A capacitor C is fully charged with voltage $V_{0}$ . After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance $\frac{C}{2}$ . The energy loss in the process after the charge is distributed between the two capacitors is:

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$\frac{1}{6} C V_{0}^{2}$

$\frac{1}{4} C V_{0}^{2}$

$\frac{1}{3} C V_{0}^{2}$

$\frac{1}{2} C V_{0}^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} Heat loss = \frac{1}{2} (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) V_{0}^{2} \\ = \frac{1}{2} \times \frac{C}{3} \times V_{0}^{2} = \frac{1}{6} {CV}_{0}^{2} \end{array}$

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