Question

If in the infinite series circuit, $C=9\mu F$ and $C_1=6\mu F$ then the capacity across $AB$ is

• A. $12\mu \text{F}$
• B. $6\mu \text{F}$
• C. $3\mu \text{F}$
• D. $1.25\mu \text{F}$

Answer 1

The correct option is C $3\mu \text{F}$


$\frac{1}{C_{AB}}=\frac{1}{x}=\frac{2}{9}+\frac{1}{(6+x)}$

$\frac{1}{x}-\frac{2}{9}=\frac{1}{6+x}$

$\frac{9-2x}{9x}=\frac{1}{6+x}$

$(9-2x)(6+x)=9x$

$54+9x-12x-2x^2=9x$

$2x^2+12x-54=0$

$x^2+6x-27=0$

$(x+9)(x-3)=0$

$x=-9\mu \text{F};x=3\mu \text{F}$

$\therefore x=3\mu \text{F}$