2 ; R1

R2 =

1 1 = = 25,464.791 Ω; ωc C (2π)(2500)(2.5 × 10−9 )

R1 =

R2 25,464.791 = = 8052.62 Ω. K 3.1623

Problems

15–5

[b]

P 15.4

[a]

1 = 2π(2500) so Rf C = 63.662 × 10−6 . Rf C There are several possible approaches. Here, choose C = 22 nF. Then 63.662 × 10−6 = 2893.7 Ω. 22 × 10−9 Choose Rf = 2.7 kΩ. This gives

Rf =

ωc =

1 = 16,835.02 rad/s so fc = 2679.4 Hz. (2700)(22 × 10−9 )

To get a passband gain of 10 dB, Rf 2700 = = 853.8 Ω. 3.1623 3.1623 Choose Ri = 1 kΩ to give K = 20 log10 (2700/1000) = 8.63 dB. The resulting circuit is

Ri =

[b] % error in fc =

2679.4 − 2500 (100) = 7.2%; 2500

% error in passband gain =

8.63 − 10 (100) = −13.7%. 10

15–6

P 15.5

CHAPTER 15. Active Filter Circuits

[a] R1 =

1 1012 = = 5.85 kΩ; ωc C (2π)(40)(103 )(680)

K = 100.6 = 3.98 =

R2 ; R1

.·. R2 = 3.98R1 = 23.29 kΩ. [b]

P 15.6

[a]

1 = 2π(40,000) so R1 C = 3.98 × 10−6 . R1 C There are several possible approaches. Here, choose C = 220 pF. Then 3.98 × 10−6 = 18,085.8 Ω. 220 × 10−12 Choose R1 = 18 kΩ. This gives

R1 =

ωc =

1 (220 ×

10−12 )(18,000)

= 252.5 krad/s so fc = 40,190.6 Hz.

To get a passband gain of 12 dB, R2 = 3.98R1 = 3.98(18,000) = 71,640 Ω. Choose R2 = 68 kΩ to give a passband gain of 20 log10 (68/18) = 11.54 dB. The resulting circuit is

[b] % error in fc =

40,190.6 − 40,000 (100) = 0.48%; 40,000

% error in passband gain =

11.54 − 12 (100) = −3.83%. 12

Problems

P 15.7

[a] ωc =

1 R1 C

K=

R2 R1

so R1 =

1 1 = = 5305 Ω; ωc C 2π(300)(100 × 10−9 )

so R2 = KR1 = (5)(5305) = 26.5 kΩ.

[b] The passband gain changes but the cutoff frequency is unchanged. P 15.8

[a] 5(0.15) = 0.75 V so Vcc ≥ 0.75 V. −5jω ; [b] H(jω) = jω + 600π H(j600π) =

−5(j600π) 5 = √ / − 135◦ ; 600π + j600π 2

5 Vo = √ / − 135◦ Vi 2 [c] H(j60π) =

−5(j60π) = 0.5/ − 95.7◦ ; 600π + j60π

Vo = 0.5/ − 95.7◦ Vi [d] H(j6000π) =

so vo (t) = 74.63 cos(60πt − 95.7◦ ) mV.

−5(j6000π) = 4.98/ − 174.3◦ ; 600π + j6000π

Vo = 4.98/ − 174.3◦ Vi P 15.9

so vo (t) = 530.33 cos(600πt − 135◦ ) mV.

so vo (t) = 746.3 cos(6000πt − 174.3◦ ) mV.

Summing the currents at the inverting input node yields 0 − Vi 0 − Vo + = 0; Zi Zf .·.

Vo Vi =− ; Zf Zi

Vo Zf .·. H(s) = =− . Vi Zi

15–7

15–8

CHAPTER 15. Active Filter Circuits

P 15.10 [a] Zf =

R2 (1/sC2 ) R2 = [R2 + (1/sC2 )] R2 C2 s + 1

(1/C2 ) . s + (1/R2 C2 ) Likewise (1/C1 ) Zi = . s + (1/R1 C1 ) =

−(1/C2 )[s + (1/R1 C1 )] .·. H(s) = [s + (1/R2 C2 )](1/C1 ) =−

C1 [s + (1/R1 C1 )] . C2 [s + (1/R2 C2 )] #

"

−C1 jω + (1/R1 C1 ) [b] H(jω) = ; C2 jω + (1/R2 C2 ) −C1 H(j0) = C2 C1 [c] H(j∞) = − C2



R2 C2 R1 C1 j j

!

=



=

−R2 . R1

−C1 . C2

[d] As ω → 0 the two capacitor branches become open and the circuit reduces to a resistive inverting amplifier having a gain of −R2 /R1 . As ω → ∞ the two capacitor branches approach a short circuit and in this case we encounter an indeterminate situation; namely vn → vi but vn = 0 because of the ideal op amp. At the same time the gain of the ideal op amp is infinite so we have the indeterminate form 0 · ∞. Although ω = ∞ is indeterminate we can reason that for finite large values of ω, H(jω) will approach −C1 /C2 in value. In other words, the circuit approaches a purely capacitive inverting amplifier with a gain of |(−1/jωC2 )/(1/jωC1 )| or −C1 /C2 . P 15.11 [a] Zf =

(1/C2 ) ; s + (1/R2 C2 )

Zi = R 1 +

1 R1 = [s + (1/R1 C1 )]; sC1 s

H(s) = −

(1/C2 ) s · [s + (1/R2 C2 )] R1 [s + (1/R1 C1 )]

=−

1 s . R1 C2 [s + (1/R1 C1 )][s + (1/R2 C2 )]

Problems [b] H(jω) = −

1  R1 C2 jω +

jω 1 R1 C1



jω +

1 R2 C2

15–9

;

H(j0) = 0. [c] H(j∞) = 0. [d] As ω → 0 the capacitor C1 disconnects vi from the circuit. Therefore vo = vn = 0. As ω → ∞ the capacitor short circuits the feedback network, thus Zf = 0 and therefore vo = 0. P 15.12 [a] km =

500 = 0.125; 4000

1 (625 × 10−9 ) = 1 × 10−6 km kf L0 =

so

kf =

625 × 10−9 = 5. (0.125)(1 × 10−6 )

(0.125)(5) = 125 mH. 5

[b] Zab = jωL + Rk =

R/jωC 1 = jωL + jωC R + (1/jωC)

[jωL(1 + jωRC) + R](1 − jωRC) . (1 + jωRC)(1 − jωRC)

The denominator is purely real, so set the imaginary part of the numerator equal to 0 and solve for ω: ωL + ω 3 R2 LC 2 − ωR2 C = 0 .·.

