---
title: "8-pole motor: speed at max torque (50 N-m, j2)"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-8pole-50nm-j2"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-8pole-50nm-j2
site: BCT Engineering
---

# 8-pole motor: speed at max torque (50 N-m, j2)

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2070 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/im-8pole-50nm-j2
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

A 8-pole, 50 Hz, 3 phase induction motor develops a starting torque of 50 N-m. The rotor winding has an impedance of $(0.8 + j2)\ \Omega$ per phase. At what speed the motor will develop maximum torque and calculate the magnitude of maximum torque.

## Short answer

Synchronous speed 750 rpm. Slip at maximum torque = 0.8/2 = 0.4, speed = 450 rpm. T_max = 50 / (2×0.4/(1+0.4²)) = 72.5 N-m.

## Answer

**Given:** 8 poles, 50 Hz, $T_{st} = 50$ N-m, rotor impedance $0.8 + j2\ \Omega$ per phase at standstill, so $R_2 = 0.8\ \Omega$, $X_2 = 2\ \Omega$.

#### Synchronous speed
$$
N_s = \frac{120 f}{P} = \frac{120\times 50}{8} = 750\ \text{rpm}
$$
#### Slip and speed at maximum torque
$$
s_m = \frac{R_2}{X_2} = \frac{0.8}{2} = 0.4
$$
$$
N_m = N_s(1-s_m) = 750\times(1 - 0.4) = 450\ \text{rpm}
$$
#### Maximum torque
The torque at slip $s$ compared to maximum torque is $\frac{T}{T_{max}} = \frac{2 s s_m}{s^2 + s_m^2}$. At starting $s = 1$:
$$
\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1 + s_m^2} = \frac{2\times 0.4}{1 + 0.4^2} = 0.6897
$$
$$
T_{max} = \frac{T_{st}}{0.6897} = \frac{50}{0.6897} = 72.5\ \text{N-m}
$$

**Answer:** maximum torque occurs at **450 rpm** (slip 0.4), and $T_{max} = 72.5$ N-m.

## Check yourself

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

1. For R_2 = 0.8, X_2 = 2, s_m is:
   A. 2.5
   B. 0.4
   C. 0.8
   D. 0.2
   Answer: B. 0.8/2 = 0.4. (Key point: Slip at Tmax)

2. Speed at maximum torque (N_s = 750 rpm):
   A. 300 rpm
   B. 600 rpm
   C. 450 rpm
   D. 750 rpm
   Answer: C. 750 x 0.6 = 450. (Key point: Speed)

3. T_st/T_max for s_m = 0.4 is:
   A. 0.4
   B. 0.6897
   C. 0.5
   D. 1.0
   Answer: B. 2(0.4)/(1.16) = 0.6897. (Key point: Ratio)

4. T_max with T_st = 50 N-m is about:
   A. 34.5 N-m
   B. 50 N-m
   C. 72.5 N-m
   D. 130 N-m
   Answer: C. 50/0.6897 = 72.5. (Key point: T_max)

## Close variants set in other exams

- [8-pole motor: speed at max torque (50 N-m, j4)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-8pole-50nm-starting-torque-j4) — Electrical Machines, chapter 5, asked 4×
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