Solution of Fundamentals of Electric Circuits [4 ed.] 0077263197, 9780077263195

Citation preview

Chapter 1, Solution 1 (a) q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C

Chapter 1, Solution 2

(a) (b) (c) (d) (e)

i = dq/dt = 3 mA i = dq/dt = (16t + 4) A i = dq/dt = (-3e-t + 10e-2t) nA i=dq/dt = 1200π cos 120π t pA i =dq/dt = − e −4t (80 cos 50 t + 1000 sin 50 t ) µ A

Chapter 1, Solution 3

(a) q(t) = ∫ i(t)dt + q(0) = (3t + 1) C

(b) q(t) = ∫ (2t + s) dt + q(v) = (t 2 + 5t) mC

(c) q(t) = ∫ 20 cos (10t + π / 6 ) + q(0) = (2sin(10t + π / 6) + 1) µ C

(d)

q(t) = ∫ 10e -30t sin 40t + q(0) =

10e -30t ( −30 sin 40t - 40 cos t) 900 + 1600 = − e - 30t (0.16cos40 t + 0.12 sin 40t) C

Chapter 1, Solution 4

q = ∫ idt = ∫ =

−5 5sin 6 π t dt = cos 6π t 6π 0

5 (1 − cos 0.06π ) = 4.698 mC 6π

10

Chapter 1, Solution 5

q = ∫ idt = ∫ =

1 e dt mC = - e -2t 2

1 (1 − e 4 ) mC = 490 µC 2

Chapter 1, Solution 6

(a) At t = 1ms, i =

dq 80 = = 40 mA dt 2

(b) At t = 6ms, i =

dq = 0 mA dt

(c) At t = 10ms, i =

dq 80 = = - 20 mA 4 dt

Chapter 1, Solution 7 25A, dq  i= = - 25A, dt   25A,

2

-2t

0 ωo . 2L 2 1 4

s = −α ± α 2 − ωo2 = −20 ± 300 = −20 ± 10 3 = −2.68, − 37.32 i( t ) = A1e − 2.68t + A 2e −37.32 t di(0) i(0) = 0 = A1 + A 2 , = −2.68A1 − 37.32A 2 = −240 dt This leads to A1 = −6.928 = −A 2

(

i( t ) = 6.928 e −37.32 t − e − 268t

Since, v( t ) =

)

1 t ∫ i( t )dt + 60, we get C 0

v(t) = (60 + 64.53e-2.68t – 4.6412e-37.32t) V

= 20

Chapter 8, Solution 18.

When the switch is off, we have a source-free parallel RLC circuit.

ωo =

α < ωo

1 LC

=

1 0.25 x1

 →

α=

= 2,

1 = 0.5 2 RC

underdamped case ω d = ω o − α 2 = 4 − 0.25 = 1.936 2

Io(0) = i(0) = initial inductor current = 20/5 = 4A Vo(0) = v(0) = initial capacitor voltage = 0 V v(t ) = e −αt ( A1 cos ω d t + A2 sin ω d t ) = e −0.5αt ( A1 cos1.936t + A2 sin 1.936t ) v(0) =0 = A1

dv = e −0.5αt (−0.5)( A1 cos1.936t + A2 sin 1.936t ) + e −0.5αt (−1.936 A1 sin 1.936t + 1.936 A2 cos1.936t ) dt (V + RI o ) dv(0) ( 0 + 4) =− o =− = −4 = −0.5 A1 + 1.936 A2 dt RC 1 Thus,

 →

A2 = −2.066

v(t ) = −2.066e −0.5t sin 1.936t

Chapter 8, Solution 19.

For t < 0, the equivalent circuit is shown in Figure (a). 10 Ω

i +

+ 120V

+ −

i

−

v

L

(a)

i(0) = 120/10 = 12, v(0) = 0

C

(b)

−

v

For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = α. ωo =

1 LC

=

1 4

= 0.5 = ωd

i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t], which leads to -v(0)/L = 0 = B Hence,

i(t) = 12cos0.5t A and v = 0.5

However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin0.5t V

Chapter 8, Solution 20.

For t < 0, the equivalent circuit is as shown below. 2Ω

i

+ −

−

vC

12

+

v(0) = -12V and i(0) = 12/2 = 6A For t > 0, we have a series RLC circuit. α = R/(2L) = 2/(2x0.5) = 2 ωo = 1/ LC = 1 / 0.5x 1 4 = 2 2 Since α is less than ωo, we have an under-damped response. ωd = ωo2 − α 2 = 8 − 4 = 2

i(t) = (Acos2t + Bsin2t)e-2t i(0) = 6 = A

di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-αt di(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 – 12] = 0 Thus, B = 6 and i(t) = (6cos2t + 6sin2t)e-2t A

Chapter 8, Solution 21.

By combining some resistors, the circuit is equivalent to that shown below. 60||(15 + 25) = 24 ohms. 12 Ω

24V

6Ω

t=0

i

3H

24 Ω

+ −

+ v

At t = 0-,

−

(1/27)F

i(0) = 0, v(0) = 24x24/36 = 16V

For t > 0, we have a series RLC circuit.

R = 30 ohms, L = 3 H, C = (1/27) F

α = R/(2L) = 30/6 = 5 ωo = 1 / LC = 1 / 3x1 / 27 = 3, clearly α > ωo (overdamped response)

s1,2 = − α ± α 2 − ωo2 = −5 ± 5 2 − 3 2 = -9, -1 v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B

(1)

i = Cdv/dt = C[-Ae-t - 9Be-9t] i(0) = 0 = C[-A – 9B] or A = -9B From (1) and (2),

B = -2 and A = 18. Hence,

v(t) = (18e-t – 2e-9t) V

(2)

Chapter 8, Solution 22. α = 20 = 1/(2RC) or RC = 1/40

(1)

ωd = 50 = ωo2 − α 2 which leads to 2500 + 400 = ωo2 = 1/(LC)

Thus, LC 1/2900

(2)

In a parallel circuit, vC = vL = vR But,

iC = CdvC/dt or iC/C = dvC/dt = -80e-20tcos50t – 200e-20tsin50t + 200e-20tsin50t – 500e-20tcos50t = -580e-20tcos50t iC(0)/C = -580 which leads to C = -6.5x10-3/(-580) = 11.21 µF R = 1/(40C) = 106/(2900x11.21) = 2.23 kohms L = 1/(2900x11.21) = 30.76 H

Chapter 8, Solution 23. Let Co = C + 0.01. For a parallel RLC circuit, α = 1/(2RCo), ωo = 1/ LC o α = 1 = 1/(2RCo), we then have Co = 1/(2R) = 1/20 = 50 mF ωo = 1/ 0.5x 0.5 = 6.32 > α (underdamped) Co = C + 10 mF = 50 mF or 40 mF

Chapter 8, Solution 24. For t < 0, u(-t) 1, namely, the switch is on. v(0) = 0, i(0) = 25/5 = 5A For t > 0, the voltage source is off and we have a source-free parallel RLC circuit. α = 1/(2RC) = 1/(2x5x10-3) = 100

ωo = 1/ LC = 1 / 0.1x10 −3 = 100 ωo = α (critically damped) v(t) = [(A1 + A2t)e-100t] v(0) = 0 = A1 dv(0)/dt = -[v(0) + Ri(0)]/(RC) = -[0 + 5x5]/(5x10-3) = -5000 But,

dv/dt = [(A2 + (-100)A2t)e-100t]

Therefore, dv(0)/dt = -5000 = A2 – 0 v(t) = -5000te-100t V

Chapter 8, Solution 25. In the circuit in Fig. 8.76, calculate io(t) and vo(t) for t>0. 2Ω

30V

+ −

1H

io(t)

t=0, note this is a make before break switch so the inductor current is not interrupted.

Figure 8.78

8Ω

For Problem 8.25.

At t = 0-, vo(0) = (8/(2 + 8)(30) = 24 For t > 0, we have a source-free parallel RLC circuit. α = 1/(2RC) = ¼ ωo = 1/ LC = 1 / 1x 1 4 = 2 Since α is less than ωo, we have an under-damped response. ωd = ωo2 − α 2 = 4 − (1 / 16) = 1.9843

vo(t) = (A1cosωdt + A2sinωdt)e-αt

+ (1/4)F

vo(t)

−

vo(0) = 24 = A1 and io(t) = C(dvo/dt) = 0 when t = 0. dvo/dt = -α(A1cosωdt + A2sinωdt)e-αt + (-ωdA1sinωdt + ωdA2cosωdt)e-αt at t = 0, we get dvo(0)/dt = 0 = -αA1 + ωdA2

Thus, A2 = (α/ωd)A1 = (1/4)(24)/1.9843 = 3.024 vo(t) = (24cosωdt + 3.024sinωdt)e-t/4 volts

Chapter 8, Solution 26. s2 + 2s + 5 = 0, which leads to s1,2 =

− 2 ± 4 − 20 = -1±j4 2

i(t) = Is + [(A1cos4t + A2sin4t)e-t], Is = 10/5 = 2 i(0) = 2 = = 2 + A1, or A1 = 0 di/dt = [(A2cos4t)e-t] + [(-A2sin4t)e-t] = 4 = 4A2, or A2 = 1 i(t) = 2 + sin4te-t A

Chapter 8, Solution 27.

s2 + 4s + 8 = 0 leads to s =

− 4 ± 16 − 32 = −2 ± j2 2

v(t) = Vs + (A1cos2t + A2sin2t)e-2t 8Vs = 24 means that Vs = 3 v(0) = 0 = 3 + A1 leads to A1 = -3 dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t 0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3 v(t) = [3 – 3(cos2t + sin2t)e-2t] volts

Chapter 8, Solution 28.

The characteristic equation is s2 + 6s + 8 with roots − 6 ± 36 − 32 s1, 2 = = −4,−2 2 Hence,

i (t ) = I s + Ae −2t + Be −4t 8I s = 12

 →

i (0) = 0

 →

I s = 1.5 0 = 1.5 + A + B

(1)

di = −2 Ae − 2t − 4 Be − 4t dt di(0) = 2 = −2 A − 4 B  → 0 = 1 + A + 2 B dt Solving (1) and (2) leads to A=-2 and B=0.5.

(2)

i (t ) = 1.5 − 2e −2t + 0.5e −4t A

Chapter 8, Solution 29.

(a)

s2 + 4 = 0 which leads to s1,2 = ±j2 (an undamped circuit) v(t) = Vs + Acos2t + Bsin2t 4Vs = 12 or Vs = 3 v(0) = 0 = 3 + A or A = -3 dv/dt = -2Asin2t + 2Bcos2t dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 – 3cos2t + sin2t) V

(b)

s2 + 5s + 4 = 0 which leads to s1,2 = -1, -4 i(t) = (Is + Ae-t + Be-4t) 4Is = 8 or Is = 2 i(0) = -1 = 2 + A + B, or A + B = -3

(1)

di/dt = -Ae-t - 4Be-4t di(0)/dt = 0 = -A – 4B, or B = -A/4 From (1) and (2) we get A = -4 and B = 1 i(t) = (2 – 4e-t + e-4t) A (c)

s2 + 2s + 1 = 0, s1,2 = -1, -1 v(t) = [Vs + (A + Bt)e-t], Vs = 3. v(0) = 5 = 3 + A or A = 2 dv/dt = [-(A + Bt)e-t] + [Be-t] dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3 v(t) = [3 + (2 + 3t)e-t] V

Chapter 8, Solution 30.

s1 = −500 = −α + α 2 − ω o ,

s 2 = −800 = −α − α 2 − ω o

2

s1 + s 2 = −1300 = −2α

 →

α = 650 =

2

R 2L

Hence, L= s1 − s 2 = 300 = 2 α 2 − ω o

R 200 = = 153.8 mH 2α 2 x650 2

C=

 →

ω o = 623.45 =

1 = 16.25µF (632.45) 2 L

1 LC

(2)

Chapter 8, Solution 31.

For t = 0-, we have the equivalent circuit in Figure (a). For t = 0+, the equivalent circuit is shown in Figure (b). By KVL, v(0+) = v(0-) = 40, i(0+) = i(0-) = 1 By KCL, 2 = i(0+) + i1 = 1 + i1 which leads to i1 = 1. By KVL, -vL + 40i1 + v(0+) = 0 which leads to vL(0+) = 40x1 + 40 = 80 vL(0+) = 80 V, 40 Ω

vC(0+) = 40 V

10 Ω

i1 40 Ω

+ i

v

+

+ 50V

−

+ −

v

vL

−

10 Ω

0.5H

(a)

−

(b)

Chapter 8, Solution 32.

For t = 0-, the equivalent circuit is shown below. 2A

i +

v

−

6Ω

i(0-) = 0, v(0-) = -2x6 = -12V For t > 0, we have a series RLC circuit with a step input. α = R/(2L) = 6/2 = 3, ωo = 1/ LC = 1 / 0.04 s = − 3 ± 9 − 25 = −3 ± j4 Thus, v(t) = Vf + [(Acos4t + Bsin4t)e-3t]

50V

+ −

where Vf = final capacitor voltage = 50 V v(t) = 50 + [(Acos4t + Bsin4t)e-3t] v(0) = -12 = 50 + A which gives A = -62 i(0) = 0 = Cdv(0)/dt dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t] 0 = dv(0)/dt = -3A + 4B or B = (3/4)A = -46.5 v(t) = {50 + [(-62cos4t – 46.5sin4t)e-3t]} V

Chapter 8, Solution 33.

We may transform the current sources to voltage sources. For t = 0-, the equivalent circuit is shown in Figure (a). 10 Ω

+ −

v

10 Ω

1H

+ 30V

i

i 5Ω

−

+ v

−

30V

4F

(a)

+ −

(b)

i(0) = 30/15 = 2 A, v(0) = 5x30/15 = 10 V For t > 0, we have a series RLC circuit. α = R/(2L) = 5/2 = 2.5 ω o = 1 / LC = 1 / 4 = 0.25, clearly α > ωo (overdamped response)

s1,2 = − α ± α 2 − ω 2o = −2.5 ± 6.25 − 0.25 = -4.95, -0.05 v(t) = Vs + [A1e-4.95t + A2e-0.05t], v = 20. v(0) = 10 = 20 + A1 + A2

(1)

i(0) = Cdv(0)/dt or dv(0)/dt = 2/4 = 1/2 Hence,

½ = -4.95A1 – 0.05A2

From (1) and (2),

A1 = 0, A2 = -10.

