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(A) Initial energy of the system is $U_i = \frac{1}{2} C V_1^2 + \frac{1}{2} C V_2^2$.When the capacitors are connected in parallel,the total charge $

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

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(A) Initial energy of the system is $U_i = \frac{1}{2} C V_1^2 + \frac{1}{2} C V_2^2$.When the capacitors are connected in parallel,the total charge $Q = Q_1 + Q_2 = C V_1 + C V_2$ is redistributed.The equivalent capacitance of the system is $C_{eq} = C + C = 2C$.The common potential $V$ after connection is given by $V = \frac{Q_{total}}{C_{eq}} = \frac{C(V_1 + V_2)}{2C} = \frac{V_1 + V_2}{2}$.The final energy of the system is $U_f = \frac{1}{2} (2C) V^2 = C \left( \frac{V_1 + V_2}{2} \right)^2 = \frac{C}{4} (V_1 + V_2)^2$.The decrease in energy is $\Delta U = U_i - U_f = \frac{1}{2} C (V_1^2 + V_2^2) - \frac{1}{4} C (V_1 + V_2)^2$.$\Delta U = \frac{C}{4} [2V_1^2 + 2V_2^2 - (V_1^2 + V_2^2 + 2V_1V_2)]$.$\Delta U = \frac{C}{4} (V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4} C (V_1 - V_2)^2$.

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

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