A 5, mu F capacitor is charged fully by a 220 | vedclass

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Question

$u _{ i }=\frac{1}{2} 	imes 5 	imes 10^{-6}(220)^{2}$

Final common potential

$v=\frac{220 	imes 5+0 	imes 2.5}{5+2.5}=220 	imes \frac{2}{3}

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$u _{ i }=\frac{1}{2} 	imes 5 	imes 10^{-6}(220)^{2}$

Final common potential

$v=\frac{220 	imes 5+0 	imes 2.5}{5+2.5}=220 	imes \frac{2}{3}$

$u _{ f }=\frac{1}{2}(5+2.5) 	imes 10^{-6}\left(220 	imes \frac{2}{3}ight)^{2}$

$\Delta u = u _{ f }- u _{ i }$

$\Delta u =-403.33 	imes 10^{-4}$

$\Rightarrow-403.33 	imes 10^{-4}=\frac{ X }{100}$

$X=-4.03$

or magnitude or value of $X$ is approximate $4$

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