---
title: "Torque at 6% slip from maximum torque"
subject: "Electrical Machines"
chapter: 5
key: "semester_4/electrical_machines:im-torque-at-6-slip"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/im-torque-at-6-slip
site: BCT Engineering
---

# Torque at 6% slip from maximum torque

Electrical Machines · Chapter 5: Three Phase Induction Machines

- Asked in: 2078 Chaitra, 6 marks
- Video: [IM 12 Torque Slip Characteristics](https://www.youtube.com/watch?v=qoeOtMZzhkc) — Lectures in Electrical Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/im-torque-at-6-slip
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-5

## Question

A three phase 6 pole, 50 Hz induction motor develops a maximum torque of 30 Nm at 960 rpm. Calculate the torque produced by the motor at 6% slip. The rotor resistance per phase is $0.6\ \Omega$.

## Short answer

Ns=1000 rpm; sm=(1000-960)/1000=0.04. Using T/Tmax=2 s sm/(s²+sm²) with s=0.06 gives T/Tmax=0.9231, so T=0.9231×30=27.69 N·m.

## Answer

**Given:** 6-pole, 50 Hz, $T_{max} = 30$ N-m at 960 rpm, $R_2 = 0.6\ \Omega$, $s = 0.06$.

$$
N_s = \frac{120\times 50}{6} = 1000\ \text{rpm}
$$
Slip at maximum torque:
$$
s_m = \frac{1000-960}{1000} = 0.04
$$
(This gives $X_2 = R_2/s_m = 15\ \Omega$ at standstill, but the ratio method does not need it.)

Torque at any slip relative to maximum torque (neglecting stator impedance):
$$
\frac{T}{T_{max}} = \frac{2\,s\,s_m}{s^2 + s_m^2}
$$
$$
\frac{T}{T_{max}} = \frac{2\times 0.06\times 0.04}{0.06^2 + 0.04^2} = \frac{0.0048}{0.0052} = 0.9231
$$
$$
T = 0.9231\times 30 = 27.69\ \text{N-m}
$$

**Answer:** torque at 6 % slip $= 27.69$ N-m.

## Check yourself

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

1. N_s for 6 poles, 50 Hz is:
   A. 500 rpm
   B. 1000 rpm
   C. 1500 rpm
   D. 750 rpm
   Answer: B. 120 x 50/6 = 1000 rpm. (Key point: Synchronous speed)

2. Speed 960 rpm at max torque gives s_m:
   A. 0.06
   B. 0.4
   C. 0.04
   D. 0.96
   Answer: C. (1000 - 960)/1000 = 0.04. (Key point: Slip at Tmax)

3. T/T_max at s = 0.06 and s_m = 0.04 is:
   A. 0.6667
   B. 1
   C. 0.9231
   D. 0.5
   Answer: C. 2(0.06)(0.04)/(0.0036 + 0.0016) = 0.9231. (Key point: Torque ratio)

4. Torque at 6 % slip is about:
   A. 27.69 N-m
   B. 30 N-m
   C. 20 N-m
   D. 32.5 N-m
   Answer: A. 0.9231 x 30 = 27.69 N-m. (Key point: Result)

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