In a series RL circuit where R = 100 Ω and XL = 100 Ω at the operating frequency, which statement about circuit behavior is correct?

Introduction:
When resistance and inductive reactance are equal in a series RL circuit, a number of convenient relationships emerge that are useful for hand calculations. This question focuses on identifying the correct qualitative statement from several tempting but incorrect alternatives.

Given Data / Assumptions:

• Series RL configuration.
• R = 100 Ω and XL = 100 Ω.
• Ideal sinusoidal operation; RMS quantities implied.

Concept / Approach:
For series RL, tan φ = XL / R. When R = XL, tan φ = 1, so φ = 45°. The impedance magnitude is Z = sqrt(R^2 + XL^2) = R * sqrt(2), and the power factor PF = cos φ = cos 45° ≈ 0.707, not 1. Voltage drops depend on current and source voltage, not fixed values like 5 V unless specified.

Step-by-Step Solution:
1) Compute φ: tan φ = XL / R = 100 / 100 = 1 → φ = 45°.2) Compute Z: Z = sqrt(100^2 + 100^2) ≈ 141.4 Ω (for context).3) Compute PF: PF = cos φ = cos 45° ≈ 0.707, confirming PF ≠ 1.4) Conclude that the only universal statement from the options is that the phase angle equals 45°.

Verification / Alternative check:
The impedance triangle with equal legs confirms a 45° angle; this is a classic special case in AC circuit theory and appears in standard tables and examples.

Why Other Options Are Wrong:
Each component drops 5 V: voltage drops depend on current and source value; no 5 V given.Impedance equals 200 Ω: would be true only if R and XL added arithmetically, which they do not.Power factor equals 1: PF would be 0.707 for φ = 45°, not unity.

Common Pitfalls:
Adding resistive and reactive magnitudes linearly or assuming unity PF when R = XL. Remember that orthogonal phasors lead to 45° and PF ≈ 0.707 in this equality case.

Final Answer:
The phase angle equals 45°.