Question

In a series $L C R$ circuit, half power frequencies (frequency at which power dissipated is half the maximum power dissipated in the circuit) are found to be $90 \,rad / s$ and $40\, rad / s$. Find the value of $\frac{\omega_{0}}{10}$, where $\omega_{0}$ is the resonance frequency (in rad / s ).

Answer 1

At half power frequencies,
$Z=R \sqrt{2}$
$ \Rightarrow\left|X_{L}-X_{C}\right|=R $
$\left(\omega L-\frac{1}{\omega C}\right)^{2}=R^{2} $
$\Rightarrow \omega^{2} L^{2}+\frac{1}{\omega^{2} C^{2}}-\frac{2 L}{C}=R^{2} $
$\omega^{4} L^{2} C^{2}-\omega^{2}\left(\frac{2 L}{C}+R^{2}\right) C^{2}+1=0$
The product of roots is
$\omega_{1}^{2} \omega_{2}^{2}=\frac{1}{L^{2} C^{2}}=\omega_{1} \omega_{2}=\frac{1}{L C}=\omega_{0}^{2} $
$\Rightarrow \omega_{0}=\sqrt{\omega_{1} \omega_{2}}=60$ rad / s