Explanation

(d) $12\mu F$ and $6\mu F$ are in series and again are in parallel with $4\mu F$.
Therefore, resultant of these three will be
$= \frac{12 \times 6}{12 + 6} + 4 = 4 + 4 = 8\mu F$
This equivalent system is in series with $1\ \mu F$
Its equivalent capacitance $= \frac{8 \times 1}{8 + 1} = \frac{8}{9}\mu F$ …(i)
Equivalent of $8\mu F,\ 2\mu F$ and $2\mu F$
$= \frac{4 \times 8}{4 + 8} = \frac{32}{12} = \frac{8}{3}\mu F$ …(ii)
(i) and (ii) are in parallel and are in series with $C$
$\therefore\frac{8}{9} + \frac{8}{3} = \frac{32}{9}$ and $C_{eq} = 1 = \frac{\frac{32}{9} \times C}{\frac{32}{9} + C} \Rightarrow C = \frac{32}{23}\mu F$