Question Detail
Question
An inductor, L = 0.5H, in series with a resistor, R = 250Ω, are connected to an a.c. source 100V, 50Hz. Determine the power factor of the circuit.
Correct Answer
Option B is the correct answer.
Detailed Explanation
Given: Inductance (L) = 0.5 H, Resistance (R) = 250 Ω, Frequency (f) = 50 Hz.
First, calculate the angular frequency (ω):
ω = 2πf = 2π(50 Hz) = 100π rad/s.
Next, calculate the inductive reactance (XL):
XL = ωL = 100π * 0.5 H = 50π Ω ≈ 157.08 Ω.
Now, calculate the impedance (Z) of the series R-L circuit:
Z = √(R2 + XL2) = √(2502 + (50π)2)
Z = √(62500 + 24674.01) = √(87174.01) ≈ 295.25 Ω.
The power factor (cos φ) is given by R/Z:
Power factor = R / Z = 250 Ω / 295.25 Ω ≈ 0.8467.
The closest option is 0.85.
Hint
The power factor in an AC R-L circuit is given by the ratio of resistance to impedance (R/Z). Remember to calculate inductive reactance (XL = ωL) and impedance (Z = √(R2 + XL2)) first.