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

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

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$80$

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(C) Initially,the energy stored in the $2\,\mu F$ capacitor is:$U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} (2 	imes 10^{-6}) V^{2} = V^{2} 	imes 10^{-6} \, J$Initially,the charge stored in the $2\,\mu F$ capacitor is:$Q_{i} = C V = (2 	imes 10^{-6}) V = 2V 	imes 10^{-6} \, C$When the switch $S$ is turned to position $2,$ the charge flows and both capacitors share charges until a common potential $V_{c}$ is reached.$V_{c} = \frac{	ext{Total Charge}}{	ext{Total Capacitance}} = \frac{2V 	imes 10^{-6}}{(2 + 8) 	imes 10^{-6}} = \frac{V}{5} \, V$Finally,the energy stored in both capacitors is:$U_{f} = \frac{1}{2} (C_{1} + C_{2}) V_{c}^{2} = \frac{1}{2} (10 	imes 10^{-6}) \left(\frac{V}{5}ight)^{2} = 5 	imes 10^{-6} 	imes \frac{V^{2}}{25} = \frac{V^{2}}{5} 	imes 10^{-6} \, J$Percentage loss of energy is given by:$\Delta U \% = \frac{U_{i} - U_{f}}{U_{i}} 	imes 100 \%$$\Delta U \% = \frac{V^{2} 	imes 10^{-6} - \frac{V^{2}}{5} 	imes 10^{-6}}{V^{2} 	imes 10^{-6}} 	imes 100 \% = \left(1 - \frac{1}{5}ight) 	imes 100 \% = \frac{4}{5} 	imes 100 \% = 80 \%$

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

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