ω2 =

R2 C − L (500)2 (10−6 ) − 0.125 = = 4 × 106 . 2 2 2 −6 2 R LC 500 (0.125)(10 )

Thus, ω = 2 krad/s. [c] In the original, unscaled circuit, the frequency at which the impedance Zab is purely real is 2 ωus =

Thus,

40002 (625 × 10−9 ) − 5 = 16 × 104 . 40002 (5)(625 × 10−9 )2 ωus = 400 rad/s.

ωscaled 2000 = = 5 = kf . ωus 400

15–10

CHAPTER 15. Active Filter Circuits

P 15.13 [a] kf =

4000 = 8; 500

C0 =

20 × 10−6 = 2.5 µF; 8

L0 =

1 = 125 mH. 8

Io =

(50 − j100)k(250 + j500) (0.125) = 27.95/ − 116.565◦ mA; 250 + j500

[b]

io = 27.95 cos(4000t − 116.565◦ ) mA. The magnitude and phase angle of the output current are the same as in the unscaled circuit. P 15.14 [a] The scaled circuit, transformed into the frequency domain with ω = 25 krad/s, is

The equivalent impedance seen by the source is used to find the voltage drop across the source: Zeq = 400 + [−j200k(1000 + j200)] = 440 − j200 Ω; V = Zeq I = (440 − j200)(0.03) = 13.2 − j6 V; 1 1 .·. S = − VI∗ = − (13.2 − j6)(0.03) = −198 + j90 mVA. 2 2 Thus the real power delivered by the source in the scaled circuit is 198 mW.

Problems

15–11

[b] The unscaled circuit, transformed into the frequency domain with ω = 25 krad/s, is

Zeq = 2 + [−jk(5 + j)] = 2.2 − j1 Ω; V = Zeq I = (2.2 − j1)(0.03) = 66 − j30 mV; 1 1 .·. S = − VI∗ = − (0.066 − j0.033)(0.03) = −990 + j450 µVA. 2 2 The real power delivered by the source in the unscaled circuit is 990 µ W. Note that the ratio of the real power delivered by the source in the scaled circuit to the real power delivered by the source in the unscaled circuit is 990 × 10−6 = 200 = km . 0.198 P 15.15 [a] The scaled circuit, transformed into the frequency domain with ω = 25 krad/s, is

The equivalent impedance seen by the source is used to find the voltage drop across the source: Zeq = 2 + [−j0.01k(5 + j100)] = 2 − j0.01 Ω; V = Zeq I = (2 − j0.01)(0.03) = 60 − j0.3 mV; 1 1 .·. S = − VI∗ = − (0.06 − j0.0003)(0.03) = −900 + j4.5 µVA. 2 2 Thus the real power delivered by the source in the scaled circuit is 900 µW. [b] The scaled circuit, transformed into the frequency domain with the source frequency scaled by kf = 0.01 (ω = 250 rad/s), is

Zeq = 2 + [−jk(5 + j)] = 2.2 − j1 Ω;

15–12

CHAPTER 15. Active Filter Circuits V = Zeq I = (2.2 − j1)(0.03) = 66 − j30 mV; 1 1 .·. S = − VI∗ = − (0.066 − j0.033)(0.03) = −990 + j450 mVA. 2 2 The real power delivered by the source in the scaled circuit when the source frequency is also scaled is 990 µ W.

[c] When both the components of the circuit and the source frequency are scaled using the frequency scale factor, the real power delivered by the source is unchanged from the unscaled circuit. P 15.16 [a] km =

20 = 5; 4

5 × 10−9 .·. 100 × 10−12 = ; 5kf Lscaled =

.·. kf = 10.

5 (80) = 40 mH; 10

R2scaled = (21)(5 × 103 ) = 105 kΩ.

[b]

Vo Vo Vo − (300/s) + + = 0; 20,000 0.04s 105,000 + (1010 /s) .·.

Vo =

252s + 24 × 106 −4 256 = + . s2 + 500,000 + 40 × 109 s + 100,000 s + 400,000

Therefore, vo = (256e−400,000t − 4e−100,000t )u(t) V. P 15.17 km =

R0 1000 = = 0.2; R 5000

kf = 0.02.

Rf0 = km Rf = 0.2(25,000) = 5 kΩ;

C0 =

C 4 × 10−9 = = 1 µF. km kf (0.2)(0.02)

Problems

15–13

Va − (0.4/s) (Va − Vo )s Va s + + 6 = 0; 1000 106 10 Va s Vo −200 + = 0 so Va = Vo . 6 10 5000 s .·.

Vo =

−400 0.5/90◦ 0.5/ − 90◦ = + . s2 + 400s + 200 × 103 s + 200 − j400 s + 200 + j400

vo = [e−200t cos(400t + 90◦ )]u(t) V. P 15.18 From the solution to Problem 14.30, ωo = 106 rad/s and β = 133.33 krad/s. Compute the two scale factors: kf =

2π(250 × 103 ) ωo0 = = π/2; ωo 106

km =

1 C 2 25 × 10−9 5 = = . kf C 0 π 10 × 10−9 π

Thus, R 0 = km R =

5 (300) = 477.46 Ω; π

L0 =

km 5/π (40 × 10−6 ) = 40.53 µH. L= kf π/2

Calculate the cutoff frequencies: 0 ωc1 = kf ωc1 = (π/2)(935.56 × 103 ) = 1469.57 krad/s; 0 ωc2 = kf ωc2 = (π/2)(1068.89 × 103 ) = 1679.01 krad/s.

To check, calculate the bandwidth: 0 0 β 0 = ωc2 − ωc1 = 209.44 krad/s = (π/2)β (checks!).

