Discharging a Capacitoron 24/2/2012 & Updated on 15/8/2018
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.
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