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(C) Initially,both capacitors $A$ and $B$ are connected in parallel to a battery of potential $V$. The capacitance of each is $C$. Total initial energ

Vedclass · Accepted Answer

$3 : 5$

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(C) Initially,both capacitors $A$ and $B$ are connected in parallel to a battery of potential $V$. The capacitance of each is $C$. Total initial energy $U_i = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$. When the switch $S$ is opened,capacitor $A$ remains connected to the battery,so its potential remains $V$ and its capacitance becomes $K C$. Capacitor $B$ is disconnected,so its charge $Q = CV$ remains constant,and its capacitance becomes $K C$. Final energy in capacitor $A$: $U_{Af} = \frac{1}{2}(KC)V^2 = \frac{3}{2}CV^2$. Final energy in capacitor $B$: $U_{Bf} = \frac{Q^2}{2(KC)} = \frac{(CV)^2}{2(3C)} = \frac{CV^2}{6}$. Total final energy $U_f = U_{Af} + U_{Bf} = \frac{3}{2}CV^2 + \frac{1}{6}CV^2 = \frac{9+1}{6}CV^2 = \frac{10}{6}CV^2 = \frac{5}{3}CV^2$. The ratio of initial to final energy is $\frac{U_i}{U_f} = \frac{CV^2}{(5/3)CV^2} = \frac{3}{5}$.

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