The starting torque of an induction motor increases with increasing rotor resistance. However, we know that, with increasing rotor resistance, rotor current will reduce, and hence rotor flux will reduce. How does the torque increase then?


1 Answer 1


It is a tradeoff, hence actual wound-rotor motors expose the rotor windings so that a resistance can be added during start, and tapered off or just removed when at speed. The resistance added is actually not tremendous, usually from 1 to 5 times the DC resistance of the rotor.

The initial problem, is that at at start, a typical induction motor can draw up to 1000% rated current. For large (> 200 HP motors) this can severely stress the supply. Pullout torque (torque required to either start the rotor, or drop it out from near synchronous speed) is directly related to slip. Increasing resistance in the rotor, effectively increases the slip, which increases the torque.

As more resistance is added to the rotor, the peak torque curve is moved closer and closer to zero speed. That is the sweet spot for starting a wound rotor motor; lowest inrush current, highest slip, highest torque. Adding more resistance will reduce the available torque, as the slip begins again to decrease.

You can run Wound Rotor motors in a variable speed mode, if you can control the resistance, but that is generally very inefficient.

Consider that basically, an induction motor at standstill is a transformer, with an essentially shorted secondary. When power is applied to the stator, a voltage is induced into the rotor, which, being shorted develops the current which creates the magnetic field to pull the rotor along with the stator's rotating field.

Okay, since the induction motor is a transformer when starting (at zero speed), there are reactances to deal with. The reactance actually causes the induced voltage (and current, and generated magnetic field) to be out of phase with the stator field, generally lagging by about 90 degrees, so the magnetic interaction between the rotor and stator is fairly weak.

If pure resistance is added to the rotor circuit, the phase lag starts to grow smaller. (Note that different constructions of the rotor bars, with different resistances are used to permanently affect the torque curves of many motors). Add enough resistance, and the phase lag reduces to the point of what the motors design slip is, which corresponds to its maximum design torque.

The problem with leaving the resistance in circuit (aside from power dissipation) is that as the motor speeds up, and approaches its synchronous speed - slip, now you have advanced the rotor/stator magnetic phase, resulting in reduced torque at the shaft.

  • 1
    \$\begingroup\$ Would u please give a physical explanation of the answer ! \$\endgroup\$
    – sniperykc
    Nov 11, 2015 at 20:17

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