First slide

Redistribution of charges and concept of common potential

Question

The capacities of two conductors are $C_{1}$ and $C_{2}$ and their respective potentials are $V_{1}$ and $V_{2}$ . If they are connected by a thin wire, then the loss of energy will be given by

Easy

A

$\frac{C_{1} C_{2} (V_{1} + V_{2})}{2 (C_{1} + C_{2})}$
B

$\frac{C_{1} C_{2} (V_{1} - V_{2})}{2 (C_{1} + C_{2})}$
C

$\frac{C_{1} C_{2} {(V_{1} - V_{2})}^{2}}{2 (C_{1} + C_{2})}$
D

$\frac{(C_{1} + C_{2}) (V_{1} - V_{2})}{C_{1} + C_{2}}$

Solution

Initial energy $U_{i} = \frac{1}{2} C_{1} V_{1}^{2} + \frac{1}{2} C_{2} V_{2}^{2}$ , final energy $U_{f} = \frac{1}{2} (C_{1} + C_{2}) V^{2}$ $(W h e r e V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}})$
Hence energy loss $Δ U = U_{i} - U_{f} = \frac{1}{2} C_{1} V_{1}^{2} + \frac{1}{2} C_{2} V_{2}^{2} - \frac{(C_{1} + C_{2})}{2} {(\frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}})}^{2} = \frac{C_{1} C_{2}}{2 (C_{1} + C_{2})} {(V_{1} - V_{2})}^{2}$

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