# Effects of low voltage on an induction motor.

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I have come to accept that when an induction motor receives lower voltage than it was designed for it will pull higher current. This doesn't make any sense to me, but I have seen it stated enough times that I'm willing to believe it. Everywhere I look provides total nonsense explanations. They say things like "the motor will draw more current in an attempt to maintain its torque" as though the motor has agency and chooses how much current it draws, or "in order to provide the same power" as though the motor cares how much power it provides.

The version that would make sense to me is you have a resistance, both in the actual windings, and in magnetic impedance(?), and if you apply a lower voltage Ohm's law means that it draws a lower current. If that results in a lower power output... what does the physics of the situation care about that? It seems like it should just spin slower, or not be able to move as heavy of a load. I always thought of wattage as a rated upper limit for a motor, not a target it actively tries to hit. Can someone help me out with a proper explanation of what force is at play here that causes the amperage to increase?

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Think talking about the inertia the shaft plus any load will have.

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I tried to sketch motor torque curves (at nominal and reduced voltage) and some hypothetical load torque curve (a constant-torque load - red line in the graph).

You can see how, with reduced motor voltage, the motor will settle at lower speed (lower rpm). This means that its slip increases (the slip says how much its speed is reduced in comparison to the synchronous speed). The thing is that the current the motor draws (its impedance, if you want) very much depends on the slip. As a result, at the reduced voltage the motor will draw significantly more current (It might go above its nominal current causing overheating and stuff, despite the fact that the delivered torque is still at the motor nominal level).

As you can understand, some load curves (like that of a fan) will not cause so much problems with a decreased voltage as some other load curves might (the displayed constant-load curve is a bad one).

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When the motor is running ,the rotor has a magnetic field, and it is moving post the coils of the stater.

That moving field and coil of wire generates a voltage. It is traditionally called the "back EMF"

The voltage opposes that from the mains supply.

So, the current through the motor depends on the difference between the mains voltage and the back emf.

If the motor is  run at a low voltage it doesn't spin so fast and there is a smaller magnetic field produced by the rotor.

So the back emf is lower.

And so the difference between the mains and the back emf is higher and the motor draws more current.

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13 hours ago, Callipygous said:

I have come to accept that when an induction motor receives lower voltage than it was designed for it will pull higher current. This doesn't make any sense to me,

By this I assume you mean the supply or line voltage. Is that correct?

Do you mean something like this?

Here is part of a test on an (admittedly larger) induction motor.

Note that load current increases both above and bolw a particular operating voltage.

Now how is you mathematics ?

There is no simple answer to your question as to why, the answer lies in the way the voltages curents, impedances and rotational frequencies vary with applied voltage.
The maths of this is (literally) quite complex as all the above quantities are complex.

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Thanks for the replies.
I would be interested to hear more about the relationship between impedance and and slip. It seems like this is basically the same thing as in rush current, which I understand as far as what it does, but not so much the nuts and bolts of why and how.

John, your comments on back emf were helpful, just knowing some better keywords to search yields better results than I was getting before. But I was wondering, if the lower RPM is the result of lower line voltage, then it seems like the lower emf produced wouldn't be that much of a difference. Both line voltage and back emf went down, so wouldn't that leave them mostly in line with each other? I guess that assumes that a drop in line voltage produces an equal drop in back emf.

1 hour ago, studiot said:

By this I assume you mean the supply or line voltage. Is that correct?

Now how is you mathematics ?

I think that's what I mean? I would say yes, I mean the line voltage, but I'm not sure what other voltage there is.

As for my math, I got good grades with minimal effort up through intermediate calculus. Since then I have spent about 10-12 years in jobs where the hardest math I had to do were angle and side length calculations with sine, cosine and tangent. I think I can wrap my mind around some complicated stuff, but it might take me a while and my definition of complicated is probably different from yours.

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I think the lower input voltage leads to a reduction in both speed and magnetisation.

Both factors reduce the back emf. So it falls more than the mains voltage.

