Chapter 08-Electrical Circuit Analysis-Problem Solutions, Exercises of Electrical Circuit Analysis

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Chapter 8, Problem 1.

For the circuit in Fig. 8.62, find: (a) i 0  and v 0  , (b) di 0 ^ / dt and dv 0 ^ / dt , (c) i fand v f.

Figure 8. For Prob. 8.1.

Chapter 8, Solution 1.

(a)shown in Figure (a). At t = 0-, the circuit has reached steady state so that the equivalent circuit is

i(0-) = 12/6 = 2A, v(0-) = 12V At t = 0+, i(0+) = i(0-) = 2A , v(0+) = v(0-) = 12V (b) For t > 0, we have the equivalent circuit shown in Figure (b). v (^) L = Ldi/dt or di/dt = v (^) L/L Applying KVL at t = 0+, we obtain, vL(0+) – v(0+) + 10i(0+) = 0 vL(0+) – 12 + 20 = 0, or v (^) L(0+) = - Hence, di(0+)/dt = -8/2 = -4 A/s Similarly, i (^) C = Cdv/dt, or dv/dt = i (^) C /C i (^) C (0+) = -i(0+) = - dv(0+)/dt = -2/0.4 = -5 V/s (c) As t approaches infinity, the circuit reaches steady state. i(f) = 0 A , v(f) = 0 V

VS^ + ^6 :

(a) (b)

v 10 H (^) 

10 P F

vL 

By the current division principle, i (^) L(0-) = 60(2mA)/(60 + 20) = 1.5 mA vC (0-) = 0 At t = 0+, vC (0+) = vC (0-) = 0 i (^) L(0+) = i (^) L(0-) = 1.5 mA 80 = i (^) R (0+)(25 + 20) + vC (0-) i (^) R (0+) = 80/45k = 1.778 mA But, i (^) R = i (^) C + i (^) L 1.778 = iC (0+) + 1.5 or iC (0+) = 0.278 mA (b) vL(0+) = vC (0+) = 0 But, vL = LdiL/dt and di (^) L(0+)/dt = vL(0+)/L = 0 di (^) L(0+)/dt = 0 Again, 80 = 45iR + vC 0 = 45diR /dt + dvC /dt But, dvC (0+)/dt = i (^) C (0+)/C = 0.278 mohms/1 PF = 278 V/s Hence, diR (0+)/dt = (-1/45)dvC (0+)/dt = -278/ di (^) R (0+)/dt = -6.1778 A/s Also, i (^) R = i (^) C + i (^) L di (^) R (0+)/dt = di (^) C (0+)/dt + di (^) L(0+)/dt -6.1788 = di (^) C (0+)/dt + 0, or di (^) C (0+)/dt = -6.1788 A/s (c) As t approaches infinity, we have the equivalent circuit in Figure (b). i (^) R (f) = i (^) L(f) = 80/45k = 1.778 mA

Chapter 8, Problem 3.

Refer to the circuit shown in Fig. 8.64. Calculate: (a) i (^) L 0  , v (^) c 0  and vR 0  , (b) di (^) L 0 ^ / dt , dv (^) c 0 ^ / dt , and dv (^) R 0 ^ / dt , (c) i (^) L f, v (^) c fand vR f

Figure 8. For Prob. 8.3.

Chapter 8, Solution 3. At t = 00. But, -v-, u(t) = 0. Consider the circuit shown in Figure (a). iL(0-) = 0, and vR (0-) = R (0-) + vC (0-) + 10 = 0, or vC (0-) = -10V.

(a) At t = 0+, since the inductor current and capacitor voltage cannot change abruptly, the inductor current must still be equal to 0A , the capacitor has a voltage equal to –10V. Since it is in series with the +10V source, together they represent a direct short at t = 0+. This means that the entire 2A from the current source flows through the capacitor and not the resistor. Therefore, vR (0+) = 0 V.

Chapter 8, Problem 4. In the circuit of Fig. 8.65, find: (a) v 0  and i 0  , (b) dv 0 ^ / dt and di 0 ^ / dt , (c) v f and i f.

Figure 8.65 For Prob. 8.4.

Chapter 8, Solution 4. (a) At t = 0-, u(-t) = 1 and u(t) = 0 so that the equivalent circuit is shown in Figure (a). i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V.

Hence, i(0+) = i(0 - ) = 5A v(0+) = v(0-) = 25V 3 :

(a)

i (^) + v 

40V + 

4 A

0.1F

(b)

i (^) + vL  40V + 

0.25 H

i (^) C i (^) R

(b) i (^) C = Cdv/dt or dv(0+)/dt = iC(0+)/C For t = 0Since i and v cannot change abruptly,+, 4u(t) = 4 and 4u(-t) = 0. The equivalent circuit is shown in Figure (b).

i (^) R = v/5 = 25/5 = 5A, i(0+) + 4 =i (^) C (0+) + i (^) R (0+) 5 + 4 = iC (0+) + 5 which leads to iC (0+) = 4 dv(0+)/dt = 4/0.1 = 40 V/s

Similarly, vL = Ldi/dt which leads to di(0+)/dt = vL(0+)/L 3i(0+) + vL(0+) + v(0+) = 0 15 + vL(0+) + 25 = 0 or vL(0+) = - di(0+)/dt = -40/0.25 = -160 A/s (c) As t approaches infinity, we have the equivalent circuit in Figure (c).

i(f) = -5(4)/(3 + 5) = -2.5 A v(f) = 5(4 – 2.5) = 7.5 V

4 A

(c)

i (^) + v 

Chapter 8, Solution 5. (a) For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0). i (^) L(0-) = 0 and vC (0-) = 0. For t = 0+, 4u(t) = 4. Consider the circuit below.

Since the 4-ohm resistor is in parallel with the capacitor, i(0+) = vC (0+)/4 = 0/4 = 0 A Also, since the 6-ohm resistor is in series with the inductor, v(0+) = 6i (^) L(0+) = 0V.

(b) di(0+)/dt = d(v (^) R (0+)/R)/dt = (1/R)dvR (0+)/dt = (1/R)dvC(0+)/dt = (1/4)4/0.25 A/s = 4 A/s v = 6iL or dv/dt = 6diL/dt and dv(0+)/dt = 6diL(0+)/dt = 6vL(0+)/L = 0 Therefore dv(0+)/dt = 0 V/s (c) As t approaches infinity, the circuit is in steady-state. i(f) = 6(4)/10 = 2.4 A v(f) = 6(4 – 2.4) = 9.6 V

i

A

vC 

4A 4 : 0.25F

v 

1 H

i (^) C + vL 

i (^) L

Chapter 8, Problem 6.

In the circuit of Fig. 8.67, find: (a) v (^) R 0  and vL 0  , (b) dv (^) R 0 ^ / dt and dv (^) L 0 ^ / dt , (c) v (^) R f and vL f,

Figure 8. For Prob. 8.6.

Chapter 8, Problem 7. A series RLC circuit has R 10 kȍ, L 0. 1 mH, and C 10 PF. What type of damping is exhibited by the circuit? Chapter 8, Solution 7. (^36)

D 2^ R L^ 2 0.1 10 x 10 10 x^ x^  3 50 10 x

Z o LC 0.1 10 x  x 10 10 x  x

D! Z o o overdamped

Chapter 8, Problem 8. A branch current is described by ddt^2 i 2 t  4 didtt  10 it 0

Determine: (a) the characteristic equation, (b) the type of damping exhibited by the circuit, (c) i t given that i 0 1 and di 0 / dt 2. Chapter 8, Solution 8.

(a) The characteristic equation is s^2^  4 s  10 0 (b) s 1,2^ ^4 r^162 ^40  2 r j 2. This is underdamped case. (c ) i t ( ) ( A cos 2.45 t  B sin2.45 ) t e ^2 t di (^) ( 2 A cos 2.45 t  2 B sin2.45 t  2.45 A sin2.45 t  2.45 B cos 2.45 ) t e  2 t dt i(0) =1 = A di(0)/dt = 2 = –2A + 2.45B = –2 + 2.45B or B = 1. i(t) = {cos(2.45t) + 1.6327sin(2.45t)}e –2t^ A. Please note that this problem can be checked using MATLAB.

Chapter 8, Problem 9.

The current in an RLC circuit is described by (^) dtd^22 i  (^10) dtdi  25 i 0 If i 0 10 and di 0 / dt 0 find i t for t! 0.

Chapter 8, Solution 9.

s 2 + 10s + 25 = 0, thus s (^) 1,2 = ^10 r 210 ^10 = -5, repeated roots. i(t) = [(A + Bt)e -5t^ ], i(0) = 10 = A di/dt = [Be -5t^ ] + [-5(A + Bt)e -5t^ ] di(0)/dt = 0 = B – 5A = B – 50 or B = 50. Therefore, i(t) = [(10 + 50t)e -5t] A

Chapter 8, Problem 10.

The differential equation that describes the voltage in an RLC network is ddt^22 v  5 dvdt  4 v 0

Given that v 0 0 , dv 0 / dt 10 obtain v t.

Chapter 8, Solution 10.

s 2 + 5s + 4 = 0, thus s1,2 =