Question

A $4\mu F$ capacitor is charged by a $200V$ battery. It is then disconnected from the supply and is connected to another uncharged $2\mu F$ capacitor. During the process, loss of energy $(inJ)$ is

• A. $3.43\times 10^{-2}$
• B. $2.67\times 10^{-2}$
• C. $2.67\times 10^{-4}$
• D. $3.43\times 10^{-4}$

Answer 1

Answer: (B)

Charge stored at the capacitor
$q=C_1{V}_1=4\times 200=800\mu C$
When this capacitor is connected with a uncharged capacitor, then common potential on both capacitors
$V=\frac{C_1{V}_1+C_2{V}_2}{C_1+C_2}$
$=\frac{800+0}{4+2}=\frac{800}{6}V$
Loss in energy $=$ Initial energy $-$ Final energy
$=\frac{1}{2}{C}_1{V}_1^2-\frac{1}{2}\left(C_1+C_2\right)V^2$
$=\frac{1}{2}\times 4\times 10^{-6}\times (200{)}^2-\frac{1}{2}(4+2)\times 10^{-6}\times {\left(\frac{800}{6}\right)}^2$
$=2\times 10^{-6}\times 4\times 10^4-\frac{3\times 10^{-6}\times 64\times 10^4}{36}$
$=8\times 10^{-2}-\frac{64}{12}\times 10^{-2}$
$=8\times 10^{-2}-5.33\times 10^{-2}$
$=267\times 10^{-2}J$