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

Download Chapter 08-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Chapter 8, Problem 1. For the circuit in Fig. 8.62, find: (a) 0i and 0v , (b) dtdi /0 and dtdv /0 , (c) i and v . Figure 8.62 For Prob. 8.1. docsity.com Chapter 8, Solution 1. (a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a). 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). vL = Ldi/dt or di/dt = vL/L Applying KVL at t = 0+, we obtain, vL(0+) – v(0+) + 10i(0+) = 0 vL(0+) – 12 + 20 = 0, or vL(0+) = -8 Hence, di(0+)/dt = -8/2 = -4 A/s Similarly, iC = Cdv/dt, or dv/dt = iC/C iC(0+) = -i(0+) = -2 dv(0+)/dt = -2/0.4 = -5 V/s (c) As t approaches infinity, the circuit reaches steady state. i( ) = 0 A, v( ) = 0 V VS + 6 (a) (b) + v 10 H 10 F 6 6 + vL docsity.com Chapter 8, Problem 3. Refer to the circuit shown in Fig. 8.64. Calculate: (a) 0 L i , 0 c v and 0 R v , (b) dtdi L /0 , dtdv c /0 , and dtdv R /0 , (c) L i , c v and R v Figure 8.64 For Prob. 8.3. Chapter 8, Solution 3. At t = 0 - , u(t) = 0. Consider the circuit shown in Figure (a). iL(0 - ) = 0, and vR(0 - ) = 0. But, -vR(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. docsity.com (b) At t = 0 + , vL(0+) = 0, therefore LdiL(0+)/dt = vL(0 + ) = 0, thus, diL/dt = 0A/s, iC(0 + ) = 2 A, this means that dvC(0 + )/dt = 2/C = 8 V/s. Now for the value of dvR(0 + )/dt. Since vR = vC + 10, then dvR(0 + )/dt = dvC(0 + )/dt + 0 = 8 V/s. (c) As t approaches infinity, we end up with the equivalent circuit shown in Figure (b). iL( ) = 10(2)/(40 + 10) = 400 mA vC( ) = 2[10||40] –10 = 16 – 10 = 6V vR( ) = 2[10||40] = 16 V 40 10 (a) + vC 10V + + vR + vR iL 2A 40 10 (b) + vC 10V + docsity.com Chapter 8, Problem 4. In the circuit of Fig. 8.65, find: (a) 0v and 0i , (b) dtdv /0 and dtdi /0 , (c) v and i . 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 5 (a) i + v 40V + 4 A 0.1F 3 5 (b) i + vL 40V + 0.25 H iC iR docsity.com Chapter 8, Solution 5. (a) For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0). iL(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+) = 6iL(0+) = 0V. (b) di(0+)/dt = d(vR(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( ) = 6(4)/10 = 2.4 A v( ) = 6(4 – 2.4) = 9.6 V 6 i A + vC 0.25F 4A 4 + v 1 H + vL iC iL docsity.com Chapter 8, Problem 6. In the circuit of Fig. 8.67, find: (a) 0 R v and 0 L v , (b) dtdv R /0 and dtdv L /0 , (c) R v and L v , Figure 8.67 For Prob. 8.6. docsity.com Chapter 8, Solution 6. (a) Let i = the inductor current. For t < 0, u(t) = 0 so that i(0) = 0 and v(0) = 0. For t > 0, u(t) = 1. Since, v(0+) = v(0-) = 0, and i(0+) = i(0-) = 0. vR(0+) = Ri(0+) = 0 V Also, since v(0+) = vR(0+) + vL(0+) = 0 = 0 + vL(0+) or vL(0+) = 0 V. (1) (b) Since i(0+) = 0, iC(0+) = VS/RS But, iC = Cdv/dt which leads to dv(0+)/dt = VS/(CRS) (2) From (1), dv(0+)/dt = dvR(0+)/dt + dvL(0+)/dt (3) vR = iR or dvR/dt = Rdi/dt (4) But, vL = Ldi/dt, vL(0+) = 0 = Ldi(0+)/dt and di(0+)/dt = 0 (5) From (4) and (5), dvR(0+)/dt = 0 V/s From (2) and (3), dvL(0+)/dt = dv(0+)/dt = Vs/(CRs) (c) As t approaches infinity, the capacitor acts like an open circuit, while the inductor acts like a short circuit. vR( ) = [R/(R + Rs)]Vs vL( ) = 0 V docsity.com