Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

(QUESTION) electric motor nature

  1. Feb 14, 2012 #1
    ok, so im coming from a complete automotive perspective so when i refer to voltage applied to an electric motor (engine starter motor for example) i mean 12v direct current. typically consisting of a battery that puts out 500 cranking amps..

    the question is real simple but iv noticed auto technicians in the field have debate about it..

    following the rules of ohms law, when you close a circuit involving an electric motor, given the complete 12 volts it needs (in most automotive cases), the current output theoretically depends on the resistance of the electrical windings inside the electric motor, correct?
    Now, im not an engineer (yet), but its very obvious that when you apply a load or physical strain to an armature of a motor, it tends to draw more current, right? (Best analogy would be cranking an engine on a cold day, cause it requires more work to do so).

    i guess my question would be as simple as, when there is an increased load on the armature as its turning, is the electrical resistance in the motor DECREASING? it sounds like an odd phenomenon and i could be wrong. this sort of goes in to my confusion between physical resistance and electrical resistance. which i'v noticed in cases where you have a faulty car starter motor, or a worn cabin blower motor, over time these tend to draw more current.

    could someone enlighten me? maybe i wasnt correct in my observations, but this very phenomenon here is a fine distinction between diagnosing a bad car starter from a bad battery. from the perspective of a future mechanical engineer working in the automotive field, iv noticed that that the understanding of "difference between physical resistance of a load device and electrical resistance of that device" is diverse and widely misunderstood.
  2. jcsd
  3. Feb 14, 2012 #2
    One way to look at it is this:

    As the rotor spins, it produces a back EMF voltage proportional to how fast it is spinning. This can be shown with Lenz law and faradays law. Basically, the current into a motor will be proportional to the voltage drop between the voltage you apply to the motor and the back EMF the windings generate as it spins. Ideally, a motor with an applied voltage and no load would spin with out any current required because the back EMF could be equal to the voltage applied, however friction, winding resistance, and other factors make this not reality. This is all related to the application of conservation of energy to this system.

    When you load the motor, you are reducing how fast it spins, and from what I mention above, a reduction in motor speed reduces this back EMF voltage. This creates a larger voltage drop across the internal winding resistance, and so more current flows through the windings. The electrical resistance does not change, it is always there and the same value.

    Back EMF is often modeled as a negative feedback for motors, and so you can think of you loading the motor as generating a negative feedback that tells the motor to take more current.

    I think this is quite observant of you to notice this without having the background knowledge, so good job.
    Last edited: Feb 14, 2012
  4. Feb 14, 2012 #3
    From my understanding (someone stop me if I mess this up).

    A motor is an armature that spins because an electric current is applied to it.
    A Generator is an armature that generates an electric current because it is spinning.
    But they are physically almost the same.

    However as the motor spins, it is still acting like a generator at the same time, usually the motor windings are set up in such a way that the current created by the spinning action works against the current driving the motor to spin. This means once the motor is spinning, it takes far less current to keep it turning (the apparent resistance is much higher). However if you jam the shaft of the motor to keep it from turning, the only resistance is what is inherent in the windings, and is much lower. This leads to overheating and can damage it.

    They do this for motors that need to run a long time continuously to reduce the power they will consume.

    Starter motors are wound slightly differently in that the current they generate from spinning works with the current driving the motor forward. This means they do not benefit from the lower current once spinning, although the spinning will help cool the motor from the air movement. This means they will have enormous torque but will overheat extremely quickly if they don't get moving. They also draw huge currents compared to similar sized motors.
  5. Feb 14, 2012 #4
    If you are familiar with circuits, this drawing might help you visualize what I described.

    http://gershwin.ens.fr/vdaniel/Doc-Locale/Cours-Mirrored/Methodes-Maths/white/sdyn/s6/s6fmathm/s6fmathm.html [Broken]

    The first picture shows exactly what I'm talking about. Vc is the back EMF, and as the speed decreases from a load, it is lowered and so more current is dropped through it from the voltage source Va.
    Last edited by a moderator: May 5, 2017
  6. Feb 14, 2012 #5

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2018 Award

    bold mine, jh

    This might be as easy as clarifying some terminology.
    You've got the idea, just need to polish the wording.

    Resistance is already taken.
    Let us say instead that the electrical opposition to current flow in the motor is decreasing.

    That electrical opposition to current flow has two components:

    1. The resistive opposition to current flow which is the resistance of of the windings, simply what you'd read with an ohm-meter at standstill.
    That is fixed and it determines the current the motor will draw when stalled.

    2. The magnetic opposition to current flow which is caused by the windings spinning in the motor's magnetic field.
    That is variable, and the less the mechanical load the higher the magnetic opposition.
    The more the mechanical load the less the magnetic opposition so the higher the current.

    The "magnetic opposition" is what DP and EB correctly called "Back EMF" - a term dear to engineers but only after we study it. I think 'magnetic opposition' will help your colleagues grasp it.
    Maybe "Dynamic Magnetic Opposition ? - after all British term for a motor is "dynamo " .

    You can demonstrate with an old blower or windshield wiper motor and an ammeter - hold shaft with pliers and apply power (maybe only six volts?) observe high current then release shaft and watch current drop as motor comes to speed. You could do that with a small battery charger - it already has an ammeter.
    That ought to start a dialog.

    It is important to use proper terminology, but i dont mind using more descriptive words to paint a workable mental image in a just-us-guys discussion.
    So, in respectable circles the right word is "BACK EMF" not 'dynamic magnetic opposition',
    but using it it should open a discussion for you.

    I hope this helps !
    Last edited: Feb 14, 2012
  7. Feb 14, 2012 #6

    Ha! That's excellent - I would never imagine that!

    Living and learning. Thank you, Jim!
  8. Feb 14, 2012 #7


    User Avatar
    Gold Member

    I agree it's cool you thought of this without the background.. for sure.. even though you said you had a little.

    You're right that the motor draws more current when it starts.

    But you're wrong in saying it's BECAUSE it has to do work.

    Only a motor with a drive is smart enough for that

    For a constant voltage applied, like your 12 volts on your starter, the motor draws ONE current value (for a fixed temperature)

    As any electric motor speeds up, it develops counter emf, which bucks the source voltage, and the motor has less overall current drawn.

    That's why when you load a motor, it seems to TRY harder, really it only draws more current because the load slowed it down, reducing the counter emf.

    Excellent question...

    PS a good test, is to take your car blower motor, and attach it to a 12 volt source, and hook up an ammeter. Notice, the motor draws the same current when starting, as it does when you grab the shaft and force it to stop.

    That current is the locked rotor current, and it exhibits locked rotor torque.

    It doesn't try any harder to move your hand than it does to start moving initially. entirely because of no counter emf induced in the armature.
  9. Feb 15, 2012 #8

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2018 Award

    ""Notice, the motor draws the same current when starting,""

    if your eye is quick enough to see the needle . Starting is a fairly quick phenomenon.

    Use a needle type meter and wach starter current when cranking -
    it goes up as each cylinder comes up on compression -
    that's the RrrrRRRrrrrRRRrrrRRRrrrRRR you hear and the ammeter will be in step.
    You can hear a weak cylinder in your starter motor.

    But when working with them on the bench never try to hold back a starter motor it'll break your wrist and throw your pliers across the room.
  10. Feb 15, 2012 #9


    User Avatar
    Science Advisor
    Gold Member

    Now how would you know that? Haha. Also, when working with a starter off the vecichle don't forget that the case of the starter is ground so when you hook the hot lead don't forget that whatever the case of the starter is setting on may be completing the circuit.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook