In an LC circuit, angular frequency at resonance is $\omega$. The new angular frequency when inductance is…
- $\frac{\omega}{2 \sqrt{2}}$
- $\frac{\omega}{4 \sqrt{2}}$
- $\frac{\omega}{4}$
- $\frac{\omega}{\sqrt{2}}$
Solution
The angular frequency at resonance for an LC circuit is given by $\omega = \frac{1}{\sqrt{LC}}$.
When both inductance and capacitance are increased by factors of 4 and 8 respectively, the new inductance becomes $L' = 4L$ and the new capacitance $C' = 8C$. The new angular frequency is therefore
$\omega' = \frac{1}{\sqrt{L'C'}} = \frac{1}{\sqrt{(4L)(8C)}} = \frac{1}{\sqrt{32LC}} = \frac{1}{4\sqrt{2}\sqrt{LC}}$.
Recognizing that $\frac{1}{\sqrt{LC}} = \omega$, we substitute to obtain
$\omega' = \frac{\omega}{4\sqrt{2}}$.
The new angular frequency is thus reduced by a factor of $4\sqrt{2}$.
Asked in: MHT CET 2025 (05 May Shift 2)