---
title: "8-pole motor: speed at max torque (50 N-m, j4)"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-8pole-50nm-starting-torque-j4"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-8pole-50nm-starting-torque-j4
site: BCT Engineering
---

# 8-pole motor: speed at max torque (50 N-m, j4)

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2068 Bhadra, 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-starting-torque-j4
- 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 $(0.8 + j4)\ \Omega$ 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 = R2/X2 = 0.2, so speed = 600 rpm. Using T_st/T_max = 2s_m/(1+s_m²) = 0.3846 gives maximum torque = 130 N-m.

## Answer

**Given:** 8 poles, 50 Hz, $T_{st} = 50$ N-m, rotor impedance $0.8 + j4\ \Omega$ per phase at standstill, so $R_2 = 0.8\ \Omega$, $X_2 = 4\ \Omega$.

#### Synchronous speed
$$
N_s = \frac{120 f}{P} = \frac{120\times 50}{8} = 750\ \text{rpm}
$$
#### Slip and speed at maximum torque
$$
s_m = \frac{R_2}{X_2} = \frac{0.8}{4} = 0.2
$$
$$
N_m = N_s(1-s_m) = 750\times(1 - 0.2) = 600\ \text{rpm}
$$
#### 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 $s = 1$:
$$
\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1 + s_m^2} = \frac{2\times 0.2}{1 + 0.2^2} = 0.3846
$$
$$
T_{max} = \frac{T_{st}}{0.3846} = \frac{50}{0.3846} = 130.0\ \text{N-m}
$$

**Answer:** maximum torque occurs at **600 rpm** (slip 0.2), and $T_{max} = 130.0$ N-m.

## Check yourself

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

1. N_s for an 8-pole 50 Hz motor is:
   A. 750 rpm
   B. 1500 rpm
   C. 375 rpm
   D. 1000 rpm
   Answer: A. 120 x 50/8. (Key point: Sync speed)

2. s_m for 0.8 + j4 is:
   A. 0.8
   B. 0.2
   C. 5
   D. 0.4
   Answer: B. R_2/X_2 = 0.2. (Key point: Slip at Tmax)

3. Speed at maximum torque:
   A. 750 rpm
   B. 450 rpm
   C. 600 rpm
   D. 150 rpm
   Answer: C. 750(1 - 0.2) = 600. (Key point: Speed)

4. Using T_st/T_max = 0.3846, T_max with T_st = 50 N-m is:
   A. 130 N-m
   B. 19.2 N-m
   C. 50 N-m
   D. 250 N-m
   Answer: A. 50/0.3846 = 130. (Key point: T_max)

## Close variants set in other exams

- [8-pole motor: speed at max torque (50 N-m, j2)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-8pole-50nm-j2) — Electrical Machines, chapter 5, asked 4×
- [4-pole motor: speed at max torque (50 N-m, j2)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-4pole-50nm-starting-torque-j2) — Electrical Machines, chapter 5, asked 4×
- [8-pole motor: speed at max torque (50 Kg-m)](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aim-8pole-50kgm-starting-torque) — Electrical Machines, chapter 5, asked 4×

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