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A capacitor of capacity $2 \mu \mathrm{F}$ is charged upto a potential $14 \mathrm{~V}$ and then connected in parallel to an uncharged capacitor of capacity $5 \mu \mathrm{F}$. The final potential difference across each capacitor will be
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The correct answer is:
$4 \mathrm{~V}$
When two capacitors are connected then their common potential difference is
$\begin{aligned}
V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2} \\
\text { Given, } C_1 & =2 \mu \mathrm{F}, V_1=14 \mathrm{~V}, C_2=5 \mu \mathrm{F}, V_2=0 \mathrm{~V} \\
\therefore \quad V=\frac{2 \times 14+5 \times 0}{2+5}=4 \mathrm{~V}
\end{aligned}$
$\begin{aligned}
V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2} \\
\text { Given, } C_1 & =2 \mu \mathrm{F}, V_1=14 \mathrm{~V}, C_2=5 \mu \mathrm{F}, V_2=0 \mathrm{~V} \\
\therefore \quad V=\frac{2 \times 14+5 \times 0}{2+5}=4 \mathrm{~V}
\end{aligned}$
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