Two identical capacitors C_{1} and C_{2} of e | vedclass

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(C) Initially,capacitor $C_{1}$ is charged to a potential $V$ by the battery. The initial energy stored in $C_{1}$ is $U_{i} = \frac{1}{2} C V^{2}$.Wh

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$50$

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(C) Initially,capacitor $C_{1}$ is charged to a potential $V$ by the battery. The initial energy stored in $C_{1}$ is $U_{i} = \frac{1}{2} C V^{2}$.When the battery is disconnected and $C_{1}$ is connected in parallel with an uncharged capacitor $C_{2}$ (where $C_{1} = C_{2} = C$),charge is redistributed.The common potential $V'$ after connection is given by $V' = \frac{C_{1}V + C_{2}(0)}{C_{1} + C_{2}} = \frac{CV}{2C} = \frac{V}{2}$.The final energy stored in the system is $U_{f} = \frac{1}{2} (C_{1} + C_{2}) (V')^{2} = \frac{1}{2} (2C) (\frac{V}{2})^{2} = C \cdot \frac{V^{2}}{4} = \frac{1}{4} C V^{2}$.The loss in energy is $\Delta U = U_{i} - U_{f} = \frac{1}{2} C V^{2} - \frac{1}{4} C V^{2} = \frac{1}{4} C V^{2}$.The percentage loss of energy is $\frac{\Delta U}{U_{i}} 	imes 100 = \frac{\frac{1}{4} C V^{2}}{\frac{1}{2} C V^{2}} 	imes 100 = 50 \%$.

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