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How the power transfers across the Ideal Transformer

  1. Jun 18, 2015 #1
    Hi all,
    I've been having a conversation with a brilliant PF memeber about power transfer, but I wouldn't be surprised if my inability for comprehension has driven them to their knees.
    I don't quite remember/understand how the concept of power fits into a transformer. I used to think that it was something to do with inductance storing energy from the primary in the magnetic field, then it would go out through the secondary.
    Basically I know there is stored magnetic energy in inductance in the core, but I've been told it has nothing to do with power transfer (and it's purely Faradays law).

    First, I want to get someone to clarify this:
    I assume the inductance from Xm, is that actually the mutual inductance from both primary and secondary coils, ignoring leakage inductance, what I was thinking was something like:
    Average Energy in core = 1/2 *L*I2 = 1/2 * (Xm*jω) * (IO)2

    =1/2 * L1*(I1)2+M*I1*I2+ 1/2 *L2*(I2)2

    =1/2 * N1Φ12*I1+M*I1*I2+ 1/2 *N2Φ21*I2


    where 1 is primary, and 2 is secondary, LΦ12 is flux from primary without leakage, Φ21 is flux from secondary
    and total core Φ = Φ12 - Φ21

    Using the characteristic of:
    Where the inductance is on the primary circuit, power goes in from source, and goes back out to the source. So that power waveform isn't what's going to a secondary load.

    So Thigns I'm seeking to confirm are:
    Q1: (from the diagram) is the power of the TX going from one side to the other is equal to Ep*Is
    = Es*Ip/a
    This seems possible to me because I imagine that power is pretty much contingent on the secondary, as it's just an inductor when there is no secondary current. This should be true regardless?

    Q2: Jim said that power goes straight across, without waiting (like a sinusoid) So is that average Energy equation that I wrote above right?
    And if so, (given energy = power*time) does this mean:
    Aveage Energy in core = 1/2 * (Xm*jω) * (IO)2
    is going swishing and swoshing in and out of the core as reactive power?

    So if Inductive energy has nothing to do with power transfer, then does one need to think of power being transfered through the ideal TX as:
    At Case 1>OC sec: no power

    At Case 2>small load on secondary, not much current through secondary: Es = N d(Φ12 - Φ21)/dt

    At Case 3>large load on secondary, lots of secondary current: means lots of Φ21 pushing back, reduces the back EMF of Ep, means more current will flow through Ip, which will mean Ep*Is will increase and more flux will be produced for Φ12, which will mean that there is Es = N d(Φ12 - Φ21)/dt will increase as Is increase.

    And, that's how the power is transferred across the Ideal TX?
     
    Last edited: Jun 18, 2015
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  3. Jun 18, 2015 #2

    jim hardy

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    Bravo - you open circuit tested it in your mind .
    Indeed it's just an inductor so energy shuttles back and forth between core and source.

    Let's think here
    Energy in a magnetic field is in proportion to flux^2, see http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html
    And flux in a transformer is set by primary volts per turn.
    So for any given primary voltage the core contains a set amount of energy .That amount does not change because of energy flow from primary to secondary.

    (well, contains was a poor verb to use but it's succinct. We all realize the energy shuttles between the inductance and the source as in Tim's sinewave chart adjacent his transformer model, but had i said "cyclically exchanges with the source" i'd have lost the audience-- old jim )


    I think Tim has made remarkable progress in his mental workings of transformer action

    .
    What he wrote above was a "Eureka" moment for electrical researchers of late 1800's .
    Above discussion assumes sine waves which are a mathematical oddity (God Bless Euler and his Transcendental functions !) .
    Flyback converters are quite another story - energy does linger in those cores.
     
