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Karnataka Board PUCPUC Science 2nd PUC Class 12

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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Question

In series LCR circuit, the plot of Imax vs ω is shown in figure. Find the bandwidth and mark in the figure.

Short/Brief Note
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Solution

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 ω21 = 1.2 – 0.8 = 0.4 rad/sec.

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Chapter 7: Alternating Current - MCQ I [Page 43]

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NCERT Exemplar Physics Exemplar [English] Class 12
Chapter 7 Alternating Current
MCQ I | Q 7.18 | Page 43

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