Q148 / 150 · MHT CET 2023 (11 May Shift 1) · Physics
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
- A$\frac{1}{4} \mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)$
- B$\frac{1}{4} C\left(V_1^2+V_2^2\right)$
- Correct answerC$\frac{1}{4} C\left(V_1-V_2\right)^2$
- D$\frac{1}{4} \mathrm{C}\left(\mathrm{V}_1+\mathrm{V}_2\right)^2$
Solution
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:
$$
\begin{aligned}
\Delta \mathrm{U} & =\mathrm{U}_{\mathrm{L}}-\mathrm{U}_2 \\
& =\frac{\mathrm{C}}{2}\left(\mathrm{~V}_1^2+\mathrm{V}_2^2\right)-\frac{1}{2}(\mathrm{C}+\mathrm{C})\left(\frac{\mathrm{V}_1+\mathrm{V}_2}{2}\right)^2 \\
& =\frac{1}{4} \mathrm{C}\left(\mathrm{V}_1-\mathrm{V}_2\right)^2
\end{aligned}
$$
Hence, the correct answer is option: 'C'.