Sadiku Practice Problem Solution pdf PDF

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February 5, 2006 CHAPTER 1 P.P.1 A proton has 1 x C. Hence, 2 million protons have x x 2 x 106 3 x C P.P.1 i mA At t 0 sec, i 7 mA P.P.1 idt 1 2dt 0 2 1 2 1 2 2t dt 2t 0 ( 2 3) t 3 2 1 2 6 C P.P.1 (a) Vab V The negative sign indicates that point a is at higher potential than point b. P.P.1 (b) Vab 5...

Description

P.P.6.11

40mH in series with 20mH = 40 + 20 = 60mH 60mH in parallel with 30mH = 30 x 60/(90) = 20mH 20mH in series with 100mH = 120mH 120mH in parallel with 40mH = 40 x 120/(160) = 30mH 30mH in series with 20mH = 50mH 50mH in parallel with 50mH = 25mH Leq = 25mH

P.P.6.12

(a) i2 = i - i1 i2(0) = i(0) – i1(0) = 1.4 - 0.6 = 800 mA di (b) v1 = 6 1 = 6( 0.6)(−2)e −2 t = −7.2e −2 t dt 1 t 1 (− 7.2) − 2 t t e 0 + 0.8 i 2 = ∫0 v1 dt + i2 ( 0) = 3 3 ( −2) = -0.4 + 1.2e–2tA i = i1 + i2 = 0.4 + 1.8e–2tA (c) From (b), v 1 = –7.2e–2tV di v 2 = 8 = 8(− 2)(1.8)e− 2 t = –28.8e-2tV dt v = v1 + v2 = –36e-2tV

P.P.6.13

RC = 25 x 103 x 10 x 10-6 = 0.25 1 t 1 t 10dt mV + 0 vi ( t ) + vo (0) = − vo = − ∫ RC o 0.25 ∫o = 40t mV

P.P.6.14

RC = 10 x 103 x 2 x 10-6 = 2 x 10-2 dv d v o = −RC i = −2 x10 −2 (3 t ) dt dt vo = -60mV

P.P.6.15

dv 2o dv = 4 cos 10 t − 3 o − 2v o 2 dt dt

Using this we obtain the analog computer as shown below. We may let RC = 1s.

2V − +

C

C

R 2 2 d vo/dt

t=0 R

R

− +

R/2

− +

-dvo/dt

− +

vo R

R − +

R/3 dvo/dt

R R − + cos10t

+ −

R/4

-cos10t

d2vo/dt2

February 5, 2006

CHAPTER 7

The circuit in Fig. (a) is equivalent to the one shown in Fig. (b).

P.P.7.1

8Ω

io +

+ 12 Ω

6Ω

vo

−

1/3 F

vx −

+

+ Req

vc

v −

− (a)

(b)

R eq = 8 + 12 || 6 = 12 Ω τ = R eq C = (12)(1 / 3) = 4 s v c = v c (0) e - t τ = 30 e - t 4 = 30 e -0.25 t V

vx =

4 v c = 10 e -0.25t V 4+ 8

vx = vo + vc io = P.P.7.2

⎯ ⎯→

vo = vx − vc = -20 e -0.25t V

vo = - 2.5 e -0.25t A 8 When t < 0, the switch is closed as shown in Fig. (a). 6Ω + 24 V

+ −

vc(0)

12 Ω

4Ω

− (a)

R eq = 4 || 12 = 3 Ω

v c (0 ) =

3 (24) = 8 V 3+ 6

1/3 F

When t > 0, the switch is open as shown in Fig. (b). t=0

6Ω

24 V

+ −

3Ω

1/6 F

(b) τ = R eq C = (3)(1 / 6) = 1 / 2 s v ( t ) = v c (0 ) e - t τ = 8 e - 2 t V

1 1 1 w c (0) = Cv 2c (0) = × × 64 = 5.333J 2 2 6

P.P.7.3

This can be solved in two ways.