  4. Jun 18, 2015 #3
    I Kind of feel like this is rewarding, but I also kind of think it shouldn't be taking me this long to get my head around it, because I thought I understood it a few months ago.
    From your response I take it I was on the money?
    So, (using the aforementioned diagram) is it fair to say that the inductance L1 = Xm/jω
    sets up a base flux Φ = L1*IO/N1, which is the flux at OC.
    Then this flux stays, but get added to by additional flux when a load on the secondary causes an opposing flux Φ12 from the secondary, which reduces the Back EMF of Ep, which increase Ip. Now here's where it still gets 'sticky' for me, so the power being used transferred by the TX is Ep*Is (I assume? please correct otherwise), but when Ip increase there is a larger voltage drop on Rp, meaning that the Voltage on Xm will be less, meaning that IO will drop and the base flux (Φ = L1*IO/N1) will be less. But how can this make sense with more current running through the primary coil, I'd have thought the flux from the primary would fight back from this increase in current?
    This is why I thought there'd be a mutual inductance term involved, (see above equation).
     
    Last edited: Jun 18, 2015
  5. Jun 18, 2015 #4

    jim hardy

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    Well this is the trouble with words instead of drawings and formulas .

    I write in a choppy style.
    It's because i think in a jumbled mess and have to force myself to go back and put one step per sentence.

    well let's see.
    500px-TREQCCT.jpg
    you got that from definition of inductance, L=NΦ/I , right ?
    If by N1 you mean the number of turns in XM
    (which we can assign same number of turns as ideal TX primary at heart of this model)
    I'd say instead Φ = L1*IM / N1

    yes, OC = open circuit = no load

    ahhh so now you have connected a load, we're no longer OC i see......

    Some people sum fluxes,, myself i prefer to sum mmf's . I wont call you wrong
    just for me the total flux is the total MMF/reluctance of core
    so i'm holding my thought to see if we diverge further down the thought stream

    okay some of the ideal transformer's primary flux Φ is removed, for you by secondary Φ and for me by secondary MMF
    (Kirchoff has two laws , after all)


    yes,
    remember that the ideal transformer requires no magnetizing current
    so now we have Is>0 for first time in this thought experiment
    (look carefully where Is is in the model..)

    seems right per the model , so stay rigorous with labels
    indisputably
    slightly less yes
    slightly, yes, there are fewer volts per turn of induction but more volts per turn of IR drop
    slightly less, yes
    That primary coil is on the ideal transformer. So none of those Is amps are magnetizing amps, they're all load amps.
    That piece of logic i cannot follow.
    Primary and secondary amp-turns in an ideal transformer are exactly equal, there's no magnetizing current (infinite inductance)
    Primary amp-turns rise to match secondary amp turns.

    Perhaps this would help your thinking:
    go back to the model
    Set turns ratio = 1 so it's out of the way.
    Move X"S and R'S over to secondary side in between ES and VS.
    Observe that's where it belongs anyway,
    and draw a second Is arrow over there in secondary flowing through the secondary X"S and R'S .
    Now what voltage Ep is impressed across ideal TX's primary ?
    Why, of course, IM* XM.
    That's what is left of VP after RP and XP each have taken their bite.
    (sorry - too much shark talk on TV right now)
    It's the magnetizing current that makes counter-EMF not the load current.
    Moving secondary impedances over where they belong made that more visually obvious.
    Kirchoff rules, take him one step at a time
    He rules for volts, amps, maxwells and oersteds.
    just be methodical and rigorous

    ---------------------------------------------------------------------------------------------


    any help ?

    This took quite a while to type,
    i hope thread hasn't changed much in the interim...


    old jim
     
    Last edited: Jun 18, 2015
  6. Jun 18, 2015 #5
    Brilliant reply, and fear not the post did not change at all while you were graciously taking the time to write it. I think were on the same page with using flux or MMF, I do prefer to use flux but I think I know what you mean by net MMF over a path.
    Ok, this is what I'm not getting straight. So there are an amount of turns on the primary of the ideal TX, and the current through those turns goes up with there is a reduction on the Back EMF on those turns. So how are the primary amp-turns risng to match the secondary amp turns? (because amp-turns are like flux to my mind) Because the base flux from Φ = L1*IM / N1
    has dropped because IM has dropped. So is it like there are two sources of flux on the primary, one from Xm and one from the ideal TX:
    Φfrom Primary = L1*IM / N1 + N1*Is?

    Well faraday's law makes the back EMF doesn't it?
    Or could you say that these are both true and that using the back EMF, the magnetising current is equal to:
    IO = [N1*d(Φfrom primary - Φfrom secondary)/dt ] / (Rc || XM) ?
     
    Last edited: Jun 18, 2015
  7. Jun 18, 2015 #6

    jim hardy

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    Thinking, trying to replicate your thought process...

    back probably after lunch.

    I think i use a mental shortcut that might be the cause of our confusion.

    Flux indeed is driven by MMF which is amp-turns.
    That's true for DC, AC, and transients.

    Remember how i harp on sinewave as a mathematical oddity in that it and its derivative have the same shape? Euler and all that math....?

    Sinewave flux sin(wt) induces voltage wcos(wt),
    and at 60 hz w is constant 377 (well 376.9911184++).
    So....
    Neglecting the 90 degree phase shift from sin to cos , voltage and flux are related by the constant 377.

    You use current as starting point for flux
    i use it too for non sinewaves
    but for line frequency sinewaves i use voltage --- volts per turn = 377*flux in webers (@60 hz)
    that's a lazy man's shortcut but it sure cleans up a messy looking formula.

    here's the model with secondary impedance moved over into secondary...

    Transformer.jpg
    so what is flux in primary of ideal transformer?
    It's whatever is voltage Ep divided by 377.
    And voltage Ep is IM * XM

    That may be the root of our miscommunication i dont know.
    Surely it'd resolve with some application of Ohm and Kirchoff to our model.

    anyhow see ya later gotta get in my heart-walk while it's not raining.

    old jim
     
    Last edited: Jun 18, 2015
  8. Jun 18, 2015 #7
    Is it just convenience you can say flux is the sin and voltage is the cos?

    I'm not sure that's the root of our miscommnunication, I'm not even sure it is a 'miscommunication' as such.
    Yeah so:
    that Ep is also Farayday's Back EMF isn't it? So that would mean:
    IM * XM = N1*d(Φfrom primary - Φfrom secondary)/dt ?(was what I was saying lat post)

    You said that the primary and secondary produced the same number of amp turns in the ideal TX, (so there's no net flux or mmf in the ideal TX because they're opposing?)
    but you also said:
    But we agreed that when there is a voltage drop on the primary Rp and Xp The flux from the inductance of XM drops (assiming that is the only mechanism by which flux can be produced in the core from the primary).
    So nothing is "rising" to make the primary flux equal the secondary flux in the ideal TX, it just is.
    Surely the flux in the real world core peaks at OC and once you put any sort of load on the secondary (no matter how small) the flux in the real world core will just drop, (not remain stable due to the inreased IP like a tug of war), just drop. Only the ideal TX maintains primary and secondary matching amp turns. The extra current the primary is drawing when the secondary has a load on it makes no difference to the flux in the Ideal TX. And remembering the flux from XM has dropped a bit, the more load on the secondary, the more it drops, the less flux end of story.
    (Since flux is a function of IM not IP, it doesn't matter that IP increases, there won't be any additional flux from the primary to fight the secondary induced flux from a load, that is counter to the original OC flux amount)
    Is this how the model responds?
    [This doesn't sit well with me by the way, because I'd have thought there would be a bit of a tug of war to maintain OC flux amount under load, like IM would increase, but as far as I think we've discussed, it doesn't]

    Also, In reality, the opposing flux from the secondary would reduce the net flux in the core and so this would change the value of XM as the permeability of the core would increase. This would change the inductance and complicate the relationship that I've been talking about, because the flux is a net flux from both primary and secondary, hence the mutual coupling and L1 and L2 I was talking about before.
    So the formula for Average Energy in core = 1/2 *L*Iprimary2 = 1/2 * (XM*jω) * (IM)2 would be no use when there is a secondary load because the flux isn't the same, this formula would only be good for OC and one source of flux.
    For a secondary that has a load I was thinking the formula for average energy in the core =
    1/2 * N1Φ12*I1+M*I1*I2+ 1/2 *N2Φ21*I2


    Take your time in replying, it's 3:30am here and I might go to bed soon. Thanks
     
    Last edited: Jun 18, 2015
  9. Jun 18, 2015 #8

    The Electrician

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    Shouldn't the a^2 terms in the secondary impedances be removed?
     

