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Alfabris

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Hello.. I am curious how or why there is a higher amp draw on an induction motor during an under voltage condition. According to ohms law, isn't amperage directly proportional to voltage? I am just confused and can not seem to find the correct answer to help me understand how or why this works or is true. Doesn't make much sense to me..

Thank you for your help! It's been driving me crazy!

TOOL_5150

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volts goes down, amps goes up.

Military Veteran

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Power factor as the line votage drops so does counter emf,therefore line current goes up,synchronous motors sometimes hunt and develop a low frequency ripple on the supply which sometimes trips the breakers.

wildleg

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Hello.. I am curious how or why there is a higher amp draw on an induction motor during an under voltage condition. According to ohms law, isn't amperage directly proportional to voltage? I am just confused and can not seem to find the correct answer to help me understand how or why this works or is true. Doesn't make much sense to me..

Thank you for your help! It's been driving me crazy!
V=ir works across a resistive load

a motor is not a resistor.

for a given load on the motor, within a certain range, the power used
by the motor will be somewhat constant, so when voltage goes one way
the current will go the other way.

drsparky

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Hello.. I am curious how or why there is a higher amp draw on an induction motor during an under voltage condition. According to ohms law, isn't amperage directly proportional to voltage? I am just confused and can not seem to find the correct answer to help me understand how or why this works or is true. Doesn't make much sense to me..

Thank you for your help! It's been driving me crazy!
No, amperage is inversely proportional to voltage

sparky970

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P=E x I, so I=P/E. P, power, is equal to 746watts per HP. E=voltage. I=current. If E decreases, then I increases.

Tsmil

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drsparky said:
No, amperage is inversely proportional to voltage
Amperes when horsepower is known on a single phase motor I=746*HP/E*Eff*PF.

Current is inversely proportional but only to the limits of the designed operational voltage.

lectric_hand6855

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You are correct current is directly related to voltage, however in this situation like sparky 970 said you must use the power formula, I=P÷E. Are you a student?

JRaef

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Actually, none of this is all that simple in induction motors, there are very complex relationships involved between applied voltage, load and motor design. You actually can only make very grossly generalized assumptions about what will happen to a motor under low voltage conditions and even then, only if you first define what you mean by "low".

For example, at great risk of dragging them out of the sewer where they belong, the so called "Nola Circuit motor energy savers" do in fact reduce the voltage to the motor, which does in fact reduce the CURRENT drawn by the motor (just NOT THE ENERGY it consumes by as much as they claim). Also think about this; if current in a motor is supposed to increase as voltage decreases, why does a "Reduced Voltage Starter" reduce the current? That's why I say it is too complex to throw blanket statements at, other than to say that IN GENERAL, running a motor under full load at a voltage significantly lower than the design voltage will usually result in an increase in current.

Alfabris

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Thank you to all! Yeah it is starting to make sense that there are many factors to consider. Because if it was always inversely proportional then, and correct me if I am wrong, why does the amperage go up on an over voltage condition as well? That is why it didn't make sense. And yes I am a student, but I am a maintenance technician and I try to understand how everything works and why so I can do my job more efficiently. I've only been in the field for about 3 years, so there's been a lot of learning going on for me.

macmikeman

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I've only been in the field for about 3 years, so there's been a lot of learning going on for me.
Dig this:
I've only been in the field for about 36 years, so there's been a lot of learning going on for me.

8V71

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Thank you to all! Yeah it is starting to make sense that there are many factors to consider. Because if it was always inversely proportional then, and correct me if I am wrong, why does the amperage go up on an over voltage condition as well? That is why it didn't make sense. And yes I am a student, but I am a maintenance technician and I try to understand how everything works and why so I can do my job more efficiently. I've only been in the field for about 3 years, so there's been a lot of learning going on for me.
The windings/core/armature of a motor are designed to run at a certain voltage. If you go higher, the core will saturate and the windings will start drawing more current (more resistive I think) without an increase in motor power.

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Inductive components behave differently. If one ohmed a transformer the coils are usually a few ohms sometimes as low as half an ohm. Under ohms law, if there was no inductive magnetic component (pure resistance alone) you would have close to a short circuit. the windings would heat up and burn up or a breaker would trip. However, when the energized winding magnetizes the iron core, the counter EMF from the induction created in the first place limits the current so to speak in the primary coil. If you put load on the secondary that would result a drop in magnetic flux in the core (magnetic flux inducing current in the secondary) so more current would need to flow through the primary to reach equilibrium in the iron core. The more current on the secondary the more it weakens the field and the more current needs to flow on the primary to reach the right amount of EMF in the iron core, enough counter EMF to prevent the primary winding from drawing infinite power, or at least what we would normally see (ie the counter EMF limits the current, not the low resistance)

A motor works similarly, although its secondary winding is a rotor. But the principal of induction is the same. Ohms law doesn't apply.

A voltage reduction will increase slippage, which in turn means lower speed. The motor will try to maintain speed or at least a stronger flux by drawing more power. This will create more heat. Also keep in mind that a slower rotor means less air movement which further compliments the heat build up. Similar goes for rated voltage but increased mechanical (rotor) load. The inductive properties of the stator will change, the motor will try to keep up speed by drawing more current caused by the changes in flux.

