Q. In the circuit shown in figure-$3.345$ the capacitor of capacitance $C$ is charged to a potential difference $V$. The current in the circuit just after the closing of switch $S$ is :

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A

$\frac{V}{3 R}$

B

$\frac{3 V}{R}$

C

$\frac{V}{2 R}$

D

zero

Solution:

SJust after closing the switch, uncharged capacitor behave like short circuit and charged capacitor behave like an ideal battery so current in circuit is given as
$i=\frac{V}{3 R}$

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