(2)

v(t) = {20 – 10e-0.05t} V

Chapter 8, Solution 34.

Before t = 0, the capacitor acts like an open circuit while the inductor behaves like a short circuit. i(0) = 0, v(0) = 20 V For t > 0, the LC circuit is disconnected from the voltage source as shown below. Vx + − i

(1/16)F

(¼) H

This is a lossless, source-free, series RLC circuit. α = R/(2L) = 0, ωo = 1/ LC = 1/

1 1 + = 8, s = ±j8 16 4

Since α is less than ωo, we have an underdamped response. Therefore, i(t) = A1cos8t + A2sin8t where i(0) = 0 = A1 di(0)/dt = (1/L)vL(0) = -(1/L)v(0) = -4x20 = -80 However, di/dt = 8A2cos8t, thus, di(0)/dt = -80 = 8A2 which leads to A2 = -10 Now we have

i(t) = -10sin8t A

Chapter 8, Solution 35. At t = 0-, iL(0) = 0, v(0) = vC(0) = 8 V For t > 0, we have a series RLC circuit with a step input. α = R/(2L) = 2/2 = 1, ωo = 1/ LC = 1/ 1 / 5 =

5

s1,2 = − α ± α 2 − ω 2o = −1 ± j2 v(t) = Vs + [(Acos2t + Bsin2t)e-t], Vs = 12. v(0) = 8 = 12 + A or A = -4, i(0) = Cdv(0)/dt = 0. But dv/dt = [-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t] 0

= dv(0)/dt = -A + 2B or 2B = A = -4 and B = -2 v(t) = {12 – (4cos2t + 2sin2t)e-t V.

Chapter 8, Solution 36. For t = 0-, 3u(t) = 0. Thus, i(0) = 0, and v(0) = 20 V. For t > 0, we have the series RLC circuit shown below. 10 Ω

i

10 Ω

5H

+ 15V

+ −

2Ω

20 V

0.2 F

+ −

v

−

α = R/(2L) = (2 + 5 + 1)/(2x5) = 0.8 ωo = 1/ LC = 1/ 5x 0.2 = 1

s1,2 = − α ± α 2 − ω2o = −0.8 ± j0.6 v(t) = Vs + [(Acos0.6t + Bsin0.6t)e-0.8t] Vs = 15 + 20 = 35V and v(0) = 20 = 35 + A or A = -15 i(0) = Cdv(0)/dt = 0 But dv/dt = [-0.8(Acos0.6t + Bsin0.6t)e-0.8t] + [0.6(-Asin0.6t + Bcos0.6t)e-0.8t] 0

= dv(0)/dt = -0.8A + 0.6B which leads to B = 0.8x(-15)/0.6 = -20 v(t) = {35 – [(15cos0.6t + 20sin0.6t)e-0.8t]} V

i = Cdv/dt = 0.2{[0.8(15cos0.6t + 20sin0.6t)e-0.8t] + [0.6(15sin0.6t – 20cos0.6t)e-0.8t]} i(t) = [(5sin0.6t)e-0.8t] A

Chapter 8, Solution 37. For t = 0-, the equivalent circuit is shown below.

6Ω

6Ω

i2 6Ω

+

v(0) 30V

+ −

i1

10V

+ −

−

18i2 – 6i1 = 0 or i1 = 3i2

(1)

-30 + 6(i1 – i2) + 10 = 0 or i1 – i2 = 10/3

(2)

From (1) and (2).

i1 = 5, i2 = 5/3 i(0) = i1 = 5A -10 – 6i2 + v(0) = 0

v(0) = 10 + 6x5/3 = 20 For t > 0, we have a series RLC circuit. R = 6||12 = 4 ωo = 1/ LC = 1/ (1 / 2)(1 / 8) = 4 α = R/(2L) = (4)/(2x(1/2)) = 4

α = ωo, therefore the circuit is critically damped v(t) = Vs +[(A + Bt)e-4t], and Vs = 10

v(0) = 20 = 10 + A, or A = 10 i = Cdv/dt = -4C[(A + Bt)e-4t] + C[(B)e-4t] i(0) = 5 = C(-4A + B) which leads to 40 = -40 + B or B = 80 i(t) = [-(1/2)(10 + 80t)e-4t] + [(10)e-4t] i(t) = [(5 – 40t)e-4t] A

Chapter 8, Solution 38. At t = 0-, the equivalent circuit is as shown. 2A + i

v

−

10 Ω

i1 5Ω

10 Ω

i(0) = 2A, i1(0) = 10(2)/(10 + 15) = 0.8 A v(0) = 5i1(0) = 4V For t > 0, we have a source-free series RLC circuit. R = 5||(10 + 10) = 4 ohms ωo = 1/ LC = 1/ (1 / 3)(3 / 4) = 2 α = R/(2L) = (4)/(2x(3/4)) = 8/3 s1,2 = − α ± α 2 − ω 2o = -4.431, -0.903 i(t) = [Ae-4.431t + Be-0.903t] i(0) = A + B = 2

(1)

di(0)/dt = (1/L)[-Ri(0) + v(0)] = (4/3)(-4x2 + 4) = -16/3 = -5.333 Hence, -5.333 = -4.431A – 0.903B

(2)

From (1) and (2), A = 1 and B = 1. i(t) = [e-4.431t + e-0.903t] A

Chapter 8, Solution 39. For t = 0-, the equivalent circuit is shown in Figure (a). Where 60u(-t) = 60 and 30u(t) = 0. 30 Ω

60V

+ −

+ v − 20 Ω

(a)

30 Ω

0.5F

0.25H

20 Ω 30V

(b)

v(0) = (20/50)(60) = 24 and i(0) = 0

+ −

For t > 0, the circuit is shown in Figure (b). R = 20||30 = 12 ohms ωo = 1/ LC = 1/ (1 / 2)(1 / 4) =

8

α = R/(2L) = (12)/(0.5) = 24

Since α > ωo, we have an overdamped response. s1,2 = − α ± α 2 − ω 2o = -47.833, -0.167 v(t) = Vs + [Ae-47.833t + Be-0.167t], Vs = 30

Thus,

v(0) = 24 = 30 + A + B or -6 = A + B

(1)

i(0) = Cdv(0)/dt = 0 But,

dv(0)/dt = -47.833A – 0.167B = 0 B = -286.43A

From (1) and (2),

(2)

A = 0.021 and B = -6.021

v(t) = 30 + [0.021e-47.833t – 6.021e-0.167t] V

Chapter 8, Solution 40. At t = 0-, vC(0) = 0 and iL(0) = i(0) = (6/(6 + 2))4 = 3A For t > 0, we have a series RLC circuit with a step input as shown below. i

14 Ω

0.02 F

2H +

6Ω

v

−

24V

12V

+ −

ωo = 1/ LC = 1/ 2 x 0.02 = 5

α = R/(2L) = (6 + 14)/(2x2) = 5

+ −

Since α = ωo, we have a critically damped response. v(t) = Vs + [(A + Bt)e-5t], Vs = 24 – 12 = 12V v(0) = 0 = 12 + A or A = -12 i = Cdv/dt = C{[Be-5t] + [-5(A + Bt)e-5t]} i(0) = 3 = C[-5A + B] = 0.02[60 + B] or B = 90 Thus, i(t) = 0.02{[90e-5t] + [-5(-12 + 90t)e-5t]} i(t) = {(3 – 9t)e-5t} A

Chapter 8, Solution 41. At t = 0-, the switch is open. i(0) = 0, and v(0) = 5x100/(20 + 5 + 5) = 50/3 For t > 0, we have a series RLC circuit shown in Figure (a). After source transformation, it becomes that shown in Figure (b). 4Ω

10 H

5A

20 Ω

5Ω

10 µF

i +

20V

+ −

(a) ωo = 1/ LC = 1/ 1x1 / 25 = 5 α = R/(2L) = (4)/(2x1) = 2

s1,2 = − α ± α 2 − ω 2o = -2 ± j4.583 Thus,

1H

v(t) = Vs + [(Acosωdt + Bsinωdt)e-2t], where ωd = 4.583 and Vs = 20 v(0) = 50/3 = 20 + A or A = -10/3

v

(b)

−

0.04F

i(t) = Cdv/dt = C(-2) [(Acosωdt + Bsinωdt)e-2t] + Cωd[(-Asinωdt + Bcosωdt)e-2t] i(0) = 0 = -2A + ωdB B = 2A/ωd = -20/(3x4.583) = -1.455 i(t) = C{[(0cosωdt + (-2B - ωdA)sinωdt)]e-2t} = (1/25){[(2.91 + 15.2767) sinωdt)]e-2t} i(t) = {0.7275sin(4.583t)e-2t} A

Chapter 8, Solution 42. For t = 0-, we have the equivalent circuit as shown in Figure (a). i(0) = i(0) = 0, and v(0) = 4 – 12 = -8V 4V

− +

1Ω

5Ω

+ −

6Ω

12V + v(0) −

− +

i

1H +

12V

v

−

0.04F

(a) (b) For t > 0, the circuit becomes that shown in Figure (b) after source transformation. ωo = 1/ LC = 1/ 1x1 / 25 = 5 α = R/(2L) = (6)/(2) = 3

s1,2 = − α ± α 2 − ω 2o = -3 ± j4 Thus,

v(t) = Vs + [(Acos4t + Bsin4t)e-3t], Vs = -12 v(0) = -8 = -12 + A or A = 4

i = Cdv/dt, or i/C = dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t] i(0) = -3A + 4B or B = 3 v(t) = {-12 + [(4cos4t + 3sin4t)e-3t]} A

Chapter 8, Solution 43.

For t>0, we have a source-free series RLC circuit.

α=

R = 2αL = 2 x8 x0.5 = 8Ω

 →

R 2L

ω d = ω o 2 − α 2 = 30

ωo =

1 LC

ω o = 900 − 64 = 836

 →

 →

C=

ω oL 1

2

=

1 = 2.392 mF 836 x0.5

Chapter 8, Solution 44.

α=

R 1000 = = 500, 2L 2 x1

ωo > α

 →

ωo =

1 LC

=

1 100 x10

−9

= 10 4

underdamped.

Chapter 8, Solution 45. ωo = 1/ LC = 1/ 1x 0.5 =

2

α = R/(2L) = (1)/(2x2x0.5) = 0.5

Since α < ωo, we have an underdamped response. s1,2 = − α ± α 2 − ω 2o = -0.5 ± j1.323 Thus,

i(t) = Is + [(Acos1.323t + Bsin1.323t)e-0.5t], Is = 4 i(0) = 1 = 4 + A or A = -3 v = vC = vL = Ldi(0)/dt = 0

di/dt = [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t] di(0)/dt = 0 = 1.323B – 0.5A or B = 0.5(-3)/1.323 = -1.134

Thus,

i(t) = {4 – [(3cos1.323t + 1.134sin1.323t)e-0.5t]} A

Chapter 8, Solution 46.

For t = 0-, u(t) = 0, so that v(0) = 0 and i(0) = 0. For t > 0, we have a parallel RLC circuit with a step input, as shown below. +

i 8mH

v

5µF

−

2 kΩ 6mA

α = 1/(2RC) = (1)/(2x2x103 x5x10-6) = 50 ωo = 1/ LC = 1/ 8x10 3 x 5x10 −6 = 5,000

Since α < ωo, we have an underdamped response. s1,2 = − α ± α 2 − ωo2 ≅ -50 ± j5,000 Thus,

i(t) = Is + [(Acos5,000t + Bsin5,000t)e-50t], Is = 6mA i(0) = 0 = 6 + A or A = -6mA v(0) = 0 = Ldi(0)/dt

di/dt = [5,000(-Asin5,000t + Bcos5,000t)e-50t] + [-50(Acos5,000t + Bsin5,000t)e-50t] di(0)/dt = 0 = 5,000B – 50A or B = 0.01(-6) = -0.06mA Thus,

i(t) = {6 – [(6cos5,000t + 0.06sin5,000t)e-50t]} mA

Chapter 8, Solution 47.

At t = 0-, we obtain,

iL(0) = 3x5/(10 + 5) = 1A

and vo(0) = 0. For t > 0, the 20-ohm resistor is short-circuited and we have a parallel RLC circuit with a step input. α = 1/(2RC) = (1)/(2x5x0.01) = 10 ωo = 1/ LC = 1/ 1x 0.01 = 10

Since α = ωo, we have a critically damped response. s1,2 = -10 i(t) = Is + [(A + Bt)e-10t], Is = 3

Thus,

i(0) = 1 = 3 + A or A = -2 vo = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t] vo(0) = 0 = B – 10A or B = -20 Thus, vo(t) = (200te-10t) V

Chapter 8, Solution 48.

For t = 0-, we obtain i(0) = -6/(1 + 2) = -2 and v(0) = 2x1 = 2. For t > 0, the voltage is short-circuited and we have a source-free parallel RLC circuit. α = 1/(2RC) = (1)/(2x1x0.25) = 2 ωo = 1/ LC = 1/ 1x 0.25 = 2

Since α = ωo, we have a critically damped response. s1,2 = -2 Thus,

i(t) = [(A + Bt)e-2t], i(0) = -2 = A v = Ldi/dt = [Be-2t] + [-2(-2 + Bt)e-2t] vo(0) = 2 = B + 4 or B = -2 Thus, i(t) = [(-2 - 2t)e-2t] A and v(t) = [(2 + 4t)e-2t] V

Chapter 8, Solution 49.

For t = 0-, i(0) = 3 + 12/4 = 6 and v(0) = 0. For t > 0, we have a parallel RLC circuit with a step input. α = 1/(2RC) = (1)/(2x5x0.05) = 2 ωo = 1/ LC = 1/ 5x 0.05 = 2

Since α = ωo, we have a critically damped response. s1,2 = -2 i(t) = Is + [(A + Bt)e-2t], Is = 3

Thus,

i(0) = 6 = 3 + A or A = 3 v = Ldi/dt or v/L = di/dt = [Be-2t] + [-2(A + Bt)e-2t] v(0)/L = 0 = di(0)/dt = B – 2x3 or B = 6 Thus, i(t) = {3 + [(3 + 6t)e-2t]} A

Chapter 8, Solution 50.

For t = 0-, 4u(t) = 0, v(0) = 0, and i(0) = 30/10 = 3A. For t > 0, we have a parallel RLC circuit.

