Q112 / 125 · BITSAT 2013 · Physics
Two identical capacitors, have the same capacitance $\mathrm{C}$. One of them is charged to potential $\mathrm{V}_{1}$ and the other to $\mathrm{V}_{2}$. 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$\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}$
- D$\frac{1}{4} C\left(V_{1}+V_{2}\right)^{2}$
Solution
Initial energy of combined system
$$
\mathrm{U}_{1}=\frac{1}{2} \mathrm{CV}_{1}^{2}+\frac{1}{2} \mathrm{CV}_{2}^{2}
$$
Final common potential, $\mathrm{V}=\frac{\mathrm{V}_{1}+\mathrm{V}_{2}}{2},$
Final energy of system,
$$
\mathrm{U}_{2}=2 \times \frac{1}{2} \mathrm{C}\left(\frac{\mathrm{V}_{1}+\mathrm{V}_{2}}{2}\right)^{2}
$$
Hence loss of energy $=\mathrm{U}_{1}-\mathrm{U}_{2}$
$$
=\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}
$$
Hence, the correct answer is option: 'C'.