Question

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. $\frac{C_{1} C_{2}}{2\left(C_{1}+C_{2}\right)} U_{0}$

Answer 1

Answer: (A)

Loss of energy during sharing
$=\frac{C_{1} C_{2}\left(V_{1}-V_{2}\right)^{2}}{2\left(C_{1}+C_{2}\right)}$
In the equation, put $V_{2}=0,\, V_{1}=V_{0}$
$\therefore $ Loss of energy
$=\frac{C_{1} C_{2} V_{0}^{2}}{2\left(C_{1}+C_{2}\right)}$
$=\frac{C_{2} U_{0}}{C_{1}+C_{2}}$
$\left[\because U_{0}=\frac{1}{2} C_{1} V_{0}^{2}\right]$