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(A) $1$. The two $2\,\mu F$ capacitors are in parallel. Their equivalent capacitance is $C_1 = 2 + 2 = 4\,\mu F$.$2$. This $C_1$ is in series with the

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$\frac{32}{23}\,\mu F$

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(A) $1$. The two $2\,\mu F$ capacitors are in parallel. Their equivalent capacitance is $C_1 = 2 + 2 = 4\,\mu F$.$2$. This $C_1$ is in series with the $8\,\mu F$ capacitor. Their equivalent is $C_2 = \frac{4 	imes 8}{4 + 8} = \frac{32}{12} = \frac{8}{3}\,\mu F$.$3$. The $6\,\mu F$ and $12\,\mu F$ capacitors are in series. Their equivalent is $C_3 = \frac{6 	imes 12}{6 + 12} = \frac{72}{18} = 4\,\mu F$.$4$. This $C_3$ is in parallel with the $4\,\mu F$ capacitor. Their equivalent is $C_4 = 4 + 4 = 8\,\mu F$.$5$. This $C_4$ is in series with the $1\,\mu F$ capacitor. Their equivalent is $C_5 = \frac{8 	imes 1}{8 + 1} = \frac{8}{9}\,\mu F$.$6$. Now, $C_2$ and $C_5$ are in parallel. Their equivalent is $C_6 = C_2 + C_5 = \frac{8}{3} + \frac{8}{9} = \frac{24 + 8}{9} = \frac{32}{9}\,\mu F$.$7$. Finally, $C$ is in series with $C_6$. Given the total equivalent capacitance $C_{eq} = 1\,\mu F$, we have $\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C_6}$.$8$. $1 = \frac{1}{C} + \frac{9}{32} \Rightarrow \frac{1}{C} = 1 - \frac{9}{32} = \frac{23}{32}$.$9$. Therefore, $C = \frac{32}{23}\,\mu F$.

The figure shows a network of capacitors where the numbers indicate capacitances in microfarads $(\mu F)$. The value of capacitance $C$ if the equivalent capacitance between points $A$ and $B$ is to be $1\,\mu F$ is:

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