A capacitor C is fully charged with voltage V | vedclass

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Question

(A) Initial charge on capacitor $C$ is $Q = CV_{0}$.Initial energy stored in the capacitor is $U_{i} = \frac{1}{2}CV_{0}^{2}$.When connected in parall

Vedclass · Accepted Answer

$0.166$

Vedclass · Answer

(A) Initial charge on capacitor $C$ is $Q = CV_{0}$.Initial energy stored in the capacitor is $U_{i} = \frac{1}{2}CV_{0}^{2}$.When connected in parallel,the common potential $V$ is given by $V = \frac{Q_{total}}{C_{total}} = \frac{CV_{0}}{C + C/2} = \frac{CV_{0}}{3C/2} = \frac{2}{3}V_{0}$.The final energy stored in the system is $U_{f} = \frac{1}{2}(C + C/2)V^{2} = \frac{1}{2}(\frac{3C}{2})(\frac{2}{3}V_{0})^{2} = \frac{1}{2}(\frac{3C}{2})(\frac{4}{9}V_{0}^{2}) = \frac{1}{3}CV_{0}^{2}$.The energy loss is $\Delta U = U_{i} - U_{f} = \frac{1}{2}CV_{0}^{2} - \frac{1}{3}CV_{0}^{2} = \frac{1}{6}CV_{0}^{2}$.Since $\frac{1}{6} \approx 0.166$,the correct option is $A$.

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