Question

A condenser of capacitance $C_1$ is charged to $V_0$ volt. The energy stored in it is $U_0$. It is connected in parallel to another uncharged condenser of capacitance $C_2$. The energy loss in the process is :

• A. $\frac{C_1{C}_2{U}_0}{2(C_1+C_2)}$
• B. ${\left(\frac{C_1-C_2}{C_1+C_2}\right)}^2{U}_0$
• C. $\frac{C_1{U}_0}{C_1+C_2}$
• D. $\frac{C_2{U}_0}{C_1+C_2}$

Answer 1

Answer: (D)

$\frac{C_2{U}_0}{C_1+C_2}$

The common potential $V_c=\frac{Q_1+Q_2}{C_1+C_2}=\frac{C_1{V}_0+0}{C_1+C_2}=\frac{C_1{V}_0}{C_1+C_2}$
As they are in parallel so equivalent capacitance $C_{eq}=C_1+C_2$
Energy loss $=\frac{1}{2}{C}_1{V}_0^2-\frac{1}{2}{C}_{eq}{V}_c^2=U_0-\frac{1}{2}(C_1+C_2)(\frac{C_1{V}_0}{C_1+C_2}{)}^2=U_0-\frac{C_1{U}_0}{C_1+C_2}=\frac{C_2{U}_0}{C_1+C_2}$