JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 24)
A capacitor of capacitance $$\mathrm{C}$$ and potential $$\mathrm{V}$$ has energy $$\mathrm{E}$$. It is connected to another capacitor of capacitance $$2 \mathrm{C}$$ and potential $$2 \mathrm{~V}$$. Then the loss of energy is $$\frac{x}{3} \mathrm{E}$$, where $$x$$ is _______.
Answer
2
Explanation
$$\begin{aligned}
& \text { Energy loss }=\frac{1}{2} \frac{C_1 C_2}{C_1+C_2}\left(V_1-V_2\right)^2 \\
& =\frac{2}{3} \cdot E \\
& \therefore x=2
\end{aligned}$$
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