# Discharging a Capacitor

Let us connect a charged capacitor of capacitance C farad in series with a resistor of resistance R ohms. We then short circuit this series combination through switching on the push switch as shown. As soon as the capacitor is short circuited, it starts discharging.

Let us assume, the voltage of the capacitor at fully charged condition is V volt. As soon as the capacitor is short circuited, the discharging current of the circuit would be, - V / R ampere. But after the instant of switching on that is at t = ^{+}0 the current through the circuit is
As per Kirchhoff’s’ Voltage Law, we get,
Integrating both sides, we get,
Where, A is the constant of integration and, at t = 0, v = V,
After, calculating value of A, we get,
We know form, KVL of the circuit,
If we plot these discharging current and voltage in graph, we get,
Hence the capacitor current exponentially reaches to zero from its initial value and the capacitor voltage reaches exponentially to zero from its initial value during discharging.

S | SHAH RA commented on 05/05/2018Very informative |