Question

# A capacitor $C$ is fully charged with voltage ${V}_{0}$. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance C/2. The energy loss in the process after the charge is distributed between the two capacitors is:

A

$\frac{1}{2}C{{V}_{0}}^{2}$

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B

$\frac{1}{4}C{{V}_{0}}^{2}$

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C

$\frac{1}{3}C{{V}_{0}}^{2}$

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D

$\frac{1}{6}C{{V}_{0}}^{2}$

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Solution

## The correct option is D $\frac{1}{6}C{{V}_{0}}^{2}$Step 1. Given data:Capacitance of fully charged capacitor is $C$ and voltage is ${V}_{0}$.Capacitance of uncharged capacitor is $\frac{C}{2}$Step 2. Solving for energy loss:Since the two capacitors are connected in parallel, using ${C}_{p}={C}_{1}+{C}_{2}$The capacity of new combination is ${C}_{f}=\left(C+\frac{C}{2}\right)=\frac{3C}{2}$Since the total charge $\left(Q=CV\right)$ is constant, we have${C}_{f}{V}_{f}=C{V}_{0}$$\frac{3C}{2}{V}_{f}=C{V}_{0}$ $⇒{V}_{f}=\frac{2}{3}{V}_{0}$The initial energy is ${U}_{0}=\frac{1}{2}C{{V}_{0}}^{2}$The energy of new configuration is, ${U}_{f}=\frac{1}{2}\frac{3C}{2}{\left(\frac{2{V}_{0}}{3}\right)}^{2}$$⇒{U}_{f}=\frac{1}{2}\frac{2}{3}C{{V}_{0}}^{2}$The Energy loss is, ${U}_{0}-{U}_{f}=\left(1-\frac{2}{3}\right)\frac{1}{2}C{{V}_{0}}^{2}$$⇒{U}_{0}-{U}_{f}=\frac{1}{3}.\frac{1}{2}C{{V}_{0}}^{2}=\frac{1}{6}C{{V}_{0}}^{2}$Hence, option D is correct

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