Power factor of a series RL: A 1.5 kΩ resistor in series with an inductor of 2.2 kΩ reactance is driven by an 18 V AC source. What is the circuit power factor?

Introduction / Context:
Power factor (PF) indicates how effectively current contributes to real power in AC circuits. In a series RL circuit, PF equals cos(theta), where theta is the angle of the impedance. Knowing PF is vital for power delivery and minimizing losses.

Given Data / Assumptions:

• R = 1.5 kΩ = 1500 Ω.
• XL = 2.2 kΩ = 2200 Ω.
• Series RL; voltage magnitude is 18 V (value not needed to compute PF).

Concept / Approach:
For a series RL, PF = cos(theta) = R / |Z|, where |Z| = sqrt(R^2 + XL^2). This follows from the right-triangle relationship of the impedance phasor diagram.

Step-by-Step Solution:
|Z| = sqrt( R^2 + XL^2 )|Z| = sqrt( 1500^2 + 2200^2 )|Z| = sqrt( 2.25e6 + 4.84e6 ) = sqrt( 7.09e6 ) ≈ 2663.8 ΩPF = R / |Z| = 1500 / 2663.8 ≈ 0.5634 ≈ 0.564

Verification / Alternative check:
Angle method: theta = arctan( XL / R ) = arctan( 2200 / 1500 ) ≈ arctan(1.4667) ≈ 55.7°. Then PF = cos(55.7°) ≈ 0.564, matching the above result.

Why Other Options Are Wrong:

• 564, 6.76, 55.7: These are dimensionless numbers linked to intermediate steps (e.g., degrees) but not valid for PF, which must lie between 0 and 1.

Common Pitfalls:
Using R/XL instead of R/|Z|, or misinterpreting angle in degrees as PF. Remember PF is the cosine of the phase angle and must be ≤ 1.

Final Answer:
0.564