Series combination of inductors: A 2 mH, a 3.3 mH, and a 0.2 mH inductor are connected in series. What is the total inductance of the series path?

Introduction / Context:
For uncoupled inductors, series connection leads to simple arithmetic addition of inductances. This is analogous to series resistors and is frequently used to realize non-standard inductance values by combining available parts.

Given Data / Assumptions:

• L1 = 2 mH, L2 = 3.3 mH, L3 = 0.2 mH.
• Series connection; mutual coupling neglected.
• All values expressed in millihenries for easy arithmetic.

Concept / Approach:
Use L_total(series) = L1 + L2 + L3. No conversions are needed since all are in mH. Ensure you do not mistakenly compute a parallel equivalent.

Step-by-Step Solution:

Verification / Alternative check:
If mutual coupling were present with tight coupling (not given here), results could change due to induced voltages. Since the problem omits coupling, pure addition applies—consistent with standard textbook treatment.

Why Other Options Are Wrong:

• 55 mH: Off by a factor of 10.
• less than 5.5 mH / less than 0.2 mH: Incorrect because series values add; they cannot yield less than the largest single inductor.

Common Pitfalls:

• Accidentally treating the connection as parallel and applying reciprocal formulas.

Final Answer:
5.5 mH