RL series circuit — phase angle when resistance dominates: In a series R–L AC circuit, if the resistance R is larger than the inductive reactance X_L (i.e., R > X_L), what is the approximate range of the circuit phase angle φ between source voltage and current?

Difficulty: Easy

Correct Answer: between 0° and 45°

Explanation:


Introduction / Context:
In AC circuit analysis, the phase angle φ between voltage and current in a series R–L network indicates how inductive the circuit is. Recognizing how φ changes as the resistance R and the inductive reactance X_L vary is essential for power factor correction, load characterization, and transformer/motor drive applications. This question asks for the qualitative range of φ when R dominates, i.e., when R > X_L.


Given Data / Assumptions:

  • Series R–L circuit driven by a sinusoidal source.
  • Resistance R and inductive reactance X_L = 2 * pi * f * L.
  • Phase angle φ defined as the angle by which voltage leads current in an R–L series circuit (current lags by φ).
  • Linear, time-invariant, steady-state conditions.


Concept / Approach:
The impedance of a series R–L circuit is Z = R + j X_L. The phase angle is φ = arctan(X_L / R). When R > X_L, the ratio X_L / R is less than 1, which directly implies 0° < φ < 45°. As R increases relative to X_L, φ approaches 0°, meaning the circuit becomes more resistive with improved power factor. Conversely, if X_L dominates (X_L > R), φ approaches 90° (strongly inductive).


Step-by-Step Solution:

Write φ = arctan(X_L / R) for a series R–L circuit.Use the given condition R > X_L ⇒ X_L / R < 1.Since arctan of a positive number smaller than 1 lies between 0° and 45°, conclude 0° < φ < 45°.Interpretation: the circuit is inductive (positive φ), but only mildly so because resistance dominates.


Verification / Alternative check:
Try numeric examples. If R = 10 Ω and X_L = 5 Ω, φ = arctan(0.5) ≈ 26.6°, which indeed lies between 0° and 45°. If R approaches X_L (e.g., both 10 Ω), φ ≈ 45°. If R ≫ X_L, φ tends to 0°, reflecting a nearly resistive load.


Why Other Options Are Wrong:

  • less than 0° (capacitive lead): A negative φ indicates capacitive behavior, which is not possible with only R and L in series.
  • exactly 45°: Occurs only when R = X_L, not when R > X_L.
  • between 45° and 90°: Requires X_L > R (inductance dominating).


Common Pitfalls:
Confusing magnitude dominance (R vs X_L) with sign of φ; forgetting that inductors cause current to lag (positive φ) while capacitors cause current to lead (negative φ); mixing up series versus parallel cases.


Final Answer:
between 0° and 45°.

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