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Kirchhoff's Junction Law

In the circui...

Question

In the circuit, $C_{1} = 2 C_{2}$ . 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

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\frac{2 V}{R}

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\infty

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\frac{V}{2 R}

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Solution

The correct option is D $\frac{V}{2 R}$
Uncharged capacitor behaves as a short circuit just after closing the switch.
But, just after closing the switch a charged capacitor behaves as a battery of e.m.f $E$ which is equal to the potential of the capacitor.

$\Rightarrow V = \frac{q_{0}}{C_{1}}$

So, the current flowing through the circuit,

$I = \frac{q_{0}}{C_{1} (R + R)} = \frac{q_{0}}{2 R C_{1}} = \frac{V}{2 R}$

Hence, option $(d)$ is correct.

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Kirchhoff's Junction Law

Standard XII Physics

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In the circuit, 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

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