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(C) The $5 \ \mu F$ and $1 \ \mu F$ capacitors are in parallel,so their equivalent capacitance is $C_p = 5 \ \mu F + 1 \ \mu F = 6 \ \mu F$.This $6 \

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$32$

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(C) The $5 \ \mu F$ and $1 \ \mu F$ capacitors are in parallel,so their equivalent capacitance is $C_p = 5 \ \mu F + 1 \ \mu F = 6 \ \mu F$.This $6 \ \mu F$ capacitor is in series with the $3 \ \mu F$ capacitor. The equivalent capacitance of this branch is $C_{branch} = \frac{3 	imes 6}{3 + 6} = \frac{18}{9} = 2 \ \mu F$.The total charge flowing through this branch is $Q = C_{branch} 	imes V = 2 \ \mu F 	imes 24 \ V = 48 \ \mu C$.Let $q_1$ be the charge on the $5 \ \mu F$ capacitor and $q_2$ be the charge on the $1 \ \mu F$ capacitor. Since they are in parallel,the potential difference across them is the same: $\frac{q_1}{5} = \frac{q_2}{1}$,which implies $q_1 = 5q_2$.Since $q_1 + q_2 = 48 \ \mu C$,we have $5q_2 + q_2 = 48 \ \mu C$,so $6q_2 = 48 \ \mu C$,which gives $q_2 = 8 \ \mu C$.The energy stored in the $1 \ \mu F$ capacitor is $U = \frac{q_2^2}{2C} = \frac{(8 \ \mu C)^2}{2 	imes 1 \ \mu F} = \frac{64}{2} = 32 \ \mu J$.

In the circuit shown,the energy stored in the $1 \ \mu F$ capacitor is ...... $\mu J$.

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