Question

A parallel-plate condenser of capacitance $C$ is connected to a battery and is charged to potential $v$. Another condenser of capacity $2C$ is connected to another battery and is charged to potential $2V$. The charging batteries are removed and now the condensers are connected in parallel in such a way that the positive plate of one is connected to the negative plate of another. The final energy of this system is

• A. $25C\frac{V^2}{6}$
• B. $9C\frac{V^2}{2}$
• C. $3C\frac{V^2}{2}$
• D. $0$

Answer 1

The correct option is C $3C\frac{V^2}{2}$

After connecting the common potential is $V_c=\frac{C_2{V}_2+C_1{V}_1}{C_1+C_2}$
$V_c=\frac{4CV-CV}{C+2C}=V$
Negative sign has come because they are connected in parallel in such a way that the positive plate of one is connected to negative plate of another.
The final energy of this system is $U=\frac{1}{2}(2C+C)V_c^2=\frac{3}{2}C{V}^2$