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(D) $1$. First,consider the right-most part of the circuit. The $12\,\mu F$ and $6\,\mu F$ capacitors are in series. Their equivalent capacitance is $

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

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(D) $1$. First,consider the right-most part of the circuit. The $12\,\mu F$ and $6\,\mu F$ capacitors are in series. Their equivalent capacitance is $C_1 = \frac{12 	imes 6}{12 + 6} = \frac{72}{18} = 4\,\mu F$.$2$. This $C_1$ is in parallel with the $4\,\mu F$ capacitor. Their equivalent is $C_2 = 4 + 4 = 8\,\mu F$.$3$. This $C_2$ is in series with the $1\,\mu F$ capacitor. Their equivalent is $C_3 = \frac{8 	imes 1}{8 + 1} = \frac{8}{9}\,\mu F$.$4$. Now consider the bottom-left part. The two $2\,\mu F$ capacitors are in parallel. Their equivalent is $C_4 = 2 + 2 = 4\,\mu F$.$5$. This $C_4$ is in series with the $8\,\mu F$ capacitor. Their equivalent is $C_5 = \frac{4 	imes 8}{4 + 8} = \frac{32}{12} = \frac{8}{3}\,\mu F$.$6$. The branches containing $C_3$ and $C_5$ are in parallel. Their equivalent is $C_6 = C_3 + C_5 = \frac{8}{9} + \frac{8}{3} = \frac{8 + 24}{9} = \frac{32}{9}\,\mu F$.$7$. Finally,$C_6$ is in series with capacitor $C$. Given the total equivalent capacitance is $1\,\mu F$,we have $1 = \frac{C_6 	imes C}{C_6 + C} = \frac{\frac{32}{9} 	imes C}{\frac{32}{9} + C}$.$8$. Solving for $C$: $1 = \frac{32C}{32 + 9C} \implies 32 + 9C = 32C \implies 23C = 32 \implies C = \frac{32}{23}\,\mu F$.

In the following circuit,the resultant capacitance between $A$ and $B$ is $1\,\mu F$. Then the value of $C$ is

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