Parallel inductance: Four coils of 75 µH, 40 µH, 25 µH, and 15 µH are connected in parallel. Estimate the total inductance.

Introduction / Context:
Parallel inductors combine like resistors in parallel: the reciprocal of the equivalent inductance equals the sum of the reciprocals. This configuration reduces inductance and increases current-handling capability.

Given Data / Assumptions:

• L1 = 75 µH, L2 = 40 µH, L3 = 25 µH, L4 = 15 µH.
• All four inductors are in parallel.
• Ideal components; negligible coupling and series resistance.

Concept / Approach:

Use 1/L_eq = Σ(1/L_i). After summing reciprocals, invert to find L_eq. Keep units in microhenries throughout to avoid conversion errors.

Step-by-Step Solution:

Verification / Alternative check:

L_eq must be less than the smallest individual value (15 µH). The computed 6.9 µH satisfies this sanity check.

Why Other Options Are Wrong:

14 µH is too large for a four-branch parallel. 2.2 µH would require much larger reciprocal sum. 155 µH incorrectly suggests series addition.

Common Pitfalls:

Arithmetic mistakes in reciprocals; forgetting to invert the final sum; mixing µH with mH.

Final Answer:

6.9 µH