EECE 210 Homework 14 with Solution, Fall 2014

Homework 14 P12.1.9 The switch in Figure P12.1.9 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: At 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. At t = 0+, 2iC + IL0 = 0, so that iC = - IL0/2 = -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. P12.2.12 The switch in Figure P12.2.12, is moved to position ‘b’ at t = 0 after 0.5 F 3 k 0.1 H 2 k iC + – vC 3 mA iC iL ix 6 k 1 k t = 0 Figure P12.1.9 500  t = 0 + – vC 100 A 10 nF R 2.5 mH iL a b + – vL 500  + – vO 125  t = 0 a b 4 H 100  1 F 0.1 A Figure P12.2.18 being in position a for a long time. (a) Choose R so that the response is critically damped; (b) determine the initial values vC(0+), iL(0+), and vL(0+); (c) determine vC(t)and iL(t) for t  0+. Solution: (a) For critical damping, , where Rs = R + 500 ; . Hence, R  500 . (b) When the switch is in position a, iL is 100 A and vC is (100 A)(500 ) = 50 mV. When the switch is moved to position b, iL and vC at t = 0+ remain the same. From KVL: 50 = 0.11000 + vL, so that vL = -50 mV. As , the energy initially stored in the circuit is dissipated and the current and all voltages become zero. (c) Since the circuit is critically damped, ; at t = 0+, vC(0+) = 50 mV; hence, A = 50 mV; iL = -CdvC/dt at t = 0+ = 10-8(-0A + B), or -10-4 = 10-8(-0A + B), or -104 = -21055010-3 + B, or -104 = -104 + B, so that B = 0. It follows that vC(t) = = mV. CdvC/dt: nFmV/s = 10-910-3/10-6 = 10-6  A; iL(t) = -10(-50)0.2 = A. P12.2.18 The switch In Figure P12.2.18, is moved to position ‘b’ at after being in position ‘a’ for a long time. Determine (t) for t  0+. Solution: iL(0-) = = 10 mA downwards, and vO(0-) 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, C iL R R R R I1 10 A t = 0 iC L 2I1 Figure P12.2.21 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. P12.2.21 The switch in Figure P12.2.21 Is closed at t = 0 after being opened for a long time. Determine: (a) iL(0-), iC(0-), iL(0+), iC(0+); (b) iL(t) for t  0+, assuming R = 1 , L = 1 H, and C = 1 F. Solution: (a) At t = 0-, the capacitor behaves as an open circuit and the inductor as a short circuit; vC(0-) = 0, where vC is a voltage drop in the direction of iC. The current through R next to the switch is I1 directed upwards. It follows from KVL around the mesh of three R’s, that the voltage across the source is zero, so that I1 = 10 A and IL(0-) = 2I1 = 20 A. At t = 0+, I1 = 5 A, and 2I1 = 10 A. The current through the inductor and the voltage across the capacitor do not change. It follows that iC = -10 A. (b) 0 = 1 rad/s, and  = = 1/2RC = 0.5 rad/s. . As t  , iL() = 10 A. Hence, ; At t = 0+, 20 = A + 10, so that A = 10. At t = 0+, vC = LdiL/dt = L(-A + dB) = 0; B = A/d . Hence, A. P12.3.12 The switch in Figure P12.3.12 is opened at t = 0 after being closed for a long time. Determine iL(t) and iC(t) for t  0+. Solution: At t = 0-, the capacitor acts as an open circuit and the inductor as a short circuit, It follows that iC(0-) = 0, vC(0-) = 0, iL(0-) = 2 + 10/5 = 4 A. At t = 0+, vC and iL do not change; hence, iC(0+) = -2 A. As t  , iC = 0 and iL = 2 A.  = 1/(2412.510-3) = 10 rad/s, rad/s. The circuit is therefore critically damped. It follows that A; at t = 0+, 4 = A + 2, so that A = 2 A; at t = 0+, vC = LdiL/dt = -0A + B = 0, so that B = 0A = 20 A/s. It follows that: A. vC(t) = LdiL/dt = V; iC(t) = CdvC/dt = -12.510-3160 A. iC 10 V iL 12.5 mF 5  + – Figure P12.3.12 t = 0 0.8 H 4  2 A 8  10 H Figure P12.3.16 t = 0 0.4 F 4  30 A iL + – vC t = 0 a b 5 A P12.3.16 Both switches in Figure P12.3.16 are moved at t = 0 after being in their initial positions for a long time. Determine vC(t) and iL(t) for t  0+. Solution: At t = 0-, the 30 A source establishes initial conditions in the circuit. From current division, iL(0-) = 10 A and the current through the 4  resistor is 20 A, so that vC(0-) = 80 V. These do not change at t = 0+. As t  , the capacitor acts as an open circuit and the inductor as a short circuit, so that iL() = 0. The 5 A current flows through the 8  resistor, making vC() = 40 V. When the 5 A source is set to zero, the circuit is a series circuit;  = R/2L = 8/20 = 0.4 rad/s; rad/s; the responses are underdamped, and rad/s. V; At t = 0+, 80 = A + 40, so that A = 40 V; at t = 0+, iL = -CdvC/dt = -C(-A + dB) = 10; dB = -25 + 0.440 = -9, B = -9/0.3 = -30. Hence, . iL(t) = -CdvC/dt A. P12.3.17 The switch in Figure P12.3.17 is moved at t = 0 from position ‘a’ to position ‘b’ after being in position ‘a’ for a long time. Determine vC(t) and iL(t) for t  0+. Solution: At t = 0-, the inductor acts as a short circuit and the capacitor as open circuit. iL(0-) = 6 A; The current in the upper 10  resistor is 2 A, and the voltage across it is vC(0-) = 20 V. When the switch is moved, the source current is reduced to 3 A, so that iL() = 3 A and vC() = 10 V. When the current source is set to zero, circuit reduced to a series RLC circuit having R = 20/3 . Hence,  = R/2L = 5 rad/s and rad/s. The responses are overdamped, with rad/s and rad/s. V; vC(0-) = 20 = A + B + 10, which makes A + B = 10. At t = 0+, 3 = iL(0+) + CdvC/dt, so that (1/6) dvC/dt = 1, or, A + 9B = 18; this gives A = 9 and B = 1. It follows that: V, t is in s. A, t is in s. 2/3 H + – vC Figure P12.3.17 iL t = 0 3 A 10  a b 6 A 1/6 F 10  10  P12.3.18 The switch in Figure P12.3.18 is opened at t = 0 after being closed for a long time. Determine vC(t) and iL(t) for t  0+. Solution: At t = 0-, the inductor acts a s a short circuit and the capacitance as an open circuit. The resistance in series with L is (6||12) = 4 . iL(0-) = (30 – 10)/4 = 5 A. From current division, the current in the 6  resistor on the right is 5/3 A, so that vC(0-) = 10 + 6(5/3) = 20 V. As t  , with the capacitor acting as an open circuit, iL() = 0 and vC() = 10 V. The circuit is a series circuit, with R = 4 . Hence,  = R/2L = 4/0.8 = 5 rad/s and rad/s. The responses are critically damped V; vC(0-) = 20 = A + B + 10, which makes A + B = 10. At t = 0+, iL(0+) = - CdvC/dt, so that -(0.1)(-5A + B) = 5, or, 5A + B = 50; this gives A = 10 and B = 0. It follows that: V, , t is in s A, t is in s 30 V iL 0.1 F + – Figure P12.3.18 0.4 H + – vC + – 10 V 6  6  6  t = 0