A $4 \mu \mathrm{F}$ capacitor is charged by a $200 \mathrm{~V}$ battery. It is then disconnected from the…
A $4 \mu \mathrm{F}$ capacitor is charged by a $200 \mathrm{~V}$ battery. It is then disconnected from the supply and is connected to another uncharged $2 \mu \mathrm{F}$ capacitor. During the process, loss of energy (in $\mathrm{J}$ ) is
$3.43 \times 10^{-2}$
$2.67 \times 10^{-2}$
$2.67 \times 10^{-4}$
$3.43 \times 10^{-4}$
Solution
Charge stored at the capacitor
$q=C_1 V_1=4 \times 200=800 \mu \mathrm{C}$
When this capacitor is connected with a uncharged capacitor, then common potential on both capacitors
$V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{800+0}{4+2}=\frac{800}{6} \mathrm{~V}$
Loss in energy = Initial energy - Final energy
$\begin{aligned}
= & \frac{1}{2} C_1 V_1^2-\frac{1}{2}\left(C_1+C_2\right) V^2 \\
= & \frac{1}{2} \times 4 \times 10^{-6} \times(200)^2 \\
& \quad-\frac{1}{2}(4+2) \times 10^{-6} \times\left(\frac{800}{6}\right)^2 \\
= & 2 \times 10^{-6} \times 4 \times 10^4-\frac{3 \times 10^{-6} \times 64 \times 10^4}{36} \\
= & 8 \times 10^{-2}-\frac{64}{12} \times 10^{-2} \\
= & 8 \times 10^{-2}-5.33 \times 10^{-2} \\
= & 2.67 \times 10^{-2} \mathrm{~J}
\end{aligned}$