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Back EMF in Solenoid

  1. Oct 25, 2013 #1
    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. jcsd
  3. Oct 25, 2013 #2


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    Staff: Mentor

    The back emf is proportional to di/dt, so if no current flowed there would be no back emf. Current will rise to the value where di/dt is such that back emf exactly opposes applied voltage.

    This is not much different from current in a resistance. The voltage across a resistance increases as the current increases (obeying Ohms Law) until the voltage across the resistance exactly opposes the applied voltage.
  4. Oct 26, 2013 #3
    i didn't really get this. I was comparing ''back EMF'' of a solenoid with the emf induced across a DC motor's armature. When the motor's back EMF = source voltage, no current flows. So it's like connecting a battery with opposite polarity.

    Considering a resistor directly connected across a source, once the voltage across it is equal to source, the current through it does not increase because for that there would have to be a difference in potential. However, if you connect a solenoid across a DC source, the current through it continues to increase for a certain time even though source voltage = back EMF. With no difference in potential, how is this possible?

    Also, the solenoid circuits's equation doesn't appear to be balanced:
    V(source)=Back EMF (which is equal to source) + CurrentxReactance
    when back EMF = source, the CurrentxReactance part should be zero!?
    Last edited: Oct 26, 2013
  5. Oct 26, 2013 #4


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    Staff: Mentor

    In an "ideal lossless" motor, yes. In a practical motor, there is armature resistance, so that drop has to be accounted for.

    v=L.di/dt so if the applied voltage is maintained constant, di/dt will be constant, meaning current continues to ramp up forever (in a lossless inductor).

    For sinusoidal AC, in the steady state, the back emf is the current x reactance term.
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