What is the sharpness of resonance and obtain an equation for the $Q$ factor?

(N/A) The resonance frequency is independent of resistance $R$,but the sharpness of resonance is inversely proportional to $R$.
Sharpness of resonance $= \frac{\omega_{0}}{2 \Delta \omega}$
$= \frac{\omega_{0} L}{R}$
Here,$\frac{\omega_{0} L}{R}$ is defined as the quality factor $Q$ of the circuit.
$\therefore$ Quality factor $Q = \frac{\omega_{0} L}{R}$
$\therefore Q = \frac{\text{Resonant frequency}}{\text{Bandwidth}} = \frac{\omega_{0}}{\omega_{2} - \omega_{1}}$
$\therefore 2 \Delta \omega = \frac{\omega_{0}}{Q}$,where $\omega_{2}$ and $\omega_{1}$ are the frequencies at which the current amplitude is $\frac{1}{\sqrt{2}}$ of $(I_{\text{rms}})_{\max}$.
Thus,a larger value of $Q$ implies a smaller bandwidth $(2 \Delta \omega)$ and a sharper resonance.
Using $\omega_{0}^{2} = \frac{1}{LC}$,we can also write:
$Q = \frac{\omega_{0} L}{R} = \frac{\omega_{0}}{R} \times \frac{1}{\omega_{0}^{2} C} = \frac{1}{\omega_{0} R C}$
If the resonance is less sharp,the maximum current is lower,and the circuit remains near resonance for a larger range of frequencies $\Delta \omega$,resulting in poor tuning. Therefore,a less sharp resonance implies lower selectivity,and a higher $Q$ factor implies higher selectivity.