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Question: Answered & Verified by Expert
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 
PhysicsAlternating CurrentJEE MainJEE Main 2020 (06 Sep Shift 2)
Solution:
2400 Upvotes Verified Answer
The correct answer is: 400

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×4×102mF

=4003πμF

N=400

 

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