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(C) $1$. Initially,the capacitor is charged by battery $\varepsilon_1$. The initial energy stored is $U_i = \frac{1}{2} C \varepsilon_1^2$ and the ini

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$\frac{C}{2}{\left( {{\varepsilon _1} + {\varepsilon _2}} \right)^2}$

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(C) $1$. Initially,the capacitor is charged by battery $\varepsilon_1$. The initial energy stored is $U_i = \frac{1}{2} C \varepsilon_1^2$ and the initial charge is $Q_i = C \varepsilon_1$.$2$. When the switch is shifted to position $2$,the capacitor is connected to battery $\varepsilon_2$ with reversed polarity. The final energy stored is $U_f = \frac{1}{2} C \varepsilon_2^2$ and the final charge is $Q_f = -C \varepsilon_2$.$3$. The change in charge is $\Delta Q = Q_f - Q_i = -C \varepsilon_2 - C \varepsilon_1 = -C(\varepsilon_1 + \varepsilon_2)$. The magnitude of charge that flows through the battery is $|\Delta Q| = C(\varepsilon_1 + \varepsilon_2)$.$4$. The work done by the battery $\varepsilon_2$ is $W_b = \varepsilon_2 \cdot |\Delta Q| = C \varepsilon_2 (\varepsilon_1 + \varepsilon_2)$.$5$. According to the work-energy theorem for the circuit,the heat generated $H$ is given by $H = W_b - (U_f - U_i)$.$6$. Substituting the values: $H = C \varepsilon_2 (\varepsilon_1 + \varepsilon_2) - [\frac{1}{2} C \varepsilon_2^2 - \frac{1}{2} C \varepsilon_1^2] = C \varepsilon_1 \varepsilon_2 + C \varepsilon_2^2 - \frac{1}{2} C \varepsilon_2^2 + \frac{1}{2} C \varepsilon_1^2 = \frac{1}{2} C (\varepsilon_1^2 + 2 \varepsilon_1 \varepsilon_2 + \varepsilon_2^2) = \frac{1}{2} C (\varepsilon_1 + \varepsilon_2)^2$.

Initially,switch $S$ is connected to position $1$ for a long time as shown in the figure. The net amount of heat generated in the circuit after it is shifted to position $2$ is:

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