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Q.
In the following circuit, the resultant capacitance between $A$ and $B$ is $1\, \mu F$. Then value of $C$ is
Electrostatic Potential and Capacitance
Solution:
$12\, \mu F$ 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 $C_{e q}=1=\frac{\frac{32}{9} \times C}{\frac{32}{9}+C}$
$\Rightarrow C=\frac{32}{23} \mu F$
