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. $\frac{C_1{C}_2}{2(C_1+C_2)}{U}_0$

Answer 1

The correct option is A $\frac{C_2}{C_1+C_2}{U}_0$

Initial energy, $U_i=U_0=\frac{1}{2}{C}_1{V}_0^2$
Initial charge of the system is $Q_i=C_1{V}_0$.
After connecting another capacitor the final charge $Q_f=(C_1+C_2)V_c$ where $V_c=$ common potential after connecting.
As total charge is conserved , $Q_i=Q_f\Rightarrow C_1{V}_0=(C_1+C_2)V_c$
$\therefore V_c=\frac{C_1{V}_0}{C_1+C_2}$
Final energy , $U_f=\frac{1}{2}(C_1+C_2)V_c^2=\frac{1}{2}(C_1+C_2){\left(\frac{C_1{V}_0}{C_1+C_2}\right)}^2=\frac{C_1{U}_0}{C_1+C_2}$
Energy dissipation $=U_i-U_f=U_0-\frac{C_1{U}_0}{C_1+C_2}=\frac{C_2{U}_0}{C_1+C_2}$