What causes an AC motor to pull locked rotor amps?

[fourthindiana]

TL;DR

What causes an AC motor to pull locked rotor amps when the AC motor first turns on?

When an alternating current motor first turns on, what causes an alternating current motor to pull locked rotor amps for approximately one second after first turning on as opposed to just pulling the same amount of amps the entire time?

What causes the amperage on an AC motor to drop after pulling locked rotor amps for approximately one second?

 

[AZFIREBALL]

1. Because the rotor IS locked!
2. Because it is rotating.

 

[Tom.G]

(semi-) Long version:
When a conductor (wire) is moving relative to a magnetic field, there is a voltage induced in the wire. The relative movement can be either the physical motion of the wire, or varying of the magnetic field strength from either motion or due to it being created from an AC current. This is the basis for an electrical generator.

When an electric motor first has power applied to it, the rotor is not rotating (moving). As the rotor gains speed, due to the relative motion, a voltage is induced in the windings. The Polarity of this induced voltage is the opposite polarity of the applied voltage.

With opposite polarities, the two voltages subtract from each other, yielding a reduced effective voltage to the windings. With the lower effective voltage, the current decreases. The faster the rotor spins, the higher the induced voltage.

The term Locked Rotor means that the rotor is not turning. If power is applied when the rotor is physically stopped from turning (locked), then the locked-rotor current will continue to flow, overhaeting and eventually destroying the motor.

Hope this helps.

Cheers,
Tom

 

[sophiecentaur]

Summary: What causes an AC motor to pull locked rotor amps when the AC motor first turns on?

What causes the amperage on an AC motor to drop after pulling locked rotor amps for approximately one second?

Where did you read that precise statement? To me, it is far too simplistic and doesn't describe reality.

At switch on, the locked rotor current will flow and it will go to a minimum once the motor is up to speed. The current will be related, inversely, to the motor speed. If you use a DMM to measure the current, the response delay could miss the fact that the current drops steadily during acceleration - which presumably takes about one second. An analogue ammeter would show the time profile of the change better (or a better, faster data-logging digital instrument).

 

[Rive]

Where did you read that precise statement? To me, it is far too simplistic and doesn't describe reality.

I think it is just a rule of thumb, saying that you should size your fuse response with this one second in mind. No textbook (worth citing) would say something that vague (I hope).

 

[essenmein]

You can see locked rotor amps drop quite a bit if sustained, this is due to the rapid heating of the winding's and copper's fairly large tcr.

 

[timmeister37]

I

(semi-) Long version:
When a conductor (wire) is moving relative to a magnetic field, there is a voltage induced in the wire. The relative movement can be either the physical motion of the wire, or varying of the magnetic field strength from either motion or due to it being created from an AC current. This is the basis for an electrical generator.

When an electric motor first has power applied to it, the rotor is not rotating (moving). As the rotor gains speed, due to the relative motion, a voltage is induced in the windings. The Polarity of this induced voltage is the opposite polarity of the applied voltage.

With opposite polarities, the two voltages subtract from each other, yielding a reduced effective voltage to the windings. With the lower effective voltage, the current decreases. The faster the rotor spins, the higher the induced voltage.

The term Locked Rotor means that the rotor is not turning. If power is applied when the rotor is physically stopped from turning (locked), then the locked-rotor current will continue to flow, overhaeting and eventually destroying the motor.

Hope this helps.

Cheers,
Tom

I searched locked rotor amps and came across this thread from last year. TomG's explanation seemed to have satisfied the OP (who apparently had his account disabled), but I'm not sure I entirely understand this.

Based on TomG's explanation of locked rotor amps, here is my understanding of the cause of locked rotor amps in an Alternating Current motor:

When an electric motor has power first applied to it, the rotor is not moving. As the rotor gains speed, a voltage is induced in the windings. The polarity of this induced voltage is the opposite polarity of the applied voltage.

With opposite polarities, the two voltages subtract from each other, yielding a reduced effective voltage to the windings. With the lower effective voltage, the current decreases. The faster the rotor spins, the higher the induced voltage.

When an electric motor has power first applied to it, there is no induced voltage in the windings. Therefore, when an electric motor has power first applied to it, there is no induced voltage to subtract from the applied voltage. Therefore, there is higher effective voltage when an electric motor has power first applied to the electric motor than the effective voltage when the rotor reaches top speed. Therefore, there is more current when the electric motor has power first applied to it than the amount of current when the rotor reaches top speed. This extra current that exists when the electric motor has power first applied to it is locked rotor amps.

 

[essenmein]

II searched locked rotor amps and came across this thread from last year. TomG's explanation seemed to have satisfied the OP (who apparently had his account disabled), but I'm not sure I entirely understand this.

Based on TomG's explanation of locked rotor amps, here is my understanding of the cause of locked rotor amps in an Alternating Current motor:

When an electric motor has power first applied to it, the rotor is not moving. As the rotor gains speed, a voltage is induced in the windings. The polarity of this induced voltage is the opposite polarity of the applied voltage.

With opposite polarities, the two voltages subtract from each other, yielding a reduced effective voltage to the windings. With the lower effective voltage, the current decreases. The faster the rotor spins, the higher the induced voltage.

