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A capacitor of capacitance $C_1$ charged up to $V$ volt and then connected to an uncharged capacitor $C_2$. Then, the final potential difference across each will be:
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Verified Answer
The correct answer is:
$\frac{C_1 V}{C_1+C_2}$
$$
\begin{aligned}
& V_{\text {comma }}=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\left(\because V_2=0\right) \\
& \Rightarrow V_{\text {comma }}=\frac{C_1 V_1}{C_1+C_2}
\end{aligned}
$$
\begin{aligned}
& V_{\text {comma }}=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}\left(\because V_2=0\right) \\
& \Rightarrow V_{\text {comma }}=\frac{C_1 V_1}{C_1+C_2}
\end{aligned}
$$
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