In the circuit shown in the figure, there are | vedclass

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(A) $1$. Initially, $S_1$ is closed. Capacitor $C_1$ charges to $Q_1 = C(2V_0) = 2CV_0$ on its upper plate. $S_1$ is then opened.$2$. Next, $S_2$ is c

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$(B, D)$

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(A) $1$. Initially, $S_1$ is closed. Capacitor $C_1$ charges to $Q_1 = C(2V_0) = 2CV_0$ on its upper plate. $S_1$ is then opened.$2$. Next, $S_2$ is closed. The charge $2CV_0$ on $C_1$ redistributes between $C_1$ and $C_2$. Since both have capacitance $C$, the potential difference across both becomes $V = \frac{Q_{total}}{C_{eq}} = \frac{2CV_0}{2C} = V_0$. Thus, the charge on the upper plate of $C_1$ becomes $CV_0$ and the charge on the upper plate of $C_2$ becomes $CV_0$. $S_2$ is then opened.$3$. Finally, $S_3$ is closed. The capacitor $C_2$ is connected to a battery of potential $V_0$ with the positive terminal connected to the lower plate. Thus, the potential of the upper plate of $C_2$ becomes $-V_0$ relative to the lower plate. The charge on the upper plate of $C_2$ becomes $Q_2 = C(-V_0) = -CV_0$. The charge on $C_1$ remains $CV_0$ as it is isolated.$4$. Therefore, the charge on the upper plate of $C_1$ is $CV_0$ (Statement $B$) and the charge on the upper plate of $C_2$ is $-CV_0$ (Statement $D$).

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