Two identical capacitors have the same capacitance ' \(C\) '. One of them is charged to potential \(\mathrm{V}_1\) and other to \(\mathrm{V}_2\). The negative ends of capacitors are connected together. When positive ends are also connected, the decrease in energy of the combined system is

\(\mathrm{U}_{\mathrm{i}}=\frac{1}{2} \mathrm{CV}_1^2+\frac{1}{2} C V_2^2=\frac{1}{2} \mathrm{C}\left(\mathrm{~V}_1^2+\mathrm{V}_2^2\right)\)
When the capacitors are joined, common potential, \(\mathrm{V}=\frac{\mathrm{CV}_1+\mathrm{CV}_2}{2 \mathrm{C}}=\frac{\mathrm{V}_1+\mathrm{V}_2}{2}\)
\(\therefore \quad\) Final energy of the system,
\(\begin{aligned}
\mathrm{U}_{\mathrm{f}} & =\frac{1}{2}(2 \mathrm{C}) \mathrm{V}^2=\frac{1}{2} 2 \mathrm{C}\left(\frac{\mathrm{~V}_1+\mathrm{V}_2}{2}\right)^2 \\
& =\frac{1}{4} \mathrm{C}\left(\mathrm{~V}_1+\mathrm{V}_2\right)^2
\end{aligned}\)
\(\begin{aligned}
\therefore \quad \text { Decrease in energy } & =\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{f}} \\
& =\frac{1}{4} \mathrm{C}\left(\mathrm{~V}_1-\mathrm{V}_2\right)^2
\end{aligned}\)