Two identical capacitors have the same capacitance C. One of them is charged to potential V₁ and the other to V₂ . 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

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$\frac{1}{4} C (V_{1}^{2} - V_{2}^{2})$

$\frac{1}{4} C {(V_{1} + V_{2})}^{2}$

$\frac{1}{4} C {(V_{1} - V_{2})}^{2}$

$\frac{1}{4} C (V_{1}^{2} + V_{2}^{2})$

answer is C.

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Detailed Solution

Initial energy of a system $V_{i} = \frac{1}{2} {CV}_{1}^{2} + {CV}_{2}^{2}$
Final energy of a system $V_{f} = \frac{1}{2} (2 C) V^{2} = \frac{1}{2} (2 C) {(\frac{V_{1} + V_{2}}{2})}^{2}$
$\begin{array}{l} = \frac{1}{4} C {(V_{1} + V_{2})}^{2} \\ ΔU = U_{i} - U_{f} = \frac{1}{4} C {(V_{1} - V_{2})}^{2} \end{array}$