3A

10 Ω

+ 10 mF 6A

v

−

40 Ω

Is = 3 + 6 = 9A and R = 10||40 = 8 ohms α = 1/(2RC) = (1)/(2x8x0.01) = 25/4 = 6.25 ωo = 1/ LC = 1/ 4x 0.01 = 5

Since α > ωo, we have a overdamped response. s1,2 = − α ± α 2 − ω o2 = -10, -2.5

i 10 H

i(t) = Is + [Ae-10t] + [Be-2.5t], Is = 9

Thus,

i(0) = 3 = 9 + A + B or A + B = -6 di/dt = [-10Ae-10t] + [-2.5Be-2.5t], v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A – 2.5B or B = -4A Thus, A = 2 and B = -8 Clearly, i(t) = { 9 + [2e-10t] + [-8e-2.5t]} A Chapter 8, Solution 51.

Let i = inductor current and v = capacitor voltage. At t = 0, v(0) = 0 and i(0) = io. For t > 0, we have a parallel, source-free LC circuit (R = ∞). α = 1/(2RC) = 0 and ωo = 1/ LC which leads to s1,2 = ± jωo

v = Acosωot + Bsinωot, v(0) = 0 A iC = Cdv/dt = -i dv/dt = ωoBsinωot = -i/C

dv(0)/dt = ωoB = -io/C therefore B = io/(ωoC) v(t) = -(io/(ωoC))sinωot V where ωo =

LC

Chapter 8, Solution 52.

α = 300 =

1 2 RC

ω d = ω o 2 − α 2 = 400

From (2),

(1)  → C=

ω o = 400 2 − 300 2 = 264.575 =

1 = 285.71µF (264.575) 2 x50 x10 −3

From (1), R=

1 1 = (3500) = 5.833Ω 2αC 2 x300

1 LC

(2)

Chapter 8, Solution 53. C1 +

v1

R2

−

+ + −

vS

i1

R1

C2

i2

vo

−

i2 = C2dvo/dt

(1)

i1 = C1dv1/dt

(2)

0 = R2i2 + R1(i2 – i1) +vo

(3)

Substituting (1) and (2) into (3) we get, 0 = R2C2dvo/dt + R1(C2dvo/dt – C1dv1/dt)

(4)

Applying KVL to the outer loop produces, vs = v1 + i2R2 + vo = v1 + R2C2dvo/dt + vo, which leads to v1 = vs – vo – R2C2dvo/dt

(5)

Substituting (5) into (4) leads to, 0 = R1C2dvo/dt + R1C2dvo/dt – R1C1(dvs/dt – dvo/dt – R2C2d2vo/dt2) Hence, (R1C1R2C2)(d2vo/dt2) + (R1C1 + R2C2 +R1C2)(dvo/dt) = R1C1(dvs/dt)

Chapter 8, Solution 54.

Let i be the inductor current.

−i =

v dv + 0.5 4 dt di v = 2i + dt Substituting (1) into (2) gives

(1)

(2)

−v =

v dv 1 dv 1 d 2 v + + + 2 dt 4 dt 2 dt 2

s 2 + 2.5s + 3 = 0

 →

 →

d 2v dv + 2.5 + 3v = 0 2 dt dt

s = −1.25 ± j1.199

v = Ae −1.25t cos1.199t + Be −1.25t sin 1.199t v(0) = 2=A. Let w=1.199 dv = −1.25( Ae −1.25t cos wt + Be −1.25t sin wt ) + w(− Ae −1.25t sin wt + Be −1.25t cos wt ) dt dv(0) = 0 = −1.25 A + Bw dt

 →

B=

1.25 X 2 = 2.085 1.199

v = 2e −1.25t cos1.199t + 2.085e −1.25t sin 1.199t V

Chapter 8, Solution 55.

At the top node, writing a KCL equation produces, i/4 +i = C1dv/dt, C1 = 0.1 5i/4 = C1dv/dt = 0.1dv/dt

But,

i = 0.08dv/dt

(1)

-dv/dt = 2di/dt + 2i

(2)

v = − (2i + (1 / C 2 ) ∫ idt ) , C2 = 0.5 or

Substituting (1) into (2) gives, -dv/dt = 0.16d2v/dt2 + 0.16dv/dt 0.16d2v/dt2 + 0.16dv/dt + dv/dt = 0, or d2v/dt2 + 7.25dv/dt = 0 Which leads to s2 + 7.25s = 0 = s(s + 7.25) or s1,2 = 0, -7.25 v(t) = A + Be-7.25t

(3)

v(0) = 4 = A + B

(4)

From (1),

i(0) = 2 = 0.08dv(0+)/dt or dv(0+)/dt = 25

But,

dv/dt = -7.25Be-7.25t, which leads to,

dv(0)/dt = -7.25B = 25 or B = -3.448 and A = 4 – B = 4 + 3.448 = 7.448 Thus, v(t) = {7.45 – 3.45e-7.25t} V

Chapter 8, Solution 56.

For t < 0, i(0) = 0 and v(0) = 0. For t > 0, the circuit is as shown below. 4Ω 6Ω

i

i

0.04F + −

20

io

0.25H

Applying KVL to the larger loop,

-20 +6io +0.25dio/dt + 25 ∫ (i o + i)dt = 0

Taking the derivative, 6dio/dt + 0.25d2io/dt2 + 25(io + i) = 0

For the smaller loop,

(1)

4 + 25 ∫ (i + i o )dt = 0

Taking the derivative,

25(i + io) = 0 or i = -io

From (1) and (2)

6dio/dt + 0.25d2io/dt2 = 0 This leads to, 0.25s2 + 6s = 0 or s1,2 = 0, -24

(2)

io(t) = (A + Be-24t) and io(0) = 0 = A + B or B = -A As t approaches infinity, io(∞) = 20/10 = 2 = A, therefore B = -2 Thus, io(t) = (2 - 2e-24t) = -i(t) or i(t) = (-2 + 2e-24t) A

Chapter 8, Solution 57.

(a) Let v = capacitor voltage and i = inductor current. At t = 0-, the switch is closed and the circuit has reached steady-state. v(0-) = 16V and i(0-) = 16/8 = 2A At t = 0+, the switch is open but, v(0+) = 16 and i(0+) = 2. We now have a source-free RLC circuit. R 8 + 12 = 20 ohms, L = 1H, C = 4mF. α = R/(2L) = (20)/(2x1) = 10 ωo = 1/ LC = 1/ 1x (1 / 36) = 6 Since α > ωo, we have a overdamped response. s1,2 = − α ± α 2 − ωo2 = -18, -2 Thus, the characteristic equation is (s + 2)(s + 18) = 0 or s2 + 20s +36 = 0. (b)

i(t) = [Ae-2t + Be-18t] and i(0) = 2 = A + B To get di(0)/dt, consider the circuit below at t = 0+. i

12 Ω +

+ v

−

8Ω

(1/36)F

vL

−

1H

-v(0) + 20i(0) + vL(0) = 0, which leads to,

(1)

-16 + 20x2 + vL(0) = 0 or vL(0) = -24 Ldi(0)/dt = vL(0) which gives di(0)/dt = vL(0)/L = -24/1 = -24 A/s Hence -24 = -2A – 18B or 12 = A + 9B From (1) and (2),

(2)

B = 1.25 and A = 0.75

i(t) = [0.75e-2t + 1.25e-18t] = -ix(t) or ix(t) = [-0.75e-2t - 1.25e-18t] A v(t) = 8i(t) = [6e-2t + 10e-18t] A

Chapter 8, Solution 58.

(a) Let i =inductor current, v = capacitor voltage i(0) =0, v(0) = 4 [v(0) + Ri(0)] (4 + 0) dv(0) =− =− = − 8 V/s dt RC 0.5 (b) For t ≥ 0 , the circuit is a source-free RLC parallel circuit.

α=

1 1 = = 1, 2 RC 2 x0.5 x1

ωo =

1 LC

=

1 0.25 x1

=2

ω d = ω 2 o − α 2 = 4 − 1 = 1.732 Thus, v(t ) = e −t ( A1 cos1.732t + A2 sin 1.732t ) v(0) = 4 = A1

dv = −e −t A1 cos1.732t − 1.732e −t A1 sin 1.732t − e −t A2 sin 1.732t + 1.732e −t A2 cos1.732t dt dv(0) = −8 = − A1 + 1.732 A2  → A2 = −2.309 dt v(t ) = e − t (4 cos 1.732t − 2.309 sin 1.732t ) V

Chapter 8, Solution 59.

Let i = inductor current and v = capacitor voltage v(0) = 0, i(0) = 40/(4+16) = 2A For t>0, the circuit becomes a source-free series RLC with

α=

1 1 16 R = = 2, ω o = = = 2,  → α = ω o = 2 2L 2 x4 4 x1 / 16 LC i (t ) = Ae −2t + Bte −2t i(0) = 2 = A di = −2 Ae − 2t + Be − 2t − 2 Bte − 2t dt 1 1 di (0) = −2 A + B = − [ Ri(0) + v(0)]  → − 2 A + B = − (32 + 0), dt L 4 i (t ) = 2e −2t − 4te −2t v=

1 idt + v(0) = 32 ∫ e − 2t dt − 64 ∫ te − 2t dt = −16e − 2t ∫ C0 0 0 t

t

t

t 0

−

64 − 2t e (−2t − 1) 4

B = −4

t 0

v = 32te −2t V

Chapter 8, Solution 60.

At t = 0-, 4u(t) = 0 so that i1(0) = 0 = i2(0)

(1)

Applying nodal analysis, 4 = 0.5di1/dt + i1 + i2 Also,

i2 = [1di1/dt – 1di2/dt]/3 or 3i2 = di1/dt – di2/dt

Taking the derivative of (2), 0 = d2i1/dt2 + 2di1/dt + 2di2/dt From (2) and (3),

(2) (3) (4)

di2/dt = di1/dt – 3i2 = di1/dt – 3(4 – i1 – 0.5di1/dt) = di1/dt – 12 + 3i1 + 1.5di1/dt

Substituting this into (4), d2i1/dt2 + 7di1/dt + 6i1 = 24 which gives s2 + 7s + 6 = 0 = (s + 1)(s + 6)

Thus, i1(t) = Is + [Ae-t + Be-6t], 6Is = 24 or Is = 4 i1(t) = 4 + [Ae-t + Be-6t] and i1(0) = 4 + [A + B]

(5)

i2 = 4 – i1 – 0.5di1/dt = i1(t) = 4 + -4 - [Ae-t + Be-6t] – [-Ae-t - 6Be-6t] = [-0.5Ae-t + 2Be-6t] and i2(0) = 0 = -0.5A + 2B From (5) and (6),

(6)

A = -3.2 and B = -0.8 i1(t) = {4 + [-3.2e-t – 0.8e-6t]} A i2(t) = [1.6e-t – 1.6e-6t] A

Chapter 8, Solution 61.

For t > 0, we obtain the natural response by considering the circuit below. 1H

a 4Ω

iL

+ vC

−

0.25F

6Ω

At node a,

vC/4 + 0.25dvC/dt + iL = 0

(1)

But,

vC = 1diL/dt + 6iL

(2)

Combining (1) and (2), (1/4)diL/dt + (6/4)iL + 0.25d2iL/dt2 + (6/4)diL/dt + iL = 0 d2iL/dt2 + 7diL/dt + 10iL = 0 s2 + 7s + 10 = 0 = (s + 2)(s + 5) or s1,2 = -2, -5 Thus, iL(t) = iL(∞) + [Ae-2t + Be-5t], where iL(∞) represents the final inductor current = 4(4)/(4 + 6) = 1.6 iL(t) = 1.6 + [Ae-2t + Be-5t] and iL(0) = 1.6 + [A+B] or -1.6 = A+B diL/dt = [-2Ae-2t - 5Be-5t]

(3)

and diL(0)/dt = 0 = -2A – 5B or A = -2.5B

(4)

From (3) and (4), A = -8/3 and B = 16/15 iL(t) = 1.6 + [-(8/3)e-2t + (16/15)e-5t] v(t) = 6iL(t) = {9.6 + [-16e-2t + 6.4e-5t]} V vC = 1diL/dt + 6iL = [ (16/3)e-2t - (16/3)e-5t] + {9.6 + [-16e-2t + 6.4e-5t]} vC = {9.6 + [-(32/3)e-2t + 1.0667e-5t]} i(t) = vC/4 = {2.4 + [-2.667e-2t + 0.2667e-5t]} A

Chapter 8, Solution 62.

This is a parallel RLC circuit as evident when the voltage source is turned off. α = 1/(2RC) = (1)/(2x3x(1/18)) = 3 ωo = 1/ LC = 1/ 2x1 / 18 = 3 Since α = ωo, we have a critically damped response. s1,2 = -3 Let v(t) = capacitor voltage Thus, v(t) = Vs + [(A + Bt)e-3t] where Vs = 0 But -10 + vR + v = 0 or vR = 10 – v Therefore vR = 10 – [(A + Bt)e-3t] where A and B are determined from initial conditions.

Chapter 8, Solution 63.

R

v1

+

R vo

vs

v2 C

C

At node 1, v s − v1 dv =C 1 R dt

(1)

At node 2, dv v2 − vo =C o (2) dt R As a voltage follower, v1 = v 2 = v . Hence (2) becomes dv (3) v = v o + RC o dt and (1) becomes dv v s = v + RC (4) dt Substituting (3) into (4) gives

v s = vo + RC

dvo d 2 vo dv + RC o + R 2 C 2 dt dt dt 2

or R 2C 2

d 2 vo dv + 2 RC o + vo = v s 2 dt dt

Chapter 8, Solution 64. C2 R2 R1 vs

1 v1

C1

2

− +

vo

At node 1,

(vs – v1)/R1 = C1 d(v1 – 0)/dt or vs = v1 + R1C1dv1/dt

At node 2,

C1dv1/dt = (0 – vo)/R2 + C2d(0 – vo)/dt or

From (1) and (2), or

(1)

–R2C1dv1/dt = vo + C2dvo/dt

(2)

(vs – v1)/R1 = C1 dv1/dt = -(1/R2)(vo + C2dvo/dt) v1 = vs + (R1/R2)(vo + C2dvo/dt)

(3)

Substituting (3) into (1) produces, vs = vs + (R1/R2)(vo + C2dvo/dt) + R1C1d{vs + (R1/R2)(vo + C2dvo/dt)}/dt = vs + (R1/R2)(vo)+ (R1C2/R2) dvo/dt) + R1C1dvs/dt + (R1R1C1/R2)dvo/dt + (R12 C1C2/R2)[d2vo/dt2] Simplifying we get, d2vo/dt2 + [(1/ R1C1) + (1/ C2)]dvo/dt + [1/(R1C1C2)](vo) = - [R2/(R1C2)]dvs/dt

Chapter 8, Solution 65. At the input of the first op amp, (vo – 0)/R = Cd(v1 – 0)

(1)

At the input of the second op amp, (-v1 – 0)/R = Cdv2/dt

(2)

Let us now examine our constraints. Since the input terminals are essentially at ground, then we have the following, vo = -v2 or v2 = -vo Combining (1), (2), and (3), eliminating v1 and v2 we get, d 2 vo  1  d 2vo − 100 v o = 0 − v o = dt 2  R 2 C 2  dt 2

(3)

Which leads to s2 – 100 = 0 Clearly this produces roots of –10 and +10. And, we obtain, vo(t) = (Ae+10t + Be-10t)V At t = 0, vo(0+) = – v2(0+) = 0 = A + B, thus B = –A This leads to vo(t) = (Ae+10t – Ae-10t)V. Now we can use v1(0+) = 2V. From (2), v1 = –RCdv2/dt = 0.1dvo/dt = 0.1(10Ae+10t + 10Ae-10t) v1(0+) = 2 = 0.1(20A) = 2A or A = 1 Thus, vo(t) = (e+10t – e-10t)V It should be noted that this circuit is unstable (clearly one of the poles lies in the righthalf-plane).

