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

# 8-pole motor: speed at max torque (50 Kg-m)

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2075 Baisakh (old course), 8 marks
- Video: [(English)EM | Example 7.5 (4ed) | Slip, Power, and Torque | Y-connected Wound-Rotor Induction Motor](https://www.youtube.com/watch?v=UvyE32jle0A) — Electrical Engineering Academy
- 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-50kgm-starting-torque
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

An 8 pole, 50 Hz, three phase induction motor develops a starting torque of 50 Kg-m. The rotor 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

Ns=750 rpm; sm=R2/X2=0.8/4=0.2; speed at Tmax=750(1-0.2)=600 rpm. Tst/Tmax=2 sm/(1+sm²)=0.3846; Tmax=50/0.3846=130 kg·m (≈1275 N·m).

## Answer

**Given:** 8 poles, 50 Hz, $T_{st} = 50$ kg-m, $Z_2 = 0.8 + j4\ \Omega$ per phase, so $R_2 = 0.8\ \Omega$, $X_2 = 4\ \Omega$.

$$
N_s = \frac{120\times 50}{8} = 750\ \text{rpm}
$$
**Slip at maximum torque:**
$$
s_m = \frac{R_2}{X_2} = \frac{0.8}{4} = 0.2
$$
**Speed at maximum torque:**
$$
N_m = N_s(1 - s_m) = 750\times 0.8 = 600\ \text{rpm}
$$
**Maximum torque:** using $\frac{T_{st}}{T_{max}} = \frac{2 s_m}{1+s_m^2}$ (stator impedance neglected)
$$
\frac{T_{st}}{T_{max}} = \frac{2\times 0.2}{1 + 0.2^2} = \frac{0.4}{1.04} = 0.3846
$$
$$
T_{max} = \frac{50}{0.3846} = 130\ \text{kg-m} \quad (= 130\times 9.81 = 1275.3\ \text{N-m})
$$

**Answer:** the motor develops maximum torque at **600 rpm**, and $T_{max} = 130$ kg-m ($\approx 1275$ N-m).

## Check yourself

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

1. N_s for 8 poles, 50 Hz is:
   A. 1500 rpm
   B. 1000 rpm
   C. 600 rpm
   D. 750 rpm
   Answer: D. 120 x 50/8 = 750 rpm. (Key point: Sync speed)

2. With R_2 = 0.8 and X_2 = 4, s_m =
   A. 5
   B. 0.2
   C. 0.8
   D. 4
   Answer: B. 0.8/4 = 0.2. (Key point: Slip at Tmax)

3. Speed at maximum torque is:
   A. 150 rpm
   B. 750 rpm
   C. 600 rpm
   D. 900 rpm
   Answer: C. 750 x 0.8 = 600 rpm. (Key point: Speed at Tmax)

4. With T_st/T_max = 2 s_m/(1 + s_m^2) = 0.3846 and T_st = 50, T_max =
   A. 130 kg-m
   B. 19.2 kg-m
   C. 50 kg-m
   D. 65 kg-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, 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×
- [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×

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