Question

Two parallel plate capacitors $\text{A}$ and $\text{B}$ with capacitance $1\mu \text{F}$ and $5\mu \text{F}$ are charged separately to the same potential of $100\text{V}$. Now, the positive plate of $\text{A}$ is connected to the negative plate of $\text{B}$. Find the total loss of electrical energy in the given system.

• A. $33.4\text{mJ}$
• B. $8.35\text{mJ}$
• C. $44.67\text{mJ}$
• D. $16.7\text{mJ}$

Answer 1

The correct option is D $16.7\text{mJ}$

Energy stored in capacitor is given by

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

Initially total energy stored in capacitors

$E_i=\left[\frac{1}{2}\right({10}^{-6}){10}^4+\frac{1}{2}(5\times {10}^{-6}\left){10}^4\right]$

$\Rightarrow E_i=3\times {10}^{-2}\text{J}$

After the connection is made:


So, common potential of the above diagram is given by

$V=\frac{|C_1{V}_1-C_2{V}_2|}{C_1+C_2}$

$V=\frac{|(1\times {10}^{-6}\times 100)-(5\times {10}^{-6}\times 100)|}{(1\times {10}^{-6})+(5\times {10}^{-6})}$

$V=\frac{|100-500|}{6}=\frac{200}{3}\text{Volt}$

Final energy stored by the capacitors is given by

$E_f=\frac{1}{2}(C_1+C_2)V^2$

$E_f=\frac{1}{2}\left[\right(1\times {10}^{-6})+(5\times {10}^{-6}\left)\right](200/3{)}^2$

$\therefore E_f=\frac{4}{3}\times {10}^{-2}\text{J}$

Loss of Energy will be

$\Delta E=E_i-E_f=3\times {10}^{-2}-\frac{4}{3}\times {10}^{-2}$

$\Rightarrow \Delta E=\frac{5}{3}\times {10}^{-2}\text{J}=16.7\text{mJ}$

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

Alternate solution: Loss of energy $=\frac{1}{2}\frac{C_1{C}_2}{C_1+C_2}(V_1+V_2{)}^2=\frac{1}{2}\frac{5\times 1}{6}(200{)}^2$ $=\frac{5}{3}\times {10}^{-2}\text{J}=16.7\text{mJ}$