Two capacitors,each of value 10 mu F,are con | vedclass

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(C) For capacitors connected in parallel,the equivalent capacitance $C_{eq}$ is given by $C_{eq} = C_1 + C_2$.Given $C_1 = C_2 = 10 \ \mu F$,the equiv

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$0.4$

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(C) For capacitors connected in parallel,the equivalent capacitance $C_{eq}$ is given by $C_{eq} = C_1 + C_2$.Given $C_1 = C_2 = 10 \ \mu F$,the equivalent capacitance is $C_{eq} = 10 \ \mu F + 10 \ \mu F = 20 \ \mu F = 20 	imes 10^{-6} \ F$.The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} C_{eq} V^2$.Substituting the values $C_{eq} = 20 	imes 10^{-6} \ F$ and $V = 200 \ V$:$U = \frac{1}{2} 	imes (20 	imes 10^{-6}) 	imes (200)^2$$U = 10 	imes 10^{-6} 	imes 40000$$U = 10 	imes 10^{-6} 	imes 4 	imes 10^4$$U = 40 	imes 10^{-2} = 0.4 \ J$.

Two capacitors,each of value $10 \ \mu F$,are connected in parallel and charged to $200 \ V \ DC$. What is the energy stored in the combination in Joules?

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