Method 1:

Find R th at the inductor terminals by inserting a voltage source. 3Ω

io +

vx

−

1Ω vo = 1 V

+ −

i1

+ −

Applying mesh analysis gives Loop 1: 4i 1 − i 2 + 2v x − 1 = 0 , 10i 1 − i 2 = 1 Loop 2:

6i 2 − i1 − 2 v x = 0 7 i 2 = i1 6

From (1) and (2), 6 i o = i1 = 53

i2

5Ω

2vx

where v x = 3i 1 (1)

(2)

R th =

vo 53 = , io 6

τ=

16 L 1 = = R 53 6 53

i( t ) = 5e -53t A Method 2:

We can obtain i using mesh analysis. 3Ω i 1/6 H

+

vx

−

1Ω i2

i1 + −

2vx

Applying KVL to the loops, we obtain 1 di 1 + 4i 1 − i 2 + 2v x = 0 Loop 1: 6 dt 1 di 1 + 10i 1 − i 2 = 0 6 dt Loop 2:

6i 2 − i 1 − 2 v x = 0 7 i 2 = i1 6

Substituting (4) into (3) yields 1 di 1 7 +10i 1 − i 1 = 0 6 dt 6 di 1 or + 53 i1 = 0 dt i 1 = Ae -53t i = - i1 = Be -53t i( 0 ) = 5 = B i( t ) = 5e -53t A Therefore, i( t ) = 5e -53t A and

v x (t ) = -3i(t) = - 15e -53t V

5Ω

where v x = 3i 1 (3)

(4)

For t < 0, the equivalent circuit is shown in Fig. (a).

P.P.7.4

i(t)

12 Ω

12 Ω

5A

5Ω

8Ω

8Ω 2H

(a) (b) i (0 ) =

8 (5) = 2 A 8 + 12

For t > 0, the current source is cut off and the RL circuit is shown in Fig. (b). L 2 R eq = (12 + 8) || 5 = 20 || 5 = 4 Ω , τ= = = 0.5 Req 4 i( t ) = i (0) e- t τ = 2 e -2 t A , t > 0 P.P.7.5 For t < 0, the switch is closed. The inductor acts like a short so the equivalent circuit is shown in Fig. (a). 3Ω i i

io

1H

io 4Ω

6A

2Ω

(a) i=

4 (6 ) = 4 A , 4+ 2

4Ω

2Ω (b)

io = 2 A ,

v o = 2i = 8 V

For t > 0, the current source is cut off so that the circuit becomes that shown in Fig. (b). The Thevenin equivalent resistance at the inductor terminals is L 1 τ= = R th = ( 4 + 2) || 3 = 2 Ω , R th 2 3 (-i) - 1 - 4 -2t 8 io = = i= and v o = -2i o = e-2t e 6+ 3 3 3 3

Thus, t 0 ⎩4 e

P.P.7.6

8V t 0

⎧ 0 t 4,

t

I = 20 − 10 t

4 2

t 2

= 40 − 10 t

=0

Thus, ⎧ 0 t 4

or

I = 10 [r (t ) − 2r (t − 2 ) + r( t − 4 ) ]A

which is sketched below ∫ i dt 20

0

P.P.7.7

⎧ 2 − 2t 0 < t < 2 ⎪ i( t ) = ⎨- 6 + 2t 2 < t < 3 ⎪ 0 otherwise ⎩

2

4

t

i(t )= (2− 2t )[ u(t )− u(t − 2)] + (-6+ 2t)[ u(t - 2) - u(t - 3)] i( t ) = 2 u ( t ) − 2 t u ( t ) + 4( t − 2) u ( t − 2) − 2( t − 3) u ( t −3) i( t ) = 2 u(t )− 2 r(t ) + 4 r( t − 2) − 2 r (t − 3) A h(t ) = 4[ u (t ) − u(t − 2) ] + (6 − t ) [ u (t − 2) − u( t − 3) ] h ( t ) = 4 u ( t ) − ( t − 2) u ( t − 2) + r ( t − 6) h ( t ) = 4 u ( t ) − r ( t − 2) + r ( t − 6)

