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____________.
JEE MAIN 2024, Diffcult
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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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