# tripped breaker

#### pilotnmech

Joined Feb 26, 2005
27
This may seem like a really stupid question, but if I get a good, understandable answer I'll never have to ask it again. Why would a circuit breaker trip if you have an extension cord plugged into a compressor, but it won't trip when you plug it into the wall? It seems that the extension cord would create more resistance, and so less current would flow, (according to ohm's law), and the extension cord would heat up from the resistance, but the breaker wouldn't trip. I hear that the motor will draw the amperage it needs whether it's a high resistance or not, which seems to go against ohm's law, but I know that it must be true because the breaker does trip. Please help me to understand the reasoning behind it. Thank You very much.

#### Papabravo

Joined Feb 24, 2006
16,828
OK, I'll try. The motor wants to draw a fixed current at a fixed voltage. If the voltage available to the motor goes down, then it tries to draw MORE current and trips the breaker. The IR drop in the cable reduces the voltage availabe at the motor.

Joined Jan 10, 2006
614
Like he said.
It's a general rule when using compressors to use an air hose extension rather than a Power cord extension for exactly that reason. Not only can the voltage drop in the cable cause the motor to draw more current, it is also one of the more common reasons why compressor motors burn out.

#### pilotnmech

Joined Feb 26, 2005
27
I still don't understand how an end user ( a motor) can demand more than is available. How about some examples with ohm's law. If you have a fixed voltage, and a higher resistance, then the current should go down. 100v and 5 ohms=20A 100v and 10 ohms=10A

#### kubeek

Joined Sep 20, 2005
5,758
Originally posted by pilotnmech@Jun 6 2006, 09:49 AM
I still don't understand how an end user ( a motor) can demand more than is available. How about some examples with ohm's law. If you have a fixed voltage, and a higher resistance, then the current should go down. 100v and 5 ohms=20A 100v and 10 ohms=10A
[post=17544]Quoted post[/post]​
Remember it is AC and the motor is not a purely resistive load. That´s why it can demand more current on lower voltage.

#### thingmaker3

Joined May 16, 2005
5,084
The motor is not demanding more current than is available. It is simply drawing more current than the breaker is rated for. (At leas for the breif time required for the breaker to trip.)

#### aac

Joined Jun 13, 2005
35
Originally posted by pilotnmech@Jun 6 2006, 02:49 AM
I still don't understand how an end user ( a motor) can demand more than is available. How about some examples with ohm's law. If you have a fixed voltage, and a higher resistance, then the current should go down. 100v and 5 ohms=20A 100v and 10 ohms=10A
[post=17544]Quoted post[/post]​
A motor generates a back emf voltage depending on how fast it is spinning. The current is determined by the source voltage minus the back emf divided by the line resistance, sort of. When you increase the voltage drop by adding an extension cord, the motor spins slower. This reduces the back emf which causes more current. Even though you have increased the resistance, the 120V-back emf got bigger faster. If you lock the motor shaft, you'll get zero motor speed, zero back emf, and a lot of current. In this case, you just have the 120V divided by the line resistance. This is the locked rotor or stall current.

It isn't true that you can't use an extension cord. You just need one with a large enough gage wire for the length you need. This may not always be possible. I run a 3hp porter cable twin tank from a 50Ft 12 awg I bought just for this purpose. The lenght of wire in your wall has an impact too.

I hope this helped.

#### pilotnmech

Joined Feb 26, 2005
27
Originally posted by aac@Jun 6 2006, 06:58 AM
A motor generates a back emf voltage depending on how fast it is spinning. The current is determined by the source voltage minus the back emf divided by the line resistance, sort of. When you increase the voltage drop by adding an extension cord, the motor spins slower. This reduces the back emf which causes more current. Even though you have increased the resistance, the 120V-back emf got bigger faster. If you lock the motor shaft, you'll get zero motor speed, zero back emf, and a lot of current. In this case, you just have the 120V divided by the line resistance. This is the locked rotor or stall current.

It isn't true that you can't use an extension cord. You just need one with a large enough gage wire for the length you need. This may not always be possible. I run a 3hp porter cable twin tank from a 50Ft 12 awg I bought just for this purpose. The lenght of wire in your wall has an impact too.

