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A series combination of $n_1$ capacitors, each of value $\mathrm{C}_1$, is charged by a source of potential difference $4 \mathrm{~V}$. When another parallel combination of $n_2$ capacitors, each of value $C_2$, is charged by a source of potential difference $V$, it has the same (total) energy stored in it, as the first combination has. The value of $\mathrm{C}_2$, in terms of $\mathrm{C}_1$, is then
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The correct answer is:
$\frac{16 C_1}{n_1 n_2}$
P.D. across each capacitor is \(V_1=\frac{4 \mathrm{~V}}{\mathrm{n}_1}\)
Energy stored across the combination is \(\mathrm{U}_1=\mathrm{n}_1 \times \frac{1}{2} \mathrm{C}_1 \frac{16 \mathrm{~V}^2}{\mathrm{n}_1^2}=\frac{16 \mathrm{C}_1 \mathrm{~V}^2}{2 \mathrm{n}_1}\)
\(\begin{aligned}
& \text {Energy stored across parallel combination is } \mathrm{U}_2=\mathrm{n}_2 \times \frac{1}{2} \mathrm{C}_2 \mathrm{~V}^2\\
& \frac{16 C_1 v^2}{2 n_1}=\frac{n_2 C_1 v^2}{2} \\
& \Rightarrow C_2=\frac{16 C_1}{n_1 n_2}
\end{aligned}\)
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