# Solve this

Question:

A $5 \mu \mathrm{F}$ capacitor is charged fully by a $220 \mathrm{~V}$ supply. It is then disconnected from the supply and is connected in series to another uncharged $2.5 \mu \mathrm{F}$ capacitor. If the energy

change during the charge redistribution is $\frac{X}{100} \mathrm{~J}$ then value of $X$ to the nearest integer is

Solution:

Given, $C_{1}=5 \mu \mathrm{F}$ and $V_{1}=220$ Volt

When capacitor $C_{1}$ fully charged it is disconnected from the supply and connected to uncharged capacitor $C_{2}$. $C_{2}=2.5 \mu \mathrm{F}, V_{2}=0$

Energy change during the charge redistribution,

$\Delta U=U_{i}-U_{f}=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2}$

$=\frac{1}{2} \times \frac{5 \times 2.5}{(5+2.5)}(220-0)^{2} \mu \mathrm{J}$

$=\frac{5}{2 \times 3} \times 22 \times 22 \times 100 \times 10^{-6} \mathrm{~J}$

$=\frac{5 \times 11 \times 22}{3} \times 10^{-4} \mathrm{~J}=\frac{55 \times 22}{3} \times 10^{-4} \mathrm{~J}$

$=\frac{1210}{3} \times 10^{-4} \mathrm{~J}=\frac{1210}{3} \times 10^{-3} \mathrm{~J} \simeq 4 \times 10^{-2} \mathrm{~J}$

According to questions, $\frac{x}{100}=4 \times 10^{-2}$

$\therefore x=4$