Question

1. Find v(t) fort> 0 in the circuit in below: t=0 222 w 622 w + 10V 2 F +1 50 V Assume the switch has been open for a long time and is closed at t=0.

Answer 1

(1) clear 22 WM +) lov → At time t= 0-: just before the switch t is Closed our creuil looks like in fig 1. As switch is open long time, so od t=0-, the Crycuit reaches il steady shale → And Capacitar Ć' when DC voldoge applied becomes open circuit as shown in Jig 1. from dig 1, His very that no current flow in Grcuit V.CO- and all the voltage appears al Capacitor Fig1 At time t=0- Je Velo-) = 10 V de - When too When switch is closed I lov Velt) at t=0, circuit looks luke of Higa. figa t>O Now, Voltage across Capacitor doesn't change abruktly, so Uc (0+) - Velo-) from fig 0 Vcl0+) 1 6.2 t 50V + lov 2 -> Now al t=0 (infinile), the urcuit again goes to cs Steady oketen error and capacitor becomes open Grcuit as Camb canner

shown in Hag 3. de 6-2 2 M w from dig 3 1) lov YA H Velo) 5ov V = V 100) — 3 To Calculate vi fiq3 At t=0 we will apkey te Nodal Analysis at Node A. Apkey Nodal Analysis al A V+50 GV-60 +2V +100 V-10 =0 + =0 le 6 12 Multiplying both side by 12 and adding 8V + 40 = 0 8 V= - 40 => V= -4018 V = -5 from ean 3 260) = -5V 4 ņ the entry Now Dlandard equahon to calculade voltage aeross Capacitor is given by Velt) = [vélt) – Ve boy] e- t/kep ce + Velo) -6 We know all of eq? 6 except Req. Now since both resisters 21 and 6.2 Ghe paralle with respect to Capacitor , we will use equahon Rs + R2 Ri=92 R2 = 6.2 Req = R1xRz where ch Scanned with CamScanner

Now 12 66 Q 6+2 1) 8 Req= Req = 3/4 2 C - Putting / f giren all values inlo eg" Velt) = [ 10-1-5)] e -1/2 x 3 - 5 - 2t 15 e 5 = 5 Common taking Velt) 5(3ed4-1) :v when t>o VECA) is nothing but VIA Vlt) 513€ 27.1) v When to Am .? 11 Scanned with CamScanner