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
$=\frac{C_1{C}_2{V}_0^2}{2\left(C_1+C_2\right)}$
$=\frac{C_2{U}_0}{C_1+C_2}$
$\left[\because U_0=\frac{1}{2}{C}_1{V}_0^2\right]$