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Isn't the current, coming from the 3-phase source of the stator, constant? We don't change the source, so how does the current change? Is a voltage induced because of the magnetic field of the rotor?

This leads me to another question which might help: Does the rotor's magnetic field of an induction motor affect the stator's current value?

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Since the motor converts electrical energy to mechanical energy, the electrical input power must be equal to the mechanical power transmitted to the load plus power lost in the motor. Since the input voltage is constant, that would have to be reflected in the input current and power factor since power = voltage X current x power factor.

The mechanism by which electrical power is converted to mechanical power is explained using the equivalent circuit of the motor. Just as in a DC motor, a back EMF is generated in an AC synchronous motor. So the motor equivalent circuit is an AC back EMF generator in series with the internal impedance of the motor. The back EMF opposes the source voltage so that the stator current is proportional to the source voltage minus the back EMF divided by the internal impedance of the machine. With an AC machine, the voltage, current and impedance values are all complex numbers. The phase angle difference between the terminal voltage and the back EMF is determined by torque angle, the angle between the rotating stator and rotor magnetic fields. As the name implies, the torque angle is proportional to torque.

This leads me to another question which might help: Does the rotor's magnetic field of an induction motor affect the stator's current value?

I don't think that is related to the synchronous motor question. The induction motor rotor current is ultimately supplied by the stator, but the magnetic fields in both the stator and rotor are pretty much constant as long as the applied voltage and frequency are constant and the ratio of voltage to frequency is constant.

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  • \$\begingroup\$ Clarification for the unfamiliar reader in the phrase "With an AC machine, the voltage, current and impedance values are all complex numbers": instantaneous quantities are real numbers, while phasors are complex numbers; impedance is always a complex number (assuming we're talking about the phasor division \$\tilde{V} / \tilde{I}\$ and not about its magnitude \$V/I\$.) You can get the instantaneous quantities from the phasors and the frequency. \$\endgroup\$
    – alejnavab
    Sep 22, 2020 at 17:13
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In an electric motor, almost all types, the current is proportional to the torque.

If the load increases, in other words, the torque increases and the current will increase.

The magnetic field in a motor are related to the voltages.

The rotor produces a field that turns at the same speed as the supply. When there is no phase difference between the fields, the current is almost zero.

The phase difference between the stator and the rotor is called the torque angle. This is the load on the motor.

Changes in the magnitudes of the magnetic fields, caused by changes in the voltage magnitudes, have little effect on the torque angle. These changes mostly alter the apparent power and are therefore related to the power factor.

(This is a simple explanation and I am making it a community answer so that terminology can be improved by those that understand more than I do. :-)

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Charles Cowie explained the reason based on the equation \$P = V I \cos \theta\$. I've seen other people explain the why using the equivalent circuit. I'm basing my answer simply on Faraday's and Lenz's law. While both Charles' and my answer lead to the same conclusion, mine not only explains why in an AC motor the stator current increases when the mechanical load increases (which is the main asked question), but also why the stator current increases when the rotor is locked or braked, or why the starting current is high. The answer is the same for all of these three questions. The reason a motor draws more current at starting, is the same as why the motor draws more current for heavier loads (and why it stops.)

My explanation (answering your main question)

In the following explanation I'll assume the RMS/effective value (or the amplitude) of the applied voltage(s) is constant. A motor can do useful mechanical work when its shaft connected to its spinning coil is coupled with an external device. However, as the coil spins in the motor, the variable magnetic flux through it induces an EMF that reduces the current in the coil (if the current increased, Lenz's law would be violated.) The phrase counter-EMF is used for an EMF that tends to reduce an applied current. The counter-EMF increases in magnitude as the rotating speed of the coil increases.

  • When a motor is started, initially there's no counter-EMF and the current is high, since it is limited only by the resistance of the coil (which is low). As the coil starts to spin, the induced counter-EMF opposes the applied voltage, and so the current reduces.

  • If the mechanical load increases, the rotor stops a bit (since it's harder for the rotor to rotate a heavier load), which decreases the counter-EMF. This reduction in the counter-EMF thus increases the current in the coil, and so the power needed from the external voltage source.

As a result, the requirements of power to start a motor and to operate it under heavy loads, are greater than those to operate it under average loads.

Answering your individual questions

I think with the previous explanation I already answered your individual questions, but I'll answer them for the sake of it.

Isn't the current, coming from the 3-phase source of the stator, constant?

When the motor is starting, no, it's not constant in magnitude (RMS or amplitude). When the load increases, no. The reason for this was already explained. When the applied voltage decreases in magnitude, no (think about conservation of power and/or Ohm's law.)

We don't change the source, so how does the current change?

It changes by changing the net applied voltage. The counter-EMF opposes the applied voltage, thus reducing the net applied voltage.

Is a voltage induced because of the magnetic field of the rotor?

As far as I know, yes. Firstly, the current induced in the rotor comes from the counter-EMF induced in the rotor, which comes from the magnetic field produced by the stator. On the other hand, the counter-EMF induced in the stator that I was talking about in the previous explanation comes from the magnetic field produced by the current in the rotor.

Maybe thinking about the stator-rotor system as two magnetically coupled coils helps visualizing this. One coil (the stator) is connected to a voltage source (if single-phase) or to three voltage sources (if thee-phase), i.e. to the grid. The other coil (the rotor) is connected to a passive element or load. Now think about all of the magnetic flux that passes through the coil connected to the sources: it'll be the flux due to the current through this coil, and the flux due to the current through the other coil.

To further visualize this, look at this figure from the textbook Circuitos Eléctricos [in English, Electric Circuits] by Jesús Fraile Mora. The core is assumed to be linear, such that we can sum the individual fluxes to get the net flux. \$\Phi_{d1}\$ and \$\Phi_{d2}\$ are the leakage fluxes, they only pass through the coil indicated in the subscript; for example, \$\Phi_{d1}\$ is the leakage flux coming from coil 1 that doesn't reach the coil 2. \$\Phi_{11}\$, \$\Phi_{12}\$, \$\Phi_{21}\$ and \$\Phi_{22}\$ and the self- and mutual fluxes, in which the first digit indicates the coil that receives the flux and the second digit indicates the coil that produces that flux; for example, \$\Phi_{11} = \Phi_{d1} + \Phi_{21} \$ is the flux produced by coil 1 due to the current in the same coil 1, while \$\Phi_{21}\$ is the flux received by coil 2 that comes as a part of the flux produced by the current in coil 1. \$\Phi_{m} = \Phi_{12} + \Phi_{21}\$ is the common or mutual flux, which passes through both coils. \$\Phi_{1} = \Phi_{11} + \Phi_{12}\$ and \$\Phi_{2} = \Phi_{22} + \Phi_{21} \$ are the total fluxes that pass through coils 1 and 2, respectively, which are due to the fluxes by their current and also by the flux that comes from the other coil.

Flux linkages by two coils with flux in the same direction

This leads me to another question which might help: Does the rotor's magnetic field of an induction motor affect the stator's current value?

Yes, exactly, and that's why the stator current increases when load increases, or why the starting current in the stator is high.

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