Parallel inductors calculation: Two coils of 5 H and 100 mH are connected in parallel. What is the total equivalent inductance of the combination?

Introduction / Context:
Combining inductors is common in filter design and energy storage networks. For inductors in parallel, the equivalent value decreases, analogously to parallel resistors. This question reinforces correct use of the parallel inductance formula and careful unit handling between henry and millihenry.

Given Data / Assumptions:

• L1 = 5 H.
• L2 = 100 mH = 0.1 H.
• Coils are ideal and not magnetically coupled (no mutual inductance considered).

Concept / Approach:

For two uncoupled inductors in parallel, the equivalent inductance is L_eq = (L1 * L2) / (L1 + L2). This mirrors the product-over-sum form used for two resistors in parallel, but applies to inductors when mutual coupling is negligible.

Step-by-Step Solution:

Verification / Alternative check:

Because one inductor (0.1 H) is much smaller than the other (5 H), the parallel combination should be slightly less than 0.1 H. The result 98 mH is consistent with this expectation.

Why Other Options Are Wrong:

• 4.76 mH / 33.3 mH / 150.0 mH: Do not satisfy the parallel formula for the given values.
• 5.10 H: This is the series sum, not a parallel result.

Common Pitfalls:

• Forgetting to convert millihenry to henry before calculation.
• Confusing series and parallel formulas or accidentally adding instead of using product-over-sum.

Final Answer:

98.0 mH