Parallel inductors: what happens to the equivalent inductance when multiple inductors are connected in parallel (assume negligible coupling and ideal components)?

Introduction / Context:
Inductors, like resistors, have predictable combination rules. Designers often place inductors in parallel to decrease the overall inductance while increasing current capability. This question checks recognition of the qualitative rule without requiring numeric calculation.

Given Data / Assumptions:

• Two or more ideal inductors connected in parallel.
• Negligible mutual coupling between coils.
• Sinusoidal steady state, but the rule for inductance is frequency-independent for ideal inductors.

Concept / Approach:
For parallel inductors, 1 / L_eq = 1 / L1 + 1 / L2 + ... + 1 / Ln. Since the sum of positive reciprocals exceeds the largest single reciprocal, L_eq must be less than the smallest individual inductance. This mirrors the rule for resistors in parallel.

Step-by-Step Solution:
Write the formula: 1 / L_eq = Σ (1 / Li).Observe Σ (1 / Li) > 1 / L_min.Take reciprocals: L_eq < L_min.Therefore, equivalent inductance is less than the smallest individual inductor.

Verification / Alternative check:
Example: L1 = L2 = L. Then 1 / L_eq = 2 / L ⇒ L_eq = L / 2, which is clearly less than L, confirming the rule.

Why Other Options Are Wrong:
Sum of inductive reactances: mixes a frequency-dependent quantity; not how L combines.Sum of inductances: that is the series rule, not parallel.Source voltage divided by current: gives impedance at a given frequency, not a structural rule for L.None of the above: incorrect because (d) is correct.

Common Pitfalls:
Confusing series and parallel combination rules; forgetting that coupling (mutual inductance) can alter results if coils are closely spaced or intentionally coupled.

Final Answer:
less than the inductance value of the smallest inductor