Mutually coupled inductors in parallel

[Simon]

Yes it's another assignment question as always one that they couldn't be bothered to explain in the course material although this particular part of the question apparently I have to do the explaining.

For those unfamiliar with the subject or needing a refresher please see here:http://www.electronics-tutorials.ws/inductor/parallel-inductors.html if you scroll down to the parallel aiding inductors this is the situation I have. I've been asked to explain formula. I'm not really sure how to do this even with the information I have been given so far and unfortunately Internet like my material is full of serious explanation but not parallel, or at least not detailed parallel.

So my understanding is that the aiding inductors effectively have a phantom inductor in series with each one equivalent to the mutual inductance.

Now I don't know if it is a fluke but if I separate the two mutual inductance terms into a separate fraction I get the same result. And for parallel inductors this seems to work for aiding inductors because they minus divided by minus is a plus and therefore the equivalent inductance is that of the individual inductors plus in series the result of the two phantom inductors in parallel. This seems to make sense. But then why are the terms been subtracted from the main inductors are they not adding to the main inductors? If I compare it to the formula for opposing inductors it starts to make a bit more sense because now the mutual inductance on the bottom is being added which will have the effect of reducing the equivalent inductance as there is a bigger divisor in the equation.

But to me it is a total mystery as to how anybody came to any of these equations.

If I was to try work this out from scratch (in this case I have been given the end formula and told to demonstrate it) I would have treated each inductor as its value plus the mutual inductance:

(L1+M)*(L2+M)
-------------------
L1+L2+2M

Why would my assumption not be correct? My understanding is that coupled inductors are equivalent to 2 inductors with the mutual inductance in series with each inductor. Depending on the orientation of the inductors with respect to each other mutual inductance may be added or subtracted from the real inductance. If there are huge chunk missing from my study material or is there something bleeding obvious that I'm missing?

[Kremmen]

Well, your assumption is not correct because the unverse decided to work differently.

But check here for a more in depth tractation: http://www.electronicshub.org/inductors-in-parallel/

[T3sl4co1l]

Inductors that are completely isolated, add in series and reduce in parallel just as resistors and capacitors do (well, capacitors do it reciprocally because their impedance goes as 1/C, but, you know).

Inductors that are coupled, multiply in series (aiding or opposing), as the formula gives.  In parallel, if they are different numbers of turns, then you are creating a short circuit through the difference of turns; if equal turns (and same direction), nothing happens at all.

Parallel, well-coupled windings are easily understood: suppose you have a helix of wire on a core.  Suppose you split the wire lengthwise into two pieces, in place.  Are they not the same wire, whether measured individually or connected in parallel (restoring the original single wire)?

Tim

[orolo]

An inductor is defined by the equation:

\$ \epsilon = L \dot{I} \$

Take two coupled inductors L1 and L2, with mutual inductance (meaning a shared flux) M. If the inductors are directly coupled, you get the equations:

\$ \epsilon_1 = L_1\dot{I}_1 \ + \ M   \dot{I}_2 \$
\$ \epsilon_2 = M\dot{I}_1 \ + \ L_2  \dot{I}_2 \$

Since the flux is shared, the current in one of the inductors generates a voltage in the other inductor, and viceversa. Since the inductors are in parallel, their voltages are equal, so \$ \epsilon_1 = \epsilon_2 \$. Using that, you get:

\$ L_1\dot{I}_1 \ + \ M  \dot{I}_2 \quad = \quad  M \dot{I}_1 \ + \ L_2  \dot{I}_2 \$

hence:

\$ (L_1 - M) \dot{I}_1 = (L_2 - M) \dot{I}_2 \$

So, finally:

\$ \displaystyle \dot{I}_2 \quad = \quad \frac{L_1 - M}{L_2 - M} \dot{I}_1 \$

The equivalent inductance \$ L_T \$ verifies:

\$ \epsilon = L_T  (\dot{I}_1 + \dot{I}_2) \$

So, to obtain the equivalent inductance you apply:

\$ \displaystyle L_T = \frac{\epsilon}{\dot{I}_1 + \dot{I}_2} = \frac{ L_1\dot{I}_1 \ + \ M  \dot{I}_2 } {\dot{I}_1 + \frac{L_1 - M}{L_2 - M} \dot{I}_1} = \$

\$ \displaystyle \frac{ L_1 \dot{I}_1 + M \frac{L_1 - M}{L_2 - M} \dot{I}_1}{\dot{I}_1 + \frac{L_1 - M}{L_2 - M} \dot{I}_1} =

\frac{L_1  (L_2 - M) + M(L_1 - M)}{L_2 - M + L_1 - M} = \frac{L_1 L_2 - M^2}{L_1 + L_2 - 2M} \$

Just use the definitions and a little algebra. The coupled differential equation from the beginning is the key.

