Induction motor startup and losses calculation
Induction motor startup and losses calculation

Induction motor startup

The main objectives while starting an induction motor are:

  1. To handle high-starting current
  2. To achieve high-starting torque.

As we know, rotor resistance determines starting torque. Usually, this rotor resistance is small, giving small starting torque, but good running conditions. So, the squirrel-cage motor can run only with low-starting loads.

If the rotor resistance is increased by some means, then the slip and speed at which maximum torque occurs can be shifted. For that purpose, external resistance can be introduced in the rotor circuit, which is done in the case of slip ring or wound rotor type motors.

When power is applied to a stationary rotor, excessive current will start flowing.

This happens due to the fact that there is a transformer action between the stator winding and the rotor winding, and the rotor conductors are short-circuited. This causes heavy current flow through the rotor. If, for reducing this heavy starting current, starting voltage applied is reduced then it affects the starting torque as well.

Methods of Starting the Motor

To get everything out, the following method of starting is generally used:

  1. DOL starting
  2. Auto transformer starting
  3. Star–delta starting.

Losses Calculation

The following are the losses in an induction motor:

  1. Core loss in the stator and the rotor
  2. Stator and rotor copper losses
  3. Friction and windage loss.

Core loss is due to the main and leakage fluxes. As the voltage is assumed constant, the core loss can also be approximated as a constant. DC can measure the stator resistance. The hysteresis and eddy current loss in the conductors increase the resistance, and the effective resistance is taken at 1.2 times the DC resistance.

The rotor copper loss is calculated by subtracting the stator copper loss from the total measured loss or the rotor I2R loss. The friction and windage loss may be assumed constant, irrespective of the load.

  • Efficiency = Rotor output/stator input
  • Output = Input – Losses

Example With Calculations

Consider a three-phase 440 V, 50 Hz, six-pole induction motor. The motor takes 50 kW at 960 rpm for a certain load. Assume stator losses of 1 kW and friction and windage loss of 1.5 kW.

To determine the percentage slip, rotor copper loss, rotor output, and efficiency of the motor, perform the following function:

Percentage slip //

The synchronous speed of the motor = (50 ×120) / 6 = 6000 / 6 = 1000 rpm
Slip = (Synchronous speed – Actual speed) = 1000 – 960 = 40 rpm
Percentage slip = [(40 / 1000) × 100] = 4% = 0.04

Rotor copper loss //

Rotor input = 50 1 = 49 kW
Rotor copper loss = Rotor input × Slip = 49 × 0.04 = 1.96 kW

Rotor output //

Rotor output = Rotor input – Rotor copper loss – Friction and Windage loss
= 49 – 1.96 + 1.5
= 49 – 3.46
= 45.54 kW

Motor efficiency //

Motor efficiency = Rotor output/Motor input
= 45.54 / 50 = 0.9108
= 91.08%

Century Electric Repulsion Start Induction Motor (VIDEO)

Resource: Practical Troubleshooting of Electrical Equipment and Control Circuits – Mark Brown, Jawahar Rawtani and Dinesh Patil (Get it from Amazon)

About Author //


Edvard Csanyi

Edvard - Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV high power busbar trunking (<6300A) in power substations, buildings and industry fascilities. Designing of LV/MV switchgears.Professional in AutoCAD programming and web-design.Present on


  1. Stuart
    Dec 12, 2016

    Hi Sir, if i reduce the speed of my 60kW-6 pole from 1140 to 730rpm, what have i lost?

  2. M. S.
    Oct 15, 2016

    Very nice. Thanks so much.

  3. Buks de Vry
    Aug 29, 2015

    Informative, well done, thank you, Buks

  4. alamin
    Sep 17, 2014


  5. riz
    Aug 03, 2013

    Hi to all,

    I just want to say

    Very Very Very informative and useful stuff available here, simply

    ThE bEsT

    Mar 31, 2013

    good as an example

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