P.P.7.8

(a)

P.P.7.9

∫ (t ∞

-∞

3

+ 5 t 2 + 10) δ ( t + 3) dt = t 3 + 5 t 2 + 10 t =-3

= -27 + 45 + 10 = 28 (b)

∫

10

0

δ( t − π ) cos(3t ) dt = cos(3π) = - 1

P.P.7.10 For t < 0, the capacitor acts like an open circuit. − v( 0 ) = v (0 + ) = v( 0) = 10 6 2 (50) = -5 (10) − 2+6 6+2 3 3 1 1 R th = 2 || 6 = Ω , τ = R th C = × = 2 3 2 2

For t > 0,

v(∞) =

v( t ) = v (∞) + [ v (0) − v(∞) ] e - t τ v ( t ) = -5 + (10 + 5) e -2t v( t ) = - 5 + 15 e -2t V At t = 0.5,

v (0.5) = -5 + 15 e -1 = 518.2 mV

P.P.7.11 For t < 0, only the left portion of the circuit is operational at steady state. − v( 0 ) = v (0 + ) = v (0) = 20 , i (0 ) = 0 For t > 0, 20u ( -t) = 0 so that the voltage source is replaced by a short circuit. Transforming the current source leads to the circuit below. 10 Ω

i

10 Ω

+ v −

0.2 F

+ −

30 V

5 (30) = 10 15 10 = 5 || 10 = Ω, 3

v(∞) = R th

τ = R th C =

2 10 × 0.2 = 3 3

v( t ) = v (∞) + [ v (0) − v(∞) ] e - t τ v( t ) = 10 + ( 20 − 10) e-3t 2 v( t ) = 10 (1 + e -1.5t ) - v( t ) = -2 ( 1 + e -1.5 t ) 5 ⎧ 0 t 0 i( t ) =

P.P.7.12

⎧ 20 V t< 0 v( t) = ⎨ - 1.5t ⎩10 ( 1 + e ) V t > 0

Applying source transformation, the circuit is equivalent to the one below. i

1.5 H

t=0

5Ω

10 Ω

+ −

30 V

At t < 0, the switch is closed so that the 5 ohm resistor is short circuited. 30 i (0 − ) = i (0 ) = =3A 10 For t > 0, the switch is open. R th = 10 + 5 = 15 ,

i(∞) =

τ=

L 1.5 = = 0.1 R th 15

30 =2A 10 + 5

i( t ) = i(∞) + [ i (0) − i(∞) ] e - t τ i( t ) = 2 + (3 − 2) e -10t i( t ) = (2 + e-10t ) A, t > 0

P.P.7.13

For 0 < t < 2, the given circuit is equivalent to that shown below.

10 Ω

20 Ω i(t)

6A

15 Ω

5H

Since switch S1 is open at t = 0 − , i(0 − ) = 0 . Also, since i cannot jump, i (0) = i (0 − ) = 0 . 90 i(∞) = =2 A 15 + 10 + 20 L 5 1 R th = 45 Ω , τ = = = R th 45 9 i( t ) = i(∞) + [ i( 0) − i(∞) ] e - t τ i ( t ) = 2 + (0 − 2) e -9 t i( t ) = 2 (1 − e -9 t ) A

When switch S2 is closed, the 20 ohm resistor is short-circuited. i( 2+ ) = i (2− ) = 2 (1 − e -18 ) ≅ 2 This will be the initial current 90 i(∞) = = 3.6 A 15 + 10 5 1 R th = 25 Ω , τ = = 25 5 i( t ) = i(∞) + [ i( 2 + ) − i(∞) ] e -( t −2)

