Back EMF in Solenoid

Discussion in 'General Electronics Chat' started by QwertyXP, Oct 25, 2013.

  1. QwertyXP

    Thread Starter New Member

    Oct 18, 2013
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    Consider an ideal solenoid (no resistance, no leakage reactance etc.) connected across an AC supply. The back EMF induced in it will be exactly equal and in opposite direction to the source voltage (which means that when a certain terminal of the AC supply is positive, the side of solenoid connected with it would also be positive, and vice versa).

    My question is, how will current flow at all when the EMFs of AC source and solenoid are cancelling each other out? It's like having having a circuit with only two batteries and terminals of similar polarities shorted with each other.

    I've read quite a few explanations on the internet but have yet to fully understand what's happening here.
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    The voltage across the inductor V(a) is the same as the applied voltage because there are only two nodes in the circuit.

    The current in the inductor lags the applied voltage by 90 degrees.
     
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  3. QwertyXP

    Thread Starter New Member

    Oct 18, 2013
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    But why does current continue to increase through the solenoid even though the p.d. across solenoid is equal to source voltage? In order for current to increase, shouldn't there be a potential difference somewhere in the circuit?

    Secondly, the solenoid circuits's equation doesn't appear to be balanced:
    V(source)=Back EMF (which is equal to source) + CurrentxReactance
    when back EMF is equal to source, the CurrentxReactance part should be zero!?
     
  4. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    The potential difference is between node a and the grounded node. The potential difference is the time-dependant ac voltage supplied by the source.
    By KVL, the voltage rise across the source equals the voltage drop across the inductor. Notice in the simulation that the applied voltage is a sinosoid and so is the current through the inductor. The current wave lags the voltage wave by 90 degrees.

    V(source) = voltage across the inductor (what you call back emf) PERIOD.

    The current that flows around the loop for AC excitation is V/X, where X = 2∏fL

    For the example shown, V=170 Vpeak so I=170/(2*∏*60*1) = 0.45 Apeak
     
    Last edited: Oct 26, 2013
  5. QwertyXP

    Thread Starter New Member

    Oct 18, 2013
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    This equation seems to hold true only for peak values, and not for values at other instants of time. Isn't Ohm's Law valid at all times?
     
  6. Duane P Wetick

    Active Member

    Apr 23, 2009
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    Consider the solenoid coil an electromagnet which alternately energizes one way, and then in the next half cycle, energizes the opposite way. Enough magnetizing force must be generated such that the field in the iron core cannot collapse fast enough to allow the core to fall back to its rest position. If it does, you've created a buzzer. The reluctance of the magnetized core is what causes it to hold its position until the field reverses and then hold again.

    Cheers, DPW [ Everything has limitations...and I hate limitations.]
     
  7. MikeML

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    Oct 2, 2009
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    Yes. For AC circuits, the Reactance is the ratio between the Magnitude of the voltage and current. You can talk about the Peak values, Average values, RootMeanSquare (RMS) values, etc. However, the voltage vs time and current vs time have the waveshapes shown in Post 2.
     
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