Transformers (coils) wired in parallel

wrote:


------------------------ You are absolutely correct.
I looked at this for a particular transformer ( 100KVA, 1000/200V) and the result is that the input current was about 3200A at 200V applied with individual primary and secondary currents of 800 A and 4000A respectively as compared to 1000A/5000A short circuit currents.
Parallel mutual inductances where L1 =L2 or M^2 <<L1*L2 will work. That is 1:1 turns ratio or weak coupling is OK.
--

Don Kelly snipped-for-privacy@shawcross.ca
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It will be good if you clarify what type of transformer you are referring to , only in Auto transformer there is electrical connection between primary & secondary windings, in normal transformer the coils are not electrically connected, in the power system study transformer is modeled as such but I guess this is not the subject which you are looking for. In the Steady state condition of your circuit when your coils are connected in parallel to the AC source with fixed frequency following should be valid Leq=(L1*L2*L3)/(L2*L3+L1*L3+L1*L2) in your measurements you should distinguish between reactance Xl & inductance L, in regards to the mutual inductance, this is depend if there is any considerable magnetic coupling between them for example if they are using the same core or if you are transferring any power signal from one coil to the other , as an example if you connect all three in the same active source, you could ignore M.
Regards Mansour

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You guys are trying to make this problem much more complex than I am asking! No, it's NOT an autotransformer. No, it's NOT a power transformer. And for now, I'm not assuming ANY ferromagnetic core, not iron, not powered iron, not ferrite, NUTTIN'! Consider it an RF transformer (untuned) as in a radio. Cores can be dealt with later once I figure out what is going on here.
*I* am wiring the three coils in parallel! Take three coils. Hook one wire from each coil together. Then hook the remaining three wires together. Measure inductance between those two connections. What is the value? Catch: The coils have some mutual inductance between each other!!! Even doing this for just two coils might be enlightening!
Simple parallel inductance is no problem. It's that MUTUAL inductance that is the fly in the ointment!
Thanks for the reply,
Benj
---------------------------------- mjalali wrote:

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I think you should think of the product of current and the number of windings, "Ampere-Turns" if you like that terminology. You know have the inductance of one coil with the others open circuit. If all the other windings are identical then they have the same number of windings. All you are doing is increasing the amount of parallel conductors, but not the number of windings - thus it will have the SAME inductance as a single winding. In the case of a real set of coils it will have a lower resistance and thus a greater current capacity.
That's if I have understood what you mean by: "*I* am wiring the three coils in parallel!"
Robert
Benj wrote:

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Yes, that is the idea and the general expected result in a case where M is larger and all coils are identical! However, we are struggling here with a case where the self-inductance (size) of each coil of different and the mutual coupling of each coil to the other two is also variably and not necessarily large. Of course as I pointed out in a recent post, once you go to identical coils that are very close coupled, one expects the inductance to be nearly the same as for a single winding! (I used the idea of using stranded wire) So what you say is right but still not the general solution I'm looking for!
Benj
------------------- snipped-for-privacy@reply.nonet wrote:

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OK I accept that it's no so simple: it's a matrix calcualtion, as follows:
Imagine you have three separate circuit segments, coupled by the transformer, the self and mutual inductances are: M11, M12, M13, M22, M23, M33 (M11=L1, etc.) the circuits have currents, with the derivatives (note the convenient Newtonian dot on the top ;-) ): i1, i2, i3 and voltage across the inductor: V1, V1, V3 so you get the matrix relation:
/V1\ /M11 M12 M13\ /i1\ |V2| = |M12 M22 M23| |i2| \V3/ \M13 M23 M33/ \i3/
or, more succinctly (just like the single circuit):
V = M . i
solve for the currents:
i = M^ . V
(where ^ denotes inverse)
Now we connect them all together, in parallel, so the voltages are identical and the the individual currents in the branches may be calculated. Of course they will all be in phase because the voltage is the same for each, so it is easy to sum them to get the overall current. Then, clearly, the overall inductance is the reciprocal of the sum of all the elements of the inverse of the inductance matrix.
Is that better?
Robert
snipped-for-privacy@reply.nonet wrote:

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snipped-for-privacy@reply.nonet wrote:

Which works out to: (L1.L2.L3 - L3.M12^2 - L2.M13^2 - L1.M23^2 + 2.M12.M13.M23)/ (L1.L2 + L1.L3 + L2.L3 -2(L1.M23 + L2.M13 + L3.M12) + 2(M12.M13 + M12.M23 + M13.M23) - M12^2 - M13^2 - M23^2)
... Simple really ;-)

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On 4/25/07 2:38 AM, in article f0n7kh$uo5$ snipped-for-privacy@frank-exchange-of-views.oucs.ox.ac.uk, " snipped-for-privacy@reply.nonet"

Once it is a matrix calculation, it is simple! Excel has matrix features built in as do many hand calculators. Linear algebra is a mainstay of computer science. IIRC, in the book by Smythe, Static and Dynamic Electricity covers such matters. Smythe himself thought that such circuit issues in a book that covered much more complicated issues was out of place.
Bill -- Fermez le Bush--about two years to go.
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You got it! Simple isn't it when you sit down and think about it. It also helps to start with the correct equations as you have.
--

Don Kelly snipped-for-privacy@shawcross.ca
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