First slide

Redistribution of charges and concept of common potential

Question

A capacitor of capacity $C_{1}$ is charged to the potential of $V_{0}$ . On disconnecting from the battery, it is connected with a capacitor of capacity $C_{2}$ as shown in the adjoining figure. The ratio of energies of system of capacitors before and after the connection of switch ‘ $S$ ’ will be

Moderate

A

$(C_{1} + C_{2}) / C_{1}$
B

$C_{1} / (C_{1} + C_{2})$
C

$C_{1} C_{2}$
D

$C_{1} / C_{2}$

Solution

Given Capacity of capacitor $= C_{1}, C_{2}$
Potential $V_{1} = V_{0}$
We know Common potential $V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$
Common potential, $V = [\frac{C_{1} V_{0} + 0}{C_{1} + C_{2}}] = \frac{C_{1} V_{0}}{C_{1} + C_{2}}$
We know potential energy stored in capacitor,
$U = \frac{1}{2} C V^{2}$
Now $\frac{U_{i}}{U_{f}} = \frac{\frac{1}{2} C_{1} V_{0}^{2}}{\frac{1}{2} (C_{1} + C_{2}) V^{2}} = [\frac{C_{1} + C_{2}}{C_{1}}]$
Therefore, the correct answer is (A)

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