[The Electrician]

Once again, this very problem, no doubt also from the HNC course, has been dealt with extensively here: https://www.physicsforums.com/threads/parallel-mutual-inductances.792899/

[G0HZU]

Parallel, well-coupled windings are easily understood: suppose you have a helix of wire on a core.  Suppose you split the wire lengthwise into two pieces, in place.  Are they not the same wire, whether measured individually or connected in parallel (restoring the original single wire)?

Winding an inductor using Litz wire is a good example of this. Sometimes there can be lots of strands within the 'wire'. The obvious benefit is increased Q at lowish RF frequencies due to increased 'skin' area but the inductance is hardly affected by having all those wires in parallel.

[Simon]

Once again, this very problem, no doubt also from the HNC course, has been dealt with extensively here: https://www.physicsforums.com/threads/parallel-mutual-inductances.792899/

Thank you I will take a look although it does look like somebody bringing up their own incorrect reasoning which I will have to read through before fathoming out what they are on about. What I was actually interested in was more about understanding how this works rather than having people solve it for me. Whoever produced the assignments for this course seems to think that I have every book ever written at my fingertips or that I should be able to find information that I am otherwise paying them for for free on the Internet.

As per the discussion about graduates failing practical problems I too am starting to be faced with the dilemma do I just send in a plagiarised assignment just to get through this shit and get on with life or do I try to understand it and be an honest student. Oh I forgot employers want me to have a qualification regardless of whether or not I actually know my stuff. It just looks good to see.

[Vtile]

The small amount I yeas ago understood about this have started to fade, but IIRC one thing that "weren't there" were a practical way to give value for M. Also I can not remember were this steady state formulation (like most) or did it actually have any use in dynamic case. I think I need to refresh this subject to my head.

[Simon]

For the purposes of working out "I" is fine di/dt just makes stuff harder to read (particularly for one with ADHD and visual stress)

[rstofer]

So, Simon, how did we do on the last problem?  I sure enjoyed it and I really appreciate the introduction to Maxima.  Did we get a good grade?

Orolo's explanation of this mutual inductance problem is as good as it gets.  Nice clean math and well laid out.  Kudos!

[Simon]

Well the last problem is part of this same assignment. Yes this is why I get frustrated it does take me this long. The problem with, the last problem is that I never got the mesh and nodal analysis to agree but I resolved to just send them in once I have everything completed and let them pick the bones out of that. Given the appalling standards of education these days no doubt I will get a pass just for making an effort!

Yes the explanation from Orolo is pretty good although I'm still a little confused about where his last 2 lines come from and have been trying to write down my own version of the calculations to see that I have it clear in my head. As much as I want to pass this course just to shut everybody up I would also like to learn something from it as I'm not your average student.

[rstofer]

Well the last problem is part of this same assignment. Yes this is why I get frustrated it does take me this long. The problem with, the last problem is that I never got the mesh and nodal analysis to agree but I resolved to just send them in once I have everything completed and let them pick the bones out of that. Given the appalling standards of education these days no doubt I will get a pass just for making an effort!

Yes the explanation from Orolo is pretty good although I'm still a little confused about where his last 2 lines come from and have been trying to write down my own version of the calculations to see that I have it clear in my head. As much as I want to pass this course just to shut everybody up I would also like to learn something from it as I'm not your average student.

Next to the last line, Orolo is substituting in the value for e₁ in the numerator and for I'₂ in the denominator.  Then there is some really ugly algebra in the last line.  It takes a minute to see how the substitutions work out but they do.

I seem to get both mesh and nodal to agree but I'm only looking at Va and Vb.

wxMaxima code for Mesh Analysis

Code: [Select]

ratprint    : false$
fpprintprec : 4$

eq1  :   0 = -V1 + Z1*I1 + Z4*(I1-I2) ;
eq2  :   0 = Z4*(I2-I1) + Z2*(I2-I4) + Z5*(I2-I3) ;
eq3  :   0 = Z5*(I3-I2) + Z3*I3 + V2 ;
eq4  :   0 = Z2*(I4-I2) + V3 ;
eq5  :   VA = V1 - I1*Z1 ;
eq6  :   VB = VA - V3 ;
eq7  :   Z1 = 2 ;
eq8  :   Z2 = -5*%i ;
eq9  :   Z3 = 4 ;
eq10 :   Z4 = -5*%i ;
eq11 :   Z5 = 4*%i ;
eq12 :   V1 = 120 ;
eq13 :   V2 = 120*%i ;
eq14 :   V3 = 14.14*%i + 14.14 ;

res  : solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10,eq11,eq12,eq13,eq14])$