τ

i( t ) = 3.6 + (2 − 3.6) e -5( t− 2) i( t ) = 3.6 − 1.6 e-5( t −2)

⎧ 0 t 0 To get to v o from v, we notice that v is the potential difference between node 1 and the output terminal, i.e. 0 − vo = v ⎯ ⎯→ v o = -v vo = - 4 e -2 t V , t > 0

P.P.7.15 Let v1 be the potential at the inverting terminal. v( t ) = v (∞) + [ v (0) − v(∞) ] e- t τ where τ = RC = 100 × 10 3 × 10 -6 = 0.1 , v(0) = 0

v1 = 0 for all t v1 − v o = v

(1)

For t > 0, the switch is closed and the op amp circuit is an inverting amplifier with - 100 v o (∞) = ( 4 mV ) = -40 mV 10 From (1), v (∞) = 0 − v o (∞) = 40 mV Thus, v( t ) = 40 (1 − e-10 t ) mV v o = v 1 − v = -v vo = 40( e -10 t − 1 ) mV This is a noninverting amplifier so that the output of the op amp is ⎛ Rf ⎞ ⎟v v a = ⎜1 + R1 ⎠ i ⎝

P.P.7.16

⎛ Rf ⎞ ⎛ 40 ⎞ ⎟ v i = ⎜1 + ⎟2 u ( t ) = 6 u ( t ) vth = va = ⎜1+ ⎝ 20 ⎠ R1 ⎠ ⎝ To get R th , consider the circuit shown in Fig. (a), where R o is the output resistance of the op amp. For an ideal op amp, R o = 0 so that R th = R 3 = 10 k Ω

R3

Ro

Rth Rth

R2 (a)

τ = Rth C = 10× 10 3 × 2 × 10 -6 =

Vth

+ −

C (b)

1 50

The Thevenin equivalent circuit is shown in Fig. (b), which is a first order circuit. Hence,

vo ( t ) = 6 ( 1 − e - t τ ) u ( t )

v o ( t ) = 6 (1 − e-50t )u(t ) V

P.P.7.17 The schematic is shown in Fig. (a). Construct and save the schematic. Select Analysis/Setup/Transient to change the Final Time to 5 s. Set the Print Step slightly greater than 0 (20 ns is default). The circuit is simulated by selecting Analysis/ Simulate. In the Probe menu, select Trace/Add and display V(R2:2) as shown in Fig. (b).

(a)

(b)

P.P.7.18 The schematic is shown in Fig. (a). While constructing the circuit, rotate L1 counterclockwise through 270° so that current i(t) enters pin 1 of L1 and set IC = 10 for L1. After saving the schematic, select Analysis/Setup/Transient to change the Final Time to 1 s. Set the Print Step slightly greater than 0 (20 ns is default). The circuit is simulated by selecting Analysis/ Simulate. After simulating the circuit, select Trace/Add in the Probe menu and display I(L1) as shown in Fig. (b).

(a)

(b)

P.P.7.19

v(0) = 0 . When the switch is closed, we have the circuit shown below.

10 k Ω

R

a

+ 80 μF

9V − b

We find the Thevenin equivalent at terminals a-b. 10 (R + 4) R th = (R + 4) || 10 = R + 14 v th = v (∞) =

R+ 4 (9 ) R + 14

v( t ) = v (∞) + [ v (0) − v(∞) ] e- t τ ,

v( t ) = v (∞) ( 1 − e

-t τ

)

τ = R th C

Since v(0) = 0 , v( t) 9 (1 − e -t τ ) mA = i( t ) = R +4 R + 4 Assuming R is in kΩ, 9 ( 1 − e -t 0 τ ) ×10 -3 R + 14 R + 14 = 1− e- t 0 τ (0.12) 9 0.12R + 1.68 7.32 − 0.12R = e-t 0 τ = 1 − 9 9

