A 2,mu F capacitor is charged as shown in the | vedclass

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(D) Initial energy stored in the $2\,\mu F$ capacitor when connected to voltage $V$ is:$U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} (2 	imes 10^{-6}) V

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$80$

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(D) Initial energy stored in the $2\,\mu F$ capacitor when connected to voltage $V$ is:$U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} (2 	imes 10^{-6}) V^{2} = 10^{-6} V^{2} 	ext{ J}$.When the switch $S$ is turned to position $2$,the $2\,\mu F$ capacitor is connected in parallel with the uncharged $8\,\mu F$ capacitor.The total charge $Q$ remains conserved:$Q = C_{1} V = (2 	imes 10^{-6}) V$.The common potential $V_{	ext{common}}$ after redistribution is:$V_{	ext{common}} = \frac{Q}{C_{1} + C_{2}} = \frac{2 	imes 10^{-6} V}{2 	imes 10^{-6} + 8 	imes 10^{-6}} = \frac{2V}{10} = \frac{V}{5}$.The final energy stored in the system is:$U_{f} = \frac{1}{2} (C_{1} + C_{2}) V_{	ext{common}}^{2} = \frac{1}{2} (10 	imes 10^{-6}) \left(\frac{V}{5}ight)^{2} = 5 	imes 10^{-6} 	imes \frac{V^{2}}{25} = 0.2 	imes 10^{-6} V^{2} 	ext{ J}$.The energy dissipated is $\Delta U = U_{i} - U_{f} = 10^{-6} V^{2} - 0.2 	imes 10^{-6} V^{2} = 0.8 	imes 10^{-6} V^{2} 	ext{ J}$.The percentage of energy dissipated is:$\frac{\Delta U}{U_{i}} 	imes 100 = \frac{0.8 	imes 10^{-6} V^{2}}{10^{-6} V^{2}} 	imes 100 = 80 \%$.

$A$ $2\,\mu F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$ is.....$\%$

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