A 100 mH inductor (XL = 6 kΩ at the operating frequency) is in series with a 1 kΩ resistor and an AC source. What is the magnitude of the phase angle between the source voltage and current?

Introduction:
The phase angle in a series RL circuit indicates how far the current lags the voltage due to inductive reactance. This question verifies your ability to compute the angle from R and XL, a common task in AC circuit design and power calculations.

Given Data / Assumptions:

• Series RL circuit.
• R = 1 kΩ.
• XL = 6 kΩ (given explicitly for the operating frequency).
• Ideal sinusoidal steady state.

Concept / Approach:
For series RL, the phase angle φ (magnitude) satisfies tan φ = XL / R, since reactive and resistive drops are orthogonal. The current lags the voltage by φ for an inductive circuit.

Step-by-Step Solution:
1) Write tan φ = XL / R.2) Substitute values: tan φ = 6000 / 1000 = 6.3) Compute φ = arctan(6) ≈ 80.5°.4) Rounded to the nearest whole-number option, φ ≈ 81.0°.

Verification / Alternative check:
Use the impedance triangle: Z = sqrt(R^2 + XL^2) and cos φ = R / Z. With R = 1 kΩ and XL = 6 kΩ, Z ≈ sqrt(1^2 + 6^2) kΩ = sqrt(37) kΩ ≈ 6.083 kΩ, giving cos φ ≈ 1 / 6.083 ≈ 0.164 and φ ≈ 80.6°, consistent with 81°.

Why Other Options Are Wrong:
0.1° and 9.0°: would require XL ≪ R; not the case here.61.0°: corresponds to a smaller XL/R ratio (tan φ ≈ 1.8), not the given 6.

Common Pitfalls:
Confusing series and parallel angle relationships or using degrees vs radians incorrectly. Always compute φ with tan φ = XL / R for series RL.

Final Answer:
81.0°