Here's a question. Say I have a transformer and instead of wiring the
primary and secondary in series (like all the textbooks do) I want to

wire them in parallel. What total inductance would I measure for the
two windings both with adding and subtracting mutual coupling?
But the real problem I'm interested in would be three close spaced
wire loops in parallel. Each loop has it's own inductance L and a
given mutual inductance M to each of the other loops. What is the
value of the total inductance measured at the terminals. I suspect
it's one over the sum of all the 1/L and 1/M terms. [None of the Ls
and Ms have the same values] But I haven't been able to prove it. Any
hints?
Thanks.
Benj

Excuse me- what texts wire transformer primaries and secondaries in series?
If you are referring to "equivalent circuits" which often present primary
and secondary leakage inductances in series as seen from one side-ignoring
the relatively high mutual inductance- then go back to basic principles to
see the basis for this model and why it is a simple and useful approximation
to a better approximation, for many, in particular power, transformers.
The problem of additive/subtractive mutuals is well covered in any of the
texts that I have seen. I don't know what you have seen but it appears that
a crucial step has been missed.
If you have a multiwinding situation -total inductance as seen from one set
of terminals is not as you suspect. It does depend on what is connected to
each winding. If winding 2 and 3 are open circuited then the inductance of
winding 1 is that which would occur if windings 2,3 weren't even there.
If, as you are considering, a 3 or more winding transformer, you are getting
into a matrix model -are you ready for this at the present time? It is a bit
messier and your intuitive guess (based on parallel resistances?) is not
correct.

--

Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer

OK. I've just looked in a few old books I have here. What I'm talking
about is they start with the basic equations: You know V = L di/dt + M
di/dt etc. and then show that if you wire those two mutually coupled
coils in series you can get an inductance that is the sum of each
coil's L and either PLUS twice the mutual inductance or MINUS twice
the mutual if the mutual coupling has the dots the other way. That's
an easy problem and as you say, well-covered.

Let me clarify again. The problem I'm proposing is to take a multi-
winding transformer (or more exactly mutually coupled coils as above
each with inductance and a coupling to the others and this time hook
them all in PARALLEL! No windings are open! All windings are hooked
in parallel with all other windings. The whole thing ends up as a TWO
terminal device with a certain inductance (Just as in the series case
above). The question is what is the formula for that inductance? I've
made a guess but I haven't been able to prove it. I keep thinking that
like the series case the parallel case can't be that hard to do, but I
still can't seem to find the way to do it! Is that more clear?
Here's a practical example: Say I've got this "IF can" type
transformer. It has 3 windings. If I wire those three windings in
parallel and hook that circuit to an inductance meter, what inductance
would I read? It's obvious the self-inductance of each coil add
together like typical parallel inductors 1/( 1/L +1/L +1/L) [I'm
ignoring subscripts here for ASCII reasons], but the serious question
is what role does the mutual inductances between the coils play in the
final value? IF M = 0 it's obvious the above formula gives the
answer. But what happens when the various Ms are not zero and you
actually have coupling between the coils? OK?
If anybody has a reference to a book that has this worked out that
would be great!
So far, I haven't found one.
Thanks in advance!
Benj

Why do you make this so convoluted?
Just google "inductors in parallel"
Here's the first hit:
http://www.play-hookey.com/dc_theory/parallel_inductors.html
Ed

------snip-------
-----------------
Unfortunately your reference is incorrect, having made a fundamental error.
It has assumed that the inductances can be represented by a parallel
combination of L1+M and L2+M.
This would be like writing KVL as
v1=v=(L1+M)di1/dt
v2=v =(L2+M)di2/dt
where in fact:
v1=v=L1di1/dt +Mdi2/dt
v2=v=Mdi1/dt +L2di2/dt
is correct.
This makes quite a difference.

