Find $C$ if the equivalent capacitance between the points $A$ and $B$ in the circuit shown is $1 \mu F$. All the capacitances are in $\mu F$.

After a brief analysis, we conclude the following information

(a) $2$ and $2$ are in parallel to give $4 \mu F$

(b)  $6$ and $12$ are in series to give $\frac{6\times 12}{6+12}=4 \mu F$

(c) This $4 \mu F$ (net capacitance of  $6$ and $12$) is in parallel combination to another $4$, so as to give is $8 \mu F$

So, using the above information, we redraw the above circuit

Further, we conclude

(d) $1$ and $8$ are in series, so we get $\frac{1\times 8}{1+8}=\frac{8}{9} \mu F$

(e) $8$ and $4$ are in series, so we get $\frac{8\times 4}{8+4}=\frac{32}{12} \mu F$

(f) $\frac{8}{9} \mu F$ and $\frac{32}{12} \mu F$ are in a parallel combination to give net capacitance $\frac{8}{9}+\frac{32}{12}=\frac{32}{9}\mu F$

Finally  $\frac{32}{9} \mu F$ and $C$ are in series between $A$ and $B$, so as to give $1 \mu F$. So, the final equivalent circuit is shown in figure.

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{32}{9}}C}{{\displaystyle \frac{32}{9}}+C}=1 \\ \Rightarrow 32C=32+9C \\ \Rightarrow 23C=32 \\ \Rightarrow C=\frac{32}{23} \mu F \end{gathered}$$