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The capacitor $C_{1}$ in the figure initially carries a charge $q_{0}$ . When the switch $S_{1}, S_{2}$ are open capacitor $C_{1}$ is connected to a resistor R and a second capacitor $C_{2}$ , which initially does not carry any charge.
What is heat lost in resistor after a long time of closing the switch?

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$\frac{q_{0}^{2}}{2} [\frac{C_{2}}{C_{1} (C_{1} + C_{2})}]$

$\frac{q_{0}^{2}}{2} [\frac{C_{2}}{(C_{1} + C_{2})}]$

$\frac{q_{0}^{2}}{2} [\frac{1}{C_{1} (C_{1} + C_{2})}]$

None of these

answer is C.

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Detailed Solution

$U_{i} = \frac{1}{2} \frac{q_{0}^{2}}{C_{1}} and U_{f} = \frac{1}{2} \frac{q_{0}^{2}}{C_{1} + C_{2}}$

Heat lost in the resistor $= U_{i} - U_{f} = \frac{q_{0}^{2}}{2} [\frac{C_{2}}{C_{1} (C_{1} + C_{2})}]$

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