Series RL sensitivity: If the frequency is reduced by one-half while the resistance is doubled, how does the total impedance of a series RL circuit change?

Introduction / Context:
This question examines how the impedance of a series RL circuit responds when both frequency and resistance are changed in opposite directions. It tests understanding of how resistive and inductive parts combine vectorially rather than arithmetically.

Given Data / Assumptions:

• A series RL circuit with resistance R and inductive reactance XL.
• Frequency is halved ⇒ XL, which depends on frequency, is reduced.
• Resistance is doubled at the same time.
• Assume ideal inductor (no winding resistance) and sinusoidal steady state.

Concept / Approach:
Impedance magnitude for a series RL circuit is Z = sqrt(R^2 + XL^2), where XL = 2 * pi * f * L. If f is halved, XL becomes XL/2. If R is doubled, R becomes 2R. The new impedance is Znew = sqrt((2R)^2 + (XL/2)^2). Whether Znew is larger or smaller than the original depends on the relative sizes of R and XL.

Step-by-Step Solution:

Verification / Alternative check:
Try numerical examples: R = 10 Ω, XL = 100 Ω ⇒ Zorig ≈ 100.5 Ω; Znew ≈ 50.0 Ω (halves). Conversely, R = 100 Ω, XL = 10 Ω ⇒ Zorig ≈ 100.5 Ω; Znew ≈ 200.1 Ω (nearly doubles).

Why Other Options Are Wrong:

• 'doubles' / 'halves' / 'remains constant': Each may occur only for specific R–XL ratios, not universally.
• 'increases by one-fourth': Not supported by the impedance formula.

Common Pitfalls:

• Adding R and XL directly instead of using vector (root-sum-square) combination.
• Forgetting that XL depends on frequency while R does not.

Final Answer:
cannot be determined without values