---
title: "4-pole IM: speed at maximum torque"
subject: "Electrical Machine"
chapter: 5
key: "semester_4/electrical_machine:im-4pole-starting-torque-50nm-max-speed"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/im-4pole-starting-torque-50nm-max-speed
site: BCT Engineering
---

# 4-pole IM: speed at maximum torque

Electrical Machine · Chapter 5: Three Phase Induction Machines

- Asked in: 2072 Asoj
- Video: [Numerical problems on 3-phase induction motors: Maximum Torque & Corresponding slip](https://www.youtube.com/watch?v=14-83DFgA4Q) — Gargi Basu
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/im-4pole-starting-torque-50nm-max-speed
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5

## Question

A 4-pole, 50 Hz, 3 phase induction motor develops a starting torque of 50 N-m. The rotor winding has an impedance of (0.8+j2) Ω per phase at stand still. At what speed the motor will develop maximum torque and calculate magnitude of the maximum torque.

## Short answer

Maximum torque occurs at slip s_m = R₂/X₂ = 0.4, speed 900 rpm. Magnitude T_max = 50 × (0.8²+2²)/(2×2×0.8) = 72.5 N·m.

## Answer

**Given:** 4-pole, 50 Hz; Z₂ = R₂ + jX₂ = 0.8 + j2 Ω/phase at standstill; starting torque T_st = 50 N·m. Stator impedance is neglected.

N_s = 120 × 50 / 4 = 1500 rpm.

#### Slip and speed at maximum torque
s_m = R₂ / X₂ = 0.8 / 2 = 0.4

N_m = N_s (1 − s_m) = 1500 × (1 − 0.4) = **900 rpm**

#### Maximum torque
With the supply voltage fixed (the same E₂ in both cases), the starting and maximum torques are

T_st = k E₂² R₂ / (R₂² + X₂²), T_max = k E₂² / (2X₂), where k = 3/ω_s.

```text
 T_max / T_st = (R2² + X2²) / (2 X2 R2)
              = (0.64 + 4) / (2 × 2 × 0.8)
              = 4.64 / 3.2 = 1.45
 T_max = 1.45 × 50 = 72.5 N·m
```
(Check using T/T_max = 2 s_m/(1 + s_m²) = 0.8/1.16 = 0.690 at s = 1, which gives 50/0.690 = 72.5 N·m.)

**Answer: maximum torque occurs at 900 rpm (slip 0.4), and its magnitude is 72.5 N·m.**

## Check yourself

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

1. How is the slip at maximum torque found?
   A. sm = X2/R2 = 2.5
   B. sm = R2/X2 = 0.8/2 = 0.4
   C. sm = R2 x X2
   D. sm = 1 - R2
   Answer: B. Maximum torque occurs when R2/s = X2, i.e. sm = R2/X2. (Key point: Slip at maximum torque)

2. What speed corresponds to sm = 0.4 for a 4-pole 50 Hz motor?
   A. 600 rpm
   B. 1500 rpm
   C. 1100 rpm
   D. 900 rpm
   Answer: D. N = Ns(1 - sm) = 1500 x 0.6 = 900 rpm. (Key point: Speed at maximum torque)

3. How is Tmax obtained from Tst?
   A. Tmax/Tst = (R2^2+X2^2)/(2 X2 R2) = 1.45
   B. Tmax = Tst x X2/R2
   C. Tmax = Tst/2
   D. Tmax = Tst x sm
   Answer: A. Using the same supply, the ratio is 4.64/3.2 = 1.45, so Tmax = 72.5 N·m. (Key point: Maximum torque)

## Close variants set in other exams

- [8-pole IM: speed at maximum torque](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aim-8pole-starting-torque-50nm-max-speed) — Electrical Machine, chapter 5, asked 3×

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