Q 12-02-164JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
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:
Answer: (B) $\dfrac32CV^{2}$
Charges are $CV$ and $4CV$; with opposite plates joined the net charge is $4CV - CV = 3CV$ on a total capacitance $3C$, so the common voltage is $V$.
$$U = \frac12(3C)V^{2} = \frac32CV^{2}$$
Solution by Sreeraj P, M.Sc Physics