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प्रश्न
In series LCR circuit, the plot of Imax vs ω is shown in figure. Find the bandwidth and mark in the figure.
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उत्तर
We know that bandwidth = (ω2 – ω1).
Where ω1 and ω2 are two frequencies where the current amplitude of LCR circuit becomes `12sqrt(12)` times (i.e., Irms) the value of current is maximum at resonant frequency.
`I = E_0/sqrt(2) = 1/sqrt(2)` = 0.707 Amp
From graph ω1 and ω2 at 0.707A current is 0.8 and 1.2 rad/sec.
So Bandwith ω2.ω1 = 1.2 – 0.8 = 0.4 rad/sec.
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