Q. A condenser of capacity $C_{1}$ is charged to a potential $V_{0}.$ The electrostatic energy stored in it is $U_{0}.$ It is connected to another uncharged condenser of capacity $C_{2}$ in parallel. The energy dissipated in the process is
A) $\frac{C_{2}}{C_{1} + C_{2}}U_{0}$
B) $\frac{C_{1}}{C_{1} + C_{2}}U_{0}$
C) $\left( \frac{C_{1} - C_{2}}{C_{1} + C_{2}} \right)U_{0}$
D) $\frac{C_{1}C_{2}}{2(C_{1} + C_{2})}U_{0}$
Explanation
(a) Loss of energy during sharing $= \frac{C_{1}C_{2}\left( V_{1} - V_{2} \right)^{2}}{2(C_{1} + C_{2})}$
In the equation, put $V_{2} = 0,\ V_{1} = V_{0}$
$\therefore$ Loss of energy $= \frac{C_{1}C_{2}V_{0}^{2}}{2(C_{1} + C_{2})}$
$= \frac{C_{2}U_{0}}{C_{1} + C_{2}}\ \ \ \left\lbrack \because U_{0} = \frac{1}{2}C_{1}V_{0}^{2} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.