--

Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer

Well, DUH!
Man I HATE it when I've neglected to do the obvious thing first!!!!!
OK, NOW, I've done it!
But sorry, no cigar! Your reference has a nifty formula and I presume
since it's on the Internet is has to be true....Only it's not!
They assume that the mutual Inductance simply adds on to the self-
inductance as it does in a series case. In series inductors the
current through one inductor has to be the same as the current through
any other inductor. But in parallel that is no longer true. This means
if you look at the equations, that you cannot group L + M as a
factor.
I looked through all the crap in the search and while there is a lot
of good stuff there that I should have examined first <sigh>,
nevertheless a derivation of parallel inductors with mutual coupling
from basic equations: I = I1+I2 and V=L1 dI1/dt + M dI2/dt = L2 dI2/dt
+MdI1/dt. that ends up with L total = ? (expression with Ls and M) is
not among the search that I could find.
Thanks for getting my mind right, though!
Benj

---------------------
Ok, the way I would approach it is to write the equations for each coil
v1=L11pi1 +L12pi2 +L13pi3....
v2= L21pi1 +L22pi2 +L23pi3....
as a family of simultaneous equations where L11, L22 etc are the self
inductances and L12, L21 etc are the mutuals (The Lij values include the
polarity sign ) and pi =did/t
Since the coils are in parallel, v1 =v2 =v3 ...
and you can solve for pi1, pi2, pi3 etc Easy to see the form in matrix
notation.
The sum of the currents is the total current so you know know pi total and v
so assuming a single Lequivalent. v/(pi total) =Lequivalent
for two inductors this would give Leq ={L11 +L22 -2M}/{L11*L22 -M^2} where
M=L12=L21
For three it will be a quite a bit messier. Usually in cases of multiple
windings it is easiest to use the basic equations as above -in matrix form-
as that would be quite general. All you are adding is the constraint that
each coil has the same voltage. It usually is not worth the bother of
working out the parallel equivalent for more than 2 or 3 coils unless you
have numerical data and a program to crunch the matrix.

--

Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer

Uh, oh! I see that none of my recent posts are appearing here. Gosh,
ain't Google Groups great! Guess I'll have to try later and try to re-
write what I said. <sigh> I HATE this work-eating feature of computers
and half-baked software!
Just a moment.
Benj
(testing)

OK. My investigation to this point shows that Google is actually
posting my efforts but is not showing it as posted. In other words you
can't read anything I've said IN Google. However, if you have a
"real" newsreader, the recent stuff is there.
The bad news is that I haven't got my "free" news service working
properly yet and therefore cannot post there with a "spoof" email
address. The bottom line is I'm sort of cut off right now.
But I will make a comment or two. Ahem. There is a copy of Terman
right on my shelf here and it was one of the FIRST places I looked.
Blanket statements about how "easy" it is to derive all this or that
Terman "talks about" mutual coupling are total BS. If it's so damn
easy then how about YOU just whip it out for us all, OK? We will be
forever grateful! As you go on in life, you will one day discover
that the most difficult questions to answer are the ones that seem the
simplest to ask! I would hardly call the question of parallel coil
connections "irrelevant"! I guess all electrical questions are
irrelevant. You will recall that the FIRST thing I did was ask if
anyone had a reference to a book that had this problem worked out.
(That would be "worked out" not "talks about mutual inductance"!) Zero
references from Salmon Egg or anyone else!
I have indeed derived the equations shown in Wikipedia (I believe the
signs are wrong) but as I've indicated it's not the MATH that is the
problem it is the reality of it! The problem has to do with an
equivalent circuit for the situation. Any math can be defined valid
but if the answers don't match reality it isn't of much electrical
value. And the Wikipedia result does appear to agree at least in part
with measured data. (The Wikipedia circuit assumes that the changing
current in one leg results in a series voltage induced in the other
leg)
By the way, the other comment someone made about a power transformer
with two different windings is VERY interesting! Yes, that would be
similar to what we are talking about although I'm not exactly sure how
two different voltage windings would fight each other. I mean if you
have a 10 volt winding and you hook it to a 15 volt winding that
essentially makes the 10 volt winding have a 5 volt input, right?
Which would be reflected back to the primary which then alters the
over all output! I'm not sure it actually would smoke the thing. But
it might. It's not wise to bet against smoke! If we hooked tow
outputs in parallel OUT of phase, it's clear that this would cause the
primary to appear with no inductance or as a piece of wire. This is a
result apparent in the parallel inductor thing. And yeah, that might
result in some heavy currents.
Since 'm sort between a rock and a hard place right now, I'll have to
back off until my posts are no longer banned on Google or I fix my
terranews setup. But I will keep an eye out for comments.

