JEE Main 2020 — Alternating Current Question with Solution
From: JEE Main 2020 (Online) 6th September Evening Slot
Question
In a series LR circuit, power of 400W 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 F, then value of n is __________.
Enter your answer
Show full solutionCorrect answer: 400
Correct answer
400
Step-by-step explanation
Given, power factor of LR circuit,
cos = 0.8 = =
We know,
Power, P =
400 =
Z = 125
R = 0.8 125 = 100
As Z2 =
XL = 75
In 2nd case given.
Power factor = 1
that means XL = XC (Resonance condition)
XL =
75 =
C =
Also given, C = F
n = 400
cos = 0.8 = =
We know,
Power, P =
400 =
Z = 125
R = 0.8 125 = 100
As Z2 =
XL = 75
In 2nd case given.
Power factor = 1
that means XL = XC (Resonance condition)
XL =
75 =
C =
Also given, C = F
n = 400
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