Question

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

The correct option is D $\frac{32}{23}\mu F$

Since capacitor of capacitance $6\mu F$ and $12\mu F$ are in series and their equivalent is in parallel with $4\mu F$, hence their equivalent will be $4+\frac{6\times 12}{6+12}=8\mu F$
Simillarly, both $2\mu F$ capacitors are in parallel connections, therefore their equivalent capacitance will be $2+2=4\mu F$


Since capacitor $8\mu F$ and $4\mu F$ are in series, their equivalent resistance will be $\frac{8\times 4}{8+4}=\frac{8}{3}\mu F$
Simillarly equivalent of $1\mu F$ and $8\mu F$ will be $\frac{8\times 1}{8+1}=\frac{8}{9}\mu F$
Now, $\frac{8}{3}\mu F$ and $\frac{8}{9}\mu F$ will be in parallel and their equivalent will be $\frac{8}{3}+\frac{8}{9}=\frac{32}{9}\mu F$​​​​​​​​​​​​​​
Since total equivalent resistance is $1\mu F$, therefore equivalent capacitance of $C$ and $\frac{32}{9}\mu F$​​​​​​​​​​​​​​ will be $1\mu F$.
$\Rightarrow \frac{C\times 32/9}{C+32/9}=1$
$\Rightarrow C=\frac{32}{23}\mu F$