A capacitor of capacitance $C_{o}$ is charged to a potential $V_{o}$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S$ , the common potential across the two capacitors becomes $V$ . The capacitance $C$ is given by:

\frac{C_{0} (V_{0} - V)}{V_{0}}

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\frac{C_{0} ({V - V}_{0})}{V_{0}}

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\frac{C_{0} ({V + V}_{0})}{V_{0}}

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\frac{C_{0} (V_{0} - V)}{V}

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Solution

The correct option is D $\frac{C_{0} (V_{0} - V)}{V}$
When the switch $S$ is closed, the common potential across the two capacitors becomes $V$ , so charge on isolated plates remains same.
$∴ C_{0} V_{0} = (C_{0} + C) V$
$\Rightarrow C V = C_{0} V_{0} - C_{0} V$
$\Rightarrow C = \frac{C_{0} (V_{0} - V)}{V}$

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