Ncert Exercises & Solutions

In the series LCR circuit, the plot of $I_{\max} versus ω$ is shown in the figure. Find the bandwidth and mark in the figure.

Hint: The bandwidth depends on the frequencies corresponding to

\frac{1}{\sqrt{2}}

times value of the maximum current.

Step 1: Consider the diagram.

Bandwidth =

ω_{2} - ω_{1}

Where

ω_{1} and ω_{2}

corresponds to frequencies at which magnitude of the current is

\frac{1}{\sqrt{2}}

times of maximum value.

Step 2:

I_{rms} = \frac{I_{\max}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \approx 0.7 A

Clearly from the diagram, the corresponding frequencies are 0.8 rad/s an 1.2 rad /s.

∆ ω = Bandwidth = 1.2 - 0.8 = 0.4 rad / s