At least that's how I remember it. Someone correct me if Im wrong its been a while:laughing:

lectric_hand6855

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Thank you to all! Yeah it is starting to make sense that there are many factors to consider. Because if it was always inversely proportional then, and correct me if I am wrong, why does the amperage go up on an over voltage condition as well? That is why it didn't make sense. And yes I am a student, but I am a maintenance technician and I try to understand how everything works and why so I can do my job more efficiently. I've only been in the field for about 3 years, so there's been a lot of learning going on for me.
Keep up the hard work and critical thinking. In this field you will never stop learning!

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Actually, none of this is all that simple in induction motors, there are very complex relationships involved between applied voltage, load and motor design. You actually can only make very grossly generalized assumptions about what will happen to a motor under low voltage conditions and even then, only if you first define what you mean by "low".

For example, at great risk of dragging them out of the sewer where they belong, the so called "Nola Circuit motor energy savers" do in fact reduce the voltage to the motor, which does in fact reduce the CURRENT drawn by the motor (just NOT THE ENERGY it consumes by as much as they claim). Also think about this; if current in a motor is supposed to increase as voltage decreases, why does a "Reduced Voltage Starter" reduce the current? That's why I say it is too complex to throw blanket statements at, other than to say that IN GENERAL, running a motor under full load at a voltage significantly lower than the design voltage will usually result in an increase in current.
As a general rule, poly phase induction motors are the most efficient when the voltage is reduced just to the point of some slippage. Usually at that point current will go down, however after a point current goes up. No slippage means the motor is on overvoltage. However, we are assuming that the load remains constant. Varying load, especially one where speed can not vary makes for the most inefficient motors. Often then the motor is oversized and over voltage to the point where there is no slippage, and at that point power factor is less than 85%. Most motors when under an overvoltage condition will just saturate and make more heat. This is especially true if the frequency goes down without the voltage. Finite details aside, a 50 Hz 415 volt motor will run without overheating at 480 volt 60Hz since the Volts/hertz saturation ratio is about the same but not vice versa. A 120 volt motor will burn up at 50hz, but if the voltage is dropped down to say 95 volts it will not saturate and overheat. However, speed of course will be slower and available torque will also be different.

Tsmil

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As a general rule, poly phase induction motors are the most efficient when the voltage is reduced just to the point of some slippage. Usually at that point current will go down, however after a point current goes up. No slippage means the motor is on overvoltage. However, we are assuming that the load remains constant. Varying load, especially one where speed can not vary makes for the most inefficient motors. Often then the motor is oversized and over voltage to the point where there is no slippage, and at that point power factor is less than 85%. Most motors when under an overvoltage condition will just saturate and make more heat. This is especially true if the frequency goes down without the voltage. Finite details aside, a 50 Hz 415 volt motor will run without overheating at 480 volt 60Hz since the Volts/hertz saturation ratio is about the same but not vice versa. A 120 volt motor will burn up at 50hz, but if the voltage is dropped down to say 95 volts it will not saturate and overheat. However, speed of course will be slower and available torque will also be different.
Wow!

dmxtothemax

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Dont forget re actance also plays a part too !

A motor not running has a very low resisance = high current,
Once motor moving re actance comes into play,
thus current decreases.

Less voltage in means less magnetic interaction,
which equals less impedance, which equals less resistance,
less resistance means more current.

Motors are not a linear load !

I've tried to simplify this a lot,
So it's easier to understand,
It's complex.

So all you technical podantics,
don't have a cow man.

It would take a very long time to explain it perfectily technically.

Aussielec

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Hello.. I am curious how or why there is a higher amp draw on an induction motor during an under voltage condition. According to ohms law, isn't amperage directly proportional to voltage? I am just confused and can not seem to find the correct answer to help me understand how or why this works or is true. Doesn't make much sense to me..

Thank you for your help! It's been driving me crazy!
When you decrease voltage in an induction motor you decrease the torque applied to the rotor, so depending on the motor load the current draw could increase or decrease.

lets say the motor is fully loaded in which case decreasing voltage would most likely cause the rotor to slow down. What happens now is that the speed/angle of stator field that moves across the rotor bars increases as the mechanical speed of the rotor decreases. This results in a larger magnetic field created within the rotor. This magnetic field opposes the stator field reducing the total net flux within the motor, which in turn increases the current flowing through the stator winding.

Likewise if the rotor is close to synchronous speed the magnetic field created by the rotor is minimal, which provides very little opposition to your stator field, so current flowing through your stator winding will decrease.

Auselect

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Aussielec said:
When you decrease voltage in an induction motor you decrease the torque applied to the rotor, so depending on the motor load the current draw could increase or decrease. lets say the motor is fully loaded in which case decreasing voltage would most likely cause the rotor to slow down. What happens now is that the speed/angle of stator field that moves across the rotor bars increases as the mechanical speed of the rotor decreases. This results in a larger magnetic field created within the rotor. This magnetic field opposes the stator field reducing the total net flux within the motor, which in turn increases the current flowing through the stator winding. Likewise if the rotor is close to synchronous speed the magnetic field created by the rotor is minimal, which provides very little opposition to your stator field, so current flowing through your stator winding will decrease.
Nice handle man...

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