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
Medium
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(c) Initial energy ${U_i} = \frac{1}{2}{C_1}V_1^2 + \frac{1}{2}{C_2}V_2^2$, Final energy ${U_f} = \frac{1}{2}({C_1} + {C_2}){V^2}$ (where $V = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1}{C_2}}})$
Hence energy loss $\Delta U = {U_i} - {U_f} = \frac{{{C_1}{C_2}}}{{2({C_1} + {C_2})}}{({V_1} - {V_2})^2}$
Hence energy loss $\Delta U = {U_i} - {U_f} = \frac{{{C_1}{C_2}}}{{2({C_1} + {C_2})}}{({V_1} - {V_2})^2}$
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