How electric motors consume energy?

  • Thread starter Charles123
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  • #51
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So the voltage drop would be solely due to resistance in the wires of the motor?
 
  • #52
we say motor consumes power or energy in contrast with transformer,that transformer does not consume power it utilisese it. While motor which works on the same principle of transformer consumes power.This is because in motor work is to be done to rotate the rotor so that much amount of energy gets consume in motor and hence the rating of motor is in kilowatt while rating of transformer is in KVA.
hope it may help you .
 
  • #53
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So the voltage drop would be solely due to resistance in the wires of the motor?

In a stalled motor, I would say yes.

If you broke apart the motor and threw away the casing/ magnets/ armature/ everything except the wire...the effect on the voltage drop would probably not be much different.
 
  • #54
CWatters
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So the voltage drop would be solely due to resistance in the wires of the motor?

In a stalled motor yes. In the wire and the brushes if it's a brushed motor.

In a running motor there are other sources of loss. See page one of this thread.
 
  • #55
sophiecentaur
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"What about the explanations you already got did you not understand?" - The only explanation given that would happen in an ideal motor is back emf, that if you read the first posts was actually suggested by me, while asking if it was it, all the others are loses related to the functioning of a real motor. What I mean is that it has to be a voltage drop in the motor, or as I said before you could have all the motors you wanted in series, and they would all perform the same way.
CWatters, I understand how electrical energy, using electromagnetism produces mechanical work. That is not my question, I you read the first posts you will see that I am not questioning how an electrical motor works. I am asking how is that the electrical energy going in the motor, and use to produce mechanical work (and heat in a real motor), is no longer available in the wires going out of the motor.
Thank you all
Regards

This shows that you still haven't got this right. Which wire is the one "coming out of the motor"? 'Positive and negative' are only arbitrary names, given to them. The energy is used in the motor (the potential drop across the terminals). No energy is needed for the current to pass through the wires, either the one from the positive or the negative terminal of the battery.
If one or both of the wires were resistive, then energy would be transferred in them and there would be less potential available for the motor.
It's the same situation when you connect three resistors in series, according to their resistance, different proportions of the supply volts are dropped across each one. This sharing out is established in the first nanosecond or so after the supply is switched on.
 
  • #56
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(Let the supply voltage=V and the motor resistance=R)

When the motor is turned on it starts to turn and because the coil is rotating in a magnetic field it acts like a generator and a back emf (E) builds up.As it picks up speed E increases and the current (I) decreases.E reaches a maximum value which depends on the speed reached which ,in turn, depends on how heavily the motor is loaded.We can write:

V-E=IR

Multiplying by I we can write:

VI=EI+I^2R

VI=input power,EI=mechanical output power(work done per second by the motor)and
I^2R=power losses due to electrical heating.

(This is a simplified treatment and ignores other losses)
 
  • #57
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Dadface, thank you for your exposition.
Thank you all for your inputs/answers.
Regards
 
  • #58
1) heat produced by current flowing through the resistance of the armature
2) eddy current in the iron core
3) hysteris losses in the iron
4) resistance at the commutator contacts
5) windage and bearing friction.....larger motors have a cooling fan
6) distortion of the applied magnetic field due to the armature field
 

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