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Two capacitors of capacitances C and 2C are charged to potential differences V and 2V respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is

Asked in JEE Main 5th Sept 1st Shift 2020 · Redistribution of charge

Answer: (2) 3/2CV²

Step-by-step solution

Initial charges: q₁=CV and q₂=(2C)(2V)=4CV.

Joining positive to negative makes the charges oppose, so the net charge is q=4CV-CV=3CV.

The capacitors end up in parallel, so the total capacitance is C+2C=3C and the common potential is V_f=(3CV)/(3C)=V.

U_f=1/2(3C)V²=3/2CV².

(The initial store was 9/2CV², so 3CV² is dissipated in the connecting wires.)

Why the other options are wrong

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