slove 1, 2, and 3 CONSIDER THE FOLLOWING CAPACITOR CHARGING AND DISCHARGING TRAN

Question

slove 1, 2, and 3

CONSIDER THE FOLLOWING CAPACITOR CHARGING AND DISCHARGING TRANSIENT CIRCUIT DIAGRAM AND ANSWER THESE QUESTIONS IN THE GIVEN ORDER FIND Ii, VCAP(t) AND VCAP(.2) AND PLOT VCAP(T) DURING 0 t .2[s] FIND VCAP(t), VCAP(.6) AND PLOT VCAP(t) DURING .2 T .6 [s] FIND I2, VCAP(t) AND VCAP(1.0) AND PLOT VCAP(t) DURING .6 T 1.0[s] FIND iCAP(t), iCAP(.2) AND PLOT iCAP(T) DURING 0 t .2[s] FIND iCAP(t) DURING .2 t .6[s] AND PLOT iCAP(t). DURING 0 t .2[s] SW-1 = ON, SW-2 = OFF DURING .2 t .6[s] SW-1 = OFF, SW-2 = OFF DURING .6 t 1.0[s] SW-1 = OFF, SW-2 = ON

Explanation / Answer

for RC network Vc = V e^-(t/tou) where tou= Req*C 1)when sw-1 is on and sw-2 is off R2 is opened so i) tou = R1*C =5*10^3*20*10^-6 =100msec =0.1 sec V =30v ii) Vc = 30 e^-(t/0.1) = 30 e^-10t iii) Vc(0.2) = 30 e^-10*0.2 =30e^-2 =4.06V gragh is exponentially decaying from 30 v to 4.06v as time is varying from 0 to 0.2 2) when sw-1 is opened and sw-2 isopened after t=0.2 now voltage source is opened and R2 also opened Vc(t) is constant Vc(t) = 4.06 V graph is constant with 4.06V as time is varying from 0.2 to 0.6 3) when sw-1 is opened and sw-2 isopened after t=0.6 voltage source is opened and R2 is closed now capacitor acts as voltage source with 4.06v for R2 tou = R2*C =10*10^3*20*10^-6 =200msec =0.2sec Vc(t) = -Ve^-t/tou = -4.06*e^-t/0.2 = -4.06 e^-5t negative sign for satisfying kvl Vc(1) = -4.06*e^-5 = -0.0273V graph is exponentially increasing from -4.06 to -0.2 from 0.6 to 1

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