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JEE Mains · Physics · STD 12 -Alternating current
  • JEE Mains 2021
  • Medium
  • Numerical

A resonance circuit having inductance and resistance \(2 \times 10^{-4} H\) and \(6.28 \Omega\) respectively oscillates at \(10\, MHz\) frequency. The value of quality factor of this resonator is ......... \([\pi=3.14]\)

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Answer & Solution

Correct Answer

(A) \(2000\)

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Detailed explanation

Given : \(\quad L =2 \times 10^{-4} H\)

\(R =6.28 \Omega\)

\(f =10 MHz =10^{7} Hz\)

Since quality factor,

\(Q =\omega_{0} \frac{ L }{ R }=2 \pi f \frac{ L }{ R }\)

\(\therefore Q =2 \pi \times 10^{7} \times \frac{2 \times 10^{-4}}{6.28}\)

\(Q =2 \times 10^{3}=2000\)

\(\therefore\) Ans. is 2000

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