Series RL comparison: At a fixed component set, how should frequency be changed so that the resistor's RMS voltage drop is less than the inductor's RMS voltage drop?

Introduction / Context:
In a series RL circuit, the voltage division between the resistor and inductor depends on reactance, which varies with frequency. This question checks practical intuition about how to make the inductor's drop dominate the resistor's drop.

Given Data / Assumptions:

• Series RL with fixed R and L.
• AC steady state with variable frequency f.
• VR = I * R and VL = I * XL with XL = 2 * pi * f * L.

Concept / Approach:
For a given current I, VR is independent of frequency (proportional to R), whereas VL grows linearly with frequency (proportional to XL). To make VL > VR, increase f so that XL > R in magnitude when scaled by the same current.

Step-by-Step Solution:

Verification / Alternative check:
At very low f (approaching DC), XL → 0 and VR dominates. At higher f, XL ≫ R, so VL dominates. This matches lab observation of frequency-dependent phasor drops.

Why Other Options Are Wrong:

• 'decreased': Lowers XL and makes VR dominate.
• 'doubled': Doubling does increase VL but the best generic instruction is simply 'increase frequency'; a specific factor is not required.
• 'not a factor': Frequency is the key factor here.
• 'reduced to DC': Makes VL → 0.

Common Pitfalls:

• Assuming VR also depends on frequency in a pure R–L series circuit.

Final Answer:
increased