Series RL phase behavior: When the resistor voltage becomes less than the inductor voltage in a series RL circuit, how does the overall phase angle change?

Introduction / Context:
Phase angle in a series RL circuit indicates how far the current lags the source voltage. It depends on the ratio of inductive reactance to resistance, and it can be inferred by comparing the voltage drops across the inductor and the resistor.

Given Data / Assumptions:

• Series RL circuit.
• Observed: VR < VL.
• Ideal linear components.

Concept / Approach:
In series RL, VR:VL equals R:XL because the same current flows through both. If VR < VL, then R < XL, which means the impedance angle theta = arctan(XL / R) is larger. A larger ratio XL/R yields a larger theta (more lag).

Step-by-Step Reasoning:
Given VR < VL ⇒ R < XLtheta = arctan(XL / R)If XL/R increases, then theta increasesTherefore, the phase angle increases

Verification / Alternative check:
Phasor diagram: the vertical component (inductive) becomes longer than the horizontal (resistive). The angle between the impedance vector and the real axis grows, confirming an increased phase angle.

Why Other Options Are Wrong:

• Decreases / not affected: Contradicts theta = arctan(XL/R).
• Cannot be determined: It can be determined directly from the voltage ratio in series.

Common Pitfalls:
Confusing series and parallel intuition or thinking VL and VR can be compared arithmetically without considering their phase relationship with current. In series, magnitudes directly reflect R and XL because current is common.

Final Answer:
increases