---
title: "4-pole motor: speed at max torque (50 N-m, j2)"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-4pole-50nm-starting-torque-j2"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-4pole-50nm-starting-torque-j2
site: BCT Engineering
---

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

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2072 Asoj
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/im-4pole-50nm-starting-torque-j2
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/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 $(0.8 + j2)\ \Omega$ per phase at stand still. At what speed will the motor develop maximum torque and calculate magnitude of the maximum torque.

## Short answer

Synchronous speed 1500 rpm. Slip at max torque = 0.8/2 = 0.4, speed = 900 rpm. Maximum torque = 50 / 0.6897 = 72.5 N-m.

## Answer

**Given:** 4 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
$$
N_s = \frac{120 f}{P} = \frac{120\times 50}{4} = 1500\ \text{rpm}
$$
#### Slip and speed at maximum torque
$$
s_m = \frac{R_2}{X_2} = \frac{0.8}{2} = 0.4
$$
$$
N_m = N_s(1-s_m) = 1500\times(1 - 0.4) = 900\ \text{rpm}
$$
#### 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 **900 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. N_s for 4 poles, 50 Hz is:
   A. 750 rpm
   B. 1000 rpm
   C. 3000 rpm
   D. 1500 rpm
   Answer: D. 120 x 50/4 = 1500 rpm. (Key point: Sync speed)

2. With s_m = 0.4, speed at max torque:
   A. 900 rpm
   B. 600 rpm
   C. 1500 rpm
   D. 1100 rpm
   Answer: A. 1500 x 0.6 = 900. (Key point: Speed at Tmax)

3. T_max with T_st = 50 N-m and ratio 0.6897:
   A. 34.5 N-m
   B. 130 N-m
   C. 72.5 N-m
   D. 50 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×
- [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×
- [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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