Current taken by induction motor during undervoltage

The induction motor coming up to speed plays a large role and this involves rotor resistance and slip. Now remember when an induction motor is started the slip is equal to one and since the parallel combination between the magnetizing branch of the stator and the rotor circuit will be close to zero because of the extremely low impedance of the rotor circuit hence high starting current (the all so famous 6× full load current).

Now when the motor is at the full speed the slip is anywhere between 3% to 5% which means the rotor resistance is multiplied by 33 or 20 respectively. So when the parallel combination of the magnetizing branch and rotor circuit is taken the impedance increases hence lower motor circuit.

Big thing this involves is that if the mechanically driven load still remains the same. If the load does remain the same then the statement holds true that current will increase. A induction motor is a constant power machine meaning this is the equation that hold the answer P = V * I cos(theta). So since power is constant and if voltage drops then current will rise and vice versa holds true if voltage rises then current will drop. This may result in rotor winding overheating.

One of the issues of low/under voltage supply to a motor is the whole Volts per Hertz phenomenon. A motor is also a constant flux machine. Therefore increasing the voltage without making speed regulation adjustment (frequency) may result in stator core overheating. Now in this case supplying the motor with low/under voltage without adjusting the speed results in low flux conditions. Now one will think if I apply a lower voltage without adjusting speed that it will be a less fluxing condition and the stator core runs cooler. Again it depends.... if the load on the motor remains at the same level and the Volts per Hertz changes then the speed torque curve of the motor changes and may not be able to supply the sufficient electrical torque to sustain rotational motion/overcome inertial requirement of the mechanical load.

Leave your comment