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JEE Mains · Physics · STD 12 - Electric potential and capacitance
  • JEE Mains 2024
  • Hard
  • Numerical

A capacitor of capacitance \(\mathrm{C}\) and potential \(\mathrm{V}\) has energy \(E\). It is connected to another capacitor of capacitance \(2 \mathrm{C}\) and potential \(2 \mathrm{~V}\). Then the loss of energy is \(\frac{x}{3} E\), where \(\mathrm{x}\) is _______.

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Answer & Solution

Correct Answer

(C) \(2\)

Step-by-step Solution

Detailed explanation

Energy loss \(=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2\)

\( =\frac{2}{3} \cdot E \)

\( \therefore x=2\)

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