Parallel inductors check: Inductors of 5 mH, 4.3 mH, and 0.6 mH are connected in parallel. Qualitatively, the total inductance will be…

Introduction / Context:
Understanding how inductors combine helps in quick sanity checks without full calculations. For parallel inductors, the equivalent inductance is always less than the smallest branch value, analogous to resistors in parallel.

Given Data / Assumptions:

• L1 = 5 mH, L2 = 4.3 mH, L3 = 0.6 mH.
• All inductors are in parallel.
• Assume ideal behavior and no coupling.

Concept / Approach:

The rule for parallel inductors is 1/L_eq = Σ(1/L_i). Because reciprocals add, L_eq must be less than the smallest L_i. Here, the smallest is 0.6 mH; therefore, L_eq < 0.6 mH.

Step-by-Step Solution:

Verification / Alternative check:

Numerically, 1/L_eq ≈ 0.2 + 0.2326 + 1.6667 ≈ 2.099, so L_eq ≈ 0.476 mH, confirming it is less than 0.6 mH.

Why Other Options Are Wrong:

Values near 9.9 mH suggest series addition, not parallel. 'Greater than 5 mH' contradicts the parallel rule.

Common Pitfalls:

Mixing up series and parallel rules; forgetting that parallel reduces the equivalent inductance.

Final Answer:

less than 0.6 mH