Solution Manual for Fundamentals of Electric Circuits 3rd Sadiku PDF

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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) i = dq/dt = 3 mA (b) i = dq/dt = (16t + 4) A (c) i =...

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Solution Manual for Fundamentals of Electric Circuits 3rd Sadiku Muin Kashem

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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)

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

q(t) = ∫ 10e -30t sin 40t + q(0) =

Chapter 1, Solution 4

q = ∫ idt = ∫ =

10

−5 5sin 6 π t dt = cos 6π t 6π 0

5 (1 − cos 0.06π ) = 4.698 mC 6π

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 0, we have a source-free RLC circuit. α = R/(2L) = (40 + 60)/5 = 20 and ωo =

1 LC

=

1 −3

10 x 2.5

ωo = α leads to critical damping i(t) = [(A + Bt)e-20t], i(0) = 0 = A di/dt = {[Be-20t] + [-20(Bt)e-20t]}, but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24] Hence,

B = -9.6 or i(t) = [-9.6te-20t] A

Chapter 8, Solution 17.

i(0) = I0 = 0, v(0) = V0 = 4 x15 = 60 di(0) 1 = − (RI0 + V0 ) = −4(0 + 60) = −240 dt L 1 1 ωo = = = 10 LC 1 1 4 25 R 10 α= = = 20, which is > 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

v

C

− (b)

For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = α. 1

ωo =

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

12 + −

−

vC

+

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

+ (1/27)F

v

− At t = 0-,

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. 1H

2Ω

30V

+ −

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 i

vC(0+) = 40 V

10 Ω

i1 40 Ω

+

+

+

v

50V

−

+ −

v

vL

−

10 Ω

−

0.5H

(a)

50V

(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]

+ −

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

i

i

+ 30V

+ −

1H

+

v

5Ω

−

v

10 Ω

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.

+

i2

6Ω

6Ω

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 10 Ω v

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

0.02 F

2H +

6Ω

v

14 Ω −

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). 10 H 4Ω 5A

20 Ω

5Ω

1H

i +

10 µF 20V

+ −

0.04F

v −

(a)

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

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

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

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 + −

i

1H

+ v(0)

+ − +

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 2L

R = 2αL = 2 x8 x0.5 = 8Ω

 →

ω d = ω o 2 − α 2 = 30 ωo =

1 LC

ω o = 900 − 64 = 836

 →

 →

C=

1

ω oL 2