-----------
You were told, by me and by "donotreply", how to do this.
Have you bothered to go back to the basic equations?
By the way, look up autotransformers. You seem to have some big holes in
your education.

--

Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer

Yes, this is what I've been trying to do only I'm not seeing the form
in matrix notation! In fact I wasn't trying matrix notation, but it's
a good idea especially when going to more coils.

This is the sort of thing I was looking for! Having the answer should
make my efforts to make the derivation a bit easier! I do believe,
however that your formula for Lequiv. is upside down, right?

Thanks. This is a great hint. Only I wish there were some place it
were all worked out for 3 or more coils (in matrix notation!), so I
wouldn't have to deal with the math (which always drives me nuts!) But
I do feel like I'm getting closer!
Benj

I want to bounce this idea off you guys. I can't figure out how to get
the above formula! I wonder what the equivalent circuit is? So
consider this.
Take a peer at the OSU lesson below.
http://www.physics.ohio-state.edu/~gan/teaching/summer04/Lec2.pdf
It's a very simple and straight-forward derivation of inductors in
parallel easily extended to N inductors in parallel. Now consider
this. I'm proposing an equivalent circuit for two inductors with
mutual coupling as two inductors in parallel AND where both are in
parallel with an ideal transformer (with no self-inductance in the
legs) of coupling M. If we look at the inductance of this device we
find that the same derivation holds even though the voltage in each
leg is due to the current in the OPPOSITE leg. But that doesn't matter
to the derivation! What I've done is separate the self-inductance part
out from the mutual inductance part in a parallel model! The bottom
line is I end up with the formula for total parallel inductance that I
had "guessed" at before!
In other words 1/L = 1/L1 +1/L2 + 1/M + 1/M
The idea is easily extended to many coils where it becomes a sum of 1/
L terms and a sum of all 1/M terms between all coils!
OK. Why is this wrong? Is it wrong? Why can't I make an equivalent
circuit like this separating out the mutual inductance from the self-
inductance? Inquiring minds want to know!
Thanks.
Benj

Oh man! I can't believe this! I mean people have been building and
measuring and fooling around with coils since the 19th freakin'
CENTURY and here in the 21 century it seems nobody has bothered to
figure out the case of parallel inductors with mutual coupling and put
it in a textbook somewhere!!!
So far we have three candidates. The Don Kelly formula below, the
Wikipedia formula which Don and I agreed was wrong, and my "guess" as
a sum of one over all Ls and Ms.
Don Kelly wrote:

So the question is which formula does God approve of? In other words
which one fits reality?
Lets look at Don's in the case of two identical closely coupled
coils. In that case L1=L2 and M~= L. We see Don's formula has two
cases of M. One where the denominator is zero and the other where it
equals 4L. However in both cases the numerator is equal to zero!!!
This means that we have one case with zero inductance and another
indeterminate case.
A zero case is always expected because when mutual inductance
subtracts from the self-inductance we have a non-inductive coil with
zero inductance. But the 0/0 is not able to make sense in this
case.
Lets next look at "my" formula 1/L = 1/L1+1/L2+ 1/M+1/M
Here if L1=L2 and M=L we find that L eq. = L/4
Finally lets look at the wikipedia formula:
1/L = 1/(L1+M) + 1/(L2+M)
In this case, 1/L = 2/2L =1/L or L eq = L In other words the
inductance of the parallel combination of two closely coupled coils
should pretty much equal the inductance of one of the single coils
before being hooked in parallel with a second one!
So what does God say, or as close as we can get to God which is
Grover's book on inductance! We look at stranded wire and lo: We
find that 10 meters of 2mm diameter wire has an inductance of 18.307
microHenries. And TWO parallel wires like that have an inductance of
16.454 microHenries! Not 1/4 not 1/2 but very nearly the SAME value!
Whoa! It seems the Wikipedia is actually the one giving the answer
that agrees with reality!
It seems that the correct equivalent circuit is where an ideal
mutually coupled transformer is inserted in series with the bottom of
the two inductors. This is as opposed to my circuit where I put the
transformer coils across the input voltage and Kelly's circuit as a T
network with an M inductance in the bottom leg.
Hmmmm.
Comments?
Benj