15–14

CHAPTER 15. Active Filter Circuits

P 15.19 From the solution to Problem 14.36, ωo = 8 × 106 rad/s and β = 500 krad/s. Calculate the scale factors: kf =

ωo0 500 × 103 = = 0.0625; ωo 8 × 106

km =

0.0625(50 × 10−6 ) kf L0 = = 0.005. L 625 × 10−6

Thus, R0 = km R = (0.005)(80,000) = 400 Ω;

C0 =

Calculate the bandwidth: β 0 = kf β = (0.0625)(500 × 103 ) = 31,250 rad/s. To check, calculate the quality factor: Q=

8 × 106 ωo = = 16; β 500 × 103

ωo0 500 × 103 Q = 0 = = 16 (checks). β 31,250 0

P 15.20 [a] From Eq 15.1 we have H(s) =

−Kωc s + ωc

where K =

R2 , R1

ωc =

1 ; R2 C

−K 0 ωc0 .·. H 0 (s) = s + ωc0 where K 0 =

R20 R10

ωc0 =

1 R20 C 0

.

By hypothesis R10 = km R1 ; and C 0 =

R20 = km R2 ,

C . It follows that kf km

K 0 = K and ωc0 = kf ωc , therefore H 0 (s) =

−Kkf ωc −Kωc . =s s + kf ωc + ωc kf

C 25 × 10−12 = = 80 nF. km kf (0.005)(0.0625)

Problems

[b] H(s) =

−K . s+1 −K

[c] H 0 (s) =  s  kf

=

+1

−Kkf . s + kf

P 15.21 [a] From Eq. 15.4 H(s) = ωc =

R2 −Ks where K = and s + ωc R1

1 ; R1 C

−K 0 s R20 0 where K = .·. H 0 (s) = s + ωc0 R10 and ωc0 =

1 R10 C 0

.

By hypothesis R10 = km R1 ;

R20 = km R2 ;

C0 =

C . km kf

It follows that K 0 = K and ωc0 = kf ωc ; .·. H 0 (s) =

[b] H(s) =

−K(s/kf ) −Ks = s . s + kf ωc + ω c kf

−Ks . s+1 −K(s/kf )

[c] H 0 (s) = 

s kf



+1

=

−Ks . s + kf

P 15.22 For the RC circuit H(s) =

s Vo = ; Vi s + (1/RC)

R0 = km R;

C0 =

C ; km kf

RC 1 .·. R0 C 0 = = ; kf kf H 0 (s) =

1 = kf . R0 C 0

s s (s/kf ) = = . s + (1/R0 C 0 ) s + kf (s/kf ) + 1

15–15

15–16

CHAPTER 15. Active Filter Circuits

For the RL circuit s H(s) = ; s + (R/L) R0 = km R;

km L ; kf

L0 =

R0 R = kf = kf ; 0 L L 

H 0 (s) =



s (s/kf ) = . s + kf (s/kf ) + 1

P 15.23 For the RC circuit H(s) =

(1/RC) Vo = ; Vi s + (1/RC)

R0 = km R;

C0 =

C ; km kf

C 1 1 .·. R0 C 0 = km R = RC = . km kf kf kf 1 = kf . R0 C 0 (1/R0 C 0 ) kf H (s) = = ; 0 0 s + (1/R C ) s + kf 0

H 0 (s) =

1 . (s/kf ) + 1

For the RL circuit R0 = km R;

L0 =

H(s) = km L; kf

R0 km R R = km = k f = kf ; 0 L L L kf 



H 0 (s) =

(R0 /L0 ) kf ; = 0 0 s + (R /L ) s + kf

H 0 (s) =

1 . (s/kf ) + 1

R/L s + R/L

so

Problems

15–17

(R/L)s βs = 2 . + (R/L)s + (1/LC) s + βs + ωo2 For the prototype circuit ωo = 1 and β = ωo /Q = 1/Q. For the scaled circuit

P 15.24 H(s) =

H 0 (s) =

s2

(R0 /L0 )s s2 + (R0 /L0 )s + (1/L0 C 0 )

where R0 = km R; L0 =

km C L; and C 0 = ; kf kf km

R0 km R R .·. = km = k f = kf β. 0 L L L kf 



kf2 1 kf km = = kf2 ; = km L0 C 0 LC LC kf Q0 =

ωo0 kf ωo = = Q. 0 β kf β

Therefore the Q of the scaled circuit is the same as the Q of the unscaled circuit. Also note β 0 = kf β. 



.·. H 0 (s) =

s2 +

kf s Q   kf s+ Q

 

H 0 (s) =  2 s kf

P 15.25 [a] L = 1 H; R=

1 Q

+

1 Q

s kf



kf2

;



s kf



.

+1

C = 1 F;

1 1 = = 0.04 Ω. Q 25

[b] kf = 100,000;

km =

3600 = 90,000. 0.04

Thus, R0 = (0.04)(90,000) = 3.6 kΩ; L0 = C0 =

90,000 (1) = 0.9 H; 100,000 1 1 = nF = 0.11 nF. 4 × 10 ) 9

(105 )(9

15–18

CHAPTER 15. Active Filter Circuits

[c]

P 15.26 [a] By hypothesis, LC = 1; Thus, C=

1 1 = F. L Q

[b] H(s) =

s2

H(s) =

(R/L)s ; + (R/L)s + (1/LC)

s2

(1/Q)s . + (1/Q)s + 1

[c] In the prototype circuit, R = 1 Ω;

L = 20 H;

.·. km = 5000;

C = 0.05 F;

kf = 50,000.

Thus R0 = 5 kΩ; L0 =

5000 (20) = 2 H; 50,000

C0 =

0.05 = 0.2 × 10−9 = 0.2 nF. (5000)(50,000)

[d]

[e] H 0 (s) =  H 0 (s) =

s 50,000

s2

1 20 2



+

s 50,000 1 20





s 50,000



; +1

2500s . + 2500s + 25 × 108

P 15.27 [a] Using the first prototype, ωo = 1 rad/s;

C = 1 F;

L = 1 H;

R = 16 Ω;

Problems 80,000 = 5000; 16

km =

kf = 80,000.

Thus, R0 = 80 kΩ; C0 =

L0 =

5 (1) = 62.5 mH; 80

1 = 2.5 nF. 400 × 106

Using the second prototype ωo = 1 rad/s; L=

C = 16 F;

1 = 6.25 mH; 16

km = 80,000;

R = 1 Ω;

kf = 80,000.

Thus, R0 = 80 kΩ; C0 =

L0 =

80 (6.25) = 6.25 mH; 80

16 = 2.5 nF. 64 × 108

[b]

P 15.28 [a] For the circuit in Fig. P15.28(a) 1 s+ Vo s2 + 1 s   H(s) = = = . 1 1 2+ 1 s+1 Vi s +s+ Q Q s For the circuit in Fig. P15.28(b) Qs + Qs Vo H(s) = = Vi 1 + Qs + Qs = H(s) =

Q(s2 + 1) ; Qs2 + s + Q s2 + 1 s2 +

  1 Q

s+1

.