I hope this helped.
[post=17549]Quoted post[/post]​

#### pilotnmech

Joined Feb 26, 2005
27
Originally posted by aac@Jun 6 2006, 06:58 AM
A motor generates a back emf voltage depending on how fast it is spinning. The current is determined by the source voltage minus the back emf divided by the line resistance, sort of. When you increase the voltage drop by adding an extension cord, the motor spins slower. This reduces the back emf which causes more current. Even though you have increased the resistance, the 120V-back emf got bigger faster. If you lock the motor shaft, you'll get zero motor speed, zero back emf, and a lot of current. In this case, you just have the 120V divided by the line resistance. This is the locked rotor or stall current.

It isn't true that you can't use an extension cord. You just need one with a large enough gage wire for the length you need. This may not always be possible. I run a 3hp porter cable twin tank from a 50Ft 12 awg I bought just for this purpose. The lenght of wire in your wall has an impact too.

I hope this helped.
[post=17549]Quoted post[/post]​

#### windoze killa

Joined Feb 23, 2006
605
Originally posted by pilotnmech@Jun 6 2006, 06:49 PM
I still don't understand how an end user ( a motor) can demand more than is available. How about some examples with ohm's law. If you have a fixed voltage, and a higher resistance, then the current should go down. 100v and 5 ohms=20A 100v and 10 ohms=10A
[post=17544]Quoted post[/post]​
I will have a go and will try to but into end user language. the formulas you used are correct but you are looking at them in the wrong way.

Where you have the current changing with a resistance change can also be looked at in another way. At the motor end, the windings will still try to draw say 100V at 10A now because the resistance in the cord has increased there is a larger voltage drop across the cord. Because of this the motor trys to draw more current to maintain the 100V at the motor. Hence to breaker pops.

Hope that makes a bit more sense. I am sure I will be corrected if I need to be.

#### pilotnmech

Joined Feb 26, 2005
27
You are saying the reactance is lessened because the voltage has dropped, correct? If that is true that makes sense. I was told that the motor will still draw the same current and spin at the same speed, but I couldn't understand how. The motor turns slower, and the reactance is less, so it draws more current, right?

#### aac

Joined Jun 13, 2005
35
Originally posted by pilotnmech@Jun 7 2006, 02:57 AM
You are saying the reactance is lessened because the voltage has dropped, correct? If that is true that makes sense. I was told that the motor will still draw the same current and spin at the same speed, but I couldn't understand how. The motor turns slower, and the reactance is less, so it draws more current, right?
[post=17584]Quoted post[/post]​
Not quite. I am saying the back emf changes not the reactance. Whenever a motor is spinning it is also generating a voltage. Let's say I have 120V at the wall and the motor windings have .5 Ohms impedence. When the motor is at running speed let's say it is drawing 15 Amps then the motor is generating 112.5 volts. That is how we get the 15 Amps, 120V-112.5V/0.5 Ω = 15A. When it was first started, it was standing still so it generated 0V instead of 112.5V and the current would be much higher, like 120/.5. But the motor sees enought voltage so it starts to turn and generate the back emf. It quickly gets up to speed and generates 112.5 reducing the current. If we added 100ft of 18 awg wire we would add about 0.6 Ω of resistance. When the motor is standing still we have a large current limited by the 0.6 Ω resistance and 0.5 Ω motor impedance. These don't just add of course but you can see the motor will see a much smaller voltage. Because of this the motor starts turning more slowly or not at all. If it doesnt turn it doesn't generate the back emf and the current stays high until the breaker opens. This is called the inrush current. You have it with and without the extension. It just doesn't go away with the extension.

