Question 47 of 180

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 :

A$\dfrac{25}{6}C{V}^{2}$B$\dfrac{3}{2}C{V}^{2}$CzeroD$\dfrac{9}{2}C{V}^{2}$

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Answer: (B) $\dfrac{3}{2}C{V}^{2}$

Charge on $C=CV$
Charge on $2C=\left(2C\right)2V$

From charge conservation, $2C(2V)-CV=(C+2C)V'$
Common potential, $V'=V$
Final energy, ${U}_{f}=\left(\dfrac{1}{2}C{V}^{2}+\dfrac{1}{2}\times 2C{V}^{2}\right)=\dfrac{3}{2}C{V}^{2}$