MHT CET · Physics · Capacitance
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
is
- A \(\frac{1}{4}\)
- B \(\frac{1}{6}\)
- C \(\frac{1}{3}\)
- D \(\frac{1}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
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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