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:
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:
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°.
Discussion & Comments