Question

A 127 pF capacitor in parallel of a 100k resistor is connected to a voltage source such that vc(t)=12V, t<0.
The voltage source is removed when t =0. Calculate the energy stored in the capacitor at t equal to
(a) 0
(b) 200 ms

Answer 1

a.

At t=0, the capacitor will be at steady-state and store

Energy stored  $E=\frac{1}{2}CV^{2}=\frac{1}{2}*127*10^{-12}*12^{2}=9.144nJ$

b. As soon as the voltage source is removed from the capacitor, the capacitor starts discharging through the resistor R in an exponential manner

$V_{C}(t)=12e^{-\frac{t}{RC}}=12e^{\frac{-200*10^{-3}}{127*10^{-12}*100*10^{3}}}=0J$

Energy =$E=\frac{1}{2}CV^{2}=\frac{1}{2}*127*10^{-12}*0^{2}=0J$

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