Inductive reactance calculation in AC circuits: What is the inductive reactance X_L of a 0.1 H coil operating at a frequency of 1000 Hz?

Introduction / Context:This problem tests your ability to compute inductive reactance in alternating-current (AC) circuit analysis. Inductive reactance expresses how strongly an inductor resists a change in current at a given frequency, and it grows with both inductance and frequency. Mastery of this concept is essential for filter design, impedance matching, and understanding current limits in AC systems.

Given Data / Assumptions:

• Inductance L = 0.1 H.
• Frequency f = 1000 Hz.
• Assume an ideal inductor (no winding resistance, no core losses).
• Use standard formula for inductive reactance in ohms.

Concept / Approach:The inductive reactance is calculated using X_L = 2 * pi * f * L. Because X_L is proportional to frequency, doubling f doubles X_L. This linear relationship is a cornerstone of frequency-domain circuit behavior for inductors.

Step-by-Step Solution:Write the formula: X_L = 2 * pi * f * L.Insert values: X_L = 2 * pi * 1000 * 0.1.Simplify: 1000 * 0.1 = 100 → X_L = 2 * pi * 100 = 200 * pi.Numerical value: 200 * pi ≈ 200 * 3.1416 ≈ 628.32 ohms.Rounded to the nearest whole number: 628 Ohm.

Verification / Alternative check:Quick estimation: 2 * 3.14 * 1000 * 0.1 = 628. Confirmed by calculator and back-of-the-envelope methods; both yield approximately 628 Ω.

Why Other Options Are Wrong:(a) 100 Ω ignores the factor 2 * pi. (b) “Less than 1 Ω” contradicts the strong frequency dependence. (d) 444 Ω is an arbitrary miscalculation. (e) Incorrect since 628 Ω is attainable from the standard formula.

Common Pitfalls:Dropping the factor of pi, mixing units (mH vs H), or using frequency in kHz without converting to Hz correctly can all lead to wrong answers.

Final Answer:628 Ohm.