Question

In the circuit shown in Figure, $C_1=2C_2$. Initially, capacitor $C_1$, is charged to a potential of $V$. The current in the circuit just after the switch $S$ is closed is

• A. $0$
• B. $2V/R$
• C. $\infty$
• D. $V/2R$

Answer 1

The correct option is D $V/2R$

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, the current in the circuit just after closing switch is $I=\frac{V}{R+R}=\frac{V}{2R}$