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

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(D) Initial energy stored in the $2 \ \mu F$ capacitor is $U_i = \frac{1}{2} C_1 V^2 = \frac{1}{2} 	imes 2 	imes 10^{-6} 	imes V^2 = 10^{-6} V^2 \

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$80$

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(D) Initial energy stored in the $2 \ \mu F$ capacitor is $U_i = \frac{1}{2} C_1 V^2 = \frac{1}{2} 	imes 2 	imes 10^{-6} 	imes V^2 = 10^{-6} V^2 \ 	ext{J}$.When the switch is turned to position $2$,the charge $Q = C_1 V = 2 	imes 10^{-6} V$ is shared between the $2 \ \mu F$ and $8 \ \mu F$ capacitors connected in parallel.The common potential $V_f$ is given by $V_f = \frac{Q}{C_1 + C_2} = \frac{2 	imes 10^{-6} V}{2 	imes 10^{-6} + 8 	imes 10^{-6}} = \frac{2V}{10} = 0.2V$.The final energy stored in the system is $U_f = \frac{1}{2} (C_1 + C_2) V_f^2 = \frac{1}{2} 	imes (10 	imes 10^{-6}) 	imes (0.2V)^2 = 5 	imes 10^{-6} 	imes 0.04 V^2 = 0.2 	imes 10^{-6} V^2 \ 	ext{J}$.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 (in $\%$)

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