Solving for an unknown series inductor — three uncoupled inductors are in series. If L1 = 120 mH, L2 = 70 mH, and the measured total inductance is 340 mH, what is the value of L3?

Introduction / Context:
Determining an unknown component value from a measured total is common in repair and prototyping. With series inductors that are not magnetically coupled, the total inductance is just the arithmetic sum. Given two known values and the total, the third can be obtained by subtraction. This mirrors the way series resistors add.

Given Data / Assumptions:

• L1 = 120 mH, L2 = 70 mH, L3 = ?
• Total series inductance L_total = 340 mH.
• Inductors are uncoupled (negligible mutual inductance).
• Ideal components for the computation; parasitics do not affect the sum appreciably.

Concept / Approach:
For uncoupled inductors in series, L_total = L1 + L2 + L3. Solve for the unknown: L3 = L_total − (L1 + L2). If coils were tightly coupled, the ±2M term could alter the equivalent inductance; however, in most practical series aggregates with spacing or orthogonal orientation, M is small and can be ignored for first-order calculations.

Step-by-Step Solution:

Verification / Alternative check:
Measure reactance at a test frequency f: X_L,total ≈ 2 * pi * f * 0.340 H. Separately summing X_L of L1, L2, and the computed L3 will yield the same total within measurement tolerance, confirming the calculated L3.

Why Other Options Are Wrong:

• 220 mH, 295 mH, 315 mH: These do not satisfy L1 + L2 + L3 = 340 mH and would give totals that are too large.

Common Pitfalls:
Accidentally using the parallel formula (reciprocals) or mixing units (mH vs H). Always check that the configuration is truly series and that units are consistent.

Final Answer:
150 mH for L3.