Question

Two capacitors of capacitances $\mathrm{C}$ and $2\mathrm{C}$ are charged to potential differences $\mathrm{V}$ and $2\mathrm{V}$, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is:

• A. zero
• B. $\frac{9}{2}{\mathrm{CV}}^2$
• C. $\frac{25}{6}{\mathrm{CV}}^2$
• D. $\frac{3}{2}{\mathrm{CV}}^2$

Answer 1

The correct option is D

$\frac{3}{2}{\mathrm{CV}}^2$

Step 1. Given data

Two capacitors of capacitances $\mathrm{C}$ and $2\mathrm{C}$ are charged to potential differences $\mathrm{V}$ and $2\mathrm{V}$, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other.

We have to find the final energy.

Step 2. Formula to be used

According to the law of conservation of charges, the final total charge will be the same as that of the earlier charge. Let $Q_1,Q_2$ be the charges before connecting the capacitor and $Q_1^{\text{'}},Q_2^{\text{'}}$ be the charges after connecting in parallel

So,

${\mathrm{Q}}_1+{\mathrm{Q}}_2={\mathrm{Q}}_1^{\text{'}}+{\mathrm{Q}}_2^{\text{'}}$

$=\mathrm{Q}$

Since,

$\mathrm{Q}=\mathrm{CV}$

Here, $\mathrm{C}$ is capacitance, and $\mathrm{V}$ is potential and $\mathrm{Q}$ is charge.

$\frac{{\mathrm{Q}}_1^{\text{'}}}{{\mathrm{Q}}_2^{\text{'}}}=\frac{{\mathrm{C}}_1\mathrm{V}}{{\mathrm{C}}_2\mathrm{V}}$

$=\frac{{\mathrm{C}}_1}{{\mathrm{C}}_2}$

Step 3. Find the final energy.

The net charges stored between two capacitors are,

$\mathrm{Q}\text{'}={\mathrm{Q}}_1-{\mathrm{Q}}_2$

$={\mathrm{C}}_2{\mathrm{V}}_2-{\mathrm{C}}_1{\mathrm{V}}_1$

$=\left(2\right)\left(2\mathrm{V}\right)-\mathrm{CV}$

$=3\mathrm{CV}$

When they are connected in parallel to each other.

We get,

${\mathrm{C}}_{\mathrm{net}}=\mathrm{C}+2\mathrm{C}$

$=3\mathrm{C}$

So,

$\mathrm{V}\text{'}=\frac{\mathrm{Q}\text{'}}{{\mathrm{C}}_{\mathrm{net}}}$

$=\frac{3\mathrm{CV}}{3\mathrm{C}}$

$=\mathrm{V}$

The energy of the capacitor is,

$\mathrm{E}=\frac{1}{2}{\mathrm{CV}}^2$

$=\frac{1}{2}\left(3\mathrm{C}\right){\mathrm{V}}^2$

The final energy of this configuration is $\frac{3}{2}{\mathrm{CV}}^2$

Hence, option $\mathrm{D}$ is correct answer.