Induction motor with current draw and no load

Discussion in 'General Electronics Chat' started by Graysonrobert, Aug 11, 2014.

  1. Graysonrobert

    Thread Starter New Member

    Aug 7, 2014
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    Can someone explain in simple terms how an induction motor with no load can draw a current
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Mechanical losses: Friction, windage
    Electrical losses: heating, magnetic

    An unloaded induction motor draws almost as much current as a loaded one. It is just that the current lags the voltage, so electric bill is smaller...
     
  3. crutschow

    Expert

    Mar 14, 2008
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    Not totally true. The magnetizing current to run an unloaded motor is much less than the real current to power it at full load. You can estimate the relative value of reactive current to real current from the loaded PF of a typical induction motor of about 0.85 to 0.9.
     
  4. MikeML

    AAC Fanatic!

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    But the magnitude of the motor current doesn't change much loaded to unloaded.
    Good discussion here.
     
  5. MaxHeadRoom

    Expert

    Jul 18, 2013
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    On switch on, an induction motor is a transformer with a shorted turn secondary, the current reduction is dependent on the ability of how close the rotor can approach or near synchronism, where rotor current would be zero, but unachievable in a normal induction motor.
    Anything that affects the approach to sync (e.g.load) increases the primary current.
    Max.
     
    Last edited: Aug 11, 2014
  6. crutschow

    Expert

    Mar 14, 2008
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    Yes, but I think the discussion concludes that the reactive (magnetizing) current doesn't change much from no load to full load, not that the total current doesn't change. So, for example, with a PF of .9 and a total current of 10A under load you would have 9A of real current and 4.36A of reactive current, which should also be close to the no-load reactive current. If the PF dropped to say 0.35 at no-load, then the reactive current would still be about 4.36A but the total current would drop to 5.4A with about 3.2A of real current (if my vector math is correct).
     
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