Question

5KR MM 10K2 V (4) 40nF 340k RA lokh 75V + 100V 1+0 -> 1.(t) + Ca for The switch in the above circut has been at long while. When't=o, it moves to position b'instantaneously. Determine (Fortz ot) ① Volt) @ io (+)

Answer 1

This is a problem of Electric Circuits whose solution has been provided below.

5kr a (oun T=0 7.lt + (ок 40kh loov Vo(t) – 4onf 75V Solution At t-o (Just before Switching reached Capacitor has Open arcuit. steedy state & hence acts as Skr W a Using valtape dinder sale, across lok is, Volo- 75 voltage V10 = 75% lok LOK+SK & V10 = 5ov = 75xlo 15 V. (6-) = 50v At t-ot (Just after switching) Capacitor opposes in energy hence ads shost circuit in series with, sudden change through it as Voo-) = 5ov

5KR W b Tokn To Cot) 50v JokR & •MOKR I loov 750(A Here 7 Volot) = Volo-) = 50v & i. (ot) can be obtained hode voltage at b sov. i Cot) = 50-100 =-5mA 50 lok LOK i Cot) = -5mA At t=00 Clong time after switching b LOKR Wo Capacitor has again reached steedy state & hence acts as open circuit Using potential divider rule, voltage across 4akriss чокл Vuo = 100x 40k 40k+lok Vuo = toox 40 = dov 50 IV. (A) = Por 10V -2MA io (0) = -2mA - 100 50k

RC time constant of the Now, we need to find the Circuit to 2= Reg Cep. circuit passive by shoot circuiting all the voltage sources open circuiting To obtain make the current sources. b lokh b >> w Gonf. Req Gonf 40KR = lokxЧок Tok+40k = 400k 5€ =8K Sec .. 7 = PK x 40 h = = 8xlo 3 x 40x/0-7 sec 320x10-6 sec. We know that for an the general RC circuit, expression of voltage (07) current is - f(t) = f (00) + (f (0) - fonete -+/2 Vo (t) = V. (①) + +[V:(04) - V. (]e V. (t) = Po + [50 - do]e -1/320x10-6 Volt) 80 Ans => -3125+ 30e V

-t/ Similarly, 1.(+) = 1.c.) + [5.664) [io (ot) – 1. (x)] => 1.(+) = -2 +[-5-(-2)] e - 31257 MA => 9.00 - 2-3e-3125+ mA An