A $5\, \mu F$ capacitor is charged fully by a $220\,V$ supply. It is then disconnected from the supply and is connected in series to another uncharged $2.5\;\mu F$ capacitor. If the energy change during the charge redistribution is $\frac{ X }{100} \;J$ then value of $X$ to the nearest integer is$.....$
JEE MAIN 2020, Medium
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$u _{ i }=\frac{1}{2} \times 5 \times 10^{-6}(220)^{2}$
Final common potential
$v=\frac{220 \times 5+0 \times 2.5}{5+2.5}=220 \times \frac{2}{3}$
$u _{ f }=\frac{1}{2}(5+2.5) \times 10^{-6}\left(220 \times \frac{2}{3}\right)^{2}$
$\Delta u = u _{ f }- u _{ i }$
$\Delta u =-403.33 \times 10^{-4}$
$\Rightarrow-403.33 \times 10^{-4}=\frac{ X }{100}$
$X=-4.03$
or magnitude or value of $X$ is approximate $4$
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