---
title: "Reluctances and air gap flux of two-winding core"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:two-winding-core-reluctance-airgap-flux"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/two-winding-core-reluctance-airgap-flux
site: BCT Engineering
---

# Reluctances and air gap flux of two-winding core

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2076 Baisakh, 8 marks
- Video: [EE213 - 03 - Analysis of magnetic circuits - example](https://www.youtube.com/watch?v=HRGsgPCZFXg) — MAFarooqi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/two-winding-core-reluctance-airgap-flux
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

A magnetic circuit with a single air gap is shown in fig below. The core dimensions are: Cross-sectional area Ac = 1.8×10⁻³ m², Mean core length lc = 0.6 m, Gap length g = 2.3×10⁻³ m, N1 = 83 turns, N2 = 20. Assume that the core relative permeability of 2000. i) Calculate the reluctance of the core and that of the gap Rg. For a current of i1 = 1.5 A and i2 = 1.25 A calculate ii) The total flux φ in the air gap. [Figure: rectangular core; winding N1 (current i1) on the left limb and winding N2 (current i2) on the right limb; a single air gap in the bottom yoke near the right limb.]

## Short answer

Core reluctance = 1.33×10⁵ AT/Wb, gap reluctance = 1.02×10⁶ AT/Wb. Summing the two MMFs (149.5 AT) and dividing by total reluctance gives air-gap flux = 0.130 mWb.

## Answer

Both windings are on the same single-loop core, so their mmfs add (assuming they are wound to aid each other) and drive one flux through the core and gap in series. Gap area is taken equal to core area (fringing neglected).

#### i) Reluctances

$$
\begin{aligned}
R_c &= \frac{l_c}{\mu_0\mu_r A_c} = \frac{0.6}{4\pi\times10^{-7}\times2000\times1.8\times10^{-3}} = 1.326\times10^{5}\ \text{AT/Wb} \\
R_g &= \frac{g}{\mu_0 A_c} = \frac{2.3\times10^{-3}}{4\pi\times10^{-7}\times1.8\times10^{-3}} = 1.017\times10^{6}\ \text{AT/Wb}
\end{aligned}
$$

Total reluctance: $R = R_c + R_g = 1.149\times10^{6}$ AT/Wb.

#### ii) Flux in the air gap

$$
\begin{aligned}
F &= N_1 i_1 + N_2 i_2 = 83\times1.5 + 20\times1.25 = 124.5 + 25 = 149.5\ \text{AT} \\
\phi &= \frac{F}{R} = \frac{149.5}{1.149\times10^{6}} = 1.30\times10^{-4}\ \text{Wb}
\end{aligned}
$$

Gap flux density: $B_g = \phi/A_c = 1.30\times10^{-4}/1.8\times10^{-3} = 0.0723$ T.

**Answer:** $R_c = 1.33\times10^5$ AT/Wb, $R_g = 1.02\times10^6$ AT/Wb, $\phi = 0.130$ mWb (B = 0.072 T).

(If the two windings opposed each other, $F = 124.5 - 25 = 99.5$ AT and $\phi = 0.0866$ mWb.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why are the core and gap reluctances added?
   A. They carry the same flux, so they are in series
   B. They are in parallel
   C. They have equal areas only
   D. They have equal lengths
   Answer: A. One single-loop core carries the same flux through core and gap, so R = R_c + R_g. (Key point: Series path)

2. Which is correct for the reluctances?
   A. R_c = 1.33e5 and R_g = 1.02e6 AT/Wb
   B. R_c = 1.02e6 and R_g = 1.33e5
   C. They are equal
   D. R_g is zero
   Answer: A. The gap has no mu_r factor so R_g is about 8 times R_c even though it is much shorter. (Key point: Gap reluctance size)

3. With the windings aiding, the total mmf is:
   A. 83x1.5 - 20x1.25 = 99.5 AT
   B. 83x1.5 + 20x1.25 = 149.5 AT
   C. 1.5 + 1.25 = 2.75 AT
   D. 83 + 20 = 103 AT
   Answer: B. Aiding mmfs add: 124.5 + 25 = 149.5 AT. (Key point: Total mmf)

4. The flux is:
   A. 0.130 mWb
   B. 1.30 mWb
   C. 13.0 mWb
   D. 0.0130 mWb
   Answer: A. 149.5 / 1.149e6 = 1.30e-4 Wb. (Key point: Flux)

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