Question

An infiinite circuit is connected across an $ac$ source (connecting terminals $A$ and $B$) of angular frequency $\omega$ as shown. If the circuit is purely inductive, then $\omega =\frac{p}{\sqrt{LC}}$. The value of $p$ is

Answer 1

As circuit is infinite , the equivalent reactance between $AB$ and $CD$ be considered as $Z$


Now as $Z$ and $X_C$ are in parallel effective impedence between them is $Z^{\prime }=\frac{Z{X}_c}{Z+X_c}$

Equivalent impedence between $AB$ is $Z=\omega L+Z^{\prime }$

$\therefore Z=\omega L+\frac{Z{X}_c}{Z+X_c}=\omega L+\frac{Z}{\frac{\omega C}{Z+\frac{1}{\omega C}}}$
$\Rightarrow Z(Z+\frac{1}{\omega C})=Z\omega L+\frac{L}{C}+\frac{Z}{\omega C}$
$\Rightarrow Z^2=Z\omega L+\frac{L}{C}$
On solving $Z=\frac{\omega LC\pm \sqrt{(\omega LC{)}^2-4LC}}{2C}$
For $Z$ to be purely inductive
$(\omega LC{)}^2-4LC=0\Rightarrow \omega =\frac{2}{\sqrt{LC}}$