In a series $LCR$ circuit,the plot of $I_{max}$ vs $\omega$ is shown in the figure. Find the bandwidth and mark it in the figure.

(N/A) The bandwidth of a series $LCR$ circuit is defined as the frequency range between the two half-power frequencies,$\omega_{1}$ and $\omega_{2}$,where the current amplitude drops to $\frac{1}{\sqrt{2}}$ times its maximum value.
Given $I_{max} = 1.0 \ A$,the current at half-power frequencies is $I = \frac{I_{max}}{\sqrt{2}} = \frac{1.0}{1.414} \approx 0.707 \ A$.
From the provided graph,at $I = 0.707 \ A$,the corresponding angular frequencies are $\omega_{1} = 0.8 \ rad/s$ and $\omega_{2} = 1.2 \ rad/s$.
The bandwidth $\Delta \omega$ is given by:
$\Delta \omega = \omega_{2} - \omega_{1}$
$\Delta \omega = 1.2 \ rad/s - 0.8 \ rad/s = 0.4 \ rad/s$.