In the following circuit, the resultant capacitance between $A$ and $B$ is $1\,\mu F$. Then value of $C$ is
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(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 ${C_{eq}} = 1 = \frac{{\frac{{32}}{9} \times C}}{{\frac{{32}}{9} + C}}$
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 ${C_{eq}} = 1 = \frac{{\frac{{32}}{9} \times C}}{{\frac{{32}}{9} + C}}$
$==>$ $C = \frac{{32}}{{23}}\,\mu F$
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