Question

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

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

Answer 1

Answer: (C)

$\frac{3C{V}^2}{2}$

When the condensers are connected in such a way that the positive plate of one is connected to negative plate of another then the total charge $Q=4CV-CV=3CV$

Now, let it is distributed as shown in figure below, potential across the capacitors is same

So,$\frac{q}{2C}=\frac{3CV-q}{C}\Rightarrow q=2CV$

therefore , charge of one capacitor (C) is $Q_1=3CV-2CV=CV,$ charge of other capacitor $\left(2C\right)$ is $Q_2=2CV$

Total potential energy

$=\frac{Q_1^2}{2C_1}+\frac{Q_2^2}{2C_2}=\frac{C^2{V}^2}{2C}+\frac{4C^2{V}^2}{2\times 2C}=\frac{3C{V}^2}{2}$