Chapter 8, Solution 66. C2

vS

R2

R1

2

+ –

1

vo R4

C1

Note that the voltage across C1 is

R3

v2 = [R3/(R3 + R4)]vo

This is the only difference between this problem and Example 8.11, i.e. v = kv, where k = [R3/(R3 + R4)].

At node 1, (vs – v1)/R1 = C2[d(v1 – vo)/dt] + (v1 – v2)/R2 vs/R1 = (v1/R1) + C2[d(v1)/dt] – C2[d(vo)/dt] + (v1 – kvo)/R2

(1)

At node 2, (v1 – kvo)/R2 = C1[d(kvo)/dt] or

v1 = kvo + kR2C1[d(vo)/dt]

(2)

Substituting (2) into (1), vs/R1 = (kvo/R1) + (kR2C1/R1)[d(vo)/dt] + kC2[d(vo)/dt] + kR2C1C2[d2(vo)/dt2] – (kvo/R2) + kC1[d(vo)/dt] – (kvo/R2) + C2[d(vo)/dt] We now rearrange the terms. [d2(vo)/dt2] + [(1/C2R1) + (1/ R2C2) + (1/R2C1) – (1/ kR2C1)][d(vo)/dt] + [vo/(R1R2C1C2)] = vs/(kR1R2C1C2) If R1 = R2 10 kohms, C1 = C2 = 100 µF, R3 = 20 kohms, and R4 = 60 kohms, k = [R3/(R3 + R4)] = 1/3 R1R2C1C2 = 104 x104 x10-4 x10-4 = 1 (1/C2R1) + (1/ R2C2) + (1/R2C1) – (1/ kR2C1) = 1 + 1 + 1 – 3 = 3 – 3 = 0 Hence,

[d2(vo)/dt2] + vo = 3vs = 6, t > 0, and s2 + 1 = 0, or s1,2 = ±j vo(t) = Vs + [Acost + B sint], Vs = 6 vo(0) = 0 = 6 + A or A = –6 dvo/dt = –Asint + Bcost, but dvo(0)/dt = 0 = B

Hence,

vo(t) = 6(1 – cost)u(t) volts.

Chapter 8, Solution 67. At node 1, d ( v1 − v o ) d ( v 1 − 0) v in − v1 = C1 + C2 dt dt R1

At node 2,

C2

(1)

− vo d ( v 1 − 0) 0 − v o dv1 , or = = dt R2 dt C2R 2

(2)

From (1) and (2), v in − v1 = − v1 = v in +

v dv R 1C1 dv o − R 1 C1 o − R 1 o R2 dt C 2 R 2 dt

v dv R 1C1 dv o + R 1 C1 o + R 1 o R2 dt C 2 R 2 dt

(3)

C1 R2 R1 vin

C2 1

v1

2 0V

− +

vo

From (2) and (3), −

d 2 v o R 1 dv o vo dv dv R C dv o = 1 = in + 1 1 + R 1 C1 + R 2 dt C2R 2 dt dt C 2 R 2 dt dt 2 d 2 vo vo 1  1 1  dv o 1 dv in   + =− + + 2 R 2  C1 C 2  dt C1 C 2 R 2 R 1 R 1C1 dt dt

But C1C2R1R2 = 10-4 x10-4 x104 x104 = 1 1 R2

 1 1  2 2  =  = 4 =2 + −4  C1 C 2  R 2 C1 10 x10

d 2 vo dv dv + 2 o + v o = − in 2 dt dt dt

Which leads to s2 + 2s + 1 = 0 or (s + 1)2 = 0 and s = –1, –1 Therefore,

vo(t) = [(A + Bt)e-t] + Vf

As t approaches infinity, the capacitor acts like an open circuit so that Vf = vo(∞) = 0 vin = 10u(t) mV and the fact that the initial voltages across each capacitor is 0 means that vo(0) = 0 which leads to A = 0. vo(t) = [Bte-t] dv o = [(B – Bt)e-t] dt dv o (0+ ) v (0+ ) =− o =0 dt C2R 2

From (2), From (1) at t = 0+,

dv (0+) dv o (0+ ) 1 1− 0 = −C 1 o which leads to =− = −1 dt dt C1 R 1 R1

Substituting this into (4) gives B = –1 Thus,

v(t) = –te-tu(t) V

(4)

Chapter 8, Solution 68. The schematic is as shown below. The unit step is modeled by VPWL as shown. We insert a voltage marker to display V after simulation. We set Print Step = 25 ms and final step = 6s in the transient box. The output plot is shown below.

Chapter 8, Solution 69. The schematic is shown below. The initial values are set as attributes of L1 and C1. We set Print Step to 25 ms and the Final Time to 20s in the transient box. A current marker is inserted at the terminal of L1 to automatically display i(t) after simulation. The result is shown below.

Chapter 8, Solution 70.

The schematic is shown below.

After the circuit is saved and simulated, we obtain the capacitor voltage v(t) as shown below.

Chapter 8, Solution 71. The schematic is shown below. We use VPWL and IPWL to model the 39 u(t) V and 13 u(t) A respectively. We set Print Step to 25 ms and Final Step to 4s in the Transient box. A voltage marker is inserted at the terminal of R2 to automatically produce the plot of v(t) after simulation. The result is shown below.

Chapter 8, Solution 72.

When the switch is in position 1, we obtain IC=10 for the capacitor and IC=0 for the inductor. When the switch is in position 2, the schematic of the circuit is shown below.

When the circuit is simulated, we obtain i(t) as shown below.

Chapter 8, Solution 73. (a)

For t < 0, we have the schematic below. When this is saved and simulated, we obtain the initial inductor current and capacitor voltage as iL(0) = 3 A and vc(0) = 24 V.

(b) For t > 0, we have the schematic shown below. To display i(t) and v(t), we insert current and voltage markers as shown. The initial inductor current and capacitor voltage are also incorporated. In the Transient box, we set Print Step = 25 ms and the Final Time to 4s. After simulation, we automatically have io(t) and vo(t) displayed as shown below.

Chapter 8, Solution 74. 5Ω

10 Ω + 20 V -

2F

4H

Hence the dual circuit is shown below. 2H

20A

0.1 Ω

4F

0.2 Ω

Chapter 8, Solution 75. The dual circuit is connected as shown in Figure (a). It is redrawn in Figure (b). 0.1 Ω 12V

10 Ω

+ −

12A

24A

0.5 F 24V 0.25 Ω

4Ω

10 H

10 H

10 µF

(a) 0.1 Ω 2F 0.5 H

24A

12A

0.25 Ω

(b)

+ −

Chapter 8, Solution 76. The dual is obtained from the original circuit as shown in Figure (a). It is redrawn in Figure (b). 0.1 Ω

0.05 Ω

1/3 Ω

10 Ω

20 Ω

60 A

120 A

+ −

– +

60 V

4H

30 Ω

120 V

+ −

2V

1F

1H

2A

4F

(a) 0.05 Ω

60 A 0.1 Ω

120 A

1H

1/4 F 2V

(b)

1/30 Ω + −

Chapter 8, Solution 77. The dual is constructed in Figure (a) and redrawn in Figure (b). – +

5A 2Ω

5V 1/3 Ω

1/2 Ω 1F

1Ω

1/4 H

1H

3Ω

1Ω

1/4 F

12V

+ −

12 A

(a)

1Ω

2Ω

1/3 Ω

1/4 F

12 A 1H

5V

+ −

(b)

Chapter 8, Solution 78. The voltage across the igniter is vR = vC since the circuit is a parallel RLC type. vC(0) = 12, and iL(0) = 0. α = 1/(2RC) = 1/(2x3x1/30) = 5 ωo = 1 / LC = 1 / 60 x10 −3 x1 / 30 = 22.36

α < ωo produces an underdamped response. s1, 2 = −α ± α 2 − ω o2 = –5 ± j21.794

vC(t) = e-5t(Acos21.794t + Bsin21.794t)

(1)

vC(0) = 12 = A dvC/dt = –5[(Acos21.794t + Bsin21.794t)e-5t] + 21.794[(–Asin21.794t + Bcos21.794t)e-5t]

(2)

dvC(0)/dt = –5A + 21.794B But,

dvC(0)/dt = –[vC(0) + RiL(0)]/(RC) = –(12 + 0)/(1/10) = –120

Hence,

–120 = –5A + 21.794B, leads to B (5x12 – 120)/21.794 = –2.753

At the peak value, dvC(to)/dt = 0, i.e., 0

= A + Btan21.794to + (A21.794/5)tan21.794to – 21.794B/5 (B + A21.794/5)tan21.794to = (21.794B/5) – A

tan21.794to = [(21.794B/5) – A]/(B + A21.794/5) = –24/49.55 = –0.484 Therefore,

21.7945to = |–0.451|

to = |–0.451|/21.794 = 20.68 ms

Chapter 8, Solution 79. For critical damping of a parallel RLC circuit,

α = ωo

 →

1 = 2 RC

1 LC

Hence,

C=

0.25 L = = 434 µF 2 4 x144 4R

Chapter 8, Solution 80.

t1 = 1/|s1| = 0.1x10-3 leads to s1 = –1000/0.1 = –10,000 t2 = 1/|s2| = 0.5x10-3 leads to s1 = –2,000 s1 = −α − α 2 − ωo2 s 2 = −α + α 2 − ω o2

s1 + s2 = –2α = –12,000, therefore α = 6,000 = R/(2L) L = R/12,000 = 60,000/12,000 = 5H s 2 = −α + α 2 − ωo2 = –2,000 α − α 2 − ωo2 = 2,000 6,000 − α 2 − ωo2 = 2,000 α 2 − ωo2 = 4,000

α2 – ωo2 = 16x106

ωo2 = α2 – 16x106 = 36x106 – 16x106 ωo = 103 20 = 1 / LC

C = 1/(20x106x5) = 10 nF

Chapter 8, Solution 81. t = 1/α = 0.25 leads to α = 4

But,

C = 1/(2αR) = 1/(2x4x200) = 625 µF

α 1/(2RC) or,

ωd = ωo2 − α 2

ωo2 = ωd2 + α 2 = (2π4x10 3 ) 2 + 16 ≅ (2π4 x10 3 0 2 = 1/(LC) This results in L = 1/(64π2x106x625x10-6) = 2.533 µH Chapter 8, Solution 82.

For t = 0-, v(0) = 0. For t > 0, the circuit is as shown below. R1

a +

+ C1

vo

−

R2

v

−

C2

At node a, (vo – v/R1 = (v/R2) + C2dv/dt vo = v(1 + R1/R2) + R1C2 dv/dt 60 = (1 + 5/2.5) + (5x106 x5x10-6)dv/dt 60 = 3v + 25dv/dt v(t) = Vs + [Ae-3t/25] where

3Vs = 60 yields Vs = 20

v(0) = 0 = 20 + A or A = –20 v(t) = 20(1 – e-3t/25)V

Chapter 8, Solution 83.

i = iD + Cdv/dt

(1)

–vs + iR + Ldi/dt + v = 0

(2)

Substituting (1) into (2), vs = RiD + RCdv/dt + Ldi/dt + LCd2v/dt2 + v = 0 LCd2v/dt2 + RCdv/dt + RiD + Ldi/dt = vs d2v/dt2 + (R/L)dv/dt + (R/LC)iD + (1/C)di/dt = vs/LC

Chapter 9, Solution 1. ω = 103 rad/s

(a)

angular frequency

(b)

frequency

f =

(c)

period

T =

ω = 159.2 Hz 2π

1 = 6.283 ms f

Since sin(A) = cos(A – 90°), vs = 12 sin(103t + 24°) = 12 cos(103t + 24° – 90°) vs in cosine form is vs = 12 cos(103t – 66°) V

(d)

vs(2.5 ms) = 12 sin((10 3 )(2.5 × 10 -3 ) + 24°) = 12 sin(2.5 + 24°) = 12 sin(143.24° + 24°) = 2.65 V

(e)

Chapter 9, Solution 2.

(a)

amplitude = 8 A

(b)

ω = 500π = 1570.8 rad/s

(c)

f =

(d)

ω = 250 Hz 2π

Is = 8∠-25° A Is(2 ms) = 8 cos((500π )(2 × 10 -3 ) − 25°) = 8 cos(π − 25°) = 8 cos(155°) = -7.25 A

Chapter 9, Solution 3.

(a) (b) (c)

4 sin(ωt – 30°) = 4 cos(ωt – 30° – 90°) = 4 cos(ωt – 120°) -2 sin(6t) = 2 cos(6t + 90°)

-10 sin(ωt + 20°) = 10 cos(ωt + 20° + 90°) = 10 cos(ωt + 110°)

Chapter 9, Solution 4.

(a) (b)

v = 8 cos(7t + 15°) = 8 sin(7t + 15° + 90°) = 8 sin(7t + 105°)

i = -10 sin(3t – 85°) = 10 cos(3t – 85° + 90°) = 10 cos(3t + 5°)

Chapter 9, Solution 5.

v1 = 20 sin(ωt + 60°) = 20 cos(ωt + 60° − 90°) = 20 cos(ωt − 30°) v2 = 60 cos(ωt − 10°)

This indicates that the phase angle between the two signals is 20° and that v1 lags v2.

