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Ac generator and dc motor

  1. Feb 28, 2012 #1
    When I put a current into a dc motor, there will be a turning effect. But I thought as this happens, won't there the an induced current that flows in the opposite direction? Bt Fleming's Right hand rule, the current will flow in the other direction. So I'm quite confused about this.

    Similarly, when I have an ac generator.won't there be a force acting opposite to your motion? By Fleming's Left Hand rule? So I'm quite confused about this too.

    Lastly, when I turn the handler of the coil in the ac generator, eg with 10N. Will the force solely act at one side of three armature or will the force be divided into 2 then they will act at the 2 ends and one is upwards the other downwards?

    Thanks for the help! :)
     
  2. jcsd
  3. Mar 1, 2012 #2

    Philip Wood

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    There will be an opposing emf, εback, induced. So there are two emfs acting, the external emf, εsupply and εback. There can only be one current, I, though. It is determined (using Kirchhoff's second law) by

    εsupply - εback = Ir,

    in which r is the resistance of the motor measured across the brushes.

    This ignores complications such as field windings.
     
  4. Mar 1, 2012 #3
    But will the emf back be lesser than the external emf? Thanks for the help!
     
  5. Mar 1, 2012 #4

    Philip Wood

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    Yes. Otherwise the current would be flowing backwards through the power supply, and, if it consisted of rechargeable batteries, you'd be recharging them, at the same time as getting mechanical work from the motor.
     
  6. Mar 2, 2012 #5
    Thanks for the help! BTW then when I turn the ac motor'a handle, is the force only acting on one of the armature or will the forces be divided by 2 and the force will be divided by 2? Thanks for the help.
     
  7. Mar 2, 2012 #6

    Philip Wood

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    You're asking about an a.c. generator, I assume. Are you taking it to be a simple rectangular coil being turned in a uniform field? In that case, for most orientations of the coil, both pairs of opposite sides will experience equal and opposite BIL sinθ forces on their two sides. For one pair, BC and DA, the forces will simply cancel. For the other pair, AB and CD, the forces, FAB and FCD, although equal and opposite, don't act in the same straight line, and will constitute a 'couple' giving rise to the clockwise torque that opposes your turning of the axle.
     

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    Last edited: Mar 2, 2012
  8. Mar 2, 2012 #7
    Oh ok I get it. But also, won't there be opposing forces due to the new induced current and also the magnetic field? So by Fleming's left hand rule, there's an opposing force. So how so do they turnl

    Thanks for the help.
     
    Last edited: Mar 2, 2012
  9. Mar 3, 2012 #8

    Philip Wood

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    The current I've shown in the diagram, and which gives rise to the forces I've been talking about, IS the induced current! Remember, this is a generator: you (or some external agency) has to turn the coil – using a torque which has to be at least as large as the opposing torque due to the BIL sinθ forces due to the induced current.

    [If the generator is 'open circuit' coil not connected to any load, but terminals not connected to anything, an emf will still be induced as you turn the coil, but there will be no current, nor any opposing forces.]
     
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