Inductive reactance and current: A sinewave voltage is applied across an inductor. If the frequency of the applied voltage is decreased, what happens to the current through the inductor (steady state)?

Introduction / Context:
Inductors oppose changes in current via inductive reactance X_L, which depends on frequency. Understanding how current varies with frequency for a fixed applied voltage is essential in filter design, AC analysis, and impedance matching.

Given Data / Assumptions:

• Pure inductor with inductance L.
• Applied sinusoidal voltage of fixed amplitude.
• Steady-state AC (no transients considered).

Concept / Approach:
Inductive reactance: X_L = 2 * π * f * L. Current magnitude with fixed voltage is I = V / X_L. Thus, as frequency f decreases, X_L decreases linearly, so current increases correspondingly. This inverse relationship holds for ideal inductors without series resistance considered.

Step-by-Step Solution:

Verification / Alternative check:
Numerical example: Let V = 10 V, L = 10 mH. At 1 kHz: X_L ≈ 62.8 Ω → I ≈ 159 mA. At 100 Hz: X_L ≈ 6.28 Ω → I ≈ 1.59 A, showing the increase clearly (ideal case).

Why Other Options Are Wrong:

• is decreased: Opposite of the reactance effect.
• does not change: Would only be true if X_L were constant, which it is not.
• momentarily goes to zero: Transient behavior is not part of steady-state AC analysis.

Common Pitfalls:

• Confusing inductors with capacitors (where current increases with frequency in a different way).

Final Answer:
is increased