EECE 210 Homework 13 with Solution, Fall 2013

Homework 13 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) vO; (b) iO. Ans. (a) V, t is in s; (b) A. Solution: (a) Just before the switch is moved, IL0 = -8 A. Just after the switch is moved iL stays the same and VL0 = 24 + 28 = 40 V. VLS = 0, and the time constant is 0.2/2 = 0.1 s. P11.1.11 The switch in Figure P11.1.11 is moved to position ‘b’ at t = 0, after being in position ‘a’ for a long time. Determine vC for t ≥ 0. Solution: Just before the switch is moved, VC0 = -30 V; VCS = -20 V. The time constant is 50103210-6 = 0.1 s. Hence, V, t is in s. 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 1-13 0.2 H + – 10  2  + – vO iO Figure P11.1.5 24 V t = 0 8 A a b + – Figure P11.1.11 20 V 20 k + – 10 V a t = 0 b 30 k 5 k 2 F – + vC + – vL Figure P11.1.14 2  5  t = 0 1 A iL 4 A 2 H a b  0, (a) vL; (b) iL. Solution: (a) Just before the switch is moved, IL0= 4 A. Just after the switch is moved, iL remains the same and the 1 A source current flows through the 5  resistance, so VL0 = 5 V; VLS = 0, and the time constant is 2/5 = 0.4 s. Hence, vL = 0 + (5 – 0)= V. (b) IL0= 4 A and ILS = 5 A. Hence, A. P11.1.16 In Figure P11.1.16, is infinite for , while the capacitor is charging. When becomes zero, instantly discharging the capacitor, and immediately becomes infinite again, so that the capacitor starts charging, and the cycle is repeated. Determine the frequency of oscillation. Solution: Let t  0 be the time when the capacitor is fully discharged and R becomes infinite. For t  0, the capacitor charges toward 6 V with  = 100.1 = 1 ms. Hence, vc  6(1 – e-t) V, where t is in ms and vc  3 V. When vc  3 V, e-t  0.5, or t  ln2 ms; hence, f  = 1.44 kHz. P11.1.19 The switch in Figure P11.1.19, is moved to position ‘b’ after being in position ‘a’ for a long time. Determine for Solution: After the switch has been in position ‘a’ for a long 2-13 10 k 0.1 F + – vC R 6 V Figure P11.1.16 + – + – vO 3  24 V 5  10  2 H 12 V – + t = 0 i 2i a b Figure P11.1.19 + – + – + – VO 5  10  12 V – + I 2I + – IL time, the circuit will be as shown; A. From KVL, VO0 = 12 +2I = 16.8 V. Just after the switch is moved, IL and hence VO stays the same. Hence this value of VO is the initial value, as assumed. After the switch has been in position ‘b’ for a long time, the circuit will be as shown. V = 5I, so that IS = , and IL = . Moreover, V = 5I = 10IL – 2I. Solving for I , I = A, IL = A, and VOS = V. To determine the effective resistance seen by L, L is replaced by a test source VT, with the 24 V source set to zero. The circuit becomes as shown. , From KVL: VT = , which gives , so that s. It follows that vO = V, t  0 s. P11.1.24 Both switches in Figure P11.1.24 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. 3-13 5  10  – + 2I VT IT I 3  + – 12 k 60 V 6 k 16 V t = 1 s S2 t = 0 S1 24 k 100 F + – vC Figure P11.1.24 + – + – 6 k 24 k + – VCS mA 5 mA 12 k 3 8 P11.2.1 The switch in Figure P13.1.11 is moved to position ‘b’ at t = 0 after being in position ‘a’ for a long time. Determine ix, t  0. Solution: After the switch has been in position ‘a’ for a long time, the currents through the capacitors are zero, so that the voltage across the capacitors is, from voltage division, 4 V. After the switch is moved to position ‘b’, the circuit becomes effectively as shown. IC0 = mA; ICS = 0;  = 0.3(6||12) = 1.2 ms. Since , it follows that mA, where t is in ms. P11.2.5 Both switches are opened at t = 0 after being closed for a long time. Determine iO for t  0. Solution: Just before t = 0, the capacitor currents are zero, and the current through the 2 k and 18 k resistors is 258/20 = 10 mA. Just after t = 0, the capacitor voltages, and hence, iO, remain the same. The final value of iO is zero. The two capacitances inn series are 300150/450 = 100 nF. The time constant is 210310010-9 = 210-4 s  0.2 ms. It follows that mA, where t is in ms. 4-13 12 V 6 k 0.1 F 0.2 F 12 k ix t = 0 a b Figure P11.2.1 + – 12 k ix 6 k 0.3 F iC t = 0 t = 0 18  8 k Figure P11.2.5 2 k 25 mA 300 nF 150 nF iO