Question:
A \(5\ \mu\mathrm{F}\) capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged \(2.5\ \mu\mathrm{F}\) capacitor. If the energy change during the charge redistribution is \(\frac{X}{100}\) J then value of \(X\) to the nearest integer is ______.
A \(5\ \mu\mathrm{F}\) capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged \(2.5\ \mu\mathrm{F}\) capacitor. If the energy change during the charge redistribution is \(\frac{X}{100}\) J then value of \(X\) to the nearest integer is ______.
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Charge redistribution means the two capacitors share charge until the potentials are equal; the loss is \(\frac{1}{2}\frac{C_1C_2}{C_1+C_2}V^2\).
Updated On: Nov 5, 2026
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Correct Answer: 4
Solution and Explanation
Initial energy: \(U_i=\frac12 C_1V^2=\frac12\times5\times10^{-6}\times220^2=0.121\) J.
After connecting the charged capacitor to the uncharged one (charges redistribute, common potential): \(V_f=\frac{C_1V}{C_1+C_2}=\frac{5\times220}{7.5}=146.67\) V.
\(U_f=\frac12(C_1+C_2)V_f^2=\frac12\times7.5\times10^{-6}\times(146.67)^2\approx0.0807\) J.
Energy change \(=U_i-U_f\approx0.0403\) J \(=\frac{4.03}{100}\) J.
So \(X\approx4.03\), nearest integer 4.
Answer: 4
After connecting the charged capacitor to the uncharged one (charges redistribute, common potential): \(V_f=\frac{C_1V}{C_1+C_2}=\frac{5\times220}{7.5}=146.67\) V.
\(U_f=\frac12(C_1+C_2)V_f^2=\frac12\times7.5\times10^{-6}\times(146.67)^2\approx0.0807\) J.
Energy change \(=U_i-U_f\approx0.0403\) J \(=\frac{4.03}{100}\) J.
So \(X\approx4.03\), nearest integer 4.
Answer: 4
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