Two identical capacitors have the same capacitance ' \(\mathrm{C}\) '. One of them is charged to a potential \(V_1\) and the other to \(V_2\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

Initial energy of the combined system
\(
\mathrm{U}_1=\frac{1}{2} \mathrm{CV}_1^2+\frac{1}{2} \mathrm{CV}_2^2=\frac{\mathrm{C}}{2}\left(\mathrm{~V}_1^2+\mathrm{V}_2^2\right)
\)
On joining the two condensers in parallel, common potential,
\(
\mathrm{V}=\frac{\mathrm{V}_1+\mathrm{V}_2}{2}
\)
\(\therefore \quad\) Final energy of the combined system:
\(
\mathrm{U}_2 \equiv \frac{1}{2}(\mathrm{C}+\mathrm{C})\left(\frac{\mathrm{V}_1+\mathrm{V}_2}{2}\right)^2 \text {. }
\)
\(\therefore \quad\) - Decrease in energy will be:
\(
\begin{aligned}
\Delta \mathrm{U} & =\mathrm{U}_{\mathrm{L}}-\mathrm{U}_2 \\
& =\frac{\mathrm{C}}{2}\left(\mathrm{~V}_1^2+\mathrm{V}_2^2\right)-\frac{1}{2}(\mathrm{C}+\mathrm{C})\left(\frac{\mathrm{V}_1+\mathrm{V}_2}{2}\right)^2 \\
& =\frac{1}{4} \mathrm{C}\left(\mathrm{V}_1-\mathrm{V}_2\right)^2
\end{aligned}
\)