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In series $\mathrm{LCR}$ circuit, the plot of $\mathrm{I}_{\max }$ versus $\omega$ is shown in figure. Find the bandwidth and mark in the figure.
Solution:
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Verified Answer
Let $\omega_1$ and $\omega_2$ are two frequencies at which magnitude of current is $\frac{1}{\sqrt{2}}$ times of maximum value for LCR circuit. i.e., $\quad \mathrm{I}_{\mathrm{ms}}=\frac{\mathrm{I}_{\max }}{\sqrt{2}}=\frac{1}{\sqrt{2}} \approx 0.7 \mathrm{Amp}$
The value of current is maximum at resonant frequency.
From the above diagram $\omega_1$ and $\omega_2$ at $0.7 \mathrm{amp}$ the corresponding frequencies are $0.8 \mathrm{rad} / \mathrm{s}$ and $1.2 \mathrm{rad} / \mathrm{s}$.
So,
For the given diagram,
Bandwidth $(\Delta \omega)=\omega_2-\omega_1=1.2-0.8$
$=0.4 \mathrm{rad} / \mathrm{sec}$
The value of current is maximum at resonant frequency.
From the above diagram $\omega_1$ and $\omega_2$ at $0.7 \mathrm{amp}$ the corresponding frequencies are $0.8 \mathrm{rad} / \mathrm{s}$ and $1.2 \mathrm{rad} / \mathrm{s}$.
So,
For the given diagram,
Bandwidth $(\Delta \omega)=\omega_2-\omega_1=1.2-0.8$
$=0.4 \mathrm{rad} / \mathrm{sec}$
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