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(A) $1$. The circuit consists of three parallel branches connected between points $A$ and $B$.$2$. The top branch has two $4\, \mu F$ capacitors in se

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$8$

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(A) $1$. The circuit consists of three parallel branches connected between points $A$ and $B$.$2$. The top branch has two $4\, \mu F$ capacitors in series. Their equivalent capacitance is $C_1 = \frac{4 	imes 4}{4 + 4} = 2\, \mu F$.$3$. The middle branch has a single $4\, \mu F$ capacitor,so $C_2 = 4\, \mu F$.$4$. The bottom branch has two $4\, \mu F$ capacitors in series. Their equivalent capacitance is $C_3 = \frac{4 	imes 4}{4 + 4} = 2\, \mu F$.$5$. Since these three branches are in parallel,the total equivalent capacitance is $C_{AB} = C_1 + C_2 + C_3 = 2 + 4 + 2 = 8\, \mu F$.

The equivalent capacitance between $A$ and $B$ is......$\mu F$.

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