The figure shows a network of capacitors where the number indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 $$\mu F$$ is:

Observe the section containing the $$6\ \mu F$$, $$12\ \mu F$$, and $$4\ \mu F$$ capacitors.

$$C_{s1} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\ \mu F$$

$$C_{p1} = 4 + 4 = 8\ \mu F$$

This $$8\ \mu F$$ equivalent is in series with the $$1\ \mu F$$ capacitor at the top.

$$C_{\text{right}} = \frac{1 \times 8}{1 + 8} = \frac{8}{9}\ \mu F$$

Now, look at the branch containing the $$8\ \mu F$$ and the two $$2\ \mu F$$ capacitors.

$$C_{p2} = 2 + 2 = 4\ \mu F$$

$$C_{\text{left}} = \frac{8 \times 4}{8 + 4} = \frac{32}{12} = \frac{8}{3}\ \mu F$$

$$C_{\text{parallel}} = C_{\text{left}} + C_{\text{right}} = \frac{8}{3} + \frac{8}{9}$$

$$C_{\text{parallel}} = \frac{24 + 8}{9} = \frac{32}{9}\ \mu F$$

$$\frac{1}{C_{AB}} = \frac{1}{C} + \frac{1}{C_{\text{parallel}}}$$

$$1 = \frac{1}{C} + \frac{9}{32}$$

$$C = \frac{32}{23}\ \mu F$$