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JEE Main 2020 (6 Sep, Shift 1) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2020 (6 Sep, Shift 1)

Q102JEE Main 2020 · 6 SEP · SHIFT 1Easy

Resonance, Bandwidth & Q Factor

An AC circuit has R= 100 Ω\Omega, C = 2 μ\muF and L = 80 mH, connected in series. The quality factor of the circuit is:

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Answer: (b)

  1. Q=1RLCQ = \frac{1}{R}\sqrt{\frac{L}{C}}
  2. Q=110080×10−32×10−6=11004×104=200100Q = \frac{1}{100}\sqrt{\frac{80\times10^{-3}}{2\times10^{-6}}} = \frac{1}{100}\sqrt{4\times10^{4}} = \frac{200}{100}
  3. So the answer is (b) 2.
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