Inductors in series — two ideal inductors are connected end-to-end (series). If L1 = 0.20 mH and L2 = 0.40 mH, what is the total equivalent inductance seen by the source?

Introduction / Context:Series combinations of inductors appear in filter chokes, input EMI networks, and energy-storage chains. Knowing how to combine inductances quickly is a bread-and-butter skill in electronics. This problem asks for the equivalent inductance when two ideal inductors are connected in series, using values that add neatly to 0.60 mH.

Given Data / Assumptions:

• L1 = 0.20 mH (ideal; no winding resistance or coupling losses).
• L2 = 0.40 mH (ideal).
• Series connection; mutual coupling is negligible (k ≈ 0) unless otherwise stated.
• Small-signal or DC equivalent; we are only aggregating L values, not reactances at a specific frequency.

Concept / Approach:For uncoupled inductors in series, total inductance is the sum of individual inductances: L_total = L1 + L2. This mirrors resistors in series. If there were significant mutual coupling, a ±2M term would appear (M is mutual inductance), but in most layout situations with separated coils or orthogonal cores, M is small and the simple sum applies.

Step-by-Step Solution:

Verification / Alternative check:At a test frequency f, the net reactance is X_L = 2 * pi * f * L_total. If each coil has reactances X1 and X2 measured separately, you will observe X_total ≈ X1 + X2, confirming the series rule in practice (ignoring small parasitics and coupling).

Why Other Options Are Wrong:

• 0.40 mH or 0.20 mH: These correspond to individual values, not the sum.
• 0.80 mH: Would require an additional 0.20 mH or constructive coupling that adds effective inductance; not the given case.

Common Pitfalls:Confusing series and parallel formulas (parallel uses reciprocals), forgetting that tight coupling can modify the sum, and mixing units (µH vs mH). Always confirm units before adding.

Final Answer:0.60 mH is the total inductance for 0.20 mH and 0.40 mH in series (uncoupled).