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(C) The circuit consists of a $4 \mu F$ capacitor in series with a parallel combination of $1 \mu F$ and $5 \mu F$ capacitors. First,calculate the equ

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$24$

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(C) The circuit consists of a $4 \mu F$ capacitor in series with a parallel combination of $1 \mu F$ and $5 \mu F$ capacitors. First,calculate the equivalent capacitance of the parallel combination: $C_p = 1 \mu F + 5 \mu F = 6 \mu F$. Now,the circuit simplifies to a $4 \mu F$ capacitor in series with a $6 \mu F$ capacitor connected across a $10 V$ battery. The equivalent capacitance $C_{eq}$ of the series combination is given by $\frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12}$,so $C_{eq} = 2.4 \mu F$. The total charge $Q$ drawn from the battery is $Q = C_{eq} 	imes V = 2.4 \mu F 	imes 10 V = 24 \mu C$. Since the $4 \mu F$ capacitor is in series with the rest of the circuit,the charge on it is equal to the total charge $Q = 24 \mu C$.

For the circuit shown in the figure,the charge on the $4 \mu F$ capacitor is $....... \mu C$.

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