15–19

15–20

CHAPTER 15. Active Filter Circuits 2



s 50,000  2 s + 15 50,000 2

[b] H 0 (s) =



+1 s 50,000 8



+1

s + 25 × 10 . s2 + 10,000s + 25 × 108

=

P 15.29 For the scaled circuit H 0 (s) =

L0 =

.·.

s2 +





s+

s2 +

R0 L0

km L; kf



1 L0 C 0



C0 =



;

1 L0 C 0

C km kf ;

kf2 1 = ; L0 C 0 LC

R0 = km R;

R0 R · .. = kf . 0 L L 



It follows then that s2 + H 0 (s) = s2 +



  R L



=  2 s kf



kf s + 2

kf2 LC

+



 

s kf

s kf

+

kf2 LC

R L

1 LC





+



1 LC



= H(s)|s=s/kf . P 15.30 For the circuit at the bottom of Fig. 14.31

H(s) =

s2 + s2 +

s RC



1 LC

+





1 LC

.

It follows that H 0 (s) =

s2 + L01C 0 s2 + R0sC 0 + L01C 0

where R0 = km R;

L0 =

km L; kf

Problems C0 =

.·.

C ; km kf

kf2 1 = . L0 C 0 LC

1 kf = 0 0 RC RC. s2 + H 0 (s) = s2 +



=  2 s kf



kf2 LC



kf RC



s kf

+



 kf2 LC

s+

2 1 RC

+

1 LC



s kf



+

1 LC

= H(s)|s=s/kf . P 15.31 For prototype circuit (a): H(s) =

Vo Q Q = ; 1 = Vi Q + s2s+1 Q + s+ 1 s

H(s) =

s2 + 1 Q(s2 + 1)   = . Q(s2 + 1) + s s2 + Q1 s + 1

For prototype circuit (b): H(s) = =

Vo 1 = Vi 1 + (s(s/Q) 2 +1) s2 + 1 s2 +

  1 Q

s+1

.

P 15.32 ωo = 50,000 rad/s; β = 300,000 rad/s; .·. ωc2 − ωc1 = 300,000; √

ωc1 ωc2 = ωo = 50,000.

15–21

15–22

CHAPTER 15. Active Filter Circuits

Solve for the cutoff frequencies: ωc1 ωc2 = 25 × 108 ; ωc2 =

25 × 108 ; ωc1

25 × 108 .·. − ωc1 = 300,000, ωc1 or ωc21 + 300,000ωc1 − 25 × 108 = 0; ωc1 = 8113.88 rad/s; .·. ωc2 = 300,000 + 8113.88 = 308,113.88 rad/s. Thus, fc1 = 1291.4 Hz

and

fc2 = 49,037.85 Hz;

ωc2 =

1 = 308,113.88; RL CL

RL =

1 = 21.64 Ω; (308,113.88)(150 × 10−9 )

ωc1 =

1 = 8113.88; RH CH

RH =

1 = 821.64 Ω. (8113.88)(150 × 10−9 )

P 15.33 [a] Hhp =

−s ; s+1

0 .·. Hhp =

−s ; s + 4000π

1 = 4000π; RH CH Hlp =

kf = 4000π;

−1 ; s+1

.·. RH = kf = 16,000π;

−16,000π 0 .·. Hlp = ; s + 16,000π

106 = 3.98 kΩ; (4000π)(0.02)

Problems 1 = 16,000π; RL CL

.·. RL =

15–23

106 = 994.72 Ω; (16,000π)(0.02)

Kωc2 = 0.8K; ωc1 + ωc2

|H(jωo )| =

20 log10 (0.8K) = 10;

√ .·. K = 1.25 10;

√ Rf .·. = 1.25 10; Ri Ri = 10 kΩ;

√ Rf = 12.5 10 = 39.53 kΩ.

s 16,000π −39.53 −63,248s · · = 2 [b] H 0 (s) = . s + 4000π s + 16,000π 10 s + 20,000πs + 64 × 106 π 2 √ [c] ωo = ωc1 ωc1 = 8000π rad/s; H 0 (jωo ) =

(16,000π)(j8000π) 39.53 · (4000π + j8000π)(16,000π + j8000π) 10

= (0.8)(3.953) = 3.16 = [d] 20 log10 |H 0 (jωo )| = 20 log10

√

√

10.

10 = 10 dB.

[e]

Note that because this is not a broadband filter, the cutoff frequencies, identified where the magnitude is down 3 dB from the maximum magnitude, are actually 1.4 kHz and 11.3 kHz. P 15.34 ωo = 2π(5000) rad/s; β = 2π(30,000) rad/s;

K = 4; C = 250 nF;

15–24

CHAPTER 15. Active Filter Circuits

β = ωc2 − ωc1 = 60,000π; ωo =

√

ωc1 ωc2 = 10,000π.

Solve for the cutoff frequencies: .·. ωc21 + 60,000πωc1 − (10,000π)2 = 0; ωc1 = 5098.1 rad/s; ωc2 = 60,000π + ωc1 = 193,593.7 rad/s; ωc1 =

1 ; RL CL

.·. RL =

1 (250 ×

10−9 )(5098.1)

= 784.6 Ω;

1 = ωc2 ; RH CH RH =

1 (250 ×

10−9 )(193,593.7)

= 20.7 Ω;

Rf = 4; Ri If Ri = 1 kΩ P 15.35 [a] ωc1 =

Rf = 4Ri = 4 kΩ.

