**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 $\frac{\mathrm{X}}{100} \mathrm{~J}$ then value of

$\mathrm{X}$ to the nearest integer is

**Solution:**

$\mathrm{u}_{\mathrm{i}}=\frac{1}{2} \times 5 \times 10^{-6}(220)^{2}$

Final common potential

$\mathrm{v}=\frac{220 \times 5+0 \times 2.5}{5+2.5}=220 \times \frac{2}{3}$

$\mathrm{u}_{\mathrm{f}}=\frac{1}{2}(5+2.5) \times 10^{-6}\left(220 \times \frac{2}{3}\right)^{2}$

$\Delta \mathrm{u}=\mathrm{u}_{\mathrm{f}}-\mathrm{u}_{\mathrm{i}}$

$\Delta \mathrm{u}=-403.33 \times 10^{-4}$

$\Rightarrow-403.33 \times 10^{-4}=\frac{\mathrm{X}}{100}$

or magnitude or value of $\mathrm{X}$ is approximate 4