When an electric motor has power first applied to it, there is no induced voltage in the windings. Therefore, when an electric motor has power first applied to it, there is no induced voltage to subtract from the applied voltage. Therefore, there is higher effective voltage when an electric motor has power first applied to the electric motor than the effective voltage when the rotor reaches top speed. Therefore, there is more current when the electric motor has power first applied to it than the amount of current when the rotor reaches top speed. This extra current that exists when the electric motor has power first applied to it is locked rotor amps.

This seems wrong. Let's take a PM or externally excited machine, induction machines are different beasts.

The induced voltage in this case has nothing to do with applied voltage. The only thing that determines induced voltage is rotor flux* and rotor speed. If you rotate such a machine, you will see volts generated with nothing connected to the windings.

*Stator current can and does affect rotor flux - see field weakening for example.

So now to get this machine to do something (produce torque) you have to get currents flowing in the stator to produce the torque you want. So now to control this current you apply a voltage that is slightly higher or slightly lower, and at a different phase angle, to the induced voltage. This is this voltage differential you are talking about. Note this phase angle is generally maximum +/-90deg for full field weakening, not opposite polarity.

Locked rotor means that there is no rotating field. Ie, no back EMF. Now the locked rotor current is determined purely by the DC resistance of the stator winding and the applied voltage, note that if the rotor is stationary, then for a synchronous machine, the stator field is also stationary, ie DC.

Induction machines do NOT have a stationary stator field with a locked rotor if you are trying to produce torque. So now the locked rotor current is determined by a complicated calculation including the rotor cage.

 

[timmeister37]

This seems wrong. Let's take a PM or externally excited machine, induction machines are different beasts.

The induced voltage in this case has nothing to do with applied voltage. The only thing that determines induced voltage is rotor flux* and rotor speed. If you rotate such a machine, you will see volts generated with nothing connected to the windings.

*Stator current can and does affect rotor flux - see field weakening for example.

So now to get this machine to do something (produce torque) you have to get currents flowing in the stator to produce the torque you want. So now to control this current you apply a voltage that is slightly higher or slightly lower, and at a different phase angle, to the induced voltage. This is this voltage differential you are talking about. Note this phase angle is generally maximum +/-90deg for full field weakening, not opposite polarity.

Locked rotor means that there is no rotating field. Ie, no back EMF. Now the locked rotor current is determined purely by the DC resistance of the stator winding and the applied voltage, note that if the rotor is stationary, then for a synchronous machine, the stator field is also stationary, ie DC.

Induction machines do NOT have a stationary stator field with a locked rotor if you are trying to produce torque. So now the locked rotor current is determined by a complicated calculation including the rotor cage.

essenmein, i am not a physicist or an engineer. Your explanation just went over my head. What specifically about my understanding of this is wrong?

 

[Guineafowl]

As the rotor gains speed, a voltage is induced in the windings.

The only thing that determines induced voltage is rotor flux* and rotor speed.

I don’t quite see the contention here...

The induced voltage in this case has nothing to do with applied voltage.

...nor what this argument is countering. @timmeister37 seems to have got the gist of why a stalled motor draws more current than a spinning one.

 

[essenmein]

I don’t quite see the contention here...

...nor what this argument is countering. @timmeister37 seems to have got the gist of why a stalled motor draws more current than a spinning one.

Some of the statements are correct, however the thing that sparked my response are these:

The polarity of this induced voltage is the opposite polarity of the applied voltage.

With opposite polarities, the two voltages subtract from each other, yielding a reduced effective voltage to the windings. With the lower effective voltage, the current decreases. The faster the rotor spins, the higher the induced voltage.

The first statement implies that the voltage is induced as a response to the applied voltage, other wise how could it know about polarity? This is not correct.

Then for a DC brushed machine, the induced voltage, or BEMF is the same polarity as the applied voltage if its spinning due to that applied voltage.

Eg a 12V brushed machine will induce maybe 10V at some speed and load. Then the voltage differential is the difference between the 12V supply and the 10V BEMF, ie 2V. This 2V across the DC resistance of the winding produces the current flow (and therefore torque).

If the rotor is stopped, no BEMF, and therefore all of the 12V is impressed on the DC resistance, ie much higher current.

 

[essenmein]

The other case would be free spinning at no load, where the BEMF equals the applied voltage, which means no current flows (there is always some friction load, bearings windage etc, so this is more a theoretical case).

 

[timmeister37]

I don’t quite see the contention here...

...nor what this argument is countering. @timmeister37 seems to have got the gist of why a stalled motor draws more current than a spinning one.

I did not see what that argument was countering either.

 

[Guineafowl]

The first statement implies that the voltage is induced as a response to the applied voltage, other wise how could it know about polarity? This is not correct.

Ok, agreed. The opposite polarity thing is a bit problematic. How about, for the level the OP is seeking, the applied voltage causes the rotor to spin in such a direction that its induced voltage opposes the applied, reducing the current flowing?

