The capacities of two conductors are C_1 and | vedclass

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(C) Initial energy $U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2$. When connected,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2

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$\frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}$

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(C) Initial energy $U_i = \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2$. When connected,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$. Final energy $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} (C_1 + C_2) \left( \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \right)^2 = \frac{(C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)}$. Energy loss $\Delta U = U_i - U_f = \left( \frac{1}{2} C_1 V_1^2 + \frac{1}{2} C_2 V_2^2 \right) - \frac{(C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)}$. Simplifying this expression: $\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}$.

The capacities of two conductors are $C_1$ and $C_2$ and their respective potentials are $V_1$ and $V_2$. If they are connected by a thin wire,then the loss of energy will be given by

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