Question

A capacitor of $4\mu \text{F}$ charged at $50\text{V}$ is connected with another capacitor of $2\mu \text{F}$ charged at $100\text{V}$, in such a way that plates with similar charges are connected together. Before joining and after joining, the total energy will be

• A. $1.5\times {10}^{-2}\text{J}\&1.33\times {10}^{-2}\text{J}$
• B. $1.33\times {10}^{-2}\text{J}\&1.5\times {10}^{-2}\text{J}$
• C. $1.5\times {10}^{-2}\text{J}\&2.67\times {10}^{-2}\text{J}$
• D. $2.67\times {10}^{-2}\text{J}\&1.5\times {10}^{-2}\text{J}$

Answer 1

The correct option is A $1.5\times {10}^{-2}\text{J}\&1.33\times {10}^{-2}\text{J}$

Energy stored in a capacitor is given by

$E=\frac{1}{2}C{V}^2$

Total initial energy will be

$E_i=(\frac{1}{2}\times 4\times {10}^{-6}\times {50}^2)+(\frac{1}{2}\times 2\times {10}^{-6}\times {100}^2)$

$\Rightarrow E_i=1.5\times {10}^{-2}\text{J}$

Let charge on the plates of the capacitors be $q_1$ and $q_2$ respectively when connected together. In parallel, the potential difference across them will be the same.

So,

$V=\frac{q_1}{C_1}=\frac{q_2}{C_2}$

$\Rightarrow \frac{q_1}{4}=\frac{q_2}{2}$

$\Rightarrow q_1=2q_2$

Now,
$q_1+q_2=C_1{V}_1+C_2{V}_2=(4\times 50)+(2\times 100)=400\mu \text{C}$

$\Rightarrow 2q_2+q_2=400\mu \text{C}$

$\Rightarrow q_2=\frac{400}{3}\mu \text{C}$ and $q_1=\frac{800}{3}\mu \text{C}$

So, common potential will be

$V=\frac{q_1}{C_1}=\frac{800}{3\times 4}=\frac{200}{3}\text{Volt}$

Total final energy will be

$E_f=\frac{1}{2}(C_1+C_2)V^2=\frac{1}{2}(2+4)\times {10}^{-6}\times (\frac{200}{3}{)}^2$

$\Rightarrow E_f=1.33\times {10}^{-2}\text{J}$

Hence, option $\left(a\right)$ is correct.