JEE Main 2020PhysicsAlternating CurrentHardNumerical

JEE Main 2020 — Alternating Current Question with Solution

JEE Main 2020 (06 Sep Shift 2)

Question

In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as n3π μF, then value of n is 

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Show full solutionCorrect answer: 400
Correct answer
400

Step-by-step explanation

P=Vm. in cosϕ

400=250×1m×0.8

irms=2A

1m2.R=P

4×R=400

⇒ R=100Ω.

cosϕ=RR2+XL2

1002+XL2=1000.82

1002+XL2=1000.82

XL=75Ω

Power factor is unity

XC=XL=75

1ω=75

⇒ C=175×2H×50=17500πF

3π×2500

=13π×4×102 mF

=4003πμF

N=400

 

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About this question

This is a previous-year question from JEE Main 2020, covering the Alternating Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.