# A capacitor C is fully charged with voltage

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 $\frac{C}{2}$. The energy loss in the process after the charge is distributed between the two capacitors is :

1. (1) $\frac{1}{2} C V_{0}^{2}$

2. (2) $\frac{1}{3} C V_{0}^{2}$

3. (3) $\frac{1}{4} C V_{0}^{2}$

4. (4) $\frac{1}{6} C V_{0}^{2}$

Correct Option: , 4

Solution:

(4)

When two capacitors with capacitance $C_{1}$ and $C_{2}$ at potential $V_{1}$ and $V_{2}$ connected to each other by wire, charge begins to flow from higher to lower potential till they acquire common potential. Here, some loss of energy takes place which is given by.

Heat loss, $H=\frac{C_{1} C_{2}}{2\left(C_{1}+C_{2}\right)}\left(V_{1}-V_{2}\right)^{2}$

In the equation, put $V_{2}=0, V_{1}=V_{0}$

$C_{1}=C, C_{2}=\frac{C}{2}$

Loss of heat $=\frac{C \times \frac{C}{2}}{2\left(C+\frac{C}{2}\right)}\left(V_{0}-0\right)^{2}=\frac{C}{6} V_{0}^{2}$

$H=\frac{1}{6} C V_{0}^{2}$

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