1 = 2000π rad/s; RL CL

RL =

109 = 31.83 kΩ; (2000π)(5)

ωc2 =

1 = 10,000π rad/s; RH CH

109 RH = = 6.37 kΩ; (10,000π)(5) 

20 log10

Rf Ri



= 10;

Choose Ri = 31.83 kΩ;

.·. Rf =

√

10Ri .

then Rf = 100.66 kΩ

Problems [b]

[c] H(s)LP =

−2000π −1 = ; s/kf + 1 s + 2000π

H(s)HP = −

−s/kf −s = ; s/kf + 1 s + 10,000π

√ Rf = − 10; Ri

H(s) = =

√

2000π s 10 + s + 2000π s + 10,000π

√

s2 + 4000πs + 20 × 106 π 2 10 . (s + 2000π)(s + 10,000π)

"

#

#

"

√ ωc1 ωc2 = 20 × 106 π 2 √ √ = 1000π 20 = 2000π 5 rad/s;

[d] ωo =

√

√ # j4000π(2000π 5) √ √ H(jωo ) = 10 (2000π + j2000π 5)(10,000π + j2000π 5) √ √ √ √ j2 5 10 j2 5 10 √ √ = √ = (1 + j 5)(5 + j 5) j6 5 √ 10 = = 1.05. 3 √

"

[e] 20 log10 |H(jωo )| = 20 log10 1.05 = 0.46 dB.

15–25

15–26

CHAPTER 15. Active Filter Circuits 

1−

[f ] H(jω) =





ω √ 1000 20π

2 

ω 1 + j 2000π

4 2ζ = √ ; 20



+ j √420 ·

w √ 100 20π

ω 1 + j 10,000π

2 ζ=√ ; 20



;

ζ 2 = 0.20;

√ fo = 1000 5 = 2236.07 Hz; q √ fp = fo 1 − 2ζ 2 = fo 0.6 = 1732.05 Hz;

√ ωo = 2000π 5;

AdB (fp ) = 10 log10 [4ζ 2 (1 − ζ 2 )] = 10 log10 0.64 = −1.94 dB; AdB (fo /2) = 10 log10 0.7625 = −1.18 dB; AdB (fo ) = 20 log10 2ζ = −0.97 dB. For the quadratic term, AdB = 0 when f =

P 15.36 H(s) =

Zf =

√

2fp = 2449.48 Hz.

Vo −Zf = ; Vi Zi

(1/C2 ) 1 kR2 = ; sC2 s + (1/R2 C2 )

Zi = R 1 +

sR1 C1 + 1 1 = ; sC1 sC1

−1/C2 −(1/R1 C2 )s s + (1/R2 C2 ) .·. H(s) = = s + (1/R1 C1 ) [s + (1/R1 C1 )][s + (1/R2 C2 )] s/R1 =

−Kβs . s2 + βs + ωo2

Problems

[a] H(s) =

15–27

−10,000s −10,000s −3.57(2800s) = 2 = 2 ; 5 (s + 2000)(s + 800) s + 2800s + 16 × 10 s + 2800s + (1264.91)2

ωo = 1264.91 rad/s; β = 2800 rad/s; K = −3.57. ωo [b] Q = = 0.45; β ωc1,2

β =± + 2

v u u t

β 2

!2

ωc1 = 486.8 rad/s P 15.37 [a] H(s) =

+ ωo2 = ±1400 +

√

14002 + 16 × 105 = ±1400 + 1886.8;

ωc2 = 3286.8 rad/s.

(1/sC) (1/RC) = ; R + (1/sC) s + (1/RC)

H(jω) =

(1/RC) ; jω + (1/RC)

|H(jω)| = q

|H(jω)|2 =

(1/RC) ω 2 + (1/RC)2

;

(1/RC)2 . ω 2 + (1/RC)2

[b] Let Va be the voltage across the capacitor, positive at the upper terminal. Then Va − Vin Va + sCVa + = 0. R1 R2 + sL Solving for Va yields Va =

(R2 + sL)Vin . R1 LCs2 + (R1 R2 C + L)s + (R1 + R2 )

But vo =

sLVa . R2 + sL

Therefore Vo =

sLVin ; R1 LCs2 + (L + R1 R2 C)s + (R1 + R2 )

H(s) =

R1

LCs2

sL ; + (L + R1 R2 C)s + (R1 + R2 )

15–28

CHAPTER 15. Active Filter Circuits

H(jω) =

jωL ; [(R1 + R2 ) − R1 LCω 2 ] + jω(L + R1 R2 C)

|H(jω)| = q

|H(jω)|2 =

ωL [R1 + R2 − R1 LCω 2 ]2 + ω 2 (L + R1 R2 C)2

;

ω 2 L2 (R1 + R2 − R1 LCω 2 )2 + ω 2 (L + R1 R2 C)2

ω 2 L2 = 2 2 2 4 . R1 L C ω + (L2 + R12 R22 C 2 − 2R12 LC)ω 2 + (R1 + R2 )2 [c] Let Va be the voltage across R2 positive at the upper terminal. Then Va Va − Vin + + Va sC + Va sC = 0; R1 R2 (0 − Va )sC + (0 − Va )sC + .·. Va =

0 − Vo = 0; R3

R2 Vin 2R1 R2 Cs + R1 + R2

and Va = −

Vo . 2R3 Cs

It follows directly that H(s) =

Vo −2R2 R3 Cs = ; Vin 2R1 R2 Cs + (R1 + R2 )

H(jω) =

−2R2 R3 C(jω) ; (R1 + R2 ) + jω(2R1 R2 C)

|H(jω)| = q

|H(jω)|2 =

2R2 R3 Cω (R1 + R2 )2 + ω 2 4R12 R22 C 2

4R22 R32 C 2 ω 2 . (R1 + R2 )2 + 4R12 R22 C 2 ω 2

P 15.38 n = 5: 1 + (−1)5 s10 = 0; s10 = 1/(0 + 360k)◦

so

s10 = 1; s = 1/36k ◦ .

;

Problems

k sk+1

k sk+1

0 1/0◦

5 1/180◦

1 1/36◦

6 1/216◦

2 1/72◦

7 1/252◦

3 1/108◦

8 1/288◦

4 1/144◦

9 1/324◦

Group by conjugate pairs to form denominator polynomial. (s + 1)[s − (cos 108◦ + j sin 108◦ )][(s − (cos 252◦ + j sin 252◦ )] · [(s − (cos 144◦ + j sin 144◦ )][(s − (cos 216◦ + j sin 216◦ )] = (s + 1)(s + 0.309 − j0.951)(s + 0.309 + j0.951)· (s + 0.809 − j0.588)(s + 0.809 + j0.588) which reduces to (s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1). n = 6: 1 + (−1)6 s12 = 0

s12 = −1;

15–29

15–30

CHAPTER 15. Active Filter Circuits

s12 = 1/180◦ + 360k.

