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JEE Main 2026 (5 Apr, Shift 2) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2026 (5 Apr, Shift 2)

Q8JEE Main 2026 · 5 APR · SHIFT 2Easy

Resonance, Bandwidth & Q Factor

A series LCR circuit with R=20 Ω, L=1.6 HR = 20\ \Omega,\ L = 1.6\text{ H} and C=40 μFC = 40\ \mu\text{F} is connected to a variable frequency a.c. source. The inductive reactance at resonant frequency is ‾ Ω\underline{\hspace{3em}}\ \Omega.

Numerical answer type. Work it out, then open the answer.

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Answer: 200

  1. ω0=1LC=11.6×40×10−6=164×10−6=18×10−3=125 rad/s\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{1.6 \times 40\times10^{-6}}} = \dfrac{1}{\sqrt{64\times10^{-6}}} = \dfrac{1}{8\times10^{-3}} = 125\text{ rad/s}
  2. XL=ω0L=125×1.6=200 ΩX_L = \omega_0 L = 125 \times 1.6 = 200\ \Omega
  3. So XL=200 ΩX_L = 200\ \Omega.
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