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How electric motors consume energy?

  1. Jul 20, 2012 #1
    What exactly makes the motor consume the electrical energy? I s heat dissipation higher that in a normal coil? Is it counter-electromotive force? What makes the voltage drop?
    Thank you
    regards
     
  2. jcsd
  3. Jul 20, 2012 #2
    Basically you have a force produced on wire because you have charge moving through wire in a B field. This occurs over a distance. So work is done. You get kinetic energy out of it. Of course some heat.

    Thats really the basics. Sorry if you wanted something deeper.

    Here is a nice simple illustration: http://www.walter-fendt.de/ph14e/electricmotor.htm
     
  4. Jul 20, 2012 #3
    Thank you for your answer!
    I understand how they work, how electrical energy produces mechanical energy. My question is how is the electrical energy consumed/dissipated?
    Heat, but more than if the motor wasn`t there? If so, why?
    For instance in an incandescent light bulb it's easy to see that it’s the structure of the filament itself that produces the resistance, and its due to that resistance that heat and the desired light is produced. In the electrical motor one uses the magnetic field generated by the current passing in the electromagnets to interact with the current passing in the wires wrapped around the armature making the armature move (generation of mechanical energy). But by what mechanisms does the voltage drop?
    Regards"
     
  5. Jul 20, 2012 #4
    The electrical energy is converted to Kinetic energy and heat.

    The little electric motors actually just use permanent magnets so this will make it a bit simpler maybe. The motor can be thought of as a resistor if you like. Except its not used to "control" current. The voltage drops occurs because you are indeed using Joules of energy per Coulomb of charge. The Joules of energy go into turning the motor and heat.

    The charge loses its energy and it is "regained" at the battery in a simple DC setup. We could get into the E-fields produced that drive the charge in the wire and such but I dont think that is necessary here.
     
  6. Jul 20, 2012 #5
    Oh I think I see what you are thinking:

    Say you add a resistor in series with a lightbulb. You notice that the light is dimmer. You have reduced the amount of charge that flows through the wire because you added a resistor. The resistor does give off a minor amount of heat. The light bulb has part of the circuit (the filament) that instead of turning the electrical energy into just heat, it also has properties that actually give off energy in the visible spectrum.

    Now if you just have a battery with just a resistor in a closed circuit and the resistance of the resistor is high you get very disappointed. You see nothing. And you can barely feel the resistor heat up at all. This is because the resistance is high so you dont carry as much charge thru the wire. If there is not as much charge flowing, then you cant carry as much energy assuming the battery has not been changed.

    I think its really helpful to think about the units that describe current, voltage and power to understand this in a more complete way. The units for resistance just lead to confusion.
     
  7. Jul 20, 2012 #6

    NascentOxygen

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    If the rotor was locked and not turning, then all of the electrical energy would be converted into heat in the motor´s resistances.
     
  8. Jul 20, 2012 #7
    First of all thank you for your patience and time trying to explain this to me!
    "The voltage drops occurs because you are indeed using Joules of energy per Coulomb of charge. The Joules of energy go into turning the motor and heat." - I understand this, conceptually I have no questions about it. My question is what happens in order for it to be the case?
    What I don’t understand is by witch mechanisms you dissipate electrical energy in running current trough the motor wires. See what I mean?
     
  9. Jul 20, 2012 #8
    Is it about counter-electromotive force? And why is the het dissipation higher than if there was no motor? Same motive?
     
  10. Jul 21, 2012 #9
    Charles I just wrote you a big explanation of electric fields set up in wires by batteries and related it to a gravitational field and a rock and explained work and all and this darn site kicked me off as I guess I took too long.

    And I am very tire now...

    Maybe tomorrow.
     
  11. Jul 21, 2012 #10
    That`s always unfortunate…
    Hope you find the patience to re-write it tomorrow!
    regards
     
  12. Jul 21, 2012 #11

    CWatters

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    In an ideal motor all the electrical energy is turned into mechanical energy and there are no losses. Real world motors have losses for a number of reasons depending on their type and construction.. Here are some examples..

    * Losses in the windings. The windings are typically copper and copper has resistance.
    * Various parts of the motor might be made from iron and there can be eddy current losses
    * Bearings - friction losses.
    * Brushes - electrical resistance and friction.
    * Rotors - air drag

    Consider a DC Permanant magnet motor. They typically accelerate until the back emf matches the supply voltage (or comes close). The back emf depends on the strength of the magnets and the number of turns. So the stronger the magnets the fewer turns are required. That might mean shorter, thicker and fatter wire can be used. That means copper losses are lower. So stronger magnets can help reduce copper losses and improve overall efficiency.

