EECE 210 Homework 14 with Solution, Fall 2013

Chapter 12 –Basic responses of Second-Order Circuits P12.1.1 Determine  in Figure P12.1.1 so that the response is critically damped by applying: (i) the source absorption theorem, and (ii) KVL and comparing coefficients with those of a prototypical series RLC circuit. Solution: rad/s; For critical damping, , . (i) The effective resistance of the source is vR/(vR/100) = 100. Hence, 100 + 100 = 160, which gives  = 0.6. (ii) From KVL, with vSRC = 0: + , where vR = 100i. The effective resistance is therefore ( + 1)100 . For critical damping, , ; ( + 1)100 = 160; hence,  = 0.6. P12.1.8 The switch in Figure P12.1.8 is opened at t = 0 after being closed for a long time. (a) Determine iL, vC, dvC/dt, and diL/dt just after the switch is opened. (b) Is the circuit reducible to a prototypical series LCR circuit or a parallel GCL circuit? Solution: Just before t = 0, iC = 0 and the 3 mA divides between the 3 k and 6 k resistors, so that IL0 = 33/9 = 1 mA; VC0 = 61 = 6 V. These do not change on switching. Just after t = 0, 2iC + iL = 0, so that iC = -iL/2 = -0.5 mA = C, and -1000 V/s. From KVL, VC0 + 1iC = 6IL0 + vL. Hence, vL = 6 – 0.5 – 6 = -0.5 V = L, and = = -5 A/s. P12.2.1 Given mA in Figure P12.2.1. Determine , , vL, and vC, assuming zero initial charge on the capacitor. Verify that vR + vL + vC = 0. Solution: It is seen from the expression for current that the response is critically damped; rad/s; hence H; R = 26.25400 = 5 K; = + = V; = V. As a check, . Figure P12.1.1 100  1 mH – + vSRC 125 nF – +  vR 4 mH + – vR i R i + – vC 1 F L + – vL Figure P12.2.1 0.5 F 3 k 0.1 H 2 k iC + – vC 3 mA iC iL ix 6 k 1 k t = 0 Figure P12.1.8 + – vL 0.5 H + – vC iL 2 F 5 k 2 A t = 0 a b Figure P12.2.7 P12.2.4 Given v = 0, I10 = I20 = 2 A at t = 0. Determine the total instantaneous energy in the inductors as a function of time Solution: Since v = 0 at t = 0, it means that at t = 0, i is a maximum and is given by i = 4cost A, where rad/s. The energy in the inductors is therefore J. P12.2.7 The switch in Figure P12.2.7 is moved from position ‘a’ to position ‘b’ at t = 0 after being in position ‘a’ for a long time. Determine vC and iL for t  0. Solution: rad/s, rad/s; response is overdamped. rad/s; . At t = 0, iL = 2 A, vC = 0. . At t = 0, A + B = 2. . At t = 0, . Solving for A and B gives: , and . It follows that A, where t is in ms and kV. P12.2.12 The switch in Figure P12.2.12, is moved to position ‘b’ at t = 0, with zero initial energy storage. Determine: (a) for critical damping and (b) , , and for . Solution: (a) For critical damping R  , where 0  10 krad/s; R   50 . To have the parallel resistance equal to 50,   0. 1 F 2 H 2 H i1 i2 + – v i Figure P12.2.4 1 F 100  10 mH 100  – + ix + – v 0.1A iC iL ix Figure P12.2.12 t = 0 a b (b) Just before the switch is moved, v = 0 and iL = 0. Just after the switch is moved, these remain the same and iC = 0.1 A. For t  0, ; hence, A = 0. . It follows that V/s, or 100 V/ms. Hence, v = 100te-10t V, where t is in ms; iC  A, t being in seconds, or iC  100(e-10t – 10te-10t) mA, t being in ms. Since iL + iC + mA, iL = 100 – 100(e-10t – 10te-10t) mA, t being in ms, or iL  100 –100(e-10t + 10te-10t) mA, t being in ms. As a check, A, with t in ms, or = mA. Integration gives the same answer. P12.2.14 The switch In Figure P12.2.14, is moved to position ‘b’ at after being in position ‘a’ for a long time. Determine for t  0. Solution: The initial current through the inductor is 10 mA, and the initial voltage across the capacitor is 5 V. After the switch is moved to position b, the circuit becomes a parallel circuit having C = 1F, L = 4 H, and Rp = 500 . Hence, krad/s, rad/s; V, where krad/s, krad/s and t is in ms. mA, where if vO is in V, t is ms and L is in henries, i is in mA. At t = 0, A + B = -5 and mA. This gives: V and V. It follows that V. P12.2.18 The switch in Figure P12.2.18 is opened at t = 0 after being closed for a long time. Determine vL for t  0. Solution: rad/s. rad/s. response is underdamped, with rad/s. Just before the switch is opened, vL = 0 and iL = 2 A. Just after the switch is opened, vL = 0, iL = 2 A, and iC = -2 A. . At t = 0, vL = 0, so A = 0, and . , retaining only the cosine term. At t = 0, iC = -2 A, so that B = iC/dC = V. it follows that V, where t is in ms. 500  + – vO 125  t = 0 a b 4 H 100  1 F 0.1 A Figure P12.2.14 100 F 5  10 mH 5  + – vL 10 V t = 0 + – 5  Figure P12.2.18 100 F 10 mH 5  + – vL 5  iL iC