120 × 10-6 =

or

⎞ ⎛ 9 ⎟ t 0 = τ ln ⎜ ⎝ 7.32 − 0.12R ⎠ t0 =

⎞ ⎛ 10 (R + 4) 9 ⎟ × 80 × 10 -6 × ln ⎜ ⎝ 7.32 − 0.12R ⎠ R + 14

When R = 0, ⎛ 9 ⎞ 40 × 80 × 10 -6 ⎟ = 0.04723 s × ln ⎜ t0 = ⎝ 7.32⎠ 14

4 kΩ

When R = 6 kΩ, 100 ⎛ 9 ⎞ × 80 × 10 - 6 × ln ⎜ t0 = ⎟ = 0.124 s 20 ⎝ 6.6 ⎠ The time delay is between 47.23 ms and 124 ms.

P.P.7.20

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

q = CV = ( 2 × 10 -3 )(80) = 0.16 C 1 1 CV 2 = (2 × 10 -3 )(6400) = 6.4 J 2 2 Δq 0.16 = 200 A ΔI = = Δt 0.8× 10-3 Δw 6.4 p= = 8 kW = Δt 0.8 × 10-3 Δq 0.16 = 32 s Δt = = Δ I 5 × 10-3 W=

L 500 ×10 -3 τ= = = 2.5 ms R 200 110 i(∞) = = 550 mA i( 0 ) = 0 , 200 i( t ) = 550 (1 − e - t τ ) mA

P.P.7.21

350 mA = i( t 0 ) = 550 ( 1 − e- t 0 τ ) mA 35 20 = 1 − e -t0 τ ⎯ ⎯→ e - t 0 τ = 55 55

et 0 τ =

55 20

⎛ 55 ⎞ ⎛ 55 ⎞ t 0 = τ ln ⎜ ⎟ = 2.5 ln ⎜ ⎟ ms ⎝ 20 ⎠ ⎝ 20 ⎠ t 0 = 2.529 ms

P.P.7.22

(a) (b) (c)

5L 5 × 20 ×10 -3 = = 20 ms t = 5τ = 5 R 2 1 2 1 12 ⎞ -3 ⎛ W = LI = ( 20 × 10 ) ⎜ ⎟ = 57.6 mJ ⎝5 ⎠ 2 2 ⎛ 12 5 ⎞ di ⎟ = 24 kV V = L = 20 × 10 -3 ⎜ ⎝ 2 × 10-6 ⎠ dt

February 5, 2006

CHAPTER 8 P.P.8.1 (a)

At t = 0-, we have the equivalent circuit shown in Figure (a). 10 Ω

i

+ 2Ω

v

vL

a +

24V

+ −

2Ω

vC

−

−

(a)

(b)

+ i

50mF

+ −

i(0-) = 24/(2 + 10) = 2 A, v(0-) = 2i(0-) = 4 V hence, v(0+) = v(0-) = 4V. (b)

At t = 0+, the switch is closed. L(di/dt) = vL, leads to (di/dt) = vL/L But,

vC(0+) + vL(0+) = 24 = 4 + vL(0+), or vL(0+) = 20 (di(0+)/dt) = 20/0.4 = 50 A/s C(dv/dt) = iC leading to (dv/dt) = iC/C

But at node a, KCL gives i(0+) = iC(0+) + v(0+)/2 = 2 = iC(0+) + 4/2 or iC(0+) = 0, hence (dv(0+)/dt) = 0 V/s (c) As t approaches infinity, the capacitor is replaced by an open circuit and the inductor is replaced by a short circuit. v(∞) = 24 V, and i(∞) = 12 A.

24V

P.P.8.2 (a)

At t = 0-, we have the equivalent circuit shown in (a).

