By the end of this chapter you should be able to:
Read the torque–speed curve of an induction motor and identify the starting, full-load and breakdown torques.
Explain how the operating point is set by the balance of motor torque and load torque.
Describe how the curve shifts when the voltage and frequency are varied together, and why this makes frequency control attractive.
Use the linear low-slip relation to predict the slip under changed load, rotor resistance or voltage.
Demonstrative Video
The Torque–Speed Curve
The torque of an induction motor depends on its speed, but no simple formula describes the whole range conveniently. It is easier to show the relationship as a curve.

For a typical design, the starting (locked-rotor) torque is about 1.5 times full-load torque, and the breakdown (maximum) torque is about 2.5 times.
Motor torque is electromagnetic, the product of flux and rotor current. Load torque is mechanical, force times radius.
| Condition | Torque relation | Result |
|---|---|---|
| Steady running | \(T_{motor} = T_{load}\) | Constant speed, slightly below \(N_s\) |
| Accelerating | \(T_{motor} > T_{load}\) | Speed rises |
| Decelerating | \(T_{motor} < T_{load}\) | Speed falls |
Varying Voltage and Frequency
For a squirrel-cage motor the curve depends on the stator voltage and frequency. At fixed frequency, torque is proportional to \(V^2\), and \(N_s\) is set by \(f\). In practice \(V\) and \(f\) are varied in the same proportion, which keeps the air-gap flux constant.

The shape of the curve stays the same, but its position shifts along the speed axis.
The torque depends on the slip speed \(N_s - N_r\). At constant flux, the same slip speed gives the same torque at any frequency.


The current–speed curve is V-shaped, with its minimum at \(N_s\). That minimum is the magnetising current that creates the flux. Since the flux is held constant, the magnetising current is the same at every synchronous speed.

For the example motor, at 440 V/60 Hz the locked-rotor current is 100 A and the locked-rotor torque is 75 N·m. At 110 V/15 Hz (the same V/f) the locked-rotor current falls to 75 A, while the locked-rotor torque doubles to 150 N·m. At low frequency the rotor current at standstill is closer in phase to the air-gap flux, so it produces more torque per ampere.
The Linear Low-Slip Region
Stable operation lies on the nearly straight part of the torque–slip curve between no load and breakdown. Here \(T \propto sV^2/R_2\), so the slope depends mainly on the rotor resistance: the higher \(R_2\), the larger the slip for a given torque. For two operating conditions:
Here \(R_1\) and \(R_2\) denote the rotor resistance in the two conditions.
Problem. A motor runs at 3 % slip at full load. Find the slip if the load torque falls to 80 %, the rotor resistance is doubled and the voltage falls to 90 %.
Summary
The torque–speed curve shows starting (≈1.5T), full-load and breakdown (≈2.5T) torques.
The operating point is where the motor torque equals the load torque.
At constant \(V/f\) the curve keeps its shape and shifts with \(N_s\).
At low slip, \(s \propto TR_2/V^2\).
Chapter Review
MCQ 1. If the load torque exceeds the breakdown torque, the motor
(a) speeds up (b) stalls (c) runs at \(N_s\) (d) becomes a generatorShow answer
(b) stalls.MCQ 2. At constant V/f, reducing the frequency
(a) changes the curve shape (b) shifts the curve to lower speeds (c) reduces the breakdown torque to zero (d) increases \(N_s\)Show answer
(b).MCQ 3. In the linear region, doubling the rotor resistance at the same torque
(a) halves the slip (b) doubles the slip (c) leaves the slip unchanged (d) quadruples the slipShow answer
(b) doubles the slip.
Why is the operating region right of the breakdown point stable and the region to its left unstable?
Why is the no-load current of an induction motor mainly magnetising current?