Question

“Before the switch closes”

The circuit is at steady state before the switch opens at time t-. The input to the circuit is the vol tage of the voltage source, 12 V. The output of this circuit is the voltage across the capacitor,v() Determine v() for t>0. Show al work. 10k2 12 V

Answer 1

1) At t=0- i.e before switching action network in steady state of capacitor is replaced by open circuit. N/W is replaced by using voltage division principle vL(0-) = 12 . 10k/10k+10k+10k VL(0-) = 12 times 1/2 = 4v = V(t) therefore VL(0-) = 4v = VL(0+) so voltage across capacitor does not change immediately VL(0+) means voltage across capacitor after switch is closed. At t=alpha+ i.e after long period of time switch closed, the N/W in steady state of capacitor replaced by open circuit. Here 10kohms is parallel to short circuit , so it is neglectedof N/W in can be as voltage division principle V(t) = VL(0-) = 12 times 10k/20k = 6v Rth = 10kparallel10k Rth = 5k Vinitial = Vc(0+) = 4v Vfinal = Vc(infinity) = 6v tau = Rth.C 5k times 1mu = 5milli seconds = 5sec. therefore from above V(t) is equal voltage across capacitor , so the volatage across capacitor for t geq 0 i.e given by Vc(t) = Vfinal + (Vinitial-Vfinal) times e-t/tau Vc(t) = 6+(4-6)e-t/sm = 6+(-2)e-2.0t volts therefore V(t) = Vc(t) = 6-2e-0.2t v.