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(C) Let the circuit be analyzed by identifying the series and parallel combinations.$1$. The top branch consists of two $1 \, \mu F$ capacitors in ser

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$\frac{8}{3} \, \mu F$

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(C) Let the circuit be analyzed by identifying the series and parallel combinations.$1$. The top branch consists of two $1 \, \mu F$ capacitors in series. Their equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{1} + \frac{1}{1} = 2$,so $C_1 = 0.5 \, \mu F$.$2$. However,looking at the circuit diagram,the central $1 \, \mu F$ capacitor is in parallel with the top-left $1 \, \mu F$ capacitor. Let's re-evaluate: The top-left $1 \, \mu F$ and the middle $1 \, \mu F$ are in parallel. Their equivalent is $1 + 1 = 2 \, \mu F$.$3$. This $2 \, \mu F$ combination is in series with the top-right $1 \, \mu F$ capacitor. The equivalent capacitance of this upper branch is $C_{upper} = \frac{2 	imes 1}{2 + 1} = \frac{2}{3} \, \mu F$.$4$. This entire upper branch is in parallel with the bottom $2 \, \mu F$ capacitor.$5$. The total equivalent capacitance $C_{eq} = C_{upper} + 2 = \frac{2}{3} + 2 = \frac{2 + 6}{3} = \frac{8}{3} \, \mu F$.

The equivalent capacitance between points $A$ and $B$ of the circuit shown will be

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