EECE 210 Homework 12 with Solution, Fall 2014

+ – 30 V 8 k ic + – 75 V 160 k 40 k Va Homework12 P11.1.5 The switch in Figure P11.1.5 is moved to position ‘b’ at t = 0 after being in position ‘a’ for a long time. Determine, for t 0+: (a) vC(t); (b) vO(t); (c) iO(t); (d) the total energy dissipated in the 60 k resistor as t. Solution: (a) At t = 0-, vC(0-) = 100 V; the resistance seen by C when the switch is in position ‘b’ is 32 + 240||60 = 32 + 48 = 80 k; C =801030.510-6 = 4010-3 s; 1/C = 1000/40 = 25/s. As t, vC 0. Hence, V, t is in s. (b) From voltage division, vO = 48vC/80; hence, V. (c) iO = vO/60 mA, so that mA. The energy dissipated in the 60 k resistor is W = = mJ = mJ. P11.1.14 The switch in Figure P11.1.14 is moved to position ‘b’ at t = 0 after being in position ‘a’ for a long time. Determine, for t 0+: (a) vC(t) (b) iC(t) from initial and final values as well as from the v-i relation for the capacitor. Solution: (a) vC(0-) = 4060/80 = 30 V; after the switch is moved to position ‘b’, vC() = -75160/200 = -60 V; the resistance seen by C is: 8 + 160||40 = 8 + 32 = 40 k; C = 4010-30.2510-6 = 1010-3; 1/C = 100/s; vC(t) = -60 + (30 + 60) = -60 + 90 V, t is in s. (b) The circuit at t = 0+ is a two-essential- node circuit. From KCL, = 0; this gives Va = 12 V, iC(0+) = (12 – 30)/8 = -2.25 mA. iC() = 0, so that mA. From thev-i relation for the capacitor mA. P11.1.17 The capacitor in Figure P11.1.17, was initially uncharged and the switch was 1-11 60 k + – 100 V 0.5 F 240 k 32 k 10 k a b + – + – vC vO iO Figure P11.1.5 t = 0 60 k + – 40 V 0.25 F 8 k 20 k a b + – vC iC Figure P11.1.14 + – 75 V 160 k 40 k t = 0 + 1 k 1 F vC 12 V 6 V a b c iC – Figure P11.1.17 + – in position ‘a’. At the switch is moved to position ‘b’. Determine for t ≥ 0+: (a) vC(t); (b) iC(t); (c) the energy delivered by the 12 V battery as t (d) the energy absorbed by the 6 V battery as t; (e) the energy dissipated in the resistor as t; verify this by integrating the power dissipated by iC from t = 0+ to t; (f) If after a long time, t′ = 0, the switch is moved to position c, determinevC(t) andiC(t) for t′ 0+, the energy delivered by the 6 V battery, the energy gained or lost by the capacitor, and the energy dissipated in the resistor. Solution: (a) vC(0-) = 0;. vC() = 6 V.  = 11 = 1 ms. Hence, vC = V, tis in ms; (b) iC(0+) = (12 – 6)/1 = 6 mA; iC() = 0; mA. (c) Energy delivered by the 12 V battery is = =  72 J as t. (d) Energy absorbed by the 6V battery is =36J as t. (e) As t, net energy delivered by the two batteries is 36 J. Energy stored in capacitor is (1/2)10-636 = 18 J.Energy dissipated in the resistor = 36 – 18 = 18 J= 18 J. (f) For t 0, initial value of vC is 6 V, final value is -6 V,  being the same. Hence, V; initial value of iC is -12/1 mA, final value is zero;. As , energy delivered by the battery is = 72 J; net energy lost by the capacitor is 18 – 18 = 0, energy dissipated in the resistor is J. P11.1.18 Both switches in Figure P11.1.18 have been open for a long time. is closed at and at Determine for Solution: The initial value of vC is zero. When S1 is closed, the final value of vC is V, and s. Hence, vC = V, 0 t 1 s. At t = 1 s, vC(1) = -11.21 V. When S2 is closed at t = 1s, the initial value of vC is vC(1). To find the final value, the voltage sources are transformed to current sources, as shown. The combined current source is mA, and the combined resistance is (12||6||24) = k. Hence, VCS = V. s. This gives V, t 1 s. 2-11 6 k 24 k + – VCS mA 5 mA 12 k 3 8 12 k 60 V 6 k 16 V t = 1 s S2 t = 0 S1 24 k 100 F + – vC Figure P11.1.18 + – + – 10 mA 30 k 5 k 20 nF 10 k 2 k a S1 b + – vC S2 Figure P11.1.19 t = 0 t = 50 s P11.1.19 Switch S1 in Figure P11.1.19 has been in position a for a long time, with switch S2 closed. At t = 0, S1 is moved to position b and remains in this position. S2 is opened at μs and closed again at t = 100 μs. Determine vC for: (a) 0 ≤ t ≤ 50 μs; (b) 50 ≤ t ≤ 100 μs; (c) t ≥ 100 μs. Solution: (a) With S1 in position ‘a’ and S2 closed, it follows from current division that the current in the 10 k resistor is 1030/42 = 50/7 mA; vC(0-) = 500/7 V. : The circuit is as shown; vC() = 0;  = 20(10||7) = s. Hence, vC(t) = V,t is in s. At t = 50 s, vC(50) = 38.9 V. (b) : The circuit becomes as shown. Initial value of vC is 38.9 V, and the final value is zero, = 207 = 140 s. Hence, vC(t) = V,t is in s. At t = 100 s, vC(100) = 27.2 V. (c) : The circuit becomes as in (a). The initial value of vC is vx(100), the final value is zero, and  = s. Hence, VC(t) = V, t is in s. 3-11 5 k 20 nF 2 k + – vx 5 k 20 nF 10 k 2 k + – vC S2 t 50