I am aware that a coil has inductive reactance when powered by AC besides the еlectrical resistance which is typically very low.

Considering the fact that reactance is proprotional to the inductance and supply frequency, how is that related to the load of the motor?

Let's assume the following parameters:

  • inductance 0.5 H
  • resistance 5 ohms
  • AC Supply 220V/50Hz

Further calculations show that the impedance is about 162 ohms. According to Ohm's law the current flow would be 220 volts divided by 162 ohms = 1,35 A no-load current. Now I want to understand what exactly changes in the equations above that makes the stator windings draw more current when the rotor is opposed.


2 Answers 2


The concept of back EMF works well to explain the relationship among speed, load torque and current in a DC motor, but it doesn't work as well for an induction motor. The preferred equivalent circuit is shown below.

R1 and X1 are the resistance of the stator and the part of the inductance that does not provide a useful magnetic field. Bm is the inductance that provides the stator magnetic field. Gc is a resistance that represents the losses in the stator's iron core. X2 represents the inductance of the rotor. R2/s is a variable resistance that represents the resistance of the rotor and also, most importantly, the the mechanical load.

At no load, the rotor of an induction motor turns at a synchronous speed, the speed of the motor's rotating magnetic field. The speed in revolutions per minute (RPM) is RPM = 120f/P where f is the power frequency (Hz) and P is the number of magnetic poles formed by the stator winding configuration. R can be any even number (poles are always north/south pairs). Usually P is 2 or 4, but 6 pole motors are not uncommon, higher numbers of poles are less common and usually found only in large motors.

When the motor is loaded, the operating speed decreases. The difference between the loaded speed and the synchronous speed is called slip (s). Slip is expressed as a fraction of synchronous speed. At rated load, the slip is generally 2 or 3 percent of the synchronous speed and s = 0.02 or 0.03.

When the slip is zero, R2/s in the diagram is R2/0 or infinite. Therefore, at no load, the current in the motor is determined by R1 and X1 in series with the magnetizing circuit, Gc and Xm. As the motor is loaded, the slip increases and R2/s decreases causing the motor to draw more current.

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The diagram represents one phase of a three-phase induction motor. Two similar diagrams are used to represent a single-phase motor.

  • \$\begingroup\$ I understand looking at the equivalent circuit simplifies our lives a lot and is accurate. But is the following line of reasoning correct? As load increases, 1. Slip increases, 2. Relative speed increases, 3. Induced E2 and I2 magnitude increases, 4. Opposing Rotor flux increases, decreasing the net mutual flux, 5. To compensate this drop in MMF, the stator draws an equivalent current that produces an equivalent mmf like a transformer. Thus resulting in a net increase in current drawn from stator \$\endgroup\$ Oct 22, 2021 at 5:29

Take a look at this: https://electronics.stackexchange.com/a/364242/58084

If you decrease the speed of the motor, the back-emf will decrease also, increasing the voltage across the motor resistance and reactance, and increasing the current.


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