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(C) Initial energy of the system is $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2$.When the capacitors are connected in parallel,the common potential i

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$\frac{1}{4}C(V_1 - V_2)^2$

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(C) Initial energy of the system is $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2$.When the capacitors are connected in parallel,the common potential is $V = \frac{CV_1 + CV_2}{2C} = \frac{V_1 + V_2}{2}$.Final energy of the system is $U_f = \frac{1}{2}(2C)V^2 = C \left( \frac{V_1 + V_2}{2} \right)^2 = \frac{1}{4}C(V_1 + V_2)^2$.The decrease in energy is $\Delta U = U_i - U_f = \frac{1}{2}C(V_1^2 + V_2^2) - \frac{1}{4}C(V_1 + V_2)^2$.Simplifying this,$\Delta U = \frac{1}{4}C [2V_1^2 + 2V_2^2 - (V_1^2 + V_2^2 + 2V_1V_2)] = \frac{1}{4}C(V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4}C(V_1 - V_2)^2$.

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When the positive ends are also connected,the decrease in energy of the combined system is

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