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How do I relate flux equation to transformers?

  1. Feb 12, 2012 #1
    http://imgur.com/VsDrQ,dJ2iv#1

    So according to the picture, I found the transformer voltage equation and current equation to be the following:


    Precondition:
    i1 is current in left loop
    i2 is current in 2nd loop

    Transformer Voltage Eq:

    V1/N1 = 4i2/N2


    Current Eq:

    Ni(i1) = N2(-i2)


    But since we are given that flux(t) = flux_initial sin(wt) How do i use this and relate it to the problem..? I don't get it.

    Thanks for any help.
     
  2. jcsd
  3. Feb 12, 2012 #2

    jim hardy

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    seems to me they are asking you to find V1 as function of flux? After all, flux was described as "known".


    Then you'd know V2, hence I2 and hence I1 all as functions of flux and N.
    Does the source where you got your transformer equations give you a relationship between voltage and flux?
     
  4. Feb 12, 2012 #3
    That's my question, how do i relate the voltage and flux? I am only given equation for flux in terms of t and then for transformers i only know the current equation and the voltage equation. What's the equation to relate voltage/flux or current/flux or something of that sort?
     
  5. Feb 12, 2012 #4

    jim hardy

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    Have you really not studied induction?

    Take a quick look at Faraday's law. I'd wager it is the whole point of that exercise.
    here's some google results



    http://www.google.com/url?sa=t&rct=...r4WPAg&usg=AFQjCNFEMtIAm0jJUOWdXYiV42uU5ROBpg

    http://www.engineering.com/Library/.../articleId/207/Faradays-Law-of-Induction.aspx

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html

    and if that doesn't clear it up, c'mon back.....
    probably you knew it already and just didnt make the connection.
    That's what a lot of learning is - discovering what you already know.
    hint - you're not going to get a solid number for answer, just an expression in terms of ω and [itex]\Phi[/itex]0. :smile:


    old jim
     
    Last edited by a moderator: Sep 25, 2014
  6. Feb 12, 2012 #5
    Actually I haven't even learned induction yet. I will learn it in another class in about 2 weeks from now. For some reason they didn't make that class (where you learn Faraday's law) a pre-requisite..

    I'll check it out and get back to you.
    Thanks.
     
  7. Feb 12, 2012 #6
    I looked through those links you sent me and I can't find an equation that relates voltage and flux and I'm still not sure on how to proceed with this question.
     
  8. Feb 13, 2012 #7

    jim hardy

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    """I looked through those links you sent me and I can't find an equation that relates voltage and flux """



    every one of them said e = -N * d[itex]\Phi[/itex]/dt ; [itex]\Phi[/itex] being flux

    have you had first semester calculus where they introduce derivative?
    that author used "B" for flux instead of [itex]\Phi[/itex].
    http://www.physics4kids.com/files/elec_faraday.html
     
  9. Feb 13, 2012 #8
    Voltage = d(flux)/dt for 1 turn in the coil (thats what my TA said today)

    so if i have these 2 equations:

    v(t) = 4 i2 * (n1/n2)

    and v(t) = -4 i1 ( n1^2 / n2^2)

    I set v(t) = flux initial omega cos(omega t) and then what.. ?

    i need to solve for i1, and v1..

    so is that my answer?

    v(t) = the derivative of flux which is flux w cos(wt) ?

    then plug that in the equation and solve for i 1 ?
     
  10. Feb 13, 2012 #9

    jim hardy

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    You got it - good job!

    glad you persevered.
    Careful algebra will solve for all in terms of flux, cos(wt), and turns.
    Now you have a headstart on induction for that next class.

    dont forget N.:smile:

    old jim
     
  11. Feb 13, 2012 #10
    Wow that was simple enough.. Thanks a lot for your help!!

    So my final answer I get is:


    v1(t) = -N1 flux w cos (wt)

    i1(t) = (N2)^2 flux w cos (wt)

    v1(t) / i1(t) = -N1 / (N2)^2


    Is this correct?
     
  12. Feb 14, 2012 #11

    jim hardy

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    Hi pghaff...

    well lets see... i would have expected there'd be a 4 in there someplace because that was one of the few hard facts they gave us.

    Glad you got the derivative concept. That's important, and gets real interesting when you have a non-sine wave...

    Okay let's set that derivative w * flux * cos(wt) to DFLUX just to simplify typing.

    V1 = - DFLUX * N1, as you wrote

    I2 is (volts over there) / (resistance over there) = (-DFLUX * N2) / 4
    and I1 = I2 * N2/N1

    so I1 = (-DFLUX * N2/4 ) * N2/N1,
    which = (-DFLUX * N2^2) / (4 * N1) did one of us mess up on N's?

    and V1/I1 = (-DFLUX * N1) / (-DFLUX * N2^2) / (4 * N1)
    which = (N1) / (N2^2) / (4 * N1)
    which = (4*N1^2) / (N2^2)

    which = 4 (N1/N2)^2

    they're telling us ohms are transformed by square of turns ratio.

    Check both our arithmetics and see what's right...

    now i see why teachers make us show our work......

    Thanks for your interest. Makes an old man feel less useless!

    old jim
     
  13. Feb 14, 2012 #12
    Yours is right I believe, but here:

    and I1 = I2 * N2/N1


    I believe it should be:

    I1 = I2 * -N2/N1

    Because the current i2 is flowing OUT of the transformer at the dot, so shouldnt it be Negative value?

    Thx again
     
  14. Feb 15, 2012 #13

    jim hardy

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    hmmm i worked quickly and didnt draw an arrow for current over on secondary.
    my bad - inattention to detail. I should've assigned a direction.

    i think you're right, and most important you have got the point of the exercise !

    I wish you success in your studies.

    This stuff isn't so hard, what is hard is to apply the necessary rigor and discipline.

    Form the habit of being meticulously neat in your algebra - align your equations, show every step and life will go easier. Better to work a problem once neat and slow than five times sloppy and fast producing four different answers.

    Probably you're already neat and orderly , evidenced by your catching my sign mistake.

    have fun !
     
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