    Some brushless DC Permanant magnet motors are >95% efficient.
     
    Last edited: Jul 21, 2012
  13. Jul 21, 2012 #12
    CWatters, thank you for your answer.
    "In an ideal motor all the electrical energy is turned into mechanical energy and there are no losses.", I understand this conceptually, but how does it happen? How is the electrical energy transformed? What makes the voltage drop?
    Regards
     
  14. Jul 21, 2012 #13
    Charles "you son of a gun..."

    I started thinking about this a bit more and you may have asked something beyond my capabilities and I would now apologize for seeing what you where trying to ask. See the two setups I have below that put me in quandry (that maybe you do or do not understand; MY QUANDRY)

    Case I: Battery connected to an electric motor(with permanent magnets) and you hold the electric motor not allowing it to turn. So you get the wire heating up and you read a voltage drop across the wire. The motor acts a resistor. Assume no internal resistance in the Battery. Done with experiment and now to compare it to the next.

    ok...

    Case II: Battery connected to an electric motor as above but we let in turn. We now have two sources of Pot. energy in my mind: The battery and the B-field from the permanent magnets. You do indeed get a back emf. So the voltage drop you read across the motor should not be the same as above, it should be less. Not only that the current is of course less. Not only that, but I think this may actually get into magnetic vector fields which is way out of my league. This has a lot more to it than a first thought. I would have to read up on this a lot more.

    Sorry for making this seem simple at first because it appears not to be.

    You cooked me. I must leave you to someone else to correct me or make additions or both. I have never read the phrase "electric motors consume energy" and I took this to be a big misunderstanding on your part. But in the end, you have confused me.
     
    Last edited: Jul 21, 2012
  15. Jul 21, 2012 #14
    pgardn, thank you for giving this extra input to the subject. I believe we are on the same page now!
    I "always" assumed I understood the behavior and mechanics of an electric motor, but yesterday I realized that I was missing a big point - what happens to the electric energy as it goes in and out of the motor? Something makes that energy "disappear" as it is transformed in mechanical motion/energy, otherwise you could have as much motors connected in series with no energy losses and they would all perform the same way, clearly this does not happen. it would be a miracle making infinite energy from a finite energy source. So what happens in the wires that make the electrical energy "disappear" as mechanical energy is "created"?
    "Case I: Battery connected to an electric motor(with permanent magnets) and you hold the electric motor not allowing it to turn. So you get the wire heating up and you read a voltage drop across the wire. The motor acts a resistor." - Could not agree more, but why is it so? What happens in the wire?
    "Case II: Battery connected to an electric motor as above but we let in turn. We now have two sources of Pot. energy in my mind: The battery and the B-field from the permanent magnets. You do indeed get a back emf. So the voltage drop you read across the motor should not be the same as above, it should be less. Not only that the current is of course less." - Absolutly right, but again, why? Is It only back emf?
    Best regards
     
  16. Jul 21, 2012 #15
    The basic for Case I:

    The electrical energy, as we call it, is supplied basically because there is a difference between two diff. metals ability to hold electrons. This allows a sort of gradient that allows electrons to go from one metal to another transferred often through a fluid (acid in cars) or some paste. Batteries get heavily into electrochemistry a bit much so I leave that. Now that sets up an electric field in the wire in which electrons flow from the - to the +. But we use conventional current (+ to -) primarily due to historical reasons. It works either way. Now electrons will, in any metal (not attached to a battery) be randomly drifting about in that metal (actually in a lattice), but when you apply this Electric field (you need the battery for this) the electrons drift in an overall direction in the electric field creating current. When the electrons enter an area of higher resistance the type of material usually only allows certain amount of energy in the electrons, lower than in a more conductive wire or the battery itself. The diff. in energy states allowed (this gets quantum) releases energy as heat. The diff. in the ability to hold electrons in a particular lattice results in electrons must have a certain range of energy. If you change going from one lattice to another you must lose energy. (Or gain energy but this would be like electrons walking uphill thus requiring an outside agent to do this work.) The energy is "lost" as heat.

    Those are basics. Sorry if the above is not deep enough. We could go on about exactly how gravity works and get just as confused. There comes a point when mechanistically our brains only go so far with analogies and we have to rely on math that fits experimental evidence. We just cant see the mechanism.

    Case II... got a bit complex for me. I would have to go back to read more about B-field stuff, especially in permanent magnets.

    I think the bolded above is very important.
     
