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A $10 \mu \mathrm{F}$ capacitor is charged to $10 \mathrm{~V}$ and disconnected from the battery. If another uncharged $10 \mu \mathrm{F}$ capacitor is connected across it in parallel, then the voltage across the combination will be
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Verified Answer
The correct answer is:
$5 \mathrm{~V}$
Common potential difference for a parallel combination of capacitors, is
$V_{C}=\frac{V_{1} C_{1}+V_{2} C_{2}}{C_{1}+C_{2}}$
Given,
$$
\begin{aligned}
V_{2} &=0, V_{1}=10 \mathrm{~V} \\
C_{1} &=C_{2}=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}
\end{aligned}
$$
$\begin{aligned} \therefore \quad V_{C} &=\frac{10 \times 10 \times 10^{-6}}{(10+10) \times 10^{-6}} \\ &=\frac{100}{20}=5 \mathrm{~V} \end{aligned}$
$V_{C}=\frac{V_{1} C_{1}+V_{2} C_{2}}{C_{1}+C_{2}}$
Given,
$$
\begin{aligned}
V_{2} &=0, V_{1}=10 \mathrm{~V} \\
C_{1} &=C_{2}=10 \mu \mathrm{F}=10 \times 10^{-6} \mathrm{~F}
\end{aligned}
$$
$\begin{aligned} \therefore \quad V_{C} &=\frac{10 \times 10 \times 10^{-6}}{(10+10) \times 10^{-6}} \\ &=\frac{100}{20}=5 \mathrm{~V} \end{aligned}$
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