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(A) $1$. Let the nodes be $M$,$N$,and $L$ as shown in the diagram.$2$. The two $2 \ \mu F$ capacitors between $M$ and $N$ are in parallel,so their equ

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$2/3$

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(A) $1$. Let the nodes be $M$,$N$,and $L$ as shown in the diagram.$2$. The two $2 \ \mu F$ capacitors between $M$ and $N$ are in parallel,so their equivalent capacitance is $C_{MN} = 2 \ \mu F + 2 \ \mu F = 4 \ \mu F$.$3$. Now,the $4 \ \mu F$ capacitor (from $M$ to $N$) and the $4 \ \mu F$ capacitor (from $N$ to $L$) are in series. Their equivalent capacitance $C_{ML}$ is given by $\frac{1}{C_{ML}} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$,so $C_{ML} = 2 \ \mu F$.$4$. The circuit now consists of three capacitors in series: the first $2 \ \mu F$ capacitor,the equivalent $2 \ \mu F$ capacitor $(C_{ML})$,and the last $2 \ \mu F$ capacitor.$5$. The total equivalent capacitance $C_{eq}$ is $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$.$6$. Therefore,$C_{eq} = 2/3 \ \mu F$.

Find the equivalent capacitance between $A$ and $B$ in $\mu F$.

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