k sk+1

k sk+1

0 1/15◦

6 1/195◦

1 1/45◦

7 1/225◦

2 1/75◦

8 1/255◦

3 1/105◦

9 1/285◦

4 1/135◦

10 1/315◦

5 1/165◦

11 1/345◦

Problems Grouping by conjugate pairs yields (s + 0.2588 − j0.9659)(s + 0.2588 + j0.9659)× (s + 0.7071 − j0.7071)(s + 0.7071 + j0.7071)× (s + 0.9659 − j0.2588)(s + 0.9659 + j0.2588) or (s2 + 0.5176s + 1)(s2 + 1.4142s + 1)(s2 + 1.9318s + 1). 1 = −10 log10 (1 + ω 2n ). 1 + ω 2n From the laws of logarithms we have   −10 y= ln(1 + ω 2n ). ln 10 Thus   dy −10 2nω 2n−1 = ; dω ln 10 (1 + ω 2n )

P 15.39 [a] y = 20 log10 √

x = log10 ω =

ln ω ; ln 10

.·. ln ω = x ln 10. dω 1 dω = ln 10, = ω ln 10; ω dx dx dy = dx

dy dω

!

dω dx

!

=

−20nω 2n dB/decade. 1 + ω 2n

At ω = ωc = 1 rad/s dy = −10n dB/decade. dx 1 = −10n log10 (1 + ω 2 ) [b] y = 20 log10 √ 2 n [ 1+ω ] −10n ln(1 + ω 2 ); ln 10   dy −10n 1 −20nω = . 2ω = 2 dω ln 10 1 + ω (ln 10)(1 + ω 2 ) =

As before dω = ω(ln 10); dx

dy −20nω 2 .·. = . dx (1 + ω 2 ) √ At the corner ωc = 21/n − 1 .·. ωc2 = 21/n − 1. dy −20n[21/n − 1] = dB/decade. dx 21/n

15–31

15–32

CHAPTER 15. Active Filter Circuits

[c] For the Butterworth Filter

For the cascade of identical sections

n

dy/dx (dB/decade)

n

dy/dx (dB/decade)

1

−10

1

−10

2

−20

2

−11.72

3

−30

3

−12.38

4

−40

4

−12.73

∞

−∞

∞

−13.86

[d] It is apparent from the calculations in part (c) that as n increases the amplitude characteristic at the cutoff frequency decreases at a much faster rate for the Butterworth filter. Hence the transition region of the Butterworth filter will be much narrower than that of the cascaded sections. P 15.40 [a] n ∼ =

(−0.05)(−40) ∼ = 3.32; log10 (4000/1000)

.·. n = 4. [b] Gain = 20 log10 q

1 1 + (4)8

= −10 log10 (1 + 48 ) = −48.16 dB.

P 15.41 [a] From Table 15.1: Hlp (s) =

(s +

Hhp (s) =

1)(s2

[(1/s) +

1 ; + 0.618s + 1)(s2 + 1.618s + 1)

1][(1/s)2

1 ; + 0.618(1/s) + 1][(1/s)2 + 1.618(1/s) + 1]

s5 . (s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)

Hhp (s) =

P 15.42 [a] kf = 10,000; (s/10,000)5 [(s/10,000) + 1]

0 Hhp (s) =

·

[(s/10,000)2

1 + 0.618s/10,000 + 1][(s/10,000)2 + 1.618s/10,000 + 1]

s5 = . (s + 10,000)(s2 + 6180s + 108 )(s2 + 16,180s + 108 )

Problems [b] H 0 (j10,000) = =

j(10,000)5 [10,000(j + 1)][6180(j10,000)][16,180(j10,000)]

j(10,000)2 (1 + j)(6180)(16,180)j 2

= 0.7072/ − 45◦ ; 20 log10 |H 0 (j10,000)| = −3.01 dB. (−0.05)(−25) P 15.43 [a] n ∼ = 1.79; .·. n = 2; = log10 (100/20) .·. H(s) =

s2

√ 2 = 2: C1 C2 =

1 √ . + 2s + 1 √ C1 = 2 F;

√ 1 1 = √ = 0.5 2 F; C1 2

km = 3000;

kf = 40,000π; √

C10

=

2 = 3.75 nF; (3000)(40,000π)

1 C20 = C10 = 1.88 nF; 2

R1 = R2 = 3 kΩ.

[b]

P 15.44 [a] n ∼ =

(−0.05)(−25) = 1.79; log10 (5/1)

s2 √ ; s2 + 2s + 1 √ √ 2 = 2; R2 = 2 Ω; R2

.·. n = 2;

.·. H(s) =

kf = 10,000π;

R1 =

1 1 = √ Ω; R2 2

15–33

15–34

CHAPTER 15. Active Filter Circuits 109 4000 = . (10,000π)(25) π

.·. km =

1 4000 R1 = √ · = 900.32 Ω; π 2 R2 =

√  4000  2 = 1800.63 Ω. π

[b]

P 15.45 [a] A bandpass filter. [b] fc1 = 5000 Hz; fo = Q=

q

fc2 = 20,000 Hz;

fc1 fc2 = 10,000 Hz;

ωo fo 10,000 = = = 0.67. β fc2 − fc1 15,000 s2 √ ; s2 + 2s + 1

[c] H(s)hp =

H 0 (s)hp = = H(s)lp =

(s/104 π)2 √ (s/104 π)2 + 2(s/104 π) + 1 s2 ; s2 + π 2 × 104 s + 108 π 2 √

1 √ . s2 + 2s + 1

H 0 (s)lp = =

(s/4 ×

104 π)2

1 √ + 2(s/4 × 104 π) + 1

16 × 108 π 2 ; s2 + 4π 2 × 104 s + 16 × 108 π 2 √

H(s) = H 0 (s)hp · H 0 (s)lp =

16 × 108 π 2 s2 √ . (s2 + π 2104 s + 108 π 2 )(s2 + 4π 2 × 104 s + 16 × 108 π 2 ) √

Problems [d] ωo = 20,000π rad/s = 2 × 104 krad/s; 16 × 108 π 2 (−4 × 108 π 2 ) √ (−3 × 108 π 2 + jπ 2104 (2 × 104 π))

H(s) =

×

(12 ×

108 π 2

1 √ + j4 2π104 (2 × 104 π))

−64 −64 √ √ = = 0.9412. −68 (−3 + j2 2)(12 + j8 2)

=

P 15.46 For the scaled circuit H 0 (s) =

1/(R0 )2 C10 C20 , s2 + R02C 0 s + (R0 )21C 0 C 0 1

2

1

where R0 = km R;

C10 = C1 /kf km ;

C20 = C2 /kf km .

