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(D) $1$. When $X$ is connected to $Y$,the capacitor charges to a potential difference $V = E$. The charge on the capacitor is $q_1 = CE$. The energy s

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$4$ times the energy finally stored in the capacitor

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(D) $1$. When $X$ is connected to $Y$,the capacitor charges to a potential difference $V = E$. The charge on the capacitor is $q_1 = CE$. The energy stored is $U_1 = \frac{1}{2}CE^2$.$2$. When $X$ is connected to $Z$,the capacitor is connected to cell $B$ with reversed polarity. The potential difference across the capacitor changes from $+E$ to $-E$. The final charge on the capacitor is $q_2 = -CE$.$3$. The total charge that flows through the circuit is $\Delta q = q_1 - q_2 = CE - (-CE) = 2CE$.$4$. The work done by the cell $B$ is $W = \Delta q \cdot E = (2CE) \cdot E = 2CE^2$.$5$. The change in energy stored in the capacitor is $\Delta U = U_f - U_i = \frac{1}{2}CE^2 - \frac{1}{2}CE^2 = 0$.$6$. By the work-energy theorem,$W = \Delta U + H$,where $H$ is the heat produced. Thus,$H = W = 2CE^2$.$7$. Since the final energy stored is $U_f = \frac{1}{2}CE^2$,we have $H = 4 \cdot (\frac{1}{2}CE^2) = 4U_f$. Therefore,the heat produced is $4$ times the energy finally stored in the capacitor.

In the circuit shown,the cells are ideal and of equal emfs $E$. The capacitance of the capacitor is $C$ and the resistance of the resistor is $R$. $X$ is first joined to $Y$ and then to $Z$. After a long time,the total heat produced in the resistor will be:

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