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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 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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प्रश्न

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

टीपा लिहा
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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 ω21 = 1.2 – 0.8 = 0.4 rad/sec.

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पाठ 7: Alternating Current - MCQ I [पृष्ठ ४३]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 12
पाठ 7 Alternating Current
MCQ I | Q 7.18 | पृष्ठ ४३

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