It follows that kf2 1 = ; (R0 )2 C10 C20 R2 C1 C2 2 R0 C10

=

2kf . RC1 kf2 /RC1 C2

.·. H 0 (s) = s2 +

=  2 s kf

k2

2kf s RC1

+ R2 Cf1 C2 1/RC1 C2

+

2 RC1



s kf



+

1 R 2 C1 C2

.

1 . + 0.765s + 1)(s2 + 1.848s + 1) [b] fc = 1000 Hz; ωc = 2000π rad/s; kf = 2000π;

P 15.47 [a] H(s) =

(s2

H 0 (s) =  =

1 s 2000π

(s2

2

+

0.765s 2000π

 

+1

2 s + 1.848s 2000π 2000π 6 2 2



+1

(4 × 10 π ) . + 1530πs + 4 × 106 π 2 )(s2 + 3696πs + 4 × 106 π 2 )

15–35

15–36

CHAPTER 15. Active Filter Circuits

[c] H 0 (j8000π) =

16 ; (−60 + j12.24)(−60 + j29.568)

|H 0 (j8000π)| =

16 = 3.91 × 10−3 ; (61.24)(66.89)

Gain = 20 log10 |H(j8000π)| = −48.16 dB (checks with Problem 15.40). P 15.48 [a] km = 2000; First stage: 2 = 0.765; C1

kf = 2000π. .·. C1 =

2 ; 0.765

C10 =

2 = 208.05 nF; (0.765)(2000)(2000π)

C2 =

1 0.765 = ; C1 2

C20 =

0.765 = 30.44 nF. 2(2000)(2000π)

Second stage: 2 = 1.848; C1

[b]

.·. C1 =

2 ; 1.848

C10 =

2 = 86.12 nF; (1.848)(2000)(2000π)

C2 =

1 1.848 = ; C1 2

C20 =

1.848 = 73.53 nF. 2(2000)(2000π)

Problems s2

P 15.49 H 0 (s) = s2

1 2 s+ + 2 k2 ) km R2 (C/km kf ) km R1 km R2 (C 2 /km f

.

s2

H 0 (s) = s2 + =

15–37

kf2 2kf s+ R2 C R1 R2 C 2

(s/kf )2 ! . 2 s 1 2 (s/kf ) + + R2 C kf R1 R2 C 2

P 15.50 [a] First we will design a unity gain filter and then provide the passband gain with an inverting amplifier. For the high pass section the cut-off frequency is 1000 Hz. The order of the Butterworth is n=

(−0.05)(−20) = 2.51; log10 (1000/400)

.·. n = 3. Hhp (s) =

s3 . (s + 1)(s2 + s + 1)

For the prototype first-order section, R1 = R2 = 1 Ω,

C = 1 F.

For the prototype second-order section, R1 = 0.5 Ω,

R2 = 2 Ω,

C = 1 F.

The scaling factors are, kf = 2π(1000) = 2000π; km =

109 104 = . 50(2000π) π

In the scaled first-order section, R1 = R2 =

104 (1) = 3.183 kΩ; π

C = 50nF. In the scaled second-order section, R1 = 0.5km = 1591.55 Ω; R2 = 2km = 6.366 kΩ;

15–38

CHAPTER 15. Active Filter Circuits C = 50 nF. For the low-pass section the cut-off frequency is 8000 Hz. The order of the Butterworth filter is (−0.05)(−20) .·. n = 3. n= = 2.51; log10 (20,000/8000) Hlp (s) =

1 . (s + 1)(s2 + s + 1)

For the prototype first-order section, R1 = R2 = 1 Ω,

C = 1 F.

For the prototype second-order section, R1 = R2 = 1 Ω;

C1 = 2 F;

C2 = 0.5 F.

The low-pass scaling factors are, km = 5 × 103 ;

kf = (8000)(2π) = 16,000π.

For the scaled first-order section, R1 = R2 = 5 kΩ;

C=

1 = 3.98 nF. (16,000π)(5 × 103 )

For the scaled second-order section, R1 = R2 = 5 kΩ; C1 =

2 = 7.96 nF; 8π × 107

C2 =

0.5 = 1.99 nF. 8π × 107

Gain amplifier: 20 log10 K = 10 dB,

.·. K = 3.16.

Since we are using 5 kΩ resistors in the low-pass stage, we will use Rf = 15.8 kΩ and Ri = 5 kΩ in the inverting amplifier stage.

Problems

15–39

[b]

P 15.51 [a] Unscaled high-pass stage: Hhp (s) =

s3 . (s + 1)(s2 + s + 1)

Frequency scaling factor kf = 2000π. Therefore the scaled transfer function is (s/2000π)3 0  Hhp (s) =    2 s s s +1 + 2000π + 1 2000π 2000π =

s3 . (s + 2000π)[s2 + 2000πs + 4 × 106 π 2 ]

Unscaled low-pass stage: Hlp (s) =

1 . (s + 1)(s2 + s + 1)

Frequency scaling factor kf = 16,000π. Therefore the scaled transfer function is 1  Hlp0 (s) =    2   s s s +1 + 16,000π + 1 16,000π 16,000π =

(16,000π)3 . (s + 16,000π)(s2 + 16,000πs + 256 × 106 π 2 )

Thus the transfer function for the filter is 4096 × 1010 π 3 s3 0 H 0 (s) = 10Hhp (s)Hlp0 (s) = , D1 D2 D3 D4 where D1 = s + 2000π; D2 = s + 16,000π;

15–40

CHAPTER 15. Active Filter Circuits D3 = s2 + 2000πs + 4 × 106 π 2 ; D4 = s2 + 16,000πs + 256 × 106 π 2 .

[b] At 400 Hz

ω = 800π rad/s:

D1 (j800π) = 800π(2.5 + j1); D2 (j800π) = 800π(20 + j1); D3 (j800π) = 16 × 105 π 2 (2.1 + j1.0); D4 (j800π) = 128 × 105 π 2 (19.95 + j1). Therefore D1 D2 D3 D4 (j800π) = 131,072π 6 1014 (2505.11/53◦ ); H 0 (j800π) =

(4096π 3 × 1010 )(512 × 106 π 3 )/ − 90◦ 131,072 × 1014 π 6 (2505.11/53◦ )

= 0.639/ − 143◦ ; .·. 20 log10 |H 0 (j800π)| = 20 log10 (0.639) = −3.89 dB. At f = 5000 Hz,

ω = 10,000π rad/s.

