If the switch in the following circuit opens at t = 0, find v(t) for t≥ 0 and the initial energy stored in the capacitor.
The initial energy stored in the Capacitor,
W=(1/2)CVo2
= 0.5×0.167×82
= 5.333J (Ans)
For t<0, the switch is closed.
Capacitor acts as open to DC.
Voltage across the capacitor,
v(0-)= {(12||4)×V}÷{(12||4)+6} = 8VFor t=0, the switch is opened,Voltage across the Capacitor cannot change instantaneously,So, v(0-)= v(0) = 8VAt t>0, The capacitor is discharging.Rth = (12||4) = 3 ohmC= 1/6 FTime constant, Π = Rth×C = 0.5 So, v(t) = v(0)e-t/Π = 8e-t/0.5 =8e-2t V (Ans.)
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