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(C) The capacitors are connected in series. The equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{

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$0.15$

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(C) The capacitors are connected in series. The equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{5 \mu F} + \frac{1}{10 \mu F} = \frac{2+1}{10 \mu F} = \frac{3}{10 \mu F}$$C_{eq} = \frac{10}{3} \mu F = \frac{10}{3} 	imes 10^{-6} \ F$The total energy $U$ stored in the series combination is given by:$U = \frac{1}{2} C_{eq} V^2$$U = \frac{1}{2} 	imes (\frac{10}{3} 	imes 10^{-6} \ F) 	imes (300 \ V)^2$$U = \frac{1}{2} 	imes \frac{10}{3} 	imes 10^{-6} 	imes 90000$$U = \frac{1}{2} 	imes \frac{10}{3} 	imes 10^{-6} 	imes 9 	imes 10^4$$U = \frac{1}{2} 	imes 30 	imes 10^{-2} = 15 	imes 10^{-2} \ J = 0.15 \ J$

$A$ $5 \mu F$ capacitor is connected in series with a $10 \mu F$ capacitor. When a $300 \ V$ potential difference is applied across this combination,the total energy stored in the capacitors is (in $J$)

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