Then D1 (j10,000π) = 2000π(1 + j5); D2 (j10,000π) = 10,000π(1.6 + j1); D3 (j10,000π) = 107 π 2 (−9.6 + j2); D4 (j10,000π) = 107 π 2 (15.6 + j16). H 0 (j10,000π) =

(4096 × π 3 × 1010 )(1012 π 3 )/ − 90◦ 2 × 1021 π 6 (2108.22/ − 35.35◦ )

= 9.71/ − 54.65◦ ; .·. 20 log10 |H 0 (j10,000π)| = 19.74 dB. [c] From the transfer function the gain is down 19.74 + 3.89 or 23.63 dB at 400 Hz. Because the upper cut-off frequency is eight times the lower cut-off frequency we would expect the high-pass stage of the filter to predict the loss in gain at 400 Hz. For a 3nd order Butterworth 1

GAIN = 20 log10 q

1 + (1000/400)6

= −23.89 dB.

Problems

15–41

5000 Hz is in the passband for this bandpass filter. Hence we expect the gain at 5000 Hz to nearly equal 20 dB as specified in Problem 15.50. Thus our scaled transfer function confirms that the filter meets the specifications. P 15.52 [a] At very low frequencies the two capacitor branches are open and because the op amp is ideal the current in R3 is zero. Therefore at low frequencies the circuit behaves as an inverting amplifier with a gain of R2 /R1 . At very high frequencies the capacitor branches are short circuits and hence the output voltage is zero. [b] Let the node where R1 , R2 , R3 , and C2 join be denoted as a, then (Va − Vi )G1 + Va sC2 + (Va − Vo )G2 + Va G3 = 0; −Va G3 − Vo sC1 = 0, or (G1 + G2 + G3 + sC2 )Va − G2 Vo = G1 Vi ; Va =

−sC1 Vo . G3

Solving for Vo /Vi yields H(s) = = =

= =

−G1 G3 (G1 + G2 + G3 + sC2 )sC1 + G2 G3 s2 C1 C2

−G1 G3 + (G1 + G2 + G3 )C1 s + G2 G3

−G1 G3 /C1 C2 s2 + s2 + s2

i

(G1 +G2 +G3 ) s C2 1 G2 G3 −G G2 C1 C2 h i (G1 +G2 +G3 ) s C2

G1 ; G2

bo =

G1 + G2 + G3 . C2

[c] Rearranging we see that G1 = KG2 ; G3 =

+

G2 G3 C1 C2

+

G2 G3 C1 C2

−Kbo , + b1 s + bo

where K = and b1 =

h

bo C1 C2 bo C1 = ; G2 G2 .

G2 G3 ; C1 C2

15–42

CHAPTER 15. Active Filter Circuits Since by hypothesis C2 = 1 F b1 =

G1 + G2 + G3 = G1 + G2 + G3 ; C2

bo C1 .·. b1 = KG2 + G2 + ; G2 b1 = G2 (1 + K) +

bo C1 . G2

Solving this quadratic equation for G2 we get b1 G2 = ± 2(1 + K) =

b1 ±

s

b21 − bo C1 4(1 + K) 4(1 + K)2

q

b21 − 4bo (1 + K)C1 2(1 + K)

.

For G2 to be realizable C1
0.

f (t) dt

17–11

17–12

CHAPTER 17. The Fourier Transform

[b] F{e−|t| } =

1 1 2 + = . 1 + jω 1 − jω 1 + ω2

Therefore F{e−a|t| } =

(1/a)2 . (ω/a)2 + 1

Therefore F{e−0.5|t| } =

4 , +1

4ω 2

F{e−|t| } =

ω2

2 ; +1

F{e−2|t| } = 1/[0.25ω 2 + 1], yes as “a” increases, the sketches show that f (t) approaches zero faster and F (ω) flattens out over the frequency spectrum.

P 17.15 [a] F{f (t − a)} =

Z ∞

f (t − a)e−jωt dt.

−∞

Let u = t − a, then du = dt, t = u + a, and u = ±∞ when t = ±∞. Therefore, F{f (t − a)} =

Z ∞

f (u)e−jω(u+a) du

−∞ −jωa

=e

Z ∞

f (u)e−jωu du = e−jωa F (ω).

−∞ jω0 t

[b] F{e

f (t)} =

Z ∞ −∞

f (t)e−j(ω−ω0 )t dt = F (ω − ω0 .)

ejω0 t + e−jω0 t [c] F{f (t) cos ω0 t} = F f (t) 2 (

"

#)

1 1 = F (ω − ω0 ) + F (ω + ω0 ) 2 2

.

Problems P 17.16 Y (ω) =

Z ∞ Z ∞ −∞

=

Z ∞

17–13



x(λ)h(t − λ) dλ e−jωt dt

−∞

Z ∞

−jωt

h(t − λ)e

x(λ)

−∞



dt dλ.

−∞

Let u = t − λ, du = dt, and u = ±∞, when t = ±∞. Therefore Y (ω) =

Z ∞

Z ∞

x(λ)

h(u)e

−∞

=

−jω(u+λ)

du dλ

−∞

Z ∞



−jωλ

x(λ) e

Z ∞

−∞

=



−jωu

h(u)e



du dλ

−∞

Z ∞

x(λ)e−jωλ H(ω) dλ = H(ω)X(ω).

−∞

Z ∞ 

P 17.17 F{f1 (t)f2 (t)} =

−∞

= =

.·.

∞

1Z 2π

∞

1 2π

= P 17.18 [a] f 0 (t) =

1Z 2π

1Z 2π

∞

jtu

−∞

Z ∞

−∞

−∞

−∞

F1 (u)f2 (t)e−jωt ejtu du dt Z ∞

F1 (u)

−∞

du f2 (t)e−jωt dt 



Z ∞

F1 (u)e





−∞

f2 (t)e−j(ω−u)t dt du

F1 (u)F2 (ω − u) du.

2A , τ

−τ < t < 0; 2

f 0 (t) =

2A 2A [u(t + τ /2) − u(t)] − [u(t) − u(t − τ /2)] τ τ

=

f 0 (t) =

−2A , τ

τ ; 2

0