#### billbehen

Joined May 10, 2006
39
Originally posted by aac@Jun 7 2006, 07:55 AM
Not quite. I am saying the back emf changes not the reactance. Whenever a motor is spinning it is also generating a voltage. Let's say I have 120V at the wall and the motor windings have .5 Ohms impedence. When the motor is at running speed let's say it is drawing 15 Amps then the motor is generating 112.5 volts. That is how we get the 15 Amps, 120V-112.5V/0.5 Ω = 15A. When it was first started, it was standing still so it generated 0V instead of 112.5V and the current would be much higher, like 120/.5. But the motor sees enought voltage so it starts to turn and generate the back emf. It quickly gets up to speed and generates 112.5 reducing the current. If we added 100ft of 18 awg wire we would add about 0.6 Ω of resistance. When the motor is standing still we have a large current limited by the 0.6 Ω resistance and 0.5 Ω motor impedance. These don't just add of course but you can see the motor will see a much smaller voltage. Because of this the motor starts turning more slowly or not at all. If it doesnt turn it doesn't generate the back emf and the current stays high until the breaker opens. This is called the inrush current. You have it with and without the extension. It just doesn't go away with the extension.
[post=17594]Quoted post[/post]​
Another factor is the time constant of the breaker, they trip right away on extreme overcurrent, but closer to the max (on the high side) they take a bit longer (100 milliseconds or more) so the current v. time profile can matter.

#### pilotnmech

Joined Feb 26, 2005
27
Isn't the back EMF generated called inductive reactance?

#### Papabravo

Joined Feb 24, 2006
16,828
Originally posted by pilotnmech@Jun 7 2006, 07:02 PM
Isn't the back EMF generated called inductive reactance?
[post=17619]Quoted post[/post]​
Back EMF is measured in volts. Inductive Reactance is measured in Ohms. If the units are not the same, then they can't be the same thing.

The value of inductive reactance is:
Rich (BB code):
2*pi*f*L
where f is frequency, and L is inductance in henries

#### n9xv

Joined Jan 18, 2005
329
Back emf IS inductive reactance. Back emf is measured in volts and inductive reactance is measured in ohms

- because -

the back emf is an "opposing" voltage that sets up an "opposing" current that has the electrical effect of resistance. This "opposition" to the normal flow of current is measured in ohms, just as a resistance opposes the flow of current and is measured in ohms. Its the way the inductor (motor windings) "react" to the change of current / change of time ratio - thus, an AC voltage & current.

The reason for the compressor motor tripping the breaker with the extenssion cord has nothing to do with reactance. The motor will only draw the current it requires for normal operation,

- BUT -

when the increased line resistance causes a voltage drop, the rotor is turning slower. when the rotor turns slower the operation of that motor is no longer "normal" because TOO MUCH current is being drawn at the slower (abnormal) speed. Remember, the motor draws a rated amount of current @ a rated RPM ect. IF the RPM is lowered and the voltage "relatively" unchanged, the motor will indeed draw an excessive current.

Try connecting the same motor to a variac. At a certain reduced line voltage and corresponding reduced RPM, the motor will draw an excess amount of current and trip the breaker.

Take it even further. hold the motor shaft so that it cant turn at all! Thats right! It will trip the breaker as soon as power is applied to it.

IMPORTANT! There is some emf generated even with the motor at a dead stand still. But that does'nt cause too much current. If anything that would act to reduce the current draw due to the opposition (ohmic properties) of the emf. It has nothing to do with back emf. The inrush or instantaneous current is simply too high at the lower RPM per "relatively" unchanged source. At some point on a graph, the line resistance would be high enough to cause the operating voltage to be low enough, so that the motor would no longer trip the breaker - it would just sit there and do absolutely nothing!

#### aac

Joined Jun 13, 2005
35
Originally posted by pilotnmech@Jun 7 2006, 06:02 PM
Isn't the back EMF generated called inductive reactance?
[post=17619]Quoted post[/post]​
No, as others have pointed out, the inductive reactance and back EMF are not the same. The motor has a coil of wire in its stator, the part that doesn't move. The inductance of that coil is dependent on the the number of turns in the coil and the magnetic materials it is wrapped around. Since these things don't change the inductance and inductive reactance don't change.

Do you understand that a motor and generator are the same thing? It just depends if we are taking mechanical work out or putting mechanical work in. Out is a motor, in is a generator.

Anyway whenever a motor is turning it is generating, even if it is running as a motor. If it is a motor, it will generate less then the line voltage. If it is a generator it will generate more than the line voltage. If it isn't turning at all, it generates no voltage. The more it generates as a motor the less current it needs. Picture a motor with a big old sanding disk on it. When it is just spinning it uses a certain current. Now jam a 2x4 into the disk. This will tend to slow the motor down. Less back emf is generated. More current because of the reduced emf. The motor tries to speed up. The frequency, number to turns in the coil, and magnetic materials are all the same. So the inductive reactance hasn't changed.

