Series inductors with geometric progression: Five inductors are connected in series. The smallest inductance is 8 µH, and each subsequent inductor has twice the inductance of the previous one (connected in ascending order). What is the total series inductance?

Introduction / Context:
Questions on series inductors often test whether learners remember that series inductances add directly, just like series resistances. This problem adds a twist by specifying a geometric progression (each value is twice the previous), which requires careful listing of the individual inductances before summing them. Such skills are fundamental in filter design, chokes, and energy storage planning in power electronics.

Given Data / Assumptions:

• Five inductors connected in series.
• Smallest inductance L1 = 8 µH.
• Each subsequent inductor is twice the previous: L2 = 2L1, L3 = 2L2, etc.
• Ideal components; mutual coupling ignored (no coupling coefficient specified).

Concept / Approach:
For inductors in series (without mutual coupling), the equivalent inductance is the arithmetic sum of individual inductances: L_total = L1 + L2 + L3 + L4 + L5. With a ratio of 2, the list is straightforward to generate and then sum. Keeping consistent microhenry units avoids conversion mistakes.

Step-by-Step Solution:

Verification / Alternative check:
Recognize the geometric-series pattern: sum of n terms with first term a and ratio r is S_n = a * (r^n − 1)/(r − 1). Here a = 8 µH, r = 2, n = 5: S_5 = 8*(2^5 − 1)/(2 − 1) = 8*(32 − 1) = 8*31 = 248 µH, matching the direct sum.

Why Other Options Are Wrong:

• 64 µH or 32 µH: These are intermediate single inductor values, not the series total.
• 8 H: Unit and magnitude error; result must remain in µH for these values and is far smaller than 1 H.

Common Pitfalls:

• Forgetting that inductances add in series even when values double.
• Mistaking geometric progression with exponential growth of the total; it still sums linearly.

Final Answer:
248 µH