MHT CET · Physics · Capacitance
The potential difference that must be applied across the series and parallel combination of 4 identical capacitors is such that the energy stored in them becomes the same. The ratio of potential difference in series to parallel combination is
- A \(1: 2\)
- B \(1: 4\)
- C \(4: 1\)
- D \(2: 1\)
Answer & Solution
Correct Answer
(C) \(4: 1\)
Step-by-step Solution
Detailed explanation
The equivalent capacitance for series combination,
\(C_1=\frac{C}{4}...(i)\)
\(\therefore \quad\) Potential energy, \(\mathrm{U}_1=\frac{1}{2} \mathrm{C}_1 \mathrm{~V}_1^2\)...(ii)
The equivalent capacitance for parallel combination,
\(C_2=4 C...(iii)\)
\(\therefore \quad\) Potential energy, \(U_2=\frac{1}{2} C_2 V_2^2\)...(iv)
Given that, \(\mathrm{U}_1=\mathrm{U}_2\)
\(\begin{array}{ll}
\therefore & \mathrm{C}_1 \mathrm{~V}_1^2=\mathrm{C}_2 \mathrm{~V}_2^2 \\
\therefore & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\sqrt{\frac{\mathrm{C}_2}{\mathrm{C}_1}} \\
\therefore & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\sqrt{\frac{4 \mathrm{C}}{\frac{\mathrm{C}}{4}}}=\sqrt{\frac{16}{1}} \\
\therefore & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\frac{4}{1}
\end{array}\)
...(From(ii) and (iv))
...(From (i) and (iii))
\(C_1=\frac{C}{4}...(i)\)
\(\therefore \quad\) Potential energy, \(\mathrm{U}_1=\frac{1}{2} \mathrm{C}_1 \mathrm{~V}_1^2\)...(ii)
The equivalent capacitance for parallel combination,
\(C_2=4 C...(iii)\)
\(\therefore \quad\) Potential energy, \(U_2=\frac{1}{2} C_2 V_2^2\)...(iv)
Given that, \(\mathrm{U}_1=\mathrm{U}_2\)
\(\begin{array}{ll}
\therefore & \mathrm{C}_1 \mathrm{~V}_1^2=\mathrm{C}_2 \mathrm{~V}_2^2 \\
\therefore & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\sqrt{\frac{\mathrm{C}_2}{\mathrm{C}_1}} \\
\therefore & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\sqrt{\frac{4 \mathrm{C}}{\frac{\mathrm{C}}{4}}}=\sqrt{\frac{16}{1}} \\
\therefore & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\frac{4}{1}
\end{array}\)
...(From(ii) and (iv))
...(From (i) and (iii))
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