Question

In the following circuit, the resultant capacitance between $A$ and $B$ is $1\mu F$. Then value of $C$ is

• A. $\frac{32}{11}\mu F$
• B. $\frac{11}{32}\mu F$
• C. $\frac{23}{32}\mu F$
• D. $\frac{32}{23}\mu F$

Answer 1

Answer: (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 $C_{eq}=1=\frac{\frac{32}{9}\times C}{\frac{32}{9}+C}$
$\Rightarrow C=\frac{32}{23}\mu F$