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JEE Main 2023 (31 Jan, Shift 2) · Resonance, Bandwidth & Q Factor

Resonance, Bandwidth & Q Factor question from JEE Main 2023 (31 Jan, Shift 2)

Q71JEE Main 2023 · 31 JAN · SHIFT 2Difficult

Resonance, Bandwidth & Q Factor

A series LCRLCR circuit consists of R=80 Ω,XL=100 ΩR = 80\ \Omega, X_L = 100\ \Omega, and XC=40 ΩX_C = 40\ \Omega. The input voltage is 2500 cos⁡(100πt) V\cos(100\pi t)\text{ V}. The amplitude of current, in the circuit, is _________ A.

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

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Answer: 25

  1. Z=R2+(XL−XC)2=802+602=100 ΩZ = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{80^2 + 60^2} = 100\ \Omega
  2. I0=V0Z=2500100=25 AI_0 = \frac{V_0}{Z} = \frac{2500}{100} = 25\text{ A}
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