A capacitor $C_{1}$ is charged to a potential difference $V$ . The charging battery is then removed and the capacitor is connected to an uncharged capacitor $C_{2}$ . The potential difference across the combination is

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$V (1 + \frac{C_{1}}{C_{2}})$

$V (1 + \frac{C_{2}}{C_{1}})$

$\frac{{VC}_{1}}{(C_{1} + C_{2})}$

$\frac{V C_{2}}{(C_{1} + C_{2})}$

answer is A.

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Detailed Solution

Charge $Q = C_{1} V$

Total capacity of combination (parallel)

$\begin{array}{l} C = C_{1} + C_{2} \\ P. D. = \frac{Q}{C} = \frac{C_{1} V}{C_{1} + C_{2}} \end{array}$

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