Question

Consider the infinite ladder circuit shown below.

For which angular frequency $\omega$ will the circuit behave like a pure inductance?

• A. $\frac{LC}{\sqrt{2}}$
• B. $\frac{1}{\sqrt{LC}}$
• C. $\frac{2}{\sqrt{LC}}$
• D. $\frac{2L}{\sqrt{C}}$

Answer 1

Answer: (A), (B), (C), (D)

In an infinite network, adding one more loop does not affects total impedence of circuit.

$\Rightarrow Z_{CD}=Z_{AB}$
$\Rightarrow \left(Z c\right)\text{series} L=Z$
$\Rightarrow \frac{ZX_{c}}{Z+X_{C}}+X_{L}=Z $
$\Rightarrow \frac{Z /\omega C}{Z+\frac{1}{\omega C}}=Z-\omega L$
$\Rightarrow \frac{\frac{Z}{\omega C}}{Z+\frac{1}{\omega C}}=Z-\omega L $
$\Rightarrow Z^{2}-\omega LZ-\frac{L}{C}=0 $
$\Rightarrow Z=\omega L\pm\sqrt{\omega^{2}L^{2}+\frac{4L}{C}}$
Now, $Z = \omega L$ or circuit is purelyinductive when,co
$\omega^{2}L^{2}+\frac{4L}{C}=0$
$\Rightarrow \omega^{2}=-\frac{4}{LC} $
But this is not possible or $\omega$ is imaginary.
Hence, circuit does not behaves like an inductor.