Compute total inductive reactance — two uncoupled inductors are in series and driven at f = 1.0 kHz. If L1 = 0.30 H and L2 = 0.25 H, what is the total inductive reactance as seen by the source?

Introduction / Context:
Reactance depends on both element value and operating frequency. For inductors in series, the inductances add first; their combined reactance is then X_L = 2 * pi * f * L_total. This calculation is common when sizing drive stages or estimating filter impedances at test frequencies such as 1 kHz.

Given Data / Assumptions:

• L1 = 0.30 H, L2 = 0.25 H, series and uncoupled.
• Frequency f = 1.0 kHz.
• Ideal inductors; we ignore resistance and coupling parasitics for this calculation.

Concept / Approach:
Total inductance for series inductors: L_total = L1 + L2. Then compute X_L = 2 * pi * f * L_total. Convert the answer to kilo-ohms for comparison with the options. This two-step process (sum L, then apply formula) is the reliable approach.

Step-by-Step Solution:

Verification / Alternative check:
Back-of-envelope: 2 * pi * 1k ≈ 6.28k; times 0.55 ≈ 3.45k Ω, consistent with the detailed computation. A quick SPICE AC analysis would report a magnitude near 3.46 kΩ at 1 kHz ignoring resistive losses.

Why Other Options Are Wrong:

• ≈ 4.4 kΩ: Would correspond to L_total ≈ 0.70 H at 1 kHz; our L_total is 0.55 H.
• ≈ 16.3 kΩ or ≈ 23.3 kΩ: These imply multi-henry totals or much higher frequency; not the given values.

Common Pitfalls:
Forgetting to sum L before applying the formula; mixing angular frequency ω = 2 * pi * f with f; or unit errors (using mH without converting to H). Keep track of base units to avoid kilo-factor mistakes.

Final Answer:
≈ 3.5 kΩ total inductive reactance at 1.0 kHz.