---
title: "Why rotor core loss is negligible"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:rotor-core-loss-negligible"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/rotor-core-loss-negligible
site: BCT Engineering
---

# Why rotor core loss is negligible

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2068 Bhadra, 1+3 marks; 2065 Chaitra (old course), 4 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/rotor-core-loss-negligible
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

Explain why the rotor core loss in a three phase induction motor is negligible (often neglected).

## Short answer

Rotor core loss is neglected because rotor frequency is sf, only 1–2.5 Hz at normal slip of 2–5%. Core loss depends on frequency, so rotor iron loss is a tiny fraction of stator core loss. At starting s=1 it is larger but transient.

## Answer

The core loss (hysteresis plus eddy current) in a magnetic material depends on the **frequency of the alternating flux** in it:

$$
P_h\propto f\,B_m^{1.6}, \qquad P_e\propto f^2B_m^2
$$

**Reason:**
1. In an induction motor the stator field rotates at synchronous speed, and the rotor turns at nearly the same speed. The flux therefore moves relative to the rotor at the slip speed only, so the **rotor core frequency is $f_r = sf$**.
2. At normal running the slip is only 2–5%. For a 50 Hz supply, $f_r = (0.02\ \text{to}\ 0.05)\times 50 = 1$–$2.5$ Hz.
3. The core loss is proportional to $f$ (hysteresis) and $f^2$ (eddy current), so at 1–2.5 Hz the rotor core loss is a tiny fraction of the stator core loss, which occurs at the full supply frequency of 50 Hz.

Example: at $s = 0.04$, $f_r = 2$ Hz. The rotor eddy-current loss is $(2/50)^2 = 0.0016$ of the loss at 50 Hz, i.e. 0.16%.

Hence in the equivalent circuit the core-loss branch is shown only on the stator side, and the **rotor iron loss is neglected**. The rotor copper loss $I_2^2R_2$ is the only significant rotor loss. (At starting, $s = 1$ and the rotor frequency is 50 Hz, but the rotor core loss does not matter then because the starting period is short.)

## Check yourself

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

1. What is rotor frequency during normal running?
   A. Supply frequency f
   B. s x f, only 1-2.5 Hz
   C. 2f
   D. Zero
   Answer: B. The flux moves relative to the rotor at slip speed, so fr = s f. (Key point: Rotor frequency)

2. Core loss components depend on frequency as:
   A. Hysteresis proportional to f^2, eddy to f
   B. Both independent of f
   C. Hysteresis proportional to f, eddy proportional to f^2
   D. Both proportional to 1/f
   Answer: C. Ph is proportional to f and Pe to f^2. (Key point: Loss dependence)

3. At fr = 2 Hz on a 50 Hz machine, the rotor eddy loss relative to 50 Hz value is:
   A. 4%
   B. 40%
   C. 0.4%
   D. 0.16%
   Answer: D. (2/50)^2 = 0.0016. (Key point: Eddy comparison)

4. Where is the core-loss branch shown?
   A. Stator side only
   B. Rotor side only
   C. Both equally
   D. Neither
   Answer: A. Rotor iron loss is neglected; only rotor copper loss matters. (Key point: Equivalent circuit)

## Related in Electrical Machines

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- [Delta induction motor: slip, rotor emf, turns ratio](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-delta-440v-1440rpm-rotor-emf) — Electrical Machines, chapter 5, asked 2×
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## Related in other subjects

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