Chapter 9, Solution 6.

(a)

(b)

(c)

v(t) = 10 cos(4t – 60°) i(t) = 4 sin(4t + 50°) = 4 cos(4t + 50° – 90°) = 4 cos(4t – 40°) Thus, i(t) leads v(t) by 20°.

v1(t) = 4 cos(377t + 10°) v2(t) = -20 cos(377t) = 20 cos(377t + 180°) Thus, v2(t) leads v1(t) by 170°.

x(t) = 13 cos(2t) + 5 sin(2t) = 13 cos(2t) + 5 cos(2t – 90°) X = 13∠0° + 5∠-90° = 13 – j5 = 13.928∠-21.04° x(t) = 13.928 cos(2t – 21.04°) y(t) = 15 cos(2t – 11.8°) phase difference = -11.8° + 21.04° = 9.24° Thus, y(t) leads x(t) by 9.24°.

Chapter 9, Solution 7.

If f(φ) = cosφ + j sinφ, df = -sinφ + j cos φ = j (cos φ + j sin φ) = j f (φ ) dφ

df = j dφ f

Integrating both sides ln f = jφ + ln A f = Aejφ = cosφ + j sinφ f(0) = A = 1 i.e. f(φ) = ejφ = cosφ + j sinφ Chapter 9, Solution 8.

(a)

(b)

(c)

15∠45° 15∠45° + j2 = + j2 5∠ - 53.13° 3 − j4 = 3∠98.13° + j2 = -0.4245 + j2.97 + j2 = -0.4243 + j4.97 (2 + j)(3 – j4) = 6 – j8 + j3 + 4 = 10 – j5 = 11.18∠-26.57° 8∠ - 20° (-5 − j12)(10) 8∠ - 20° 10 + + = 11.18∠ - 26.57° 25 + 144 (2 + j)(3 - j4) - 5 + j12 = 0.7156∠6.57° − 0.2958 − j0.71 = 0.7109 + j0.08188 − 0.2958 − j0.71 = 0.4151 − j0.6281 10 + (8∠50°)(13∠-68.38°) = 10+104∠-17.38° = 109.25 – j31.07

Chapter 9, Solution 9.

(3 + j4)(5 + j8) 3 + j4 = 2+ 25 + 64 5 − j8 15 + j24 + j20 − 32 = 2+ 89 = 1.809 + j0.4944

(a)

2+

(b)

4∠-10° +

1 − j2 2.236 ∠ - 63.43° = 4∠-10° + 3∠6° 3∠6°

= 4∠-10° + 0.7453∠-69.43° = 3.939 – j0.6946 + 0.2619 – j0.6978 = 4.201 – j1.392 (c)

8∠10° + 6 ∠ - 20° 7.879 + j1.3892 + 5.638 − j2.052 = 9∠80° − 4∠50° 1.5628 + j8.863 − 2.571 − j3.064 13.533∠ - 2.81° 13.517 − j0.6629 = = 5.886∠99.86° − 1.0083 + j5.799 = 2.299∠-102.67° = -0.5043 – j2.243

Chapter 9, Solution 10.

(a) z1 = 6 − j8, z 2 = 8.66 − j 5, and z 3 = −4 − j 6.9282 z1 + z 2 + z 3 = 10.66 − j19.93 (b)

z1 z 2 = 9.999 + j 7.499 z3

Chapter 9, Solution 11.

(a)

(b) (c)

z 1 z 2 = (-3 + j4)(12 + j5) = -36 – j15 + j48 – 20 = -56 + j33 z1 - 3 + j4 (-3 + j4)(12 + j5) = -0.3314 + j0.1953 = ∗ = 144 + 25 z2 12 − j5 z 1 + z 2 = (-3 + j4) + (12 + j5) = 9 + j9 z 1 − z 2 = (-3 + j4) – (12 + j5) = -15 – j z1 + z 2 9 (1 + j) - 9 (1 + j)(15 - j) - 9 (16 + j14) = = = z1 − z 2 15 2 − 12 226 - (15 + j) = -0.6372 – j0.5575

Chapter 9, Solution 12.

(a)

z 1 z 2 = (-3 + j4)(12 + j5) = -36 – j15 + j48 – 20 = -56 + j33 z1 (-3 + j4)(12 + j5) - 3 + j4 = = -0.3314 + j0.1953 ∗ = z2 144 + 25 12 − j5

(b)

z 1 + z 2 = (-3 + j4) + (12 + j5) = 9 + j9 z 1 − z 2 = (-3 + j4) – (12 + j5) = -15 – j z1 + z 2 - 9 (1 + j)(15 - j) - 9 (16 + j14) 9 (1 + j) = = = 2 2 z1 − z 2 15 − 1 226 - (15 + j) = -0.6372 – j0.5575

(c)

Chapter 9, Solution 13.

(a) (−0.4324 + j 0.4054)+ (−0.8425 − j 0.2534) = − 1.2749 + j 0.1520

(b)

50∠ − 30 o = − 2.0833 24∠150 o

(c) (2+j3)(8-j5) –(-4) = 35 +j14

Chapter 9, Solution 14.

(a)

(b)

3 − j14 = − 0.5751 + j 0.5116 − 15 + j11

(62.116 + j 231.82 + 138.56 − j80)(60 − j80) 24186 − 6944.9 = = − 1.922 − j11.55 (67 + j84)(16.96 + j10.5983) 246.06 + j 2134.7

(c) (− 2 + j 4 )

2

(260 − j120) = − 256.4 − j 200.89

Chapter 9, Solution 15.

(a)

(b)

(c)

10 + j6 2 − j3 = -10 – j6 + j10 – 6 + 10 – j15 -5 -1 + j = -6 – j11 20∠ − 30° - 4∠ - 10° = 60∠15° + 64∠-10° 16∠0° 3∠45° = 57.96 + j15.529 + 63.03 – j11.114 = 120.99 – j4.415

1− j − j 0 j 1 −j 1 j 1+ j

1− j − j j 1

0 −j

= 1 + 1 + 0 − 1 − 0 + j2 (1 − j) + j2 (1 + j) = 1 − 1 (1 − j + 1 + j) = 1 – 2 = -1

Chapter 9, Solution 16.

(a)

-10 cos(4t + 75°) = 10 cos(4t + 75° − 180°) = 10 cos(4t − 105°) The phasor form is 10∠-105°

(b)

5 sin(20t – 10°) = 5 cos(20t – 10° – 90°) = 5 cos(20t – 100°) The phasor form is 5∠-100°

(c)

4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°) The phasor form is 4∠0° + 3∠-90° = 4 – j3 = 5∠-36.87°

Chapter 9, Solution 17.

(a)

Let A = 8∠-30° + 6∠0° = 12.928 – j4 = 13.533∠-17.19° a(t) = 13.533 cos(5t + 342.81°)

(b)

We know that -sinα = cos(α + 90°). Let B = 20∠45° + 30∠(20° + 90°) = 14.142 + j14.142 – 10.261 + j28.19 = 3.881 + j42.33 = 42.51∠84.76° b(t) = 42.51 cos(120πt + 84.76°)

(c)

Let C = 4∠-90° + 3∠(-10° – 90°) = -j4 – 0.5209 – j2.954 = 6.974∠265.72° c(t) = 6.974 cos(8t + 265.72°)

Chapter 9, Solution 18.

(a) (b)

(c) (d)

v1 ( t ) = 60 cos(t + 15°)

V2 = 6 + j8 = 10∠53.13° v 2 ( t ) = 10 cos(40t + 53.13°) i1 ( t ) = 2.8 cos(377t – π/3) I 2 = -0.5 – j1.2 = 1.3∠247.4° i 2 ( t ) = 1.3 cos(103t + 247.4°)

Chapter 9, Solution 19.

(a)

(b)

(c)

3∠10° − 5∠-30° = 2.954 + j0.5209 – 4.33 + j2.5 = -1.376 + j3.021 = 3.32∠114.49° Therefore, 3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t + 114.49°) 4∠-90° + 3∠-45° = -j40 + 21.21 – j21.21 = 21.21 – j61.21 = 64.78∠-70.89° Therefore, 40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89°) Using sinα = cos(α − 90°), 20∠-90° + 10∠60° − 5∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4.699 = 6.7101 – j6.641 = 9.44∠-44.7° Therefore, 20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°) = 9.44 cos(400t – 44.7°)

Chapter 9, Solution 20. (a) V = 4∠− 60 o − 90 o − 5∠40 o = −3.464 − j 2 − 3.83 − j 3.2139 = 8.966∠ − 4.399 o v = 8.966 cos(377t − 4.399 o )

Hence,

(b) I = 10∠0 o + jω 8∠20 o − 90 o ,

ω = 5 , i.e. I = 10 + 40∠20 o = 49.51∠16.04 o

i = 49.51 cos(5t + 16.04 o )

Chapter 9, Solution 21.

(a) F = 5∠15 o − 4∠− 30 o − 90 o = 6.8296 + j 4.758 = 8.3236∠34.86 o f (t ) = 8.324 cos(30t + 34.86 o )

(b) G = 8∠ − 90 o + 4∠50 o = 2.571 − j 4.9358 = 5.565∠ − 62.49 o g (t ) = 5.565 cos(t − 62.49 o )

1 (10∠0 o + 5∠ − 90 o ), ω = 40 jω i.e. H = 0.25∠ − 90 o + 0.125∠ − 180 o = − j 0.25 − 0.125 = 0.2795∠ − 116.6 o (c) H =

h(t ) = 0.2795 cos(40t − 116.6 o )

Chapter 9, Solution 22.

dv Let f(t) = 10v(t ) + 4 − 2 ∫ v(t )dt dt −∞ 2V F = 10V + jω 4V − , ω = 5, V = 20∠ − 30 o jω t

F = 10V + j 20V − j 0.4V = (10 − j19.6)(17.32 − j10) = 440.1∠ − 92.97 o f (t ) = 440.1 cos(5t − 92.97 o )

Chapter 9, Solution 23.

(a) (b)

(c)

(d)

v(t) = 40 cos(ωt – 60°) V = -30∠10° + 50∠60° = -4.54 + j38.09 = 38.36∠96.8° v(t) = 38.36 cos(ωt + 96.8°)

I = j6∠-10° = 6∠(90° − 10°) = 6∠80° i(t) = 6 cos(ωt + 80°)

2 + 10∠-45° = -j2 + 7.071 – j7.071 j = 11.5∠-52.06° i(t) = 11.5 cos(ωt – 52.06°) I =

Chapter 9, Solution 24.

(a)

(b)

V+

V = 10∠0°, ω = 1 jω V (1 − j) = 10 10 = 5 + j5 = 7.071∠45° V= 1− j Therefore, v(t) = 7.071 cos(t + 45°)

jωV + 5V +

4V = 20∠(10° − 90°), ω = 4 jω  4 V  j4 + 5 +  = 20 ∠ - 80° j4   20∠ - 80° = 3.43∠ - 110.96° V= 5 + j3 Therefore, v(t) = 3.43 cos(4t – 110.96°)

Chapter 9, Solution 25.

(a)

(b)

2jωI + 3I = 4∠ - 45°, ω = 2 I (3 + j4) = 4∠ - 45° 4∠ - 45° 4∠ - 45° = = 0.8∠ - 98.13° I= 3 + j4 5∠53.13° Therefore, i(t) = 0.8 cos(2t – 98.13°) I + jωI + 6I = 5∠22°, ω = 5 jω (- j2 + j5 + 6) I = 5∠22° 5∠22° 5∠22° = = 0.745∠ - 4.56° I= 6 + j3 6.708∠26.56° Therefore, i(t) = 0.745 cos(5t – 4.56°) 10

Chapter 9, Solution 26. jωI + 2I +

I = 1∠0°, ω = 2 jω  1 I  j2 + 2 +  = 1 j2   1 = 0.4∠ - 36.87° I= 2 + j1.5 Therefore, i(t) = 0.4 cos(2t – 36.87°)

Chapter 9, Solution 27. jωV + 50V + 100

V = 110∠ - 10°, ω = 377 jω  j100   = 110∠ - 10° V  j377 + 50 −  377  V (380.6∠82.45°) = 110∠ - 10° V = 0.289 ∠ - 92.45° Therefore, v(t) = 0.289 cos(377t – 92.45°).

Chapter 9, Solution 28. i( t ) =

v s ( t ) 110 cos(377 t ) = = 13.75 cos(377t) A. R 8

Chapter 9, Solution 29.

Z=

1 1 = = - j 0.5 6 jωC j (10 )(2 × 10 -6 )

V = IZ = (4∠25°)(0.5∠ - 90°) = 2 ∠ - 65° Therefore

v(t) = 2 sin(106t – 65°) V.

Chapter 9, Solution 30. Z = jωL = j (500)(4 × 10 -3 ) = j2 V 60 ∠ - 65° = 30∠ - 155° I= = Z 2∠90° Therefore, i(t) = 30 cos(500t – 155°) A.

Chapter 9, Solution 31.

Thus,

i(t) = 10 sin(ωt + 30°) = 10 cos(ωt + 30° − 90°) = 10 cos(ωt − 60°) I = 10∠-60°

v(t) = -65 cos(ωt + 120°) = 65 cos(ωt + 120° − 180°) = 65 cos(ωt − 60°) Thus, V = 65∠-60°

Z=

V 65∠ - 60° = = 6.5 Ω I 10∠ - 60°

Since V and I are in phase, the element is a resistor with R = 6.5 Ω.

Chapter 9, Solution 32.

V = 180∠10°,

Z=

I = 12∠-30°,

V 180∠10° = 15∠40° = 11.49 + j 9.642 Ω = I 12∠ - 30°

ω = 2

One element is a resistor with R = 11.49 Ω. The other element is an inductor with ωL = 9.642 or

L = 4.821 H.

Chapter 9, Solution 33. 110 = v 2R + v 2L

v L = 110 2 − v 2R

v L = 110 2 − 85 2 = 69.82 V

Chapter 9, Solution 34. v o = 0 if ωL = ω=

1 ωC

1

 → ω =

(5 × 10 −3 )(2 × 10 −3 )

1 LC

= 100 rad/s

Chapter 9, Solution 35. Vs = 5∠0° jωL = j (2)(1) = j2 1 1 = = - j2 jωC j (2)(0.25) Vo =

j2 j2 Vs = 5∠0° = (1∠90°)(5∠0°) = 5∠90° 2 − j2 + j2 2 Thus, v o ( t ) = 5 cos(2t + 90°) = -5 sin(2t) V

Chapter 9, Solution 36.

