Two capacitors of capacitances $C$ and $2\, C$ are charged to potential differences $V$ and $2\, V$, 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$.....CV^2$
JEE MAIN 2020, Diffcult
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$Q _{1}= CV \quad Q _{2}=2 C \times 2 V =4 CV$
$\Rightarrow$ By conservation of charge
$q_{i}=q_{f}$
$Q_{1}+Q_{2}=q_{1}+q_{2}$
$4 CV - CV =( C +2 C ) V _{ C }$
$v _{ C }=\frac{3 CV }{3 C } \Rightarrow V$
$\Rightarrow \frac{1}{2} \times(3 C) \times V_{ c }^{2}$
$=\frac{1}{2} \times 3 C \times V ^{2}=\frac{3}{2} CV ^{2}$
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