JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 _______.
- A \(45\)
- B \(8\)
- C \(2\)
- D \(19\)
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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