Question

4. Use the circuit in UMF 5.39 but do the following (see similar example for UMF P5.35). a. Do the problem as stated. b. At t=1.25seconds, the switch is moved back to position 1. Find v(t) for t 1.25s C. Plot (using Matlab), v(t) and it for 0 St<12s *5.39 The switch in the circuit of Fig. P5.39 had been in position 1 for a long time until it was moved to position 2 at t = 0. Determine u(t) for 1 > 0, given that lo = 6 mA, Vo = 18 V, Ri = R2 = 4k2, and C = 200 /F. 11 1. R 29 Figure P5.39: Circuit for Problems 5.39 and 5.40.

Answer 1

step at t o after pushing switch from 2 to 1, capacitor wille be open. Net) Vs + --- oma Q v ok a Dior Vo ke -24 +V100) +18 = 0 V600) = 6v VC 1-25)= 48V V[00) - 6V VLL) =VCO) +[1(0)-1100] et/z vig across Capacitor. V(t)=6+/-18–6] e the vit) = 6-24e +/T] " remove all sources; and see through Capacitor. Voltage source shorted Current source opened for finding I URES ukr We are I= Reg Cep seeing (4+4) kr = 8kr > T = 8k * 200 HF = 1600x16?=106 TV (t) = 6-24 0 / 106 volte | By putting values of t, we got different voltages. For Ilt), I-Cd 200ux & [ 6-24 *+/16] - 200x106 [-24 ethro - 1 200x106 x -24 X-1 ethro 1.

Given circuit VEE 84 V = 18V 6mA = I Q LAM 4kr Soli Capacitor doesoot allow Sudden changes in Voltage. The reason is, cox = dv - Y-VE / it sudden charges occur then o at at tiba Iti-tz 6 = V 2_00) we are getting as voltage for - practically voltage too is not possible. * Hence, no sudden changes in voltage across capacitor. ve (1025 ) = Vc (1-25+)} at steady state t o corso w=0 cos S=0 X c = c = 100 Impedance -00- Hence, at steady state, capacitor) N-IR → V=I (0) acts as open I VEICO) I=O (open). Stero: Initially, switch is at "a". Ovo = 18v capacitor acts as open, as it is in steady state, before switch is changed to t yer +4 (125) * =18V tva (1925) +18=0 T no effect as no current. [vc (1025) = -187) VE (1-253) - Va (1.25+), -18V

Volks I vit)=6-24 e II) = 3 et/16mA)