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4 hours ago, Callipygous said:

Thanks for the replies.
I would be interested to hear more about the relationship between impedance and and slip. It seems like this is basically the same thing as in rush current, which I understand as far as what it does, but not so much the nuts and bolts of why and how.

John, your comments on back emf were helpful, just knowing some better keywords to search yields better results than I was getting before. But I was wondering, if the lower RPM is the result of lower line voltage, then it seems like the lower emf produced wouldn't be that much of a difference. Both line voltage and back emf went down, so wouldn't that leave them mostly in line with each other? I guess that assumes that a drop in line voltage produces an equal drop in back emf.

I think that's what I mean? I would say yes, I mean the line voltage, but I'm not sure what other voltage there is.

As for my math, I got good grades with minimal effort up through intermediate calculus. Since then I have spent about 10-12 years in jobs where the hardest math I had to do were angle and side length calculations with sine, cosine and tangent. I think I can wrap my mind around some complicated stuff, but it might take me a while and my definition of complicated is probably different from yours.

Yes I said 'complex'. and you said 'complicated'.

Whilst both mean much the same in common parlance, complex has a particular meaning in Science in general and Mathematics in particular.
In this context complex means 'made of more than one (maybe many) parts'.
So when I say the quantities are complex I mean the maths uses special numbers, called complex numebrs, which are made of two parts.
From you description I don't think you have met these?

I will have a think of how to avoid them; meanwhile

Perhaps it is a good idea to have a model to work with.

An induction motor works as follows:

There is a fixed ferrometal (magnetisable) ring with projections called the stator.
This is fixed to the chassis and does not move.
The stator contains projections called poles arranged facing inwards as shown around the center.
At the centre and mounted on the rotating mechanical output shaft is another ferromagnetic lump called the rotor.
The rotor has projections that match and face the projections (but do not touch) on the stator.

Wound around each set of projections are two separate coils, called the rotor winding adn the stator windings respectively.

These windings are connected in series to the supply voltage.
Do you know what a series connection means?
I have drawn it below the machine diagram.

The action of the motor works like this:

The alternating mains voltage in the stator windings creates an alternating magnetic field in the pole projections.
These are arranged to produce alternate N and S poles around the circle as shown.
As the magnetic field alternates at each pole, the N and S alternates at that pole (and all the others around the circle).
So there is a pattern of N and S established moving around the circle at supply frequency.

These N and S poles induce poles of the opposite sense in the facing pole projections on the rotor.

As the N S N S... pattern moves around the stator circle it draws the opposing pattern of S N S N in the rotor around with it causing rotation of the rotor with its attached shaft.

Both these poles are in the ferromagnetic material, not  the windings.

OK, you ask, so where do the rotor windings come into this?
Well the rotating magnetic fields induce a current in them which is opposite in polarity/sense to that in the stator windings.

It is the series connection of these two windings with opposing currents/voltages that prevents the whole thing spinning out of control and maintains the desired speed.

So now, in answer to your question ( but I'm not sure what other voltage there is. ) we are working with 3 voltages,  - the supply, the stator voltage and the opposing rotor voltage
and 3 current - The rotor current, the stator current and the diference between the two.

Once you have confirmed you are happy with the story so far we will explore how slip arises and the meaning of resistance, reactance as being the real and imaginary parts of a complex number.

For John Cuthber's benefit  here is the rest of the article I already posted the first page of.

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19 hours ago, studiot said:

Yes I said 'complex'. and you said 'complicated'.

Whilst both mean much the same in common parlance, complex has a particular meaning in Science in general and Mathematics in particular.
In this context complex means 'made of more than one (maybe many) parts'.
So when I say the quantities are complex I mean the maths uses special numbers, called complex numebrs, which are made of two parts.
From you description I don't think you have met these?

I have met them, but as I mentioned earlier, it has been a very long time since I have had to do anything with them.

Quote

It is the series connection of these two windings with opposing currents/voltages that prevents the whole thing spinning out of control and maintains the desired speed.

So now, in answer to your question ( but I'm not sure what other voltage there is. ) we are working with 3 voltages,  - the supply, the stator voltage and the opposing rotor voltage
and 3 current - The rotor current, the stator current and the diference between the two.