5Ω

5Ω

a iR

2A

iL 3A

+ vC

+

vR 10 μF

b −

vL

2H

−

− (a)

+ 3A

(b)

iL(0-) = -3A, vL(0-) = 0, v R(0-) = 0 At t = 0+, we have the equivalent circuit in Figure (b). At node b, iR(0+) = iL(0+) + 3, since iL(0+) = iL(0-) = -3A, iR(0+) = 0, and vR(0+) = 5iR(0+) = 0. Thus, iL(0) = –3 A, vC(0) = 0, and vR (0+) = 0. (b)

dvC(0+)/dt = iC(0+)/C = 2/0.2 = 10 V/s. To get (dvR/dt), we apply KCL to node b, i R = i L + 3, thus diR/dt = diL/dt. Since vR = 5iR, dv R/dt = 5di R/dt = 5di L/dt. But LdiL/dt = vL, diL/dt = vL/L. Hence, dvR (0+)/dt = 5vL(0+)/L. Applying KVL to the middle mesh in Figure (b), -v C(0+) + v R(0+) + v L(0+) = 0 = 0 + 0 + vR(0+), or vR (0+) = 0 Hence, dv R(0+)/dt = 0 = di L(0+)/dt; diL(0+)/dt = 0, dvC(0+)/dt = 10 V/s, dv R(0+)/dt = 0 .

(c)

As t approaches infinity, we have the equivalent circuit shown below.

5Ω 2A

iL 3A

2 = 3 + iL(∞) leads to iL(∞) = -1A vC(∞) = vR(∞) = 2x5 = 10V Thus, iL(∞) = –1 A, vC(∞) = vR(∞) = 10 V

P.P.8.3 (a)

α = R/(2L) = 10/(2x5) = 1, ω o = 1

LC = 1

5x 2x10 −2 = 10

s1,2 = − α ± α 2 − ω 2o = − 1± 1 − 100 = -1 ± j9.95. (b)

Since α < ωo , we clearly have an underdamped response.

P.P.8.4 For t < 0, the inductor is connected to the voltage source although it acts like a short circuit. i(0-) = 50/10 = 5 = i(0+) = i(0) The voltage across the capacitor is 0 = v(0-) = v(0+) = v(0). For t > 0, we have a source-free RLC circuit.

ωo = 1

LC = 1

1x

1 = 3 9

α = R/(2L) = 5/(2x1) = 2.5

Since α < ωo, we have an underdamped case. s1,2 = − α ± α 2 − ω2o = −2.5 ± 6.25 − 9 = -2.5 ± j1.6583 i(t) = e-2.5t[A1cos1.6583t + A2sin1.6583t] We now determine A 1 and A 2. i(0) = 5 = A1 di/dt = -2.5{e -2.5t[A 1cos1.6583t + A 2sin1.6583t]} + 1.6583e-2.5t [-A1 sin1.6583t + A2cos1.6583t] di(0)/dt = -(1/L)[Ri(0) + v(0)] = -2.5A 1 + 1.6583A 2 = -1[25] = -2.5(5) + 1.6583A2 A 2 = -7.5378 -2.5t

Thus, i(t) = e

P.P.8.5

[5cos1.6583t – 7.538sin1.6583t] A

α = 1/(2RC) = 1/(2x2x25x10-3) = 10 ω0= 1

LC = 1

0.4 x 25x10−3 = 10

since α = ωo , we have a critically damped response. Therefore, v(t) = [(A1 + A2t)e-10t] v(0) = 0 = A1 + A2x0 = A1, which leads to v(t) = [A 2te-10t]. dv(0)/dt = -(v(0) + Ri(0))/(RC) = -2x3/(2x25x10-3) = -120 dv/dt = [(A2 - 10A2t)e-10t] At t = 0,

-120 = A2 therefore, v(t) = (–120t)e–10t volts

P.P.8.6 For t < 0, the switch is closed. The inductor acts like a short circuit while the capacitor acts like an open circuit. Hence, i(0) = 2 and v(0) = 0. α = 1/(2RC) = 1/(2x20x4x10 -3) = 6.25 ωo = 1

