---
title: "Hysteresis and eddy losses depend on frequency"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:hysteresis-eddy-frequency-reason"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/hysteresis-eddy-frequency-reason
site: BCT Engineering
---

# Hysteresis and eddy losses depend on frequency

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2071 Bhadra, 4 marks
- Video: [Iron losses | V/ f constant |  Transformers | Lec - 19](https://www.youtube.com/watch?v=twXNJIyj--8) — Education 4u
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/hysteresis-eddy-frequency-reason
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

Give reason: Hysteresis and eddy current losses depend on the frequency of the supply system.

## Short answer

Hysteresis loss is proportional to frequency because the B-H loop area (energy/cycle) is fixed, and more cycles per second multiply the loss. Eddy loss is proportional to f^2 because induced emf (∝ f) drives currents whose I^2R loss goes as emf^2.

## Answer

Both losses grow with frequency because a higher frequency means the core is magnetised and demagnetised more times each second and the induced emf in the core is larger.

#### Hysteresis loss

The energy lost in one cycle per unit volume is the area $w_h$ of the $B$–$H$ loop. It depends on the material and $B_{max}$ only. The number of cycles per second is $f$, so

$$
P_h = w_h V f = \eta B_{max}^{1.6} f V
$$

The loss is directly proportional to $f$. At twice the frequency, the loop is traversed twice as often and the loss doubles (for the same $B_{max}$).

#### Eddy current loss

The flux in the core is $\phi = \phi_m \sin\omega t$. The emf induced in a lamination is proportional to the rate of change of flux:

$$
e \propto \frac{d\phi}{dt} = \omega\phi_m\cos\omega t \Rightarrow E_{eddy} \propto f B_{max}
$$

The eddy current is $I_e = E_{eddy}/R_{core}$, and the loss is $I_e^2 R$. So the loss varies as the square of the induced emf:

$$
P_e \propto E_{eddy}^2 \propto f^{2} B_{max}^{2}
$$

Eddy loss is therefore proportional to $f^2$. At twice the frequency the loss is four times larger.

**Note:** in a transformer on a fixed supply, $B_{max} \propto V/f$. Then $P_e$ is almost independent of $f$, while $P_h \propto f^{-0.6}$. The statement is therefore true for a given $B_{max}$.

## Check yourself

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

1. Why is hysteresis loss proportional to f?
   A. The loop area grows with f
   B. The loop is traversed f times per second, each cycle costing the same energy
   C. The resistivity rises with f
   D. B_max rises with f
   Answer: B. Energy per cycle is fixed for the material and B_max; more cycles per second give P_h proportional to f. (Key point: Hysteresis vs frequency)

2. If frequency doubles at constant B_max, by what factor does eddy loss change?
   A. 2
   B. 4
   C. Unchanged
   D. 1/2
   Answer: B. The induced emf is proportional to f, and the loss to emf squared, so P_e is proportional to f^2. (Key point: Eddy loss vs frequency)

3. Why is the induced eddy emf larger at higher frequency?
   A. The core resistance falls
   B. The flux amplitude must rise
   C. The rate of change of flux d(phi)/dt is larger
   D. The lamination thickness increases
   Answer: C. e is proportional to d(phi)/dt = omega phi_m cos(omega t), which grows with f. (Key point: Why eddy emf rises)

4. In a transformer on a fixed-voltage supply, B_max is proportional to V/f. What follows?
   A. Eddy loss is almost independent of f
   B. Eddy loss rises as f^2
   C. Hysteresis loss rises as f^2
   D. Both losses become zero
   Answer: A. P_e proportional to f^2 B_max^2 becomes nearly constant when B_max is proportional to 1/f. (Key point: Transformer on fixed supply)

## Related in Electrical Machines

- [Hysteresis and eddy current losses, reduction](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Ahysteresis-eddy-losses-reduce) — Electrical Machines, chapter 1
- [Eddy current loss and its minimization](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aeddy-current-loss-minimize) — Electrical Machines I, chapter 1
- [Short note: Eddy current loss](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Anote-eddy-current-loss) — Electrical Machines, chapter 1
- [Rotor copper loss, frequency, current and efficiency of 11 kW motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Asquirrel-cage-1450rpm-11kw-losses) — Electrical Machines I, chapter 5

## Related in other subjects

- [Reason: losses depend on frequency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aeddy-hysteresis-frequency-reason) — Electrical Machine, chapter 1
- [Hysteresis and eddy current losses; loop area](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Ahysteresis-eddy-loss-area) — Electrical Machine, chapter 1
- [Hysteresis loss and eddy current loss](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Ahysteresis-eddy-loss) — Electric Machine Design, chapter 1