Consider the infinite ladder circuit shown below.

For which angular frequency ω will the circuit behave like a pure inductance ?

Options

A. `(LC)/(sqrt(2))`
B. `(1)/(sqrt(LC))`
C. `(2)/(sqrt(LC))`
D. `(2L)/(sqrt(C ))`

Answer

C

Explanation

Let the equivalent impendence of the circuit be Z
So `Z = omega L + Z.`
Now `Z. = (ZX_(c))/(Z + X_(c))" " :. Z = omega L + ((Z xx (1)/(omega C))/(Z + (1)/(omegaC))) `
on solving we get `Z = (omega LC pm sqrt((omega LC)^(2) - 4LC))/(2C)`
for Z to be purely inductive `omega^(2) L^(2) C^(2) - 4 LC = 0 ` or `omega = (2)/( sqrt(LC))`