Series RL at changing frequency: In a series circuit with an inductor and a resistor, the frequency is initially set so that XL = R. If the frequency is increased, which relationship becomes true for the voltage drops?

Introduction / Context:
In a series RL circuit, the same current flows through both elements, and the voltage drops are proportional to their impedances. Changing frequency alters the inductor's reactance, which in turn changes the relative drops.

Given Data / Assumptions:

• Series RL; initial condition XL = R.
• Frequency is increased from that starting point.
• Sinusoidal steady state with ideal components.

Concept / Approach:

Inductive reactance XL = 2 * π * f * L. Increasing f increases XL. With the same current I in series, VL = I * XL and VR = I * R. If XL becomes greater than R, then VL exceeds VR.

Step-by-Step Solution:

Verification / Alternative check:

Impedance magnitude Z = √(R^2 + XL^2) increases; current decreases, but the ratio VL/VR = XL/R increases > 1, so VL dominates.

Why Other Options Are Wrong:

VL = VR holds only at the specific frequency where XL = R. VR ≥ VL or VL < VR would require XL ≤ R, not true after increasing f.

Common Pitfalls:

Assuming current increase with frequency; in RL, higher frequency actually increases impedance and reduces current, but the ratio of drops still favors the inductor.

Final Answer:

VL > VR