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Alternating Current question

2002 · Shift 0 · Q127
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Alternating Current question

2002 · Shift 0 · Q127

JEE MainPhysicsAlternating CurrentMCQ+4 / −1
The power factor of ACACAC circuit having resistance (R)(R)(R) and inductance (L)(L)(L) connected in series and an angular velocity ω\omegaω is
  1. A
    R/ωLR/\omega LR/ωL
  2. B
    R/(R2+ω2L2)1/2R/{\left( {{R^2} + {\omega ^2}{L^2}} \right)^{1/2}}R/(R2+ω2L2)1/2
  3. C
    ωL/R\omega L/RωL/R
  4. D
    R/(R2−ω2L2)1/2R/{\left( {{R^2} - {\omega ^2}{L^2}} \right)^{1/2}}R/(R2−ω2L2)1/2
View written solutionFree

Correct answer: B

  1. Impedance of a series RRR-LLL AC circuit

For a resistor RRR and inductor LLL in series, the inductive reactance is XL=ωLX_L = \omega LXL​=ωL

Hence, the total impedance is Z=R2+XL2=R2+(ωL)2Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (\omega L)^2}Z=R2+XL2​​=R2+(ωL)2​

  1. Definition of power factor

The power factor is cos⁡ϕ=resistanceimpedance=RZ\cos\phi = \frac{\text{resistance}}{\text{impedance}} = \frac{R}{Z}cosϕ=impedanceresistance​=ZR​

Substituting ZZZ: cos⁡ϕ=RR2+ω2L2\cos\phi = \frac{R}{\sqrt{R^2 + \omega^2 L^2}}cosϕ=R2+ω2L2​R​

  1. Match with the options
  • A: RωL\dfrac{R}{\omega L}ωLR​ — incorrect
  • B: RR2+ω2L2\dfrac{R}{\sqrt{R^2 + \omega^2 L^2}}R2+ω2L2​R​ — correct
  • C: ωLR\dfrac{\omega L}{R}RωL​ — incorrect
  • D: RR2−ω2L2\dfrac{R}{\sqrt{R^2 - \omega^2 L^2}}R2−ω2L2​R​ — incorrect
  1. Final answer

The power factor is RR2+ω2L2\boxed{\frac{R}{\sqrt{R^2 + \omega^2 L^2}}}R2+ω2L2​R​​ So, the correct option is B.

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