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

Question

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 (A) 0 (B) 2V/R (C) ∞ (D) V/2 R

Tardigrade Admin · Accepted Answer

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}

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 is

$0$

$2 V / R$

$\infty$

$V /2 R$

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,
$I = \frac{q _{0}}{C _{1} ( 2 R )} = \frac{q _{0}}{2 R C _{1}} = \frac{V}{2 R}$

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

0

2V/R

∞

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, I=C1​(2R)q0​​=2RC1​q0​​=2RV​

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 is

$0$

$2 V / R$

$\infty$

$V /2 R$

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,
$I = \frac{q _{0}}{C _{1} ( 2 R )} = \frac{q _{0}}{2 R C _{1}} = \frac{V}{2 R}$