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(B) The circuit consists of a bridge-like structure. Let us simplify it.$1$. The two $4 \mu F$ capacitors in the top branch (between $A$ and $B$) are

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$8$

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(B) The circuit consists of a bridge-like structure. Let us simplify it.$1$. The two $4 \mu F$ capacitors in the top branch (between $A$ and $B$) are in series. Their equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$, so $C_1 = 2 \mu F$.$2$. The two $4 \mu F$ capacitors in the bottom branches (connected to $C$) are also in series with each other. Their equivalent capacitance $C_2$ is $\frac{1}{C_2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$, so $C_2 = 2 \mu F$.$3$. This $C_2$ is in parallel with the middle $4 \mu F$ capacitor. Thus, the equivalent capacitance of this middle section is $C_3 = C_2 + 4 \mu F = 2 \mu F + 4 \mu F = 6 \mu F$.$4$. Finally, $C_1$ and $C_3$ are in parallel between points $A$ and $B$. Therefore, the total equivalent capacitance $C_{eq} = C_1 + C_3 = 2 \mu F + 6 \mu F = 8 \mu F$.

Find the equivalent capacitance between points $A$ and $B$ in the given figure. (in $\mu F$)

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