Here's my understanding of the induction motor, it would be really helpful if someone can verify it.

  1. 3 phase AC supply given to stator, sets up an rotating magnetic field (RMF,) \$ B \$ (discussing about the magnetic field, not the MMF.)
  2. This RMF induces EMF in each rotor conductor (\$ e = B \, l \, v \$,) this induced EMF is in phase with the RMF that is the peak of induced EMF is always under the peak \$B\$.
  3. Induced EMF causes a current flow that lags the EMF and hence the RMF, due to the inductive reactance of the rotor coil. (So the maximum current-carrying conductor at any instant is always a bit behind the max EMF induced conductor.) https://www.youtube.com/watch?v=JPn5Ou-N0b0 (to understand this lag in peak current conductor)
  4. The force on a conductor due to RMF, \$F = I \times B = I B \sinψ\$, but \$ ψ = 90 ^{\circ}\$ always, so \$F = IB\$. (Check graph link to look at force distribution on all conductors.) If the maximum current-carrying conductor is always under the maximum \$B\$, then the average force (average of forces on all conductors) is maximum, so \$F_{avg}\$ is maximum when the maximum current-carrying conductor is under a pole, but there's always a lag as mentioned, and this lag depends on the inductive reactance which is dependent on the frequency of induced current. (Also note that inductive reactance \$X_l \,\propto \,\dfrac{1}{\cos \theta}\,\,\,\,\$). During the start, induced current frequency is \$N_s\$ (maximum possible) so \$X_l\$ is maximum which implies the lag of current is maximum and therefore \$F_{avg}\$ is low. Therefore starting torque is pretty low. (Graph for \$B, I, F, F_{avg}\$ https://www.desmos.com/calculator/hh2t1tpoxa (only for high slip operation))
  5. Once the machine starts running, the relative speed decreases, so the frequency of induced current decreases, and so the reactance decreases, which implies \$F_{avg}\$ increases and thus torque increases, but remember even though the \$F_{avg}\$ is affected by the lag it is primarily dependent on the magnitude of \$I\$ itself (as \$F = BI\$ is force on each conductor.) The peak \$I\$ decreases as the motor starts running as the induced emf peak magnitude decreases, but initially, the decrease in inductive reactance (and thus the reduction in lag between, peak induced EMF and peak current, (which increases the \$F_{avg}\$) overshadows the decrease in the magnitude of \$I\$, so initially \$T \propto \dfrac{1}{slip}\$ (high slip.) Once \$T_{max}\$ occurs (near s = 5 to 10%,) the effect of reduction in reactance (and increase in PF) doesn't matter much, after \$T_{max}\$ the decrease in the magnitude of current dominates, so the torque begins to fall. So \$T \propto s\$ at low slip.

1 Answer 1


The YouTube video linked in the question provides a reasonable, but very basic explanation that avoids most of the math. More complex squirrel-cage rotor designs provide starting torque that is adequate for most purposes. Therefore slip-ring motors are not used very much. Also electronic power inverters called variable-frequency drives (VFDs) are used to provide speed control and high starting torque for difficult loads.

To get a better understanding, you need to study sources that cover the mathematical analysis of motors using equivalent circuits.

The diagram below is used to explain the operation of a three-phase induction motor. It represents one of three phases of a wye-connected motor.

enter image description here

Image from Fitzgerald, Kingsley, Umans, Electric Machinery, 4th ed

R1, X1, R2 and X2 represent the Stator and rotor resistance and reactance. Bm and Gc represent magnetizing reactance and the mechanism of power loss in the iron core. R2(1-s)/s is a variable resistance that represents the mechanism of converting electrical power to mechanical power. The slip, s is the slip / synchronous RPM. Note that the rotor circuit, is shown as being directly connected to the stator. An induction motor is like a transformer. Here it is assumed to be a 1:1 transformer and the secondary winding has been omitted.

When the motor rotor is not turning, the instant after power is applied to start the motor s = 1 and R2(1-s)/s = 0. The current is determined by R1 plus X1 plus the parallel combination of Gc, Bm and X2 + R2. The current is quite high, typically on the order of six times the motor’s full-load running current. Since the current in the magnetizing branch Im is considerably less than the motor’s full-load current most of the stator current also flows in the rotor. In actuality the effective stator to rotor turns ratio causes the rotor current to be perhaps 10 to 100 times the stator current.

The current in the rotor causes torque even though R2(1-s)/s = 0 and the electrical power into the motor is all converted to heat when the rotor is not turning. Remember mechanical power require both force and motion. No mechanical power is produced until the rotor begins to turn.

When the rotor begins to turn, slip decreases and R2(1-s)/s increases. The current decreases, but since R2(1-s)/s is no longer zero electrical power is converted to mechanical power. When the motor approaches the normal operating speed, s is about 0.03 and R2(1-s)/s is about 32 x R2. At that point most of the electrical power going to the rotor is converted to mechanical power and only about 3% is lost in R2.

See also my answers to the following:

Why does an induction motor draw more current when the load is increased?

When load increases in rotor of induction motor how does stator draws more current?

Torque-Speed Equation for Induction Motor


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.