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(A) The circuit consists of a capacitor $C_1 = 2 \mu F$ in parallel with a series combination of two capacitors $C_2 = 6 \mu F$ and $C_3 = 3 \mu F$.Fi

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$4$

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(A) The circuit consists of a capacitor $C_1 = 2 \mu F$ in parallel with a series combination of two capacitors $C_2 = 6 \mu F$ and $C_3 = 3 \mu F$.First,calculate the equivalent capacitance $C'$ of the series combination of $C_2$ and $C_3$:$\frac{1}{C'} = \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{6} + \frac{1}{3} = \frac{1+2}{6} = \frac{3}{6} = \frac{1}{2} \mu F^{-1}$Therefore,$C' = 2 \mu F$.Next,calculate the total equivalent capacitance $C_{eq}$ of the parallel combination of $C_1$ and $C'$:$C_{eq} = C_1 + C' = 2 \mu F + 2 \mu F = 4 \mu F$.Thus,the effective capacitance is $4 \mu F$.

As shown in the circuit below,the effective capacitance is . . . . . . $\mu F$.

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