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  10. Jun 18, 2015 #9

    jim hardy

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    indeed it should be - victim of click and drag.
     
  11. Jun 18, 2015 #10

    jim hardy

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    yes. Start simple. Start with flux = cos if you prefer.
    You know volts and flux have a derivative relationship e = n dΦ/dt ;;;
    and from high school precalc that d/dt of (sinwt) = wcoswt



    okay so far

    There is a net FLUX in the ideal transformer, else there'd be no voltage.
    There's no net MMF though because the ideal transformer's core is infinitely permeable.
    So fluxes add to Ep/#turns/377 (for 60 hz), and you had a different formula using IM XM and L.
    MMF's add to zero, or the to infinitesimal MMF required to push that flux through reluctance of zero .

    Very Good, Tim !
    That's an ideal transformer !
    Amp turns do NOT differ by magnetizing amp turns like a real transformer !
    And you made me aware i overuse that phrase "rise to balance".
    Awesome my friend !
    Primary and secondary fluxes add to whatever is necessary to solve your equation
    Ep = IM * XM = N1*d(Φfrom primary - Φfrom secondary)/dt

    NET mmf is zero,
    individual MMF's are individual NI 's.
    Net flux is (Ep/#turns)/377

    Maybe that's where we miscommunicate .
    In my Ideal transformer net flux is non zero, but net MMF is zero.
    How's yours doing ?




    I feel we should be playing Verdi's Triumphal March here !.


    Let's be careful here ..
    No difference to total flux, sure
    but since you sum fluxes,
    ,,,,,,
    ,,
    total flux = ( Φfrom primary - Φfrom secondary ) and both of those fluxes got larger , but they still add to (Ep/#turns)/377 in the ideal transformer

    that's so. I wasn't sure how much drop you envisioned that's why i commented about the drop being a modest one.

    Im not comfortable with that wording.
    That's why i add MMF's .
    If you add fluxes, they add to whatever induces Ep .
    If you believe Φ = MMF/Reluctance,
    you can add MMF's and figure resulting flux
    or you can figure fluxes and add them by superposition
    either one involves division by zero reluctance.
    So one must resort to another means to calculate flux.
    I like (Vp/#turns)/377, you can use IM and XM equally well i think.

    However flux is determined, and however it is calculated,
    it must agree with E = N dΦ/dt .
    Faraday and Ampere will come to an agreement.


    back to model (fixed those peaky a's that The Electrician caught)

    Transformer.jpg

    I repeat from a few paragraphs above

    I think you are addressing the non-linearity of iron's permeability.
    I've not included that at all yet because it's fine tuning of the model and should be applied after it's working smooth in one's head.

    Don't let complications into the room before we're prepped for them.
    That is a method our subconscious uses to torpedo our efforts in the hope we'll give up.

    Don't give up yet.

    Well, if Iprimary in that formula is IP in the model it won't work . You already observed IP and IM aren't the same.
    But does it look okay per model if you use IM ?
    Or if you use sum of ideal transformer's primary and secondary currents adjusted for turns ratio ( which will be magnetizing current IM edit neglecting Ic...) ?

    I'll leave that one to you. I use mutual inductance sooo very seldom that my subconscious always finds a reason to give up .

    probably a typo or three in above - i musta fixed fifty...

    old jim
     
    Last edited: Jun 18, 2015
  12. Jun 19, 2015 #11
    Although this is a difficult lengthy dicussion to have via text, I'm also having this long-running argument with this psudo-philosopher over the internet (about the Onatological Argument) and I have to say this is not only more satisfing as a topic, but you're much more fun to talk to.

