---
title: "8-pole IM: speed at maximum torque"
subject: "Electrical Machine"
chapter: 5
key: "semester_4/electrical_machine:im-8pole-starting-torque-50nm-max-speed"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/im-8pole-starting-torque-50nm-max-speed
site: BCT Engineering
---

# 8-pole IM: speed at maximum torque

Electrical Machine · Chapter 5: Three Phase Induction Machines

- Asked in: 2070 Magh, 8 marks; 2068 Bhadra, 8 marks (asked 2 times)
- Video: [Starting torque, maximum torque, full load torque & their relation of 3 phase induction motor](https://www.youtube.com/watch?v=bUo0WLRtDsQ) — Edu fun zo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/im-8pole-starting-torque-50nm-max-speed
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-5

## Question

An 8-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 what speed the motor will develop maximum torque and calculate the magnitude of maximum torque.

## Short answer

Synchronous speed is 750 rpm. Slip at maximum torque s_m = R2/X2 = 0.8/2 = 0.4, so speed = 750×(1−0.4) = 450 rpm. Maximum torque = 72.5 N·m.

## Answer

The rotor impedance per phase at standstill is Z₂ = R₂ + jX₂ = 0.8 + j2 Ω (R₂ = 0.8 Ω, X₂ = 2 Ω).

For an induction motor:
- T = K s R₂ / (R₂² + s²X₂²)
- Starting torque (s = 1): T_st = K R₂ / (R₂² + X₂²)
- Max torque: at s_m = R₂/X₂, T_max = K / (2X₂)

#### Synchronous speed
N_s = 120 f / P = 120 × 50 / 8 = 750 rpm

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

#### Speed at maximum torque
N_m = N_s (1 − s_m) = 750 × (1 − 0.4) = **450 rpm**

#### Maximum torque
Ratio:

T_max / T_st = (R₂² + X₂²) / (2 R₂ X₂) = (0.64 + 4) / (2 × 0.8 × 2) = 4.64 / 3.2 = 1.45

T_max = 1.45 × 50 = **72.5 N·m**

| Quantity | Value |
|---|---|
| N_s | 750 rpm |
| s_m | 0.4 |
| N at max torque | 450 rpm |
| T_max / T_st | 1.45 |
| T_max | 72.5 N·m |

**Answer: maximum torque of 72.5 N·m is developed at 450 rpm.**

## Check yourself

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

1. What is N_s?
   A. 750 rpm
   B. 1500 rpm
   C. 3000 rpm
   D. 375 rpm
   Answer: A. 120 × 50/8 = 750 rpm. (Key point: Synchronous speed)

2. What is s_m?
   A. 0.25
   B. 0.4
   C. 0.8
   D. 2.5
   Answer: B. R2/X2 = 0.8/2 = 0.4. (Key point: Slip at max torque)

3. What is the speed at maximum torque?
   A. 750 rpm
   B. 1050 rpm
   C. 450 rpm
   D. 300 rpm
   Answer: C. N = 750 × (1 - 0.4) = 450 rpm. (Key point: Speed at max torque)

4. What is T_max?
   A. 34.5 N·m
   B. 50 N·m
   C. 145 N·m
   D. 72.5 N·m
   Answer: D. T_max/T_st = (R2^2+X2^2)/(2R2X2) = 1.45, so 72.5 N·m. (Key point: Maximum torque)

## Close variants set in other exams

- [4-pole IM: speed at maximum torque](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Aim-4pole-starting-torque-50nm-max-speed) — Electrical Machine, chapter 5, asked 3×

## Related in Electrical Machine

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- [Speed and value of maximum torque of 2-pole motor](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aim-2pole-2ohm-8ohm-max-torque) — Electrical Machines I, chapter 5

## Related in other subjects

- [Split-phase motor winding currents at starting](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asplit-phase-starting-currents) — Electric Machines II, chapter 3
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