# Charging a Capacitor

on 24/2/2012 & Updated on 15/8/2018Let us assume the capacitor is fully uncharged. When we push the switch, the capacitor is uncharged, hence no voltage developed across the capacitor, thus the capacitor will behave as short circuit. At that moment the charge just starts accumulation in the capacitor. The current through the circuit will only be limited by resistance R.

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That initial current is V/R. Now slowly the voltage is being developed across the capacitor, and this developed voltage is in opposite of the polarity of the battery. As a result the current in the circuit gradually decreased. It will be gradually decreased to zero, when the voltage across the capacitor becomes equal and opposite of the voltage of battery. The voltage gradually increases across the capacitor during charging. Let us consider the rate of increase of voltage across the capacitor is dv/dt at any instant t. The current through the capacitor at that instant is

Applying, Kirchhoff’s Voltage Law, in the circuit at that instant, we can write,

Integrating both side we get, Now, at the time of switching on the circuit, voltage across the capacitor was zero. That means, v = 0 at t = 0. Putting these values in above equation, we get After getting the value of A, we can rewrite the above equation as, Now, we know that, This is the expression of charging current I, during process of charging. The current and voltage of the capacitor during charging is shown below.

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