    Under any other circumstance if you were to day 'the net MMF is zero' that would mean the net flux was zero, but here because we have this infintite permeability we can have a flux with an 'infinitely small MMF'?

    Is that Φfrom primary the same flux we had when the TX was OC? I'm guessing not, because you said:
    So if Φfrom primary is only a function of IM, and IM drops as the secondary current increases, then how does Φfrom primary increase too?
    If you said that Φfrom primary = Φfrom XM + N1*(IP - IO)
    Then that would make sense to me.
    But if that was the case then:
    Real core flux = Φfrom primary - Φfrom secondary
    = ΦfromXM + N1*(IP- IO) - N2*IS
    Which would mean that the net flux in a Real transformer core wouldn't change, because I assume
    N1*(IP - IO) = N2*IS
    because they're from the Ideal TX part of the model, and you said they had equal MMFs. And I think we agreed that as soon as a load is on the secondary the net flux in a real transformer core starts to drop (there's no tug of war).

    Ok, I think you're right, I'll just take baby steps at the moment, there are some questions I'm close, but not quite ready to ask you.
     
    Last edited: Jun 19, 2015
  13. Jun 19, 2015 #12

    jim hardy

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    Big ten-four good buddy.

    Okay i see the trouble.
    We use the label Φ to mean two different things and we're not shifting gears smoothly .
    Φ we are using for both a physical entity , flux , and for the magnitude of that entity which varies with transformer loading.
    We switch between OC and loaded, and don't update Φ.
    Lavoisier - be rigorous in assigning names. Take small steps not leaps.

    transformer-jpg.84964.jpg

    Open circuit: What is current through ideal transformer's primary? I see zero.
    What flux links that primary ? I see whatever is necessary to balance Ep ,
    and that will have value XM*IM/(ω *N) , N = # turns (remember we assigned same turns to XM and primary)
    So all of the flux in the ideal transformer results from the zero MMF produced by the zero primary current,
    and that flux has magnitude as determined by IM through XM
    Inductor XM has finite inductance so requires magnetizing current
    ideal transformer primary has infinite inductance so doesn't require magnetizing current.

    Break - sanity check - am i consistent ?


    Now apply load.
    Does current flow through ideal transformer's primary ? I say Yes.
    Does current flow through ideal transformer's secondary ? I say Yes.
    Do those two currents apply MMF's to core of ideal transformer ? I say Yes.

    Do those two currents create individual fluxes in the ideal transformer that add to some net flux ? In your model, yes.
    Do those two currents create individual MMFs in the ideal transformer that add to some net MMF? In my model, Yes.
    I prefer to sum MMF's then calculate flux, but we will converge i hope.
    I cannot add the mmf's and divide by reluctance because i'll get zero divided by zero.
    You cannot add fluxes because you'll have to calculate each as MMF/reluctance and you'll get infinity minus infinity.

    So what we gonna do now, bro ?


    You've added fluxes to MMF's in both of those equations. (Gosh i hope i didnt do the same thing late at night)
    To make your NI terms into fluxes you must divide them by reluctance which is zero.

    But it demonstrates that net flux in the ideal core must be same as in inductor XM.
    else they'd have different counter EMF's/

    Well sure net flux drops a little in both XM and in the ideal core, , because Ep dropped .

    Flux must induce counter EMF equal to Ep.
    So it's nonzero,
    and just how close to constant it is is determined by the real transformer's imperfections - Zp, Xm and Rc. .

    Try this on for size:
    Nonzero flux in the ideal core requires no primary current.
    So for you, primary and secondary flux totals to zero and there's a residual of whatever is set by XM at the time.
    Observe voltage divider action with RP and XP adjusts voltage at XM and Ep , hence flux at any load.
    For me, primary and secondary MMF's total to zero and the quotient of zero MMF over zero Reluctance equals whatever is set by XM.

    Do we converge at last ?
     
