Q67 / 90 · KVPY 2010 (SB/SX) · Physics
Consider the infinite ladder circuit shown below.
For which angular frequency $\omega$ will the circuit behave like a pure inductance?
For which angular frequency $\omega$ will the circuit behave like a pure inductance?
- A$\frac{\mathrm{LC}}{\sqrt{2}}$
- B$\frac{1}{\sqrt{\mathrm{LC}}}$
- Correct answerC$\frac{2}{\sqrt{L C}}$
- D$\frac{2 \mathrm{~L}}{\sqrt{\mathrm{C}}}$
Solution
Let the equivalent impedance of the circuit be $Z$
So $Z=\omega L+Z^{\prime}$
now $Z^{\prime}=\frac{Z X_{C}}{Z+X_{C}}$
$$
\therefore Z=\omega L+\left(\frac{Z \times \frac{1}{\omega C}}{Z+\frac{1}{\omega C}}\right)
$$
on ving we get $Z=\frac{\omega L C \pm \sqrt{(\omega L C)^{2}-4 L C}}{2 C}$
for $Z$ to be purely inductive
$$
\omega^{2} L^{2} C^{2}-4 L C=0 \text { or } \omega=\frac{2}{\sqrt{L C}}
$$
Hence, the correct answer is option: 'C'.