Solutions to Circuit Analysis Problems in Chapter 8, Assignments of Engineering

Download Solutions to Circuit Analysis Problems in Chapter 8 and more Assignments Engineering in PDF only on Docsity! Chapter 8, Solution 1. (a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a). + vL − 6 Ω 10 H + v − (a) 6 Ω + − 6 Ω VS 10 µF (b) 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 Chapter 8, Solution 2. (a) At t = 0-, the equivalent circuit is shown in Figure (a). 25 kΩ 20 kΩ iR + − + v − iL 60 kΩ 80V (a) 25 kΩ 20 kΩ iR + − iL 80V (b) 60||20 = 15 kohms, iR(0-) = 80/(25 + 15) = 2mA. By the current division principle, iL(0-) = 60(2mA)/(60 + 20) = 1.5 mA vC(0-) = 0 At t = 0+, vC(0+) = vC(0-) = 0 iL(0+) = iL(0-) = 1.5 mA 80 = iR(0+)(25 + 20) + vC(0-) iR(0+) = 80/45k = 1.778 mA But, iR = iC + iL 1.778 = iC(0+) + 1.5 or iC(0+) = 0.278 mA Chapter 8, Solution 16. At t = 0, i(0) = 0, vC(0) = 40x30/50 = 24V For t > 0, we have a source-free RLC circuit. α = R/(2L) = (40 + 60)/5 = 20 and ωo = LC 1 = 5.2x10 1 3− = 20 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. .iswhich,20 4 12 10 L2 R 10 25 1 4 1 1 LC 1 240)600(4)VRI( L 1 dt )0(di 6015x4V)0(v,0I)0(i o o 00 00 ω>===α ===ω −=+−=+−= ===== ( )t268t32.37 21 2121 t32.37 2 t68.2 1 2 o 2 ee928.6)t(i A928.6AtoleadsThis 240A32.37A68.2 dt )0(di,AA0)0(i eAeA)t(i 32.37,68.23102030020s −− −− −= −=−= −=−−=+== += −−=±−=±−=ω−α±α−= getwe,60dt)t(i C 1)t(v,Since t0 +∫= v(t) = (60 + 64.53e-2.68t – 4.6412e-37.32t) V 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 + − + v − t = 0 i 24 Ω 6 Ω 3 H 24V (1/27)F 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 27/1x3/1LC/1o ==ω = 3, clearly α > ωo (overdamped response) s1,2 = 222o 2 355 −±−=ω−α±α− = -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 (2) From (1) and (2), B = -2 and A = 18. Hence, v(t) = (18e-t – 2e-9t) V ωo = 1/ 310x1.0/1LC −= = 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. (1/4)F + − 8 Ω 2 Ω t=0, note this is a make before break switch so the inductor current is not interrupted. 1 H io(t) + vo(t) − 30V Figure 8.78 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/ 241x1/1LC == Since α is less than ωo, we have an under-damped response. 9843.1)16/1(422od =−=α−ω=ω vo(t) = (A1cosωdt + A2sinωdt)e-αt