 

[Tom.G]

Hey folks, some of those quotes causing contention are take out of context. Please see my post #3:

When an electric motor first has power applied to it, the rotor is not rotating (moving). As the rotor gains speed, due to the relative motion, a voltage is induced in the windings. The Polarity of this induced voltage is the opposite polarity of the applied voltage.

With opposite polarities, the two voltages subtract from each other, yielding a reduced effective voltage to the windings. With the lower effective voltage, the current decreases. The faster the rotor spins, the higher the induced voltage.

Whereas @essenmein took some statements from a single thought that spanned two paragraphs. Specifically:

The Polarity of this induced voltage is the opposite polarity of the applied voltage.

With opposite polarities, the two voltages subtract from each other, yielding a reduced effective voltage to the windings. With the lower effective voltage, the current decreases. The faster the rotor spins, the higher the induced voltage.

The first statement implies that the voltage is induced as a response to the applied voltage, other wise how could it know about polarity? This is not correct.

Me thinks we have a tempest-in-a-teapot going on here!

Cheers,
Tom

 

[Guineafowl]

Hey folks, some of those quotes causing contention are take out of context. Please see my post #3:Whereas @essenmein took some statements from a single thought that spanned two paragraphs. Specifically:

Me thinks we have a tempest-in-a-teapot going on here!

Cheers,
Tom

No dramas! Couldn’t see what Essenmein was driving at, that’s all.

It has raised a linguistic point about the meaning of ‘opposite’ in electrical terms. When the rotor spins, it induces a voltage that is the same polarity as the input, and so it acts in opposition. So the back EMF is simultaneously the same and opposite polarity.

A similar problem occurs with switches, where an ‘open’ or ‘closed’ switch will perform in the opposite manner to a door.

 

[essenmein]

Me thinks we have a tempest-in-a-teapot going on here!

Cheers,
Tom

I think this is clearly more of a sturm I am wasserglas.

 

[timmeister37]

Hey folks, some of those quotes causing contention are take out of context. Please see my post #3:Whereas @essenmein took some statements from a single thought that spanned two paragraphs. Specifically:

Me thinks we have a tempest-in-a-teapot going on here!

Cheers,
Tom

Based on my explanation of what causes an AC motor to pull locked rotor amps in post #7, do i seem to understand the gist of what causes an AC motor to pull locked rotor amps?

 

[timmeister37]

No dramas! Couldn’t see what Essenmein was driving at, that’s all.

It has raised a linguistic point about the meaning of ‘opposite’ in electrical terms. When the rotor spins, it induces a voltage that is the same polarity as the input, and so it acts in opposition. So the back EMF is simultaneously the same and opposite polarity.

I don't understand this. I thought that the back EMF is the opposite to the applied voltage. How is the back EMF simultaneously the same and opposite polarity?

 

[Tom.G]

Based on my explanation of what causes an AC motor to pull locked rotor amps in post #7, do i seem to understand the gist of what causes an AC motor to pull locked rotor amps?

Yes!

edit: (can't do quotes in an edit)

post #19 by @timmeister37 :
How is the back EMF both the same and opposite polarity?

post #16 by @Guineafowl :
A similar problem occurs with switches, where an ‘open’ or ‘closed’ switch will perform in the opposite manner to a door.


The overall explanations we supplied were somewhat confused. But yes, you managed to get the idea.

Tom

 

[Guineafowl]

I don't understand this. I thought that the back EMF is the opposite to the applied voltage. How is the back EMF simultaneously the same and opposite polarity?

Ignore R. The EMFs are opposite in terms of which way they will try to drive current. This is the way to think about your original question.

But if the EMFs are connected + to +, and - to -, would you not call this the same polarity, ie like connected to like?

Add a likely load in parallel with the motor, such as ‘motor running’ indicator lamp. You haven’t changed the EMF connections, but now you would say both are in parallel with the lamp. Draw it out and tell us what you think.

 

[artis]

back EMF cannot be 180 degrees out of phase as @essenmein already said because that would mean that there would be a potential difference and the higher the potential difference (Voltage) between two points the higher the current so as back EMF would increase current would also increase, but we know that this is not the case!

Instead as back EMF increases current through a motor decreases so apparently the Potential difference across the motor has decreased , but how can this happen if the socket in which the motor is plugged has the same voltage , say 230V as before right?
Simple, a spinning especially unloaded motor behaves a bit like a generator so the back EMF produces it's own voltage which is IN phase with the supplied voltage,
so in theory for a motor with permanent magnets in rotor if the motor was spinning and the plug was pulled from the wall the motor would still produce some 95% of the voltage it was supplied with , this voltage would decrease as the rotor RPM decreases.@timmeister37 So Imagine you have a permanent magnet motor like this and you run it from your wall socket , the motor spins, now for some reason electricity goes down and lights go dark, in the same socket you had the motor you also have a light bulb in parallel. Even though the electricity would switch off instantly , your light bulb would not go out instantly but instead glow for a while with a diminishing glow because now the motor becomes a generator , but if you started cranking it with hand (assuming you could do it fast enough) the bulb would continue shining and if you probed the wires with a scope the waveform would seem the same just like it was when the electricity was supplied from the grid.

sure in reality the frequency would change etc because your hand cranking would not be stable etc but that is a different topic.