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JEE Main 20266 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
6 April
Shift
Morning Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2026 · 6 April · Morning Shift · Physics · Alternating Current

A LCR series circuit driven with E_ rms = 90 V at frequency f_d =...

A LCR series circuit driven with Erms=90E_{rms} = 90 V at frequency fd=30f_d = 30 Hz has resistance R=80 ΩR = 80\,\Omega, an inductance with inductive reactance XL=20.0 ΩX_L = 20.0\,\Omega and capacitance with capacitive reactance XC=80.0 ΩX_C = 80.0\,\Omega. The power factor of the circuit is _______.

Options

  1. A0.80.8
  2. B0.640.64
  3. C0.90.9
  4. D0.50.5

Correct answer

A. 0.80.8

Step-by-step solution

The impedance of the LCR series circuit is given by Z=R2+(XC−XL)2Z = \sqrt{R^2 + (X_C - X_L)^2}.

Substituting the given values:

Z=802+(80−20)2Z = \sqrt{80^2 + (80 - 20)^2}

Z=802+602Z = \sqrt{80^2 + 60^2}

Z=6400+3600=10000=100 ΩZ = \sqrt{6400 + 3600} = \sqrt{10000} = 100\,\Omega

The power factor of the circuit is given by cos⁡ϕ=RZ\cos \phi = \dfrac{R}{Z}.

cos⁡ϕ=80100=0.8\cos \phi = \dfrac{80}{100} = 0.8

Answer: 0.80.8
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