On 4/24/07 6:15 PM, in article
snipped-for-privacy@n35g2000prd.googlegroups.com, "Benj"

I can't believe this disbelief! I do not understand how someone can finish a
as junior sophomore year, let alone a junior year, in an electrical
engineering program without being able to solve such (relatively
meaningless) problems. I m giving away my age by referring to "Elementary
Electric-Circuit Theory" by Richard Frazier. If that is not good enough,
Frederick Terman has described magnetic coupling in his books on radio
engineering.
Believe it or not, a technical library can be an interesting place.
Bill
-- Fermez le Bush--about two years to go.

-------
Maybe because it is a relatively trivial problem of little interest or use.
In the case of parallel wires there is some interest but there is an easy
way to deal with it.
----------

----------
Huh? Something missing in your background
Sneak up on it. L1=M+e =L2 -plug into the expression and see what happens as
e goes to 0. Result is Leq =M. Taking limits is a standard approach in such
cases. If M is -ve then the equivalent L =0.
Note that using the T equivalent circuit also gives this without having to
sneak up on the problem.
What's the big deal?
Result is Leq =M =L or Leq=0 Using the equivalent circuit also gives this
without having to sneak up on the problem .
Your "formula" apparently ignores Kirchoff's Laws (without which circuit
theory is meaningless) and is incorrect because of this.

-------
Note that the incorrect Wikipedia approach is actually correct for two coils
when L1=L2 for any M It doesn't give the correct answer when L1< >L2 except
for M=0 and cannot be extended to more than 2 coils.
-----------

---------
No kidding, Have you looked at the calculations of inductance for finite or
infinite wires and the concept of GMR or approximations needed?
>

---
Only because you have L1=L2. By the way, reality is not L1=L2=M.

----------
However the T circuit does give the expected result -so again what is the
problem. In the case of your transformer - just what is being coupled in the
two windings? Your description is questionable. The mutually coupled
windings already provide a non-ideal transformer. You can use an ideal
transformer but note that the ideal transformer has L1=L2 =M all approaching
infinity so the transformer without any load will be an open circuit. Is
this what you want? You need a load across one side of it- I suggest M.
Hell, you can have a black box with a gremlin in it and a pair of hand crank
generators and ammeters if you want -AS LONG AS THE BASIC CIRCUIT EQUATIONS
ARE SATISFIED. I suggest that you look at the these equations, given in
damn near any text, for coupled windings and make sure that they are
satisfied and also that current in coil 1 is only i1 and current in coil 2
is only i2 as in real life- then go for an equivalent circuit 1 and 2
coupled by an ideal transformer, or a T or pi model- lots of ways to skin a
cat but make sure you have the right cat to start with.
Generally they all have some form of
v1=L11di1/dt +L12di2/dt
v2=L21di1/dt +L22di2/dt
where, (with exceptions in some rotating machines) L12 =L21 =M
Look, in the library for some texts such as Krause "Analysis of Electric
Machinery" or even Fitzgerald Higgenbotham, Grabel "Basic Electrical
Engineering" (a once very popular book, updated many times-used for non-EE
students. These are now old but newer circuits and energy conversion texts
deal with this stuff. It would also be good for you to look at the magnetic
situation involved -i.e. what leads to L11 L22 and L12.
The flux relationships leading to these KVL equations are also given in a
wide variety of texts.
If you believe, as the preponderance of authors do, that these equations are
valid, then both the T circuit and the expression for equivalent inductance
follow. If you disagree with these equations- then why do you disagree? I
would like to know. If you have a problem with the approach I gave, then
question it- I welcome questioning, but do your homework- it is a trivial
problem.

--

Don Kelly snipped-for-privacy@shawcross.ca
remove the X to answer

(snip)
You'll have, effectively, a lot of shorted turns and thus no
inductance. A power transformer has very tight coupling between
windings. If you actually did what you said (BTW, dont!), the "two
terminal" result would be close to a short circuit.
Cheers,.
Roger

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