That part is throwing me off. My (limited) understanding of circuits includes a rule that in a circuit with no branches, voltage can vary from point to point, but current is the same throughout. If they are wired in series, how can they have different currents?

My understanding of slip is when there is enough load on the rotor that the force exerted by the magnetic field is not strong enough to spin the rotor at the same rotational speed as the magnetic field. So while our two fields would ideally keep up with each other, with the N's and S'es always pulling in the right direction, the rotor will fall behind some percentage of the time and either not get pushed by part of the cycle or get pulled the wrong direction. I assume learning about the different currents you referred to will illuminate how that relates to a change in current draw on the mains.

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Steinmetz equivalent circuit of induction motor has branches:

Edited by Sensei
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With an induction motor there is no electrical connection to the rotor.

The only current in it is induced by the current in the stator.

Studiot's diagram of them in series is misleading.

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• 1 month later...

I am going to attempt to put this in laymen's terms to the best of my understanding, and probably butcher it in the process. I look forward to your corrections.

We have a few carefully arranged coils of wire known as the stator. We are going to apply a voltage to it, which is going to cause current to flow. Current flowing through a conductor generates a magnetic field. Because of the nature of three phase AC power, and how we've arranged our coils, our magnetic field is rotating in a circle at 3600 RPM (dependent on who's electrical standard you are using and how you've set up your coils)  around the rotor.

A magnetic field moving relative to a conductor generates current (or maybe voltage would be more correct?) , so now we have current flowing through our rotor coils. Rotor coils with current flowing means they are generating their own magnetic field, which is stationary because the rotor is stationary. The rotating magnetic field moving past the stationary one exerts force between the two, our stator is bolted down and our rotor isn't, so the rotor begins to spin. Now the rotor has a rotating magnetic field, which is moving relative to the stationary stator coils, so it begins to generate a current in them. This current is in the opposite direction of the one produced by the mains, and at some point the forces of magnetism and friction balance out at some RPM, I'm guessing for a motor with no load, somewhere slightly below 3600 rpm.

How'd I do?

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46 minutes ago, Callipygous said:

I am going to attempt to put this in laymen's terms to the best of my understanding, and probably butcher it in the process. I look forward to your corrections.

We have a few carefully arranged coils of wire known as the stator. We are going to apply a voltage to it, which is going to cause current to flow. Current flowing through a conductor generates a magnetic field. Because of the nature of three phase AC power, and how we've arranged our coils, our magnetic field is rotating in a circle at 3600 RPM (dependent on who's electrical standard you are using and how you've set up your coils)  around the rotor.

A magnetic field moving relative to a conductor generates current (or maybe voltage would be more correct?) , so now we have current flowing through our rotor coils. Rotor coils with current flowing means they are generating their own magnetic field, which is stationary because the rotor is stationary. The rotating magnetic field moving past the stationary one exerts force between the two, our stator is bolted down and our rotor isn't, so the rotor begins to spin. Now the rotor has a rotating magnetic field, which is moving relative to the stationary stator coils, so it begins to generate a current in them. This current is in the opposite direction of the one produced by the mains, and at some point the forces of magnetism and friction balance out at some RPM, I'm guessing for a motor with no load, somewhere slightly below 3600 rpm.

How'd I do?

So do you think it would still work if the coils were wound on wooden cores and frames rather than ferrous material ?

On 10/27/2019 at 9:12 PM, John Cuthber said:

With an induction motor there is no electrical connection to the rotor.

The only current in it is induced by the current in the stator.

Studiot's diagram of them in series is misleading.

I owe you an apology as I was oversimplifying before.

John is right about the lack of physical connection, this appears in the equivalent circuit (as mentioned by sensei) and I was rather jumping the gun.

Would you like a derivation of it?

This would include the step missing from your description, which is otherwise nearly right.

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My understanding is that yes, it would work with wood, but not as well. Something about the magnetic inductance or capacitance or some such of the material?
I am not sure what you are offering with "a derivation of it" but yes please, whatever explanation you can offer of any concepts I'm missing.

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