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Maximum flux in a single phase transformer

  1. Nov 9, 2011 #1
    A single phase transformer has the following rating: 120 kVA, 2000 V/100 V, 60 Hz with 1000 primary turns.

    (a) the secondary turns
    (b) the rated primary and secondary currents
    (c) the maximum flux
    (d) given a maximum flux density of 0.25 T, the cross-sectional area of the core.

    I have answers for A and B, i'm struggling with C and D! Any help greatly appreciated!
  2. jcsd
  3. Nov 9, 2011 #2

    rude man

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    Remember emf = -Ndφ/dt?
  4. Nov 10, 2011 #3
    Could you explain that equation please?
  5. Nov 10, 2011 #4

    rude man

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    The negative of the time-rate of change of magnetic flux per winding turn, multiplied by the number of turns, equals the electromotive force applied to those turns.

    Otherwise known as Faraday's law of induction. Very famous ....
  6. Nov 10, 2011 #5
    Is dt going to be the reciprocal of the frequency? And will EMF be zero for max flux occurs when transformer is idle? I'm still unsure.
  7. Nov 10, 2011 #6
    Suppose you have a cosinusoidal voltage V=Vmax*cos(wt). What would be the equation of the flux using this EMF in Faraday's law? What is the magnitude of this flux waveform in terms of V?
  8. Nov 10, 2011 #7

    rude man

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    emf = V = -Ndφ/dt
    Oh sorry I didn't realize you were not the questioner.
  9. Nov 15, 2011 #8


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    To be fair, Greek lettered equations usually confuse me at first glance too! Clearer to state just in plain english sometimes! Here is the same again, transliterated:

    Magnetic flux (Weber) is the Volts multiplied by [integrated over] the time, divided by the turns.

    Or in 'simplified algebra';


    Bear in mind you have to take the average Volts in this formula (Av:RMS = ~1.11) [because we integrate over time], plus the flux is two sided (+ and - flux in each half cycle). (The formula above works fine if you are working with square waves only, but doesn't account for integrating the Volts over the time period being calculated for.)

    Also, the 's' is just a half cycle, so if you want to use frequency directly you get the 4.44 term, as used below (2 x 2 x 1.11).

    So, as for the cross-sectional area (T=Wb/A), this is generally approximated to the following formula:

    T(Peak Flux density, Teslas)=VRMS/{4.44 x frequency x cross-section x turns}
    Last edited: Nov 15, 2011
  10. Nov 15, 2011 #9
    This is false.
    1) The flux linkage is the integral of voltage, not volts multiplied by time. If you do the time integral of the previous voltage waveform, the magnitude of this waveform is the answer to the poster's question. Flux is simply the flux linkage divided by the number of turns.
    2) Never answer an homework type question, you have to let them work it out.
  11. Nov 15, 2011 #10


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    Well, to be pedantic, I would say the 'integral of voltage' is physically meaningless, so why use it as a concept in an educational setting? What we are looking at is the rate of change of flux is the Volts/turn at any instant. I'd says any other conclusions are just mathematical machinations after that that don't add much to comprehension of the concepts, which are difficult enough to graps before turning it into pure maths.

    I am unclear what you are saying is 'false'. We do not seem to be disagreeing.

    (I made mods to #8, to clarify on your point.)

    No-one has, yet. I've only sought to provide the basic simplified equations that practical transformer-builders use.
  12. Nov 15, 2011 #11
    I agree, but if the goal is comprehension of the concepts, then why give the formula (that only works out for a given voltage waveform)? Understanding where that formula comes from is the key in understanding, rather than "plug-and-play".

    If the question is about transformer engineering, obviously the author is familiar with integrals. The term "rate of change" is precisely the definition of the derivative. I'm not sure if there is an analog term for the integral, but perhaps it is best to talk about "surface under the waveform", instead of Volts*time. The former is only true for a DC waveform, which we would obviously not like to put on the transformer.
  13. Jan 22, 2013 #12
    I can answer your question. The formula for Maximum flux = V1= 4.44F∅N1 or V2=4.44F∅N2. Hope that does helps.
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