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
IIT 2002, Diffcult
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(c) Initial energy of the system
${U_i} = \frac{1}{2}C{V_1}^2 + \frac{1}{2}C{V_2}^2$
When the capacitors are joined, common potential $V = \frac{{C{V_1} + C{V_2}}}{{2C}} = \frac{{{V_1} + {V_2}}}{2}$
Final energy of the system
${U_f} = \frac{1}{2}(2C){V^2} = \frac{1}{2}2C\,{\left( {\frac{{{V_1} + {V_2}}}{2}} \right)^2} = \frac{1}{4}C{({V_1} + {V_2})^2}$
Decrease in energy = ${U_i} - {U_f} = \frac{1}{4}C{({V_1} - {V_2})^2}$
${U_i} = \frac{1}{2}C{V_1}^2 + \frac{1}{2}C{V_2}^2$
When the capacitors are joined, common potential $V = \frac{{C{V_1} + C{V_2}}}{{2C}} = \frac{{{V_1} + {V_2}}}{2}$
Final energy of the system
${U_f} = \frac{1}{2}(2C){V^2} = \frac{1}{2}2C\,{\left( {\frac{{{V_1} + {V_2}}}{2}} \right)^2} = \frac{1}{4}C{({V_1} + {V_2})^2}$
Decrease in energy = ${U_i} - {U_f} = \frac{1}{4}C{({V_1} - {V_2})^2}$
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