Now before the motor has started it isn't generating at all. When you start it there is a large current because of no back emf. This is usually higher than the breaker value but doesn't last long enough to open it. The extension slows the motor acceleration enough so that it the large starting current lasts long enough to pop the breaker even though this current would be less than the current without the extension. OK?

#### Erin G.

Joined Mar 3, 2005
167
Getting back to basics:

NEETS Modual 5, Introduction to Motors & Generators, defines back (counter-) EMF as described above. "The voltage generated in a coil by a moving magnetic field... in opposition (counter) to the moving field that created it...present in every motor, generator, transformer or other inductance winding whenever an alternating current flows." Since this is a "voltage generated", it is measured in volts. Since it opposes the field (voltage) that created it, it effectively cancels (reduces) some of that voltage.

More definitions,...

Inductance: The property of a circuit that tends to oppose a change in existing current flow.

Reactance: The opposition offered to the flow of an alternating current by the inductance, capacitance, or both, in any circuit.

Inductive Reactance:The opposition to the flow of an alternating current caused by the inductance of a circuit.

A little while ago (in a different topic) someone said, "Any time you have a coil, you get a free resistor". If you isolate the coil (inductor) of any motor/generator winding, solonoid coil, transformer winding, etc. and put an ohm-meter across it, you will measure a resistance value. And, ofcourse, we measure resistance values in ohms.

In other words, the coil (windings) in the motor has a resistance value that will oppose the flow of current proportional to that value. It is a value that is built into coil, does not change, and is the inductive reactance.

We can see that although counter EMF and inductive reactance are both present in all motors, they are definately not the same.

To answer the original posted question without any more technical mumbo-jumbo, the basic rule of air compressors has already been stated: Get a fat extension cord and you will probably not continue trippping the breaker. (This assumes that the extension cord you were trying to use is not damaged in any way.)

Hope this doesn't further confuse the issue

#### windoze killa

Joined Feb 23, 2006
605
Originally posted by Erin G.@Jun 9 2006, 02:18 PM

A little while ago (in a different topic) someone said, "Any time you have a coil, you get a free resistor". If you isolate the coil (inductor) of any motor/generator winding, solonoid coil, transformer winding, etc. and put an ohm-meter across it, you will measure a resistance value. And, ofcourse, we measure resistance values in ohms.

In other words, the coil (windings) in the motor has a resistance value that will oppose the flow of current proportional to that value. It is a value that is built into coil, does not change, and is the inductive reactance.
[post=17650]Quoted post[/post]​
When you measure the DC resistance of that coil which is what you are doing when you put an ohm meter across the winding it has very little to do with inductive reactance. The inductive reactance of a coil or inductor is dependent on the thickness of the wire, the tightness of the windings, the spacing of the windings, the capacitance between the windings and to a very small extent the DC resistance. The frequency of the voltage applied affects it as well.

As far as back EMF is concerned it will stretch my memory a bit. Back EMF is generated when the magnetic field that was produced by the application of a voltage colapses. If the voltage doesn't change then the magnetic field will not colapse hence no back EMF. As motors as are being discussed here are AC motors the magnetic field is constantly increasing and decreasing with the sinusiodal voltage applied.

It has been a long time since I taught this stuff so the brain is a bit rusty.

#### n9xv

Joined Jan 18, 2005
329
The DC resistance measured with an ohm-meter across a coil is NOT the reactance of that coil. Inductive reactance (6.28 * frequency * inductance) is dependant only on frequency (for a given inductance), or only on inductance (for a given frequency). The ratio of inductive reactance to DC resistance is called the "Q" of the coil. The higher the Q of the coil, the more effective it is as an inductor.

Inductive reactance is due soley to the principles of back emf and is caused by the expanding & collapsing magnetic feild due to;

∆ i / ∆ t

where,

i = current
t = time.

If there were no such thing as back emf associated with a coil, then there would be no such thing as inductive reactance offered by that coil. The reason why the current in an inductor lags the voltage applied to it is due only to the back emf. The back emf is trying to oppose the source thats producing the back emf in the first place. This opposition to the flow of current is measured in ohms and termed inductive reactance.

You cant make science conform to your understanding, instead, you must conform to the realities of the science.