---
title: "Reluctance and permeance expressions"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:reluctance-permeance-derive"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/reluctance-permeance-derive
site: BCT Engineering
---

# Reluctance and permeance expressions

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2078 Chaitra, 8 marks
- Video: [Introduction to Magnetic Circuits: MMF, Flux, Reluctance | What is a Magnetic Circuit?](https://www.youtube.com/watch?v=0egh0XBItFc) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/reluctance-permeance-derive
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

What are reluctance and permeance in the magnetic circuits? Derive their expressions in any type of magnetic circuits.

## Short answer

Reluctance S = l/(μA) is the magnetic opposition; permeance P = 1/S. Derived from Φ = BA = μA·NI/l, so Φ = F / (l/(μA)), giving S = l/(μA). For series, sum S; for parallel, sum P.

## Answer

**Reluctance** ($S$ or $\mathcal{R}$) is the opposition that a magnetic circuit offers to the setting up of flux. It is the ratio of mmf to flux, and its unit is AT/Wb (or H$^{-1}$).
**Permeance** ($P$) is the reciprocal of reluctance. It measures the ease with which flux is set up, and its unit is Wb/AT (henry).

#### Derivation for a uniform circuit

Take a core of mean length $l$, cross-section $A$ and permeability $\mu = \mu_0\mu_r$, carrying a coil of $N$ turns and current $I$.

$$
\begin{aligned}
\text{mmf } F &= NI \\
H &= \frac{NI}{l},\qquad B = \mu H = \frac{\mu NI}{l} \\
\Phi &= BA = \frac{\mu A\,NI}{l} = \frac{NI}{\dfrac{l}{\mu A}}
\end{aligned}
$$

Comparing with $\Phi = \dfrac{\text{mmf}}{\text{reluctance}}$:

$$
S = \frac{l}{\mu A} = \frac{l}{\mu_0\mu_r A}, \qquad P = \frac{1}{S} = \frac{\mu A}{l}
$$

#### For other types of circuit

- **Air gap of length $l_g$:** $S_g = \dfrac{l_g}{\mu_0 A_g}$.
- **Series circuit** (a core of several sections, or a core with a gap): the same flux flows through each part, so

$$
S = S_1 + S_2 + \dots,\qquad NI = \Phi(S_1 + S_2 + \dots)
$$

- **Parallel circuit** (for example the legs of a three-limb core): the mmf across the branches is common, so

$$
P = P_1 + P_2 + \dots \Rightarrow \frac{1}{S} = \frac{1}{S_1} + \frac{1}{S_2} + \dots
$$

- **Non-uniform section:** in general, add up the reluctances of small elements: $S = \displaystyle\int \frac{dl}{\mu A(l)}$.

**Points to note:** $S$ is small for iron (high $\mu_r$) and large for air, so even a short air gap contributes a large share of the total reluctance. Unlike resistance, reluctance does not dissipate energy; the energy used to set up the flux is stored in the magnetic field.

## Check yourself

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

1. What is reluctance?
   A. Ratio of flux to mmf
   B. Ratio of mmf to flux, unit AT/Wb
   C. Ratio of B to H
   D. Flux per unit area
   Answer: B. S = mmf/flux, the opposition to setting up flux; permeance is its reciprocal. (Key point: Definition of reluctance)

2. For a uniform core, which expression gives reluctance?
   A. mu A / l
   B. l / (mu A)
   C. A / (mu l)
   D. mu l / A
   Answer: B. From Phi = mu A NI / l, S = l/(mu A). (Key point: Reluctance formula)

3. How do reluctances of parallel magnetic paths combine?
   A. Reluctances add
   B. Permeances add
   C. Flux adds but permeance does not
   D. Mmfs add
   Answer: B. Parallel paths share a common mmf, so P = P_1 + P_2 + ... (Key point: Parallel circuits)

4. Why does a short air gap add a large share of the total reluctance?
   A. Air has high permeability
   B. Air has mu_r = 1, much lower than iron
   C. The gap area is larger
   D. Air stores energy as heat
   Answer: B. Low permeability gives high reluctance per metre, so even a short gap matters. (Key point: Air gap contribution)

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