d
(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 \(C\)
 \(\frac{8}{9} + \frac{8}{3} = \frac{{32}}{9}\) and \({C_{eq}} = 1 = \frac{{\frac{{32}}{9} \times C}}{{\frac{{32}}{9} + C}}\)

\(==>\) \(C = \frac{{32}}{{23}}\,\mu F\)