Let Z be the input impedance at the source.

10 µF

jωL = j 200 x100 x10 −3 = j 20

 →

100 mH

 →

1 1 = = − j 500 jωC j10 x10 −6 x 200

1000//-j500 = 200 –j400 1000//(j20 + 200 –j400) = 242.62 –j239.84 Z = 2242.62 − j 239.84 = 2255∠ − 6.104 o I=

60∠ − 10 o = 26.61∠ − 3.896 o mA o 2255∠ − 6.104

i = 266.1 cos(200t − 3.896 o )

Chapter 9, Solution 37. jωL = j (5)(1) = j5 1 1 = = -j jωC j (5)(0.2) Let Z1 = - j ,

Z 2 = 2 || j5 =

Then,

Ix =

(2)( j5) j10 = 2 + j5 2 + j5

Z2 I , Z1 + Z 2 s

where I s = 2∠0°

j10 j20 2 + j5 Ix = (2) = = 2.12 ∠32° j10 5 + j8 - j+ 2 + j5 Therefore,

i x ( t ) = 2.12 sin(5t + 32°) A

Chapter 9, Solution 38. 1 F  → 6

(a)

1 1 = = - j2 jωC j (3)(1 / 6)

I=

- j2 (10 ∠45°) = 4.472∠ - 18.43° 4 − j2 Hence, i(t) = 4.472 cos(3t – 18.43°) A

V = 4I = (4)(4.472∠ - 18.43°) = 17.89∠ - 18.43° Hence, v(t) = 17.89 cos(3t – 18.43°) V 1 F  → 12

(b)

3H  →

1 1 = = - j3 jωC j (4)(1 / 12)

jωL = j (4)(3) = j12

V 50∠0° = 10∠36.87° = Z 4 − j3 Hence, i(t) = 10 cos(4t + 36.87°) A I=

V=

j12 (50∠0°) = 41.6 ∠33.69° 8 + j12 Hence, v(t) = 41.6 cos(4t + 33.69°) V Chapter 9, Solution 39. Z = 8 + j5 || (- j10) = 8 + I=

( j5)(- j10) = 8 + j10 j5 − j10

V 40 ∠0° 20 = = = 3.124∠ - 51.34° Z 8 + j10 6.403∠51.34°

I1 =

- j10 I = 2 I = 6.248∠ - 51.34° j5 − j10

I2 =

j5 I = - I = 3.124∠128.66° - j5

Therefore,

i1 ( t ) = 6.248 cos(120πt – 51.34°) A i 2 ( t ) = 3.124 cos(120πt + 128.66°) A

Chapter 9, Solution 40.

(a)

For ω = 1 , 1H  →

jωL = j (1)(1) = j 1 1 0.05 F  → = = - j20 jωC j (1)(0.05) - j40 Z = j + 2 || (- j20) = j + = 1.98 + j0.802 2 − j20

V 4 ∠0° 4∠0° = = 1.872 ∠ - 22.05° = Z 1.98 + j0.802 2.136∠22.05° Hence, i o ( t ) = 1.872 cos(t – 22.05°) A Io =

(b)

For ω = 5 , 1H  →

jωL = j (5)(1) = j5 1 1 0.05 F  → = = - j4 jωC j (5)(0.05) - j4 Z = j5 + 2 || (- j4) = j5 + = 1.6 + j4.2 1 − j2

V 4∠0° 4∠0° = = 0.89∠ - 69.14° = Z 1.6 + j4 4.494∠69.14° Hence, i o ( t ) = 0.89 cos(5t – 69.14°) A Io =

(c)

For ω = 10 , 1H  → jωL = j (10)(1) = j10 1 1 0.05 F  → = = - j2 jωC j (10)(0.05) - j4 Z = j10 + 2 || (- j2) = j10 + = 1 + j9 2 − j2

V 4∠0° 4 ∠0° = = 0.4417 ∠ - 83.66° = Z 1 + j9 9.055∠83.66° Hence, i o ( t ) = 0.4417 cos(10t – 83.66°) A Io =

Chapter 9, Solution 41.

ω = 1, 1H  →

jωL = j (1)(1) = j

1F  →

1 1 = = -j jωC j (1)(1)

Z = 1 + (1 + j) || (- j) = 1 + I=

Vs 10 = , Z 2− j

- j+1 = 2− j 1

I c = (1 + j) I

V = (- j)(1 + j) I = (1 − j) I =

Thus,

(1 − j)(10) = 6.325∠ - 18.43° 2− j

v(t) = 6.325 cos(t – 18.43°) V

Chapter 9, Solution 42. ω = 200

50 µF  →

1 1 = = - j100 jωC j (200)(50 × 10 -6 )

0.1 H  →

jωL = j (200)(0.1) = j20

50 || -j100 = Vo =

(50)(-j100) - j100 = = 40 − j20 50 − j100 1 - j2

j20 j20 (60∠0°) = 17.14 ∠90° (60∠0°) = 70 j20 + 30 + 40 − j20

Thus, v o ( t ) = 17.14 sin(200t + 90°) V or

v o ( t ) = 17.14 cos(200t) V

Chapter 9, Solution 43.

ω= 2 1H  → jωL = j (2)(1) = j2 1F  → Io =

1 1 = = - j0.5 jωC j (2)(1)

j2 − j0.5 j1.5 I= 4∠0° = 3.328∠33.69° j2 − j0.5 + 1 1 + j1.5

Thus, i o ( t ) = 3.328 cos(2t + 33.69°) A

Chapter 9, Solution 44. ω = 200 10 mH  → jωL = j (200)(10 × 10 -3 ) = j2

5 mF  →

1 1 = = -j jωC j (200)(5 × 10 -3 )

Y=

3+ j 1 1 1 + + = 0.25 − j0.5 + = 0.55 − j0.4 10 4 j2 3 − j

Z=

1 1 = = 1.1892 + j0.865 Y 0.55 − j0.4

I=

6∠0° 6∠0° = = 0.96 ∠ - 7.956° 5 + Z 6.1892 + j0.865

Thus, i(t) = 0.96 cos(200t – 7.956°) A

Chapter 9, Solution 45.

We obtain I o by applying the principle of current division twice. I

I2

Z1

I2

Io -j2 Ω

Z2

(a) Z 1 = - j2 ,

(b)

Z 2 = j4 + (-j2) || 2 = j4 +

I2 =

Z1 - j10 - j2 I= (5∠0°) = Z1 + Z 2 1+ j - j2 + 1 + j3

Io =

 - j  - j10  - 10 - j2  = = -5 A I2 =  2 - j2 1 - j  1 + j  1 + 1

- j4 = 1 + j3 2 - j2

Chapter 9, Solution 46. i s = 5 cos(10 t + 40°)  → I s = 5∠40°

Let

0.1 F  →

1 1 = = -j jωC j (10)(0.1)

0.2 H  →

jωL = j (10)(0.2) = j2

Z1 = 4 || j2 =

2Ω

j8 = 0.8 + j1.6 , 4 + j2

Z2 = 3 − j

Io =

Z1 0.8 + j1.6 (5∠40°) Is = 3.8 + j0.6 Z1 + Z 2

Io =

(1.789∠63.43°)(5∠40°) = 2.325∠94.46° 3.847 ∠8.97°

Thus, i o ( t ) = 2.325 cos(10t + 94.46°) A

Chapter 9, Solution 47.

First, we convert the circuit into the frequency domain. Ix 5∠0˚ Ix =

2Ω

j4

+ −

20 Ω

-j10

5 5 5 = = = 0.4607∠52.63° − j10(20 + j4) 2 + 4.588 − j8.626 10.854∠ − 52.63° 2+ − j10 + 20 + j4 is(t) = 0.4607cos(2000t +52.63˚) A

Chapter 9, Solution 48.

Converting the circuit to the frequency domain, we get:

10 Ω

V1 30 Ω Ix

20∠-40˚

+ −

We can solve this using nodal analysis.

j20

-j20

V1 − 20∠ − 40° V1 − 0 V −0 =0 + + 1 10 j20 30 − j20 V1(0.1 − j0.05 + 0.02307 + j0.01538) = 2∠ − 40° 2∠40° = 15.643∠ − 24.29° 0.12307 − j0.03462 15.643∠ − 24.29° = = 0.4338∠9.4° 30 − j20 = 0.4338 sin(100 t + 9.4°) A

V1 = Ix ix

Chapter 9, Solution 49. Z T = 2 + j2 || (1 − j) = 2 + I

( j2)(1 − j) =4 1+ j

Ix

1Ω

j2 Ω

-j Ω

where I x = 0.5∠0° =

Ix =

j2 j2 I= I, j2 + 1 − j 1+ j 1+ j 1+ j I= Ix = j2 j4

1 2

1+ j 1+ j (4) = = 1 − j = 1.414∠ - 45° j4 j v s ( t ) = 1.414 sin(200t – 45°) V

Vs = I Z T =

Chapter 9, Solution 50. Since ω = 100, the inductor = j100x0.1 = j10 Ω and the capacitor = 1/(j100x10-3) = -j10Ω.

j10 5∠40˚

-j10

Ix 20 Ω

+ vx

−

Using the current dividing rule: − j10 5∠40° = − j2.5∠40° = 2.5∠ − 50° − j10 + 20 + j10 Vx = 20I x = 50∠ − 50°

Ix =

v x = 50 cos(100t − 50°) V

Chapter 9, Solution 51. 0.1 F  →

0.5 H  →

1 1 = = - j5 jωC j (2)(0.1) jωL = j (2)(0.5) = j

The current I through the 2-Ω resistor is Is 1 I= Is = , 1 − j5 + j + 2 3 − j4 I s = (10)(3 − j4) = 50∠ - 53.13°

where I = 10 ∠0°

Therefore, i s ( t ) = 50 cos(2t – 53.13°) A

Chapter 9, Solution 52. 5 || j5 =

j25 j5 = = 2.5 + j2.5 5 + j5 1 + j

Z1 = 10 ,

Z 2 = - j5 + 2.5 + j2.5 = 2.5 − j2.5

I2 IS

Z1

Z2

I2 =

Z1 10 4 Is = Is = I 12.5 − j2.5 5− j s Z1 + Z 2

Vo = I 2 (2.5 + j2.5)  4 8∠30° =  5 −

Is =

 10 (1 + j)  I s (2.5)(1 + j) = I 5− j s j

(8∠30°)(5 − j) = 2.884∠-26.31° A 10 (1 + j)

Chapter 9, Solution 53. Convert the delta to wye subnetwork as shown below. Z1

Io

Z2

2Ω

Z3 +

60∠ − 30 V

8Ω

o

10 Ω

-

Z Z1 =

− j 2 x4 = 0.1532 − j 0.7692, 10 − j 2

Z3 =

12 = 1.1538 + j 0.2308 10 − j 2

Z2 =

j6 x4 = −0.4615 + j 2.3077, 10 − j 2

( Z 3 + 8) //( Z 2 + 10) = (9.1538 + j 0.2308) //(9.5385 + j 2.3077) = 4.726 + j 0.6062 Z = 2 + Z 1 + 4.726 + j 0.6062 = 6.878 − j 0.163

Io =

60∠ − 30 o 60∠ − 30 o = = 8.721∠ − 28.64 o A o Z 6.88∠ − 1.3575

Chapter 9, Solution 54.

Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below. + −

Vs

+ 2Z

− Z

V1

V2

−

+

V1 = I o (1 − j) = 2 (1 − j) V2 = 2V1 = 4 (1 − j) Vs = V1 + V2 = 6 (1 − j) Vs = 8.485∠-45° V

Chapter 9, Solution 55.

12 Ω

-j20 V

I

I1

Z

I2

-j4 Ω

+ −

I1 =

Vo 4 = = -j0.5 j 8 j8

I2 =

I 1 (Z + j8) (-j0.5)(Z + j8) Z = = +j - j4 - j4 8

I = I 1 + I 2 = -j0.5 +

Z Z + j = + j0.5 8 8

- j20 = 12 I + I 1 (Z + j8)

Z j - j - j20 = 12  +  + (Z + j8)  8 2 2

+ Vo

−

j8 Ω

3 1 - 4 - j26 = Z  − j  2 2

Z=

- 4 - j26 26.31∠261.25° = = 16.64∠279.68° 3 1 1.5811∠ - 18.43° −j 2 2

Z = 2.798 – j16.403 Ω Chapter 9, Solution 56.

3H

 →

jωL = j 30

3F

 →

1 = − j / 30 jω C

1.5F

 →

1 = − j / 15 jω C

−j 15 = − j 0.06681 j 30 //( − j / 15) = j j 30 − 15 j 30 x

Z=

−j − j 0.033(2 − j 0.06681) = 6 − j 333 mΩ //(2 − j 0.06681) = − j 0.033 + 2 − j 0.06681 30

Chapter 9, Solution 57.

2H 1F

 →  →

jωL = j 2

1 =−j jω C

Z = 1 + j2 //( 2 − j) = 1 +

j2(2 − j) = 2.6 + j1.2 j2 + 2 − j

Y = 1 = 0.3171 − j0.1463 S Z

Chapter 9, Solution 58. (a)

10 mF  →

10 mH  →

1 1 = = - j2 jωC j (50)(10 × 10 -3 )

jωL = j (50)(10 × 10 -3 ) = j0.5

Z in = j0.5 + 1 || (1 − j2) 1 − j2 Z in = j0.5 + 2 − j2 Z in = j0.5 + 0.25 (3 − j) Z in = 0.75 + j0.25 Ω

(b)

0.4 H  →

0.2 H  →

1 mF  →

jωL = j (50)(0.4) = j20

jωL = j (50)(0.2) = j10 1 1 = = - j20 jωC j (50)(1 × 10 -3 )

For the parallel elements, 1 1 1 1 = + + Z p 20 j10 - j20

Z p = 10 + j10

Then,

Z in = 10 + j20 + Z p = 20 + j30 Ω

Chapter 9, Solution 59. Z eq = 6 + (1 − j2) || (2 + j4) Z eq = 6 +

(1 − j2)(2 + j4) (1 − j2) + (2 + j4)

Z eq = 6 + 2.308 − j1.5385 Z eq = 8.308 – j1.5385 Ω

Chapter 9, Solution 60. Z = (25 + j15) + (20 − j 50) //(30 + j10) = 25 + j15 + 26.097 − j 5.122 = 51.1 + j 9.878Ω

Chapter 9, Solution 61.

