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(B) To find the equivalent capacitance between $A$ and $B$,we simplify the circuit from the right side towards the left.$1$. The rightmost branch has

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(B) To find the equivalent capacitance between $A$ and $B$,we simplify the circuit from the right side towards the left.$1$. The rightmost branch has two capacitors of $1 \mu F$ and $2 \mu F$ in parallel. Their equivalent is $C_1 = 1 + 2 = 3 \mu F$.$2$. Now,this $3 \mu F$ is in series with the $3 \mu F$ capacitor on the top and the $3 \mu F$ capacitor on the bottom. The equivalent capacitance of these three in series is $\frac{1}{C_{eq1}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \Rightarrow C_{eq1} = 1 \mu F$.$3$. This $1 \mu F$ is in parallel with the $2 \mu F$ capacitor. Their equivalent is $C_2 = 1 + 2 = 3 \mu F$.$4$. Now,this $3 \mu F$ is in series with the next $3 \mu F$ capacitor on the top and the $3 \mu F$ capacitor on the bottom. The equivalent capacitance is $\frac{1}{C_{eq2}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \Rightarrow C_{eq2} = 1 \mu F$.$5$. This $1 \mu F$ is in parallel with the $2 \mu F$ capacitor. Their equivalent is $C_3 = 1 + 2 = 3 \mu F$.$6$. Finally,this $3 \mu F$ is in series with the first $3 \mu F$ capacitor on the top and the $3 \mu F$ capacitor on the bottom. The total equivalent capacitance is $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \Rightarrow C_{eq} = 1 \mu F$.

The equivalent capacitance between $A$ and $B$ in the given circuit is (in $\mu F$)

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