Chapter 8, Solution 1. (a) At t = 0-, the circuit has reached steady state so that the equivalent circuit is shown in Figure (a). 6Ω VS + − 6Ω + 6Ω + vL 10 H (b) v − 10 µF − (a) 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,. | Chapter 8 Solution 1. a At t 0- the circuit has reached steady state so that the equivalent circuit is shown in Figure a . Vs a 6 Q v À 10 pF vl i 10 H 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 -5 V s c As t approaches infinity the circuit reaches steady state. i 0 A v ro 0 V Chapter 8 Solution 2. a At t 0- the equivalent circuit is shown in Figure a . 25 kQ 20 kQ ---Wr ----VA I -iR-------- iL 80V L_î b 60 20 15 kohms Îr 0- 80 25 15 2mA. By the current division principle Îl 0- 60 2mA 60 20 mA vc 0- 0 At t 0 vc 0 vc 0- 0 Îl 0 Îl 0- mA 80 Îr 0 25 20 vc 0- Îr 0 80 45k mA But iR iC iL ic 0 or ic 0 mA b vl 0 vc 0 0 But vl LdiL dt and diL O dt vl 0 L 0 diL O dt 0 Again 8O 45iR vc O 45diR dt dvc dt But dvc O dt ic O C mohms 1 pF 278 V s Hence diR O dt -1 45 dvc O dt -278 45 diR O dt A s Also iR ic iL diR O dt dic O dt diL O dt dic O dt O or dic O dt A s c As t approaches infinity we have the equivalent circuit in Figure b . Ir Il 8O 45k mA ic z cdv dt 0. Chapter 8 Solution 3. At t O- u t O. consider the circuit shown in Figure a . iL O- O and vr 0- O. But -vr 0- vc O- 10 O or vc O- -1OV. a At t O since the inductor current and capacitor voltage cannot change abruptly the inductor current must still be equal to 0A the capacitor has a voltage equal to -10V. Since it is in series with the 1OV source together they represent a direct short at t O . This means that the entire 2A from the current source flows through the capacitor and not the resistor. Therefore vr 0 0 V. b At t O vl 0 O therefore LdiL O dt vl 0 O thus diL dt 0A s ic O 2 A this means that dvc O dt 2 c 8 V s. Now for the value of dvR O dt. Since vr vc 1O then dvR O dt dvc O dt O
