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  4. In the circuit shown in figure, C1=2 C2. Initially, capacitor C1 is charged to a potential of V. The current in the circuit just after the switch S is closed is

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Q. In the circuit shown in figure, $C_{1}=2 C_{2}$. Initially, capacitor $C_{1}$ is charged to a potential of $V$. The current in the circuit just after the switch $S$ is closed isPhysics Question Image

Current Electricity

Solution:

Uncharged capacitor behaves as a short circuit just after closing the switch. But charged capacitor behaves as the battery of emf $V$ just after closing the switch. Therefore,
$I=\frac{q_{0}}{C_{1}(2 R)}=\frac{q_{0}}{2 R C_{1}}=\frac{V}{2 R}$