Two identical capacitors have the same capacitance C. One of them is charged to a potential $v_{1}$ and the other to $v_{2}$ . If they are connected with their unlike plates together, the decrease in energy of the combined system is

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

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

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

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

answer is D.

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Detailed Solution

$initial energy U_{1} = \frac{1}{2} {CV}_{1}^{2} + \frac{1}{2} C {V_{2}}^{2} Final energy U_{2} = \frac{1}{2} {[\frac{C (V_{1} - V_{2})}{2}]}^{2} \frac{1}{C} ΔU = \frac{1}{2} \frac{C_{1} C_{2}}{C_{1} + C_{2}} {(V_{1} - V_{2})}^{2}$

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