All of the impedances are in parallel. 1 1 1 1 1 = + + + Z eq 1 − j 1 + j2 j5 1 + j3 1 = (0.5 + j0.5) + (0.2 − j0.4) + (- j0.2) + (0.1 − j0.3) = 0.8 − j0.4 Z eq

Z eq =

1 = 1 + j0.5 Ω 0.8 − j0.4

Chapter 9, Solution 62.

2 mH  →

jωL = j (10 × 10 3 )(2 × 10 -3 ) = j20 1 1 1 µF  → = = - j100 3 jωC j (10 × 10 )(1 × 10 -6 ) 50 Ω

1∠0° A

+

+

V

−

j20 Ω

Vin

−

V = (1∠0°)(50) = 50

+

-j100 Ω

Vin = (1∠0°)(50 + j20 − j100) + (2)(50) Vin = 50 − j80 + 100 = 150 − j80

Z in =

Vin = 150 – j80 Ω 1∠0°

2V

Chapter 9, Solution 63.

First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with the corresponding delta. 200 + j150 + j300 = 20 + j45 10 200 + j450 200 + j450 z2 = = 30 − j13.333, z3 = = 10 + j22.5 j15 20 z1 =

8Ω

–j12 Ω

–j16 Ω 10 Ω

z2 z1

ZT

z3

–j16 Ω

10 Ω

Now all we need to do is to combine impedances. z 2 (10 − j16) =

(30 − j13.333)(10 − j16) = 8.721 − j8.938 40 − j29.33

z3 (10 − j16) = 21.70 − j3.821

ZT = 8 − j12 + z1 (8.721 − j8.938 + 21.7 − j3.821) = 34.69 − j6.93Ω

Chapter 9, Solution 64. − j10(6 + j8) = 19 − j5Ω 6 − j2 30∠90° = −0.3866 + j1.4767 = 1.527∠104.7° A I= ZT ZT = 4 +

Chapter 9, Solution 65. Z T = 2 + (4 − j6) || (3 + j4)

ZT = 2 +

(4 − j6)(3 + j4) 7 − j2

Z T = 6.83 + j1.094 Ω = 6.917∠9.1° Ω

I=

V 120 ∠10° = = 17.35∠0.9° A Z T 6.917 ∠9.1°

Chapter 9, Solution 66. Z T = (20 − j5) || (40 + j10) =

(20 − j5)(40 + j10) 170 (12 − j) = 60 + j5 145

Z T = 14.069 – j1.172 Ω = 14.118∠-4.76°

I=

V 60∠90° = 4.25∠94.76° = Z T 14.118∠ - 4.76° I I1

I2

20 Ω

j10 Ω

+ I1 = I2 =

40 + j10 8 + j2 I= I 60 + j5 12 + j 20 − j5 4− j I= I 60 + j5 12 + j

Vab = -20 I 1 + j10 I 2

Vab

−

Vab = Vab =

- (160 + j40) 10 + j40 I+ I 12 + j 12 + j

- 150 (-12 + j)(150) I= I 12 + j 145

Vab = (12.457 ∠175.24°)(4.25∠97.76°) Vab = 52.94∠273° V

Chapter 9, Solution 67.

(a)

20 mH  →

jωL = j (10 3 )(20 × 10 -3 ) = j20 1 1 12.5 µF  → = = - j80 3 jωC j (10 )(12.5 × 10 -6 ) Z in = 60 + j20 || (60 − j80) ( j20)(60 − j80) Z in = 60 + 60 − j60 Z in = 63.33 + j23.33 = 67.494 ∠20.22°

Yin =

(b)

1 = 0.0148∠-20.22° S Z in

10 mH  → 20 µF  →

30 || 60 = 20

jωL = j (10 3 )(10 × 10 -3 ) = j10 1 1 = = - j50 jωC j (10 3 )(20 × 10 -6 )

Z in = - j50 + 20 || (40 + j10) (20)(40 + j10) Z in = - j50 + 60 + j10 Z in = 13.5 − j48.92 = 50.75∠ - 74.56°

Yin =

1 = 0.0197∠74.56° S = 5.24 + j18.99 mS Z in

Chapter 9, Solution 68. Yeq =

1 1 1 + + 5 − j2 3 + j - j4

Yeq = (0.1724 + j0.069) + (0.3 − j0.1) + ( j0.25)

Yeq = 0.4724 + j0.219 S

Chapter 9, Solution 69. 1 1 1 1 = + = (1 + j2) Yo 4 - j2 4

Yo =

4 (4)(1 − j2) = = 0.8 − j1.6 1 + j2 5

Yo + j = 0.8 − j0.6 1 1 1 = + + = (1) + ( j0.333) + (0.8 + j0.6) Yo ′ 1 - j3 0.8 − j0.6 1

Yo ′ 1

= 1.8 + j0.933 = 2.028∠27.41°

Yo ′ = 0.4932∠ - 27.41° = 0.4378 − j0.2271 Yo ′ + j5 = 0.4378 + j4.773 1 1 1 0.4378 − j4.773 = + = 0.5 + 22.97 Yeq 2 0.4378 + j4.773 1 = 0.5191 − j0.2078 Yeq

Yeq =

0.5191 − j0.2078 = 1.661 + j0.6647 S 0.3126

Chapter 9, Solution 70.

Make a delta-to-wye transformation as shown in the figure below. a Zan

Zbn

Zeq

n

Zcn

b 2Ω

Z an =

(- j10)(10 + j15) (10)(15 − j10) = 7 − j9 = 5 − j10 + 10 + j15 15 + j5

Z bn =

(5)(10 + j15) = 4.5 + j3.5 15 + j5

Z cn =

(5)(- j10) = -1 − j3 15 + j5

Z eq = Z an + (Z bn + 2) || (Z cn + 8 − j5) Z eq = 7 − j9 + (6.5 + j3.5) || (7 − j8) Z eq = 7 − j9 +

(6.5 + j3.5)(7 − j8) 13.5 − j4.5

Z eq = 7 − j9 + 5.511 − j0.2 Z eq = 12.51 − j9.2 = 15.53∠-36.33° Ω

c 8Ω -j5 Ω

Chapter 9, Solution 71.

We apply a wye-to-delta transformation.

j4 Ω Zab

b

a Zac

1Ω

Zbc

-j2 Ω

c Z ab =

2 − j2 + j4 2 + j2 = = 1− j j2 j2

Z ac =

2 + j2 = 1+ j 2

Z bc =

2 + j2 = -2 + j2 -j

j4 || Z ab = j4 || (1 − j) = 1 || Z ac = 1 || (1 + j) =

( j4)(1 − j) = 1.6 − j0.8 1 + j3

(1)(1 + j) = 0.6 + j0.2 2+ j

j4 || Z ab + 1 || Z ac = 2.2 − j0.6 1 1 1 1 = + + Z eq - j2 - 2 + j2 2.2 − j0.6

= j0.5 − 0.25 − j0.25 + 0.4231 + j0.1154

= 0.173 + j0.3654 = 0.4043∠64.66°

Z eq = 2.473∠-64.66° Ω = 1.058 – j2.235 Ω

Zeq

Chapter 9, Solution 72.

Transform the delta connections to wye connections as shown below. a j2 Ω

j2 Ω

-j18 Ω

-j9 Ω

j2 Ω

R1

R2

R3 b

- j9 || - j18 = - j6 , R1 =

(20)(20) = 8 Ω, 20 + 20 + 10

R2 =

Z ab = j2 + ( j2 + 8) || (j2 − j6 + 4) + 4 Z ab = 4 + j2 + (8 + j2) || (4 − j4) Z ab = 4 + j2 +

(8 + j2)(4 − j4) 12 - j2

Z ab = 4 + j2 + 3.567 − j1.4054 Z ab = 7.567 + j0.5946 Ω

(20)(10) = 4Ω, 50

R3 =

(20)(10) = 4Ω 50

Chapter 9, Solution 73.

Transform the delta connection to a wye connection as in Fig. (a) and then transform the wye connection to a delta connection as in Fig. (b). a j2 Ω

j2 Ω

-j18 Ω

-j9 Ω

j2 Ω

R1

R2

R3 b Z1 =

( j8)(- j6) 48 = = - j4.8 j8 + j8 − j6 j10 Z 2 = Z1 = -j4.8 ( j8)( j8) - 64 Z3 = = = j6.4 j10 j10 (2 + Z1 )(4 + Z 2 ) + (4 + Z 2 )(Z 3 ) + (2 + Z1 )(Z 3 ) = (2 − j4.8)(4 − j4.8) + (4 − j4.8)( j6.4) + (2 − j4.8)( j6.4) = 46.4 + j9.6 46.4 + j9.6 = 1.5 − j7.25 j6.4 46.4 + j9.6 = 3.574 + j6.688 Zb = 4 − j4.8 46.4 + j9.6 = 1.727 + j8.945 Zc = 2 − j4.8 Za =

(6∠90°)(7.583∠61.88°) = 07407 + j3.3716 3.574 + j12.688 (-j4)(1.5 − j7.25) - j4 || Z a = = 0.186 − j2.602 1.5 − j11.25 j6 || Z b =

j12 || Z c =

(12∠90°)(9.11∠79.07°) = 0.5634 + j5.1693 1.727 + j20.945

Z eq = ( j6 || Z b ) || (- j4 || Z a + j12 || Z c )

Z eq = (0.7407 + j3.3716) || (0.7494 + j2.5673) Z eq = 1.508∠75.42° Ω = 0.3796 + j1.46 Ω

Chapter 9, Solution 74.

One such RL circuit is shown below. 20 Ω

+

V

j20 Ω

Vi = 1∠0°

20 Ω

j20 Ω

+ Vo

−

Z

We now want to show that this circuit will produce a 90° phase shift. Z = j20 || (20 + j20) = V=

( j20)(20 + j20) - 20 + j20 = = 4 (1 + j3) 20 + j40 1 + j2

Z 1 + j3 1 4 + j12 Vi = (1∠0°) = = (1 + j) Z + 20 6 + j3 3 24 + j12

Vo =

 j  1 j20  (1 + V = 20 + j20 1 + j  3

 j j)  = = 0.3333∠90°  3

This shows that the output leads the input by 90°.

Chapter 9, Solution 75. Since cos(ωt ) = sin(ωt + 90°) , we need a phase shift circuit that will cause the output to lead the input by 90°. This is achieved by the RL circuit shown below, as explained in the previous problem.

10 Ω +

10 Ω

j10 Ω

Vi

−

j10 Ω

+ Vo

−

This can also be obtained by an RC circuit.

Chapter 9, Solution 76. Let Z = R – jX, where X = | Z |= R 2 + X 2

 →

C=

1 1 = ωC 2πfC

X = | Z |2 − R 2 = 1162 = 662 = 95.394

1 1 = = 27.81µF 2πfX 2πx 60x95.394

Chapter 9, Solution 77.

(a)

Vo =

- jX c V R − jX c i 1 1 where X c = = = 3.979 ωC (2π)(2 × 10 6 )(20 × 10 -9 )

Vo - j3.979 = = Vi 5 - j3.979

Vo = Vi

3.979

25 + 15.83

3.979

5 + 3.979 2

2

∠(-90° + tan -1 (3.979 5))

∠(-90° − 38.51°)

Vo = 0.6227 ∠ - 51.49° Vi

Therefore, the phase shift is 51.49° lagging

(b)

θ = -45° = -90° + tan -1 (X c R )

45° = tan -1 (X c R )  → R = X c =

ω = 2πf = f=

1 ωC

1 RC

1 1 = = 1.5915 MHz 2πRC (2π)(5)(20 × 10 -9 )

Chapter 9, Solution 78.

8+j6 R Z -jX

Z = R //[8 + j (6 − X )] =

R[8 + j (6 − X )] =5 R + 8 + j (6 − X )

i.e 8R + j6R – jXR = 5R + 40 + j30 –j5X Equating real and imaginary parts: 8R = 5R + 40 which leads to R=13.33Ω 6R-XR =30-5 which leads to X=4.125Ω.

Chapter 9, Solution 79.

(a)

Consider the circuit as shown. 20 Ω + Vi

−

j10 Ω

Z2

V2

40 Ω j30 Ω

Z1

V1

30 Ω j60 Ω

+ Vo

−

( j30)(30 + j60) = 3 + j21 30 + j90 ( j10)(43 + j21) Z 2 = j10 || (40 + Z1 ) = = 1.535 + j8.896 = 9.028∠80.21° 43 + j31

Z1 = j30 || (30 + j60) =

Let Vi = 1∠0° . Z2 (9.028∠80.21°)(1∠0°) Vi = 21.535 + j8.896 Z 2 + 20 V2 = 0.3875∠57.77°

V2 =

Z1 3 + j21 (21.213∠81.87°)(0.3875∠57.77°) V2 = V2 = Z1 + 40 43 + j21 47.85∠26.03° V1 = 0.1718∠113.61°

V1 =

Vo =

j60 j2 2 V1 = V1 = (2 + j)V1 30 + j60 1 + j2 5 Vo = (0.8944∠26.56°)(0.1718∠113.6°) Vo = 0.1536∠140.2° Therefore, the phase shift is 140.2° (b) (c)

The phase shift is leading. If Vi = 120 V , then Vo = (120)(0.1536∠140.2°) = 18.43∠140.2° V and the magnitude is 18.43 V.

Chapter 9, Solution 80.

200 mH  → Vo =

(a)

jωL = j (2π)(60)(200 × 10 -3 ) = j75.4 Ω

j75.4 j75.4 Vi = (120∠0°) R + 50 + j75.4 R + 50 + j75.4

When R = 100 Ω , j75.4 (75.4∠90°)(120∠0°) (120 ∠0°) = Vo = 150 + j75.4 167.88∠26.69° Vo = 53.89∠63.31° V

(b)

(c)

When R = 0 Ω , j75.4 (75.4∠90°)(120 ∠0°) (120∠0°) = Vo = 50 + j75.4 90.47 ∠56.45° Vo = 100∠33.55° V To produce a phase shift of 45°, the phase of Vo = 90° + 0° − α = 45°. Hence, α = phase of (R + 50 + j75.4) = 45°. For α to be 45°, R + 50 = 75.4 Therefore, R = 25.4 Ω

Chapter 9, Solution 81.