    Last edited: Jun 19, 2015
  14. Jun 19, 2015 #13

    jim hardy

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    I suppose if you preferred
    you could say

    "primary and secondary flux totals to whatever residual is set by XM at the time."
    That'd be closer to what a real transformer does. In the ideal transformer that takes zero primary current.
    Title of the thread was ideal transformer. We've beat it if not to death, senseless.

    Does the model work in your head now? If so the formulas will be intuitive and that beats cramming for exams..
     
  15. Jun 19, 2015 #14
    My bad, I'll cop that one, my fault, I forgot I put that it the title, sorry!
    I just read through your last two replies once, and I'm going to read through them again before I can really digest the content, but quick question before I do (to make a proper reply) which is:
    when there is a SC on the secondary, what will the net flux in the real transformer core be?
     
  16. Jun 19, 2015 #15

    jim hardy

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    Whatever is required to induce Es = IS * (jXS + RS +R(of the short circuit))

    transformer-jpg.84964.jpg

    It'll be quite low. That's how a current transformer works.
     
  17. Jun 19, 2015 #16

    jim hardy

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    Assume your short circuit is some small number Zsc of (milli)ohms.


    Flux will be whatever is required to induce ES = IS * ( Zsc + RS +jXS )


    transformer-jpg.84964.jpg

    It'll be quite low. That's how a current transformer works.
     
  18. Jun 19, 2015 #17
    M''mmm, yeah, ok, hmmm. So if you had a theoretical source Vp value and B-H curve would you still need to measure IS or ES to find that flux?

    I don't think you addressed the point (not that Ive digested your response yet) that: The flux in the real transformer, that's all coming from what's left of IM after (Ip - Io) have taken their share isn't it?
    There is no additional flux cominf from the increase of Ip - Io, is there?
     
  19. Jun 19, 2015 #18

    jim hardy

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    I dunno how the double post happened. It disappeared so i re-wrote it... and it reappeared.

    Anyhow you knew that from Ohm's law.

    transformer-jpg.84964.jpg

    Real transformer? Not the model anymore? Draw what circuit you want to address and label the currents.

    Your wording sounds as if you mean (Ip -Io) took a bite from IM instead of Ip and i dont want to assume different.

    Net MMF is a sum of all the currents , phasors are handy for that calculation.
    Flux willl be that MMF divided by reluctance of the core.
    Since you said it's a short circuited secondary the net flux will be low, in the linear region of the curve well below saturation.

    Do you know what are winding resistances and leakage reactances? That's why you'd make tests on a real transformer.
    If you already know them, there'd be no need to measure Is and Es . If you dont know them you'd have to experiment.

    .
     
  20. Jun 19, 2015 #19

    jim hardy

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    Net flux, no, the increase cancelled by the increase in Is.
     
  21. Jun 19, 2015 #20
    Well it does dosn't it? IM is higher when there's no (Ip - Io) ? Its all coming from Ip into Io until (Ip - Io) starts flowing when a load is put on the TX.

    Yeah, say you can theoretically define whatever you want to observe what will happen. I was thinking that you still didn't know Es or Is though (non linear permeability etc)

    Yeah, net flux, no, the net flux is dropping. That's the root issues I'm having is that a secondary flux starts opposing the flux caused from XM, and IM drops so the primary flux drops. Even though Ip has increased, there's no primary flux push-back from the increased Ip.
    I would have thought the increase in Ip would cause it to sort of fight back and maintain equilibrium, but since IM will just drop, there is no primary flux puch-back against the load current flux.
     
    Last edited: Jun 19, 2015
  22. Jun 19, 2015 #21
    On reading your last big post through once more:
    Hmm.
    Though this seems to contradict what I just commeted. Which was based on something I posed a few posts ago, I'll try revising how I said it as we've had a lot more time to understand where eachother are coming from:
    So although there is always a reduction in net flux when there is any load, no matter how small, caused by the voltage drop on (Rp + Xp), this is only very small. There is still a flux fight-back from the increased Ip current but it takes place in the ideal TX where the MMFs are equal.
    How's this?
    Although wouldn't the voltage drop on (Rp + Xp) have to be very very big for a significant drop in the flux of (in the real world core) to reduce IM enough to reduce the flux significantly?

