Question

In the following circuit, the resultant capacitance between $\text{A}$ and $\text{B}$ is $1\mu \text{F}$. Then the value of $C$ is $\frac{32}{x}\mu \text{F}$

Answer 1

On simplifying the given circuit,



For the given figure we can see $6\mu \text{F}$ and $12\mu \text{F}$ are in series on simplify we get,

$\because \frac{6\times 12}{6+12}=4\mu \text{F}$



Similarly simplifying the series and parallel combination of capacitors we get,


$\because C_1=\frac{8}{3}+\frac{8}{9}=\frac{8+24}{9}=\frac{32}{9}\mu \text{F}$





$\frac{1}{C_{AB}}=\frac{1}{C}+\frac{1}{C_1}$

$1=\frac{1}{C}+\frac{9}{32}$

$1-\frac{9}{32}=\frac{1}{C}$

$\frac{1}{C}=\frac{23}{32}$

$\therefore C=\frac{32}{23}\mu \text{F}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (d) is the correct answer.