Let Zx = Rx + Rx =

Z1 = R 1 ,

Z2 = R 2 +

1 , jωC 2

Z 3 = R 3 , and Z x = R x +

Z3 Z Z1 2

R3  1 1   R 2 + = jωC x R 1  jωC 2  R3 1200 R2 = (600) = 1.8 kΩ R1 400

R1  400  1 R3   1  (0.3 × 10 -6 ) = 0.1 µF → C x = C2 =  =     1200  Cx  R1   C2  R3

Chapter 9, Solution 82. Cx =

R1  100  (40 × 10 -6 ) = 2 µF Cs =   2000  R2

Chapter 9, Solution 83. Lx =

R2  500  (250 × 10 -3 ) = 104.17 mH Ls =  1200  R1

1 . jωC x

Chapter 9, Solution 84.

Let

Z1 = R 1 ||

1 Z2 = R 2 , , jωC s R1 jωC s R1 Z1 = = 1 jωR 1C s + 1 R1 + jωC s

Since Z x =

Z3 Z , Z1 2

R x + jωL x = R 2 R 3

Z 3 = R 3 , and Z x = R x + jωL x .

jωR 1C s + 1 R 2 R 3 = (1 + jωR 1C s ) R1 R1

Equating the real and imaginary components, R 2R 3 Rx = R1 ωL x =

R 2R 3 (ωR 1C s ) implies that R1 L x = R 2 R 3Cs

Given that R 1 = 40 kΩ , R 2 = 1.6 kΩ , R 3 = 4 kΩ , and C s = 0.45 µF R 2 R 3 (1.6)(4) = kΩ = 0.16 kΩ = 160 Ω R1 40 L x = R 2 R 3 C s = (1.6)(4)(0.45) = 2.88 H

Rx =

Chapter 9, Solution 85.

Let

Z1 = R 1 ,

Z2 = R 2 +

1 , jωC 2 R4 - jR 4 Z4 = = jωR 4 C 4 + 1 ωR 4 C 4 − j

Since Z 4 =

Z3 Z Z1 2

 → Z1 Z 4 = Z 2 Z 3 ,

Z 3 = R 3 , and Z 4 = R 4 ||

1 . jωC 4

 - jR 4 R 1 j   = R 3 R 2 − ωR 4 C 4 − j ωC 2  

jR 3 - jR 4 R 1 (ωR 4 C 4 + j) = R 3R 2 − 2 2 2 ω R 4C4 + 1 ωC 2

Equating the real and imaginary components, R 1R 4 = R 2R 3 2 ω R 24 C 24 + 1 (1) 2 R3 ωR 1 R 4 C 4 = 2 2 2 ω R 4 C 4 + 1 ωC 2 (2) Dividing (1) by (2), 1 = ωR 2 C 2 ωR 4 C 4 1 ω2 = R 2C2R 4C4 1 ω = 2πf = R 2C2 R 4C4 f=

1

2π R 2 R 4 C 2 C 4

Chapter 9, Solution 86. Y=

1 1 1 + + 240 j95 - j84

Y = 4.1667 × 10 -3 − j0.01053 + j0.0119 Z=

1 1000 1000 = = Y 4.1667 + j1.37 4.3861∠18.2°

Z = 228∠-18.2° Ω

Chapter 9, Solution 87. Z1 = 50 +

-j 1 = 50 + (2π)(2 × 10 3 )(2 × 10 -6 ) jωC

Z1 = 50 − j39.79

Z 2 = 80 + jωL = 80 + j (2π)(2 × 10 3 )(10 × 10 -3 ) Z 2 = 80 + j125.66

Z 3 = 100

1 1 1 1 = + + Z Z1 Z 2 Z 3 1 1 1 1 = + + Z 100 50 − j39.79 80 + j125.66 1 = 10 -3 (10 + 12.24 + j9.745 + 3.605 − j5.663) Z = (25.85 + j4.082) × 10 -3 = 26.17 × 10 -3 ∠8.97° Z = 38.21∠-8.97° Ω Chapter 9, Solution 88.

(a)

(b)

Z = - j20 + j30 + 120 − j20 Z = 120 – j10 Ω 1 1 = would cause the capacitive ωC 2πf C impedance to double, while ωL = 2πf L would cause the inductive impedance to halve. Thus, Z = - j40 + j15 + 120 − j40 Z = 120 – j65 Ω If the frequency were halved,

Chapter 9, Solution 89.  1   Z in = jωL ||  R + jωC  

 1  L  jωL  R + + jωL R jωC   C = Z in = 1  1  R + jωL +  R + jωL − jωC  ωC  Z in =

  L 1    + jωL R  R − jωL −   C ωC   1   R + ωL −  ωC 

2

2

To have a resistive impedance, Im(Z in ) = 0 . Hence,  L  1  =0 ωL R 2 −  ωL −  C  ωC 

ωR 2 C = ωL −

1 ωC

ω2 R 2 C 2 = ω2 LC − 1 ω2 R 2 C 2 + 1 L= ω2 C

(1) Ignoring the +1 in the numerator in (1), L = R 2 C = (200) 2 (50 × 10 -9 ) = 2 mH

Chapter 9, Solution 90.

Let

Vs = 145∠0° , I=

X = jωL = j (2π)(60) L = j377 L

Vs 145∠0° = 80 + R + jX 80 + R + jX

V1 = 80 I =

50 =

(80)(145) 80 + R + jX

(80)(145) 80 + R + jX

Vo = (R + jX) I =

110 =

(1) (R + jX)(145∠0°) 80 + R + jX

(R + jX)(145) 80 + R + jX

(2)

From (1) and (2), 50 80 = 110 R + jX

11  R + jX = (80)   5 R 2 + X 2 = 30976 From (1), (80)(145) 80 + R + jX = = 232 50

(3)

6400 + 160R + R 2 + X 2 = 53824

160R + R 2 + X 2 = 47424

(4)

Subtracting (3) from (4), 160R = 16448  → R = 102.8 Ω From (3), X 2 = 30976 − 10568 = 20408 X = 142.86 = 377 L  → L = 0.3789 H

Chapter 9, Solution 91. Z in = Z in =

1 + R || jωL jωC

-j jωLR + ωC R + jωL

- j ω 2 L2 R + jωLR 2 = + ωC R 2 + ω 2 L2

To have a resistive impedance, Im(Z in ) = 0 . Hence, ωLR 2 -1 + 2 =0 ωC R + ω2 L2 1 ωLR 2 = 2 ωC R + ω2 L2

R 2 + ω2 L2 C= ω2 LR 2

where ω = 2π f = 2π × 10 7 C=

9 × 10 4 + (4π 2 × 1014 )(400 × 10 −12 ) (4π 2 × 1014 )(20 × 10 − 6 )(9 × 10 4 )

C=

9 + 16π 2 nF 72π 2

C = 235 pF

Chapter 9, Solution 92. (a) Z o =

Z = Y

100∠75 o = 471.4∠13.5 o Ω o −6 450∠48 x10

(b) γ = ZY = 100∠75 o x 450∠48 o x10 −6 = 0.2121∠61.5 o

Chapter 9, Solution 93. Z = Zs + 2 ZA + ZL Z = (1 + 0.8 + 23.2) + j(0.5 + 0.6 + 18.9) Z = 25 + j20 IL =

VS 115∠0° = Z 32.02 ∠38.66°

I L = 3.592∠-38.66° A

Chapter 10, Solution 1.

ω=1

10 cos( t − 45°)  → 10∠ - 45° 5 sin( t + 30°)  → 5∠ - 60°

1H  →

1F  →

jωL = j 1 = -j jωC

The circuit becomes as shown below. 3Ω

10∠-45° V

+ −

Vo

jΩ

2 Io

+ −

5∠-60° V

Applying nodal analysis, (10∠ - 45°) − Vo (5∠ - 60°) − Vo Vo = + 3 j -j j10∠ - 45° + 15∠ - 60° = j Vo Vo = 10 ∠ - 45° + 15∠ - 150° = 15.73∠247.9° Therefore, v o ( t ) = 15.73 cos(t + 247.9°) V

Chapter 10, Solution 2. ω = 10 4 cos(10t − π 4)  → 4∠ - 45°

20 sin(10 t + π 3)  → 20 ∠ - 150°

1H  →

jωL = j10 1 1 0.02 F  → = = - j5 jωC j 0.2

The circuit becomes that shown below.

10 Ω

20∠-150° V

Vo Io

j10 Ω

+ −

4∠-45° A

-j5 Ω

Applying nodal analysis, (20∠ - 150°) − Vo Vo Vo + 4∠ - 45° = + 10 j10 - j5 20 ∠ - 150° + 4∠ - 45° = 0.1(1 + j) Vo Io =

Therefore,

Vo 2 ∠ - 150° + 4 ∠ - 45° = 2.816 ∠150.98° = j10 j (1 + j) i o ( t ) = 2.816 cos(10t + 150.98°) A

Chapter 10, Solution 3. ω= 4 2 cos(4t )  → 2∠0°

16 sin(4 t )  → 16∠ - 90° = -j16 2H  →

jωL = j8 1 1 1 12 F  → = = - j3 jωC j (4)(1 12) The circuit is shown below.

4Ω

-j16 V

+ −

-j3 Ω

Vo 1Ω

j8 Ω 2∠0° A

6Ω

Applying nodal analysis, - j16 − Vo Vo Vo +2= + 4 − j3 1 6 + j8

 - j16 1 1  V + 2 = 1 + + 4 − j3  4 − j3 6 + j8  o Vo =

3.92 − j2.56 4.682∠ - 33.15° = = 3.835∠ - 35.02° 1.22 + j0.04 1.2207 ∠1.88° v o ( t ) = 3.835 cos(4t – 35.02°) V

Therefore,

Chapter 10, Solution 4. 16 sin(4 t − 10°)  → 16∠ - 10°, ω = 4 1H  →

jωL = j4

0.25 F  → Ix

16∠-10° V

+ −

1 1 = = -j jωC j (4)(1 4) j4 Ω

0.5 Ix

(16∠ - 10°) − V1 1 V + Ix = 1 j4 2 1− j But

So,

Ix =

(16∠ - 10°) − V1 j4

3 ((16∠ - 10°) − V1 ) V = 1 j8 1− j

V1

-j Ω 1Ω

+ Vo

−

V1 =

48∠ - 10° - 1 + j4

Using voltage division, 1 48∠ - 10° Vo = V1 = = 8.232∠ - 69.04° 1− j (1 - j)(-1 + j4) v o ( t ) = 8.232 sin(4t – 69.04°) V

Therefore, Chapter 10, Solution 5.

Let the voltage across the capacitor and the inductor be Vx and we get: Vx − 0.5I x − 10∠30° Vx Vx + + =0 − j2 j3 4 (3 + j6 − j4)Vx − 1.5I x = 30∠30° but I x =

Vx = j0.5Vx − j2

Combining these equations we get: (3 + j2 − j0.75)Vx = 30∠30° or Vx = I x = j0.5

30∠30° = 4.615∠97.38° A 3 + j1.25

30∠30° 3 + j1.25

Chapter 10, Solution 6.

Let Vo be the voltage across the current source. Using nodal analysis we get: Vo − 4Vx Vo 20 −3+ = 0 where Vx = Vo 20 + j10 20 20 + j10 Combining these we get: Vo 4Vo Vo − −3+ = 0 → (1 + j0.5 − 3)Vo = 60 + j30 20 20 + j10 20 + j10 Vo =

60 + j30 20(3) or Vx = = 29.11∠–166˚ V. − 2 + j0.5 − 2 + j0.5

Chapter 10, Solution 7. At the main node, V V 120∠ − 15 o − V = 6∠30 o + + − j30 50 40 + j20  1 j 1 + +  V  40 + j20 30 50 

V=

 →

115.91 − j31.058 − 5.196 − j3 = 40 + j20

− 3.1885 − j4.7805 = 124.08∠ − 154 o V 0.04 + j0.0233

Chapter 10, Solution 8.

ω = 200, 100mH 50µF

 →  →

jωL = j200x 0.1 = j20 1 1 = = − j100 jωC j200x 50x10 − 6

The frequency-domain version of the circuit is shown below. 0.1 Vo

40 Ω

V1 6∠15

o

At node 1,

or

20 Ω

+ Vo -

Io -j100 Ω

V2 j20 Ω

V V1 V − V2 6∠15 o + 0.1V1 = 1 + + 1 20 − j100 40 5.7955 + j1.5529 = (−0.025 + j 0.01)V1 − 0.025V2

(1)

V V1 − V2 = 0.1V1 + 2 j20 40 From (1) and (2),

At node 2,

→

0 = 3V1 + (1 − j2)V2

(−0.025 + j0.01) − 0.025 V1   (5.7955 + j1.5529)   =   3 (1 − j2)  V2   0  

or

(2)

AV = B

Using MATLAB, V = inv(A)*B V2 = −110.3 + j161.09 leads to V1 = −70.63 − j127.23, V − V2 Io = 1 = 7.276∠ − 82.17 o 40

Thus,

i o ( t ) = 7.276 cos(200 t − 82.17 o ) A

Chapter 10, Solution 9. → 10 ∠0°, ω = 10 3 10 cos(10 3 t )  10 mH  →

jωL = j10

50 µF  →

1 1 = = - j20 3 jωC j (10 )(50 × 10 -6 )

Consider the circuit shown below.

20 Ω

10∠0° V

At node 1,

V1

-j20 Ω

V2

j10 Ω

Io + −

20 Ω

4 Io

30 Ω

+ Vo

−

10 − V1 V1 V1 − V2 = + 20 20 - j20 10 = (2 + j) V1 − jV2

(1)

At node 2,

V V1 − V2 V V2 = (4) 1 + , where I o = 1 has been substituted. 20 - j20 20 30 + j10 (-4 + j) V1 = (0.6 + j0.8) V2

V1 =

0.6 + j0.8 V2 -4+ j

(2)

Substituting (2) into (1) (2 + j)(0.6 + j0.8) 10 = V2 − jV2 -4+ j

V2 =

or

Vo =

170 0.6 − j26.2 30 3 170 V2 = ⋅ = 6.154 ∠70.26° 30 + j10 3 + j 0.6 − j26.2 v o ( t ) = 6.154 cos(103 t + 70.26°) V

Therefore,

Chapter 10, Solution 10.

50 mH 2µF

 →  →

jωL = j2000x50 x10 − 3 = j100,

ω = 2000

1 1 = = − j250 jωC j2000 x 2x10 − 6

Consider the frequency-domain equivalent circuit below. V1

36