LC = 1 10 x 4 x10 −3 = 5

Since α > ωo, this is an overdamped response. s1,2 = − α ± α 2 − ω o = − 6.25 ± (6.25) 2 − 25 = -2.5 and –10 Thus, v(t) = A1 e-2.5t + A2 e-10t v(0) = 0 = A 1 + A 2, which leads to A 2 = -A 1 dv(0)/dt = -(v(0) + Ri(0))/(RC) = -12.5(2x20) = -500 But, dv/dt = -2.5A1e -2.5t -10A2e-10t At t = 0, -500 = -2.5A1 -10A2 = 7.5A1 since A1 = -A2 A1 = -66.67, A2 = 66.67 Thus, v(t) = 66.67(e–10t – e–2.5t) V P.P.8.7

The initial capacitor voltage is obtained when the switch is in position a. v(0) = [2/(2 + 1)]12 = 8V

The initial inductor current is i(0) = 0. When the switch is in position b, we have the RLC circuit with the voltage source. α = R/(2L) = 10/(2x2.5) = 2

ωo = 1 LC = 1

(5 / 2) x(1 / 40) = 4

Since α < ωo, we have an underdamped case.

2 s1,2 = − α ± α − ωo = −2 ± (2) 2 − 16 = -2 ± j 3.464

Thus, v(t) = vf + [(A 1cos3.464t + A 2sin3.464t)e -2t] where vf = v(∞) = 10, the final capacitor voltage. We now impose the initial conditions to get A1 and A 2. v(0) = 8 = 10 + A1 leads to A1 = -2 The initial capacitor current is the same as the initial inductor current. i(0) = C(dv(0)/dt) = 0 therefore, dv(0)/dt = 0 But, dv/dt = 3.464[(-A1sin3.464t + A 2cos3.464t)e-2t] -2[(A1cos3.464t + A2sin3.464t)e-2t] dv(0)/dt = 0 – 2A1 + 3.464A2 , which leads to A2 = -4/3.464 = -1.1547 Thus, v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e-2t]} V i = C(dv/dt), v R = Ri = RC(dv/dt) = (1/4)dv/dt = (1/4)[(4 – 4)cos3.464t + (2x1.1547 + 2x3.464)sin3.464t]e-2t vR(t) = {[2.31sin3.464t]e-2t} V

P.P.8.8

When t < 0, v(0) = 0, i(0) = 0; for t > 0, α = 0, ωo = 1 LC = 1 0.2 x5 = 1 i(t) = is + A1 cost + A2sint = 20 + A1cost + A2 sint i(0) = 0 = 20 + A1, therefore A1 = -20 Ldi(0)/dt = v(0) = 0

But di/dt = -A1sint + A2cost At t = 0, di(0)/dt = 0 = 0 + A2 leading to i(t) = 20(1 – cost) A v(t) = Ldi/dt = 5x20sint = 100sint V

At t = 0, the switch is open so that v(0) = 0, i(0-) = 0

P.P.8.9

(1)

For t > 0, the switch is closed. We have the equivalent circuit as in Figure (a).

iC

4Ω

10 Ω

10 Ω

2A (1/20)F

i

iC

i

+

4Ω 2A

v

2H

− (b)

(a) v(0+) = 0, i(0+) = 0

(2)

-2 + iC + i = 0

(3)

From (3), i(0+) = 0 means that iC (0+) = 2, but iC (0+) = Cdv(0+)/dt which leads to dv(0+)/dt = i C(0+)/C = 2/(1/20) = 40 V/s As t approaches infinity, we have the equivalent circuit in (b). i(∞) = 2 A, v(∞) = 4i(∞) = 8V

(5)

Next we find the network response by turning off the current source as shown in Figure (c).

iC

i

10 Ω

4Ω i

+ (1/20)F

v

2H

− (c) Applying ...