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(C) Initial energy of the system is $U_1 = \frac{1}{2} CV_1^2 + \frac{1}{2} CV_2^2 = \frac{C}{2}(V_1^2 + V_2^2)$.When the capacitors are connected in

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

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(C) Initial energy of the system is $U_1 = \frac{1}{2} CV_1^2 + \frac{1}{2} CV_2^2 = \frac{C}{2}(V_1^2 + V_2^2)$.When the capacitors are connected in parallel,the common potential is $V = \frac{V_1 + V_2}{2}$.The final energy of the system is $U_2 = \frac{1}{2}(C + C)V^2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{C}{4}(V_1 + V_2)^2$.The decrease in energy is $\Delta U = U_1 - U_2$.$\Delta U = \frac{C}{2}(V_1^2 + V_2^2) - \frac{C}{4}(V_1 + V_2)^2$.$\Delta U = \frac{C}{4} [2(V_1^2 + V_2^2) - (V_1^2 + V_2^2 + 2V_1V_2)]$.$\Delta U = \frac{C}{4} [V_1^2 + V_2^2 - 2V_1V_2] = \frac{1}{4} C(V_1 - V_2)^2$.

Two identical capacitors have the same capacitance $C$. One of them is charged to a 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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