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Mmf in transformer

  1. Jun 12, 2015 #1


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    I am studying about transformer and get confused about some concept. I have learn that in an ideal simple transformer the primary has two components of current:
    • Magnetization current i.e. the open-secondary primary current.
    • Current that (ignoring small leakage inductances) finds it way (via the turns ratio) to the secondary.
    The flux in iron core is determined by primary voltage and cannot be affected by secondary coil in a transformer and this is due to the load's ampere-turns in the primary exactly cancelling the secondary ampere-turns. In textbooks that i1N1 = i2N2 since the counter mmf is opposing the mmf of the primary coil.

    I just wonder if they are exactly the same does it implies that flux will be canceled out? I know flux should not canceled out but the two mmfs are the same this make me confused.
  2. jcsd
  3. Jun 12, 2015 #2


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    Well, that's because your textbooks imagine some ideal transformer, that does not need any flux at all.

    i1*N1 = i2*N2

    i1 = i2 * N2 / N1 (which is not correct)


    I1 = i2 * N2 / N1 + i1m ( which is more correct )

    Say that I2 = 0, then i1 = im1 ( im1 is the magnitizing current ).

    The flux is not canceled, it's just balancing, so that the flux will be kept constant at any load ( by an ideal transformer ).
  4. Jun 12, 2015 #3


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    Thanks a lot. I would also like to ask is that in practical situation magnetizing current is very small compared to the total current? In addition, you said "the flux will be kept constant at any load" do you actually mean the flux is only determined by primary voltage (which is varying) and is not affected by secondary current?
  5. Jun 13, 2015 #4


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    It is "small", but not "very small".

    I1 is a function of I2:

    I1 = i2 * N2 / N1 + i1m.

    In practice L1 has an impedance ( Z1 ), so higher current in L1 will lead to a higher voltage drop ( V1 ) in L1. Now, if I1 is increased ( due to I2 ) , V1 will be increased as well: V1 = I1 * Z1.

    Now, call the back-emf in L1: E1 and the supply-voltage Vs: The current in L1 is driven by: I1 = ( Vs - E1 ) / Z1. So increasing I1, ( Vs - E1 ) must be increased, and as Vs is assumed constant, E1 must be decreased, thus the flux must be decreased: E1 = dψv/dt.
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