Frequency dependence of inductors: How does inductive reactance behave as the excitation frequency changes in a purely inductive circuit?

Introduction / Context:Inductive reactance determines how strongly an inductor resists AC. Designers use this dependence to create filters, chokes, and impedance-matching networks that behave differently over frequency. Recognizing the trend with frequency is fundamental to AC analysis.

Given Data / Assumptions:

• Ideal inductor with inductance L.
• Sinusoidal steady-state excitation at frequency f.
• No significant parasitics considered.

Concept / Approach:Inductive reactance is defined by XL = 2 * pi * f * L. Thus, for fixed inductance L, XL grows linearly with frequency. Higher frequency means larger reactance, which reduces current for a given applied AC voltage (I = V / XL in a purely inductive branch).

Step-by-Step Solution:Formula: XL = 2 * pi * f * L.Holding L constant, compare f1 and f2: if f2 > f1 then XL2 > XL1.Therefore, as frequency increases, XL increases.Conversely, lowering frequency reduces XL proportionally.

Verification / Alternative check:Plot XL versus f; the graph is a straight line through the origin with slope 2 * pi * L, confirming linear dependence.

Why Other Options Are Wrong:Decrease with frequency: opposite of the formula.Independent from frequency: contradicts XL expression.Depend on XC: capacitive reactance is unrelated to the definition of XL.

Common Pitfalls:Mixing up XL and XC. Remember XC = 1 / (2 * pi * f * C) decreases with frequency, whereas XL increases.

Final Answer:increase with frequency