Q75 / 90 · JEE Main 2024 (30 Jan Shift 1) · Physics

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

Correct answer: 2

Solution

Given, $E = \frac{1}{2} C V^{2}$

Required energy loss

$= \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2} = \frac{1}{2} \frac{C \times 2 C}{C + 2 C} {(V - 2 V)}^{2}$

$= \frac{2}{6} C V^{2} = \frac{2}{3} \cdot E$

$∴ x = 2$

Hence, the correct answer is 2.