Single-inductor reactance — in a circuit operating at f = 1.4 kHz, an inductor L2 has inductance 0.50 H. What is the inductive reactance of L2?

Introduction / Context:
Inductive reactance quantifies how strongly an inductor resists changes in current at a given frequency. This calculation is central to filter design, impedance matching, and estimating current draw from AC sources. The problem asks for X_L of a single inductor at a specified frequency.

Given Data / Assumptions:

• L2 = 0.50 H (ideal).
• Frequency f = 1.4 kHz.
• We ignore winding resistance and core losses for this calculation.

Concept / Approach:
The inductive reactance magnitude is X_L = 2 * pi * f * L. Substitute L = 0.50 H and f = 1.4 kHz to obtain a numerical value in ohms, then express in kilo-ohms for comparison with the options.

Step-by-Step Solution:

Verification / Alternative check:
Back-of-envelope: 2 * pi ≈ 6.28; 6.28 * 1.4k ≈ 8.79k; half of that (for 0.5 H) ≈ 4.40k Ω. A quick simulator would report essentially the same magnitude when plotting |Z_L| at 1.4 kHz.

Why Other Options Are Wrong:

• ≈ 3.5 kΩ: Corresponds to about 0.40 H at 1.4 kHz or 0.55 H at 1.0 kHz; not our case.
• ≈ 16.3 kΩ or ≈ 23.3 kΩ: These would require multiple henries or much higher frequency.

Common Pitfalls:
Mixing f and angular frequency ω; remember ω = 2 * pi * f. Also, avoid unit slips (H vs mH) which can introduce thousand-fold errors.

Final Answer:
≈ 4.4 kΩ is the inductive reactance of L2 at 1.4 kHz.