  17. Jul 21, 2012 #16
    I agree that the bolded is a god point, but, and thinking about case I, what happens is that you are not able to produce work, so all energy must be dissipated as heat. Simple thermodynamics tells us that. The entering in an area of higher resistance is easy to see in the example that I gave of the filament in a incandescent light bulb, but not so much in the motor. Because the reason that, and still thinking about case I, all energy is dissipated as heat, is because the rotor is not able to move. So why is that this inability to move the rotor causes a drop in electrical energy?
     
  18. Jul 22, 2012 #17
    Because the motor is no longer a motor.

    Its just essentially a long piece of wire. Long thin wire that attaches to the terminals of a battery so that current flows. Work IS still done because there is an electric field in the wire. The electric field does work on the electrons. They are forced over a distance of Efield within the wire.

    Now if you go back to the applet I showed you, current would flow as long as the black insulating strip did not touch the terminals (actually the brushes that attach to the terminals of the battery.
     
    Last edited: Jul 22, 2012
  19. Jul 22, 2012 #18
  20. Jul 22, 2012 #19
    Charles123 the simple answer to your question, per my understanding, is indeed "back emf".

    Electrons flow through the wire, creating a magnetic field which pushes the wire/ motor. If the motor turns as a result of this and there's any load on it, then that load will push back on the motor, which will push back on the magnetic field, which will push back on the electrons.
     
    Last edited: Jul 22, 2012
  21. Jul 22, 2012 #20
    Yes but that is only part of his question, case II.

    If a motor does not turn, there is no back emf.

    My final explanation from Case II was however incomplete and possibly wrong. And it remains that way. The ideas of what is doing work on charge and in which direction is still muddled in Case II imo. Maybe you could clear that up for both of us. The other thread https://www.physicsforums.com/showthread.php?t=621018 appears to be a mess with more details; last time I read it. Thanks.
     
    Last edited: Jul 22, 2012
  22. Jul 22, 2012 #21
    pgardn:
    "Because the motor is no longer a motor.

    Its just essentially a long piece of wire. Long thin wire that attaches to the terminals of a battery so that current flows." - So the fact that it is now a resistor has only to do with the characteristics of the wires?
    "Work IS still done because there is an electric field in the wire. The electric field does work on the electrons. They are forced over a distance of Efield within the wire." - you are of course right, I meant no mechanical work making the rotor turn.
    Lsos, thank you for your answer! So back emf is the answer then…
     
  23. Jul 22, 2012 #22
    What do you mean by 'back emf is the answer'.
    Do you know how to calculate back emf?
    what is your background knowledge, your profile gives no clues, what text books on this topiuc have you referred to?....were they of no help?
     
    Last edited: Jul 22, 2012
  24. Jul 22, 2012 #23
    Yep. If we ignore internal resistance in the battery to make it simple.

    Back emf...

    Charles you actually asked a question that opened up a much larger can of worms.
    For me anyway... back emf does not answer my problem with what you presented. And thanks for asking, because I still dont have it straight. And after reading about it, I think its a bit beyond what I was thinking.
     
  25. Jul 22, 2012 #24
    pgardn, but, and thinking about your case I, if you could build a motor with wires that would offer the same resistance as the wires leading to the motor, by not allowing the motor to turn electrical energy would have to be dissipated. What would be the explanation in that case?
    truesearch, I don’t understand your question. Back emf (http://en.wikipedia.org/wiki/Counter-electromotive_force) is part of the functioning of an electrical motor. Lsos wrote that back emf is responsible for the voltage drop across the motor. I am just wondering if that is the definite explanation… Please feel free to give your inputs.
     
  26. Jul 22, 2012 #25
    You can play with back emf with those little hand generators as well. You can hook two of them together, one acts as the motor, and the other supplies the voltage. I got that. Back emf is an just a name with a clear mechanism and rules. I explained what I thought it would do to the current and the voltage drops in an earlier post but...

    I dont understand which fields are doing work and at what points in time in the motor/battery setup. And that is the essence of understanding the energy situations in your very first post that I did not realize, for me. Work. I personally get myself to understand most of this stuff with fields and a reference object and what is doing work on that object through a series of events. I cannot manage that with this problem. Maybe you understand what fields are doing work on electrons, or the electrons within the lattice, during the different phases of a motor turning? If so, please give it to me.

    As for case I. I was already assuming the wires had the same resistance. Therefore there are two diff ranges of energy allowed for the electrons. 1. in the battery 2. In the wire. No motor necessary.

    The energy ranges differ for electrons in the two places mentioned. The range allowed in the battery is higher than in the wire therefore we have a conservation of energy problem. My solution, heat given off in the wire. Take note that I do not even want to get into the guts of a battery for this case, I just know it supplies a potential difference. Case I is not a problem for me. You dont have to have a motor. Its just a battery and a single wire. Thats how I am considering it.
     
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