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(D) Initially,the capacitor $C$ is connected to the battery $V_1$. The initial charge on the capacitor is $Q_i = CV_1$ and the initial energy stored i

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Dependent on $V_2$ as well as $V_1$

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(D) Initially,the capacitor $C$ is connected to the battery $V_1$. The initial charge on the capacitor is $Q_i = CV_1$ and the initial energy stored is $U_i = \frac{1}{2}CV_1^2$.When the switch is shifted to position $2$,the capacitor is connected to the battery $V_2$. The final charge on the capacitor is $Q_f = CV_2$ and the final energy stored is $U_f = \frac{1}{2}CV_2^2$.The work done by the battery $V_2$ is $W = Q_f V_2 = (CV_2)V_2 = CV_2^2$.According to the work-energy theorem,the heat generated $H$ is given by $H = W - (U_f - U_i)$.Substituting the values: $H = CV_2^2 - (\frac{1}{2}CV_2^2 - \frac{1}{2}CV_1^2) = \frac{1}{2}CV_2^2 + \frac{1}{2}CV_1^2 = \frac{1}{2}C(V_1^2 + V_2^2)$.Thus,the heat generated depends on both $V_1$ and $V_2$.

The switch $S_w$ is shifted from position $1$ to position $2$ as shown in the figure. Heat generated in the circuit is

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