    No need to mist up "fight-back fluxes from Ip" and counter fluxes from Is, they're all contained in the ideal TX and you can always say (loaded or not) that the net energy of the core is calculable from the inductance of XM?

    That aside, regardless is it fair to say that the magnetic core is working hardest at OC and the copper coils are working hardest at SC?
     
    Last edited: Jun 19, 2015
  23. Jun 19, 2015 #22

    jim hardy

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    transformer-jpg.84964.jpg

    What's the larger current ? Ip or Im ? From whom does (Ip-Io) take his bite?
    The voltage divider action between (Rp + Xp) and the rest of the circuit determines Ep.
    Ep over Xm determines Im.
    Which establishes what must be the magnitude of flux in the ideal transformer's core.
    To me you've confused cause and effect.

    You've flipflopped on me again. What's driving the system ?
    Let Zp = Rp +jSp .
    Im drops because Ip* Zp went up ,
    lowering Ep a little bit.
    So net flux went down a little bit.

    ?????? push back ? Where ? Are you acknowledging tug-of-war?

    Ip-Io flows in primary of ideal tx and Is flows in secondary.
    Since you sum fluxes, and fluxes are in proportion to currents,
    over in the ideal transformer,
    the increased primary flux is cancelled by the increased secondary flux.
    Their sum remains that number established by Im through Xm for whatever is voltage Ep.
    That's equilibrium.

    Since i sum MMF's, their MMF's add to zero which is the MMF necessary to push any amount of flux,
    namely that number established by Im through Xm,
    through zero reluctance.
    That's equilibrium.

    what is it ? fight back where ?

    It's down to elementary circuits now
    set Rp = 0, Xp = j1, Xm = j1, Rc = ∞, Vp = 1, Rs and Xs both = 0
    and solve for Vs & Is at OC
    and then with shorted secondary..


    Lavoisier :
    I think we have framed understanding anew.
     
  24. Jun 19, 2015 #23

    jim hardy

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    we crossed in the mail again....

    I'm feeling better already

    Still addressing our model ?
    If so
    Yes,
    but which current are you applying to XM (XM in the model)
    That energy has to be half LI^2 where I is IM and L is XM/ω.

    What do you think?

    That leads to the observation that the energy content of a transformer is more a function of the voltage across it than the current through it.

    Do i sense a breakthrough ?

    Search on hyperphysics magnetic field energy

    and you'll see the energy is a function of flux.
    Poynting vectors et al get involved when you want to look into transport of energy.
    Having these basics will help you apply that higher math.

    whew !

    old jim
     
  25. Jun 19, 2015 #24

    jim hardy

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    Yes.
     
  26. Jun 19, 2015 #25

    jim hardy

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    Do you have access to a test bench, variac, multimeters and small transformer ?

    If so

    measure primary current and secondary voltage as you vary primary voltage .

    Driving a transformer into saturation , feeling the core buzz and seeing the primary amps skyrocket gives one an intuitive feel for flux vs voltage. With sinewaves that is.

    Snake one turn around the core and observe .linearity of millivolts/turn vs miliamps at low excitation and flattening of the curve near saturation.

    Be aware driving it into hard saturation may distort your sinewave so much a DMM no longer reports accurately. I used a bridge rectifier and D'Arsonval ammeter....

    I had the good luck to have a 200 lb 3phase core in my garage. And a 5 amp Variac. I spent days tinkering with them.
    A smaller transformer will work fine.

    Volts per turn is a handy measure of flux at line frequency. Takes a big core to make 1v'turn, though. Can you calculate what cross section it'd take to do make that voltage at 1 Tesla peak flux density and 60 hz ??

    I learn more by simple calculations like that than i do from working out long derivations.
    I guss that's because i spent a lifetime fixing machinery . So my academic side is lacking, i do realize.

    Have fun , and fear not inductance !
     
    Last edited: Jun 19, 2015
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