આકૃતિમાં દર્શાવ્યા અનુસાર 2 mu F ના કેપેસિટરન | vedclass

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Question

Initially, the energy stored in $2\, \mu F$ capacitor is

$U_{i}=\frac{1}{2} C V^{2}=\frac{1}{2}\left(2 	imes 10^{-6}ight)\, V^{2}=V^{2} 	imes 1

Vedclass · Accepted Answer

$80$

Vedclass · Answer

Initially, the energy stored in $2\, \mu F$ capacitor is

$U_{i}=\frac{1}{2} C V^{2}=\frac{1}{2}\left(2 	imes 10^{-6}ight)\, V^{2}=V^{2} 	imes 10^{-6} \,\mathrm{J}$

Initially, the charge stored in $2 \,\mu \mathrm{F}$ capacitor is

$Q_{i}=C V=\left(2 	imes 10^{-6}ight) V=2 V 	imes 10^{-6}$ coulomb. When switch $S$ is turned to position $2,$ the charge flows and both the capacitors share charges till a common potential $V_{c}$ is reached.

$V_{C}=\frac{	ext { total charge }}{	ext { total capacitance }}=\frac{2 V 	imes 10^{-6}}{(2+8) 	imes 10^{-6}}=\frac{V}{5} 	ext { volt }$

Finally, the energy stored in both the capacitors

$U_{f}=\frac{1}{2}\left[(2+8) 	imes 10^{-6}ight]\left(\frac{V}{5}ight)^{2}=\frac{V^{2}}{5} 	imes 10^{-6}\, \mathrm{J}$

$\%$ loss of energy, $\Delta U=\frac{U_{i}-U_{f}}{U_{i}} 	imes 100 \%$

$=\frac{\left(V^{2}-V^{2} / 5ight) 	imes 10^{-6}}{V^{2} 	imes 10^{-6}} 	imes 100 \%=80 \%$

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