Series RL phase relationship: A 100 mH inductor has inductive reactance of 6 kΩ at the operating frequency. It is in series with a 1 kΩ resistor and driven by a source. What is the phase angle between source voltage and current?

Introduction / Context:
In AC analysis, series RL circuits exhibit a phase shift between the source voltage and the circuit current due to the inductor’s reactance. Estimating this angle quickly helps in phasor calculations, power factor estimation, and filter design.

Given Data / Assumptions:

• Inductive reactance XL = 6 kΩ.
• Resistance R = 1 kΩ in series.
• Sinusoidal steady-state operation.

Concept / Approach:
The phase angle φ of a series RL circuit (voltage referenced to current) is given by φ = arctan(XL / R). A larger ratio XL/R yields a larger lagging angle, since current lags voltage in inductive circuits. We compute the arctangent using the given values.

Step-by-Step Solution:
Compute ratio: XL / R = 6000 / 1000 = 6.Find angle: φ = arctan(6) ≈ 80.5 degrees (rounded).Compare to options: closest available is 81.0 degrees.Therefore, the phase angle is approximately 81.0 degrees (current lags voltage).

Verification / Alternative check:
Compute impedance magnitude: |Z| = sqrt(R^2 + XL^2) = sqrt(1^2 + 6^2) kΩ = sqrt(37) kΩ ≈ 6.083 kΩ. Then cos φ = R / |Z| = 1 / 6.083 ≈ 0.164, so φ ≈ arccos(0.164) ≈ 80.6°, consistent with arctan calculation.

Why Other Options Are Wrong:
0.1° and 9.0°: far too small; would require XL much less than R.61.0°: corresponds to a smaller XL/R ratio (about 1.8), not 6.

Common Pitfalls:
Forgetting that angle is referenced as the voltage leading the current in inductive circuits; current lags by φ in an RL series network.

Final Answer:
81.0 degrees