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The potential differences that must be applied across the parallel and series combination of 3 identical capacitors is such that the energy stored in them becomes the same. The ratio of potential difference in parallel to series combination
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The correct answer is:
$\frac{1}{3}$
Let $C$ be the capacity of each capacitor. The equivalent capacitance of three capacitors in parallel combination will be $C_{p}$ $=3 C$ and in series combination $C_{S}=\frac{C}{3}$
Let $V_{p}$ be the potential difference in parallel combination and $V_{S}$ in series combination then the energy stored in the same in the two cases.
$$
\begin{array}{l}
\therefore \frac{1}{2} C_{p} V_{p}^{2}=\frac{1}{2} C_{s} V_{s}^{2} \\
\therefore \frac{V_{p}^{2}}{V_{s}^{2}}=\frac{C_{s}}{C_{p}}=\frac{C}{3} \cdot \frac{1}{3 C}=\frac{1}{9} \\
\therefore \frac{V_{p}}{V_{s}}=\frac{1}{3}
\end{array}
$$
Let $V_{p}$ be the potential difference in parallel combination and $V_{S}$ in series combination then the energy stored in the same in the two cases.
$$
\begin{array}{l}
\therefore \frac{1}{2} C_{p} V_{p}^{2}=\frac{1}{2} C_{s} V_{s}^{2} \\
\therefore \frac{V_{p}^{2}}{V_{s}^{2}}=\frac{C_{s}}{C_{p}}=\frac{C}{3} \cdot \frac{1}{3 C}=\frac{1}{9} \\
\therefore \frac{V_{p}}{V_{s}}=\frac{1}{3}
\end{array}
$$
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