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(C) Initially,the capacitor is connected to the $200\,V$ source. The energy stored in the capacitor is:$U = \frac{1}{2} CV^{2} = \frac{1}{2} 	imes (5

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$2$

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(C) Initially,the capacitor is connected to the $200\,V$ source. The energy stored in the capacitor is:$U = \frac{1}{2} CV^{2} = \frac{1}{2} 	imes (5 	imes 10^{-6}\,F) 	imes (200\,V)^{2} = \frac{1}{2} 	imes 5 	imes 10^{-6} 	imes 40000 = 0.1\,J$.When the switch is shifted to contact $2$,the capacitor discharges through the series combination of the $300\,\Omega$ and $500\,\Omega$ resistors.The total resistance in the circuit is $R_{eq} = 300\,\Omega + 500\,\Omega = 800\,\Omega$.The total heat generated in the circuit is equal to the initial energy stored in the capacitor,$H_{total} = 0.1\,J$.The heat generated in a specific resistor is proportional to its resistance: $H_{500} = \left( \frac{R_{500}}{R_{eq}} ight) 	imes H_{total}$.$H_{500} = \left( \frac{500}{800} ight) 	imes 0.1 = \frac{5}{8} 	imes 0.1 = \frac{5}{8} 	imes \frac{1}{10} = \frac{5}{80} = \frac{1}{16}\,J$.To match the options,we express this as $\frac{2}{32}\,J$.

$A$ capacitor of capacitance $5\,\mu F$ is connected to a source of constant $emf$ of $200\,V$ for a long time. Then,the switch is shifted from contact $1$ to contact $2$. The total amount of heat generated in the $500\,\Omega$ resistance thereafter is: (in $/32\,J$)

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