results : expand(float(res))$
lngth   : length(results[1])$
sorted  : sort(results[1])$

print("")$
for i:1 thru lngth do
    print(sorted[i])$

wxMaxima code for Nodal Analysis

Code: [Select]

ratprint    : false$
fpprintprec : 4$

eq1 :   0 = -(V1-VA)/Z1 +(VA/Z4) + (VB / Z5) - (V2 - VB)/Z3;
eq2 :   VA = VB + V3;
eq3 :   V1 = 120;
eq4 :   V2 = 120*%i;
eq5 :   V3 = 14.14 + 14.14 * %i;
eq6 :   Z1 = 2;
eq7 :   Z3 = 4;
eq8 :   Z4 = 0 -5 * %i;
eq9 :   Z5 = 0 + 4 * %i;

res : solve([eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9])$
results : expand(float(res))$
lngth   : length(results[1])$
sorted  : sort(results[1])$
print("")$
for i:1 thru lngth do
    print(sorted[i])$

For some strange reason, I can't do a copy and paste of the output.  Nevertheless, for Va I get 86.38+j45.76 and for Vb 72.24+j31.62
One of the things I spent time on was getting the variables printed in sorted order with limited precision.

If you run out of things to do, you might try these in wxMaxima.

[orolo]

I'm still a little confused about where his last 2 lines come from and have been trying to write down my own version of the calculations to see that I have it clear in my head.

Sorry, I was trying to find a sweet spot between being too brief and burdening the calculation with too many details. Since we want \$ L_T \$, the idea is starting from its definition, \$ \epsilon = L_T (\dot{I_1} + \dot{I_2}) \$, and solve for \$ L_T \$, so:

\$ L_T = \displaystyle \frac{\epsilon}{\dot{I_1} + \dot{I_2}} \$

Now, since we want \$ L_T \$ to depend only on L1, L2 and M, we use all the relations derived before to elliminate \$ \epsilon, \dot{I_1} \$ and \$ \dot{I_2} \$.

I have written down the proccess in some detail, including the starting relations (within squares), and then all the substitutions and simplfications.

A good check to see if you grasped the idea is to repeat the computation for inductors coupled in reverse.

[Simon]

Thank you. My big problem is maths. You see what I just tried to do doing it myself was to reduce the top and bottom terms in terms of I1 but then I end up with everything below plus 1. I wasn't really sure if I could just cancel out to independent variables at the top and bottom. So once again my apparent understanding of electronics is hampered by maths but I suppose I can't do the maths I can't understand electronics. Oh well either way I can still sell a collection of £20 worth of components for £150.

[Simon]

Oh I see you ended up with one as well. This may take a while.

[Simon]

Right the maths makes sense. But I have no creativity with maths and would not have thought of how to push things around in order to get that result.

[rstofer]

Thank you. My big problem is maths.

Math is always a stumbling block and, unfortunately, EE is all math.

For practice, rework the same problem from the original 4 equations.  But think about where those come from.  You should be able to derive those with little effort having already seen them.  Then keep reworking the problem until you have the Algebra mastered.  That's the only thing I can think of that will help.  Practice, practice and more practice.

In your other spare time, you know, that time you waste sleeping, consider looking at the videos on Khan Academy.  I have spent a lot of time there since my grandson started college.  40+ years of Margarita brain grenades hasn't done much for my recall.  That's why I enjoy your threads!  It gets me to thinking about other stuff.

[Simon]

It's the creativity bit I miss. I'd never have thought of multiplying by:

L2-M
-----
L2-M

[rstofer]

It's the creativity bit I miss. I'd never have thought of multiplying by:

L2-M
-----
L2-M

I'm not sure I would have done that either.  But I would have put the numerator (of the 3rd eq from the bottom in the handwritten version) all over a common denominator (mutliplying the 1 by L2-M and then done the same thing with the denominator (again, multiplying through by L2-M).  Then I would have noticed that the fraction in the numerator and the fraction in the denominator had common  denominators and dumped them (this is a lot harder to write than do).  Think of 1/4 divided by 2/4.  Clearly you can dump the /4 in both cases and wind up with 1/2

I would have gotten to the same place in two steps instead of one.  We're just expanding and clearing fractions.  Clearly the neater solution was is shown in red on the notes.  I might have thought about that after I did it the hard way or, maybe not.  I'm not into post mortems.  If I get the answer, I quit looking for alternatives.

[orolo]

It's like rstofer said: practice until the calculations flow naturally. Learning consists of building and reinforcing neural pathways. Each time you try an exercise or see one solved, the associated neurons reinforce their connections. Von Neumann said you don't understand math, you just get used to it. Going through al these exercises that exasperate you so much is like starting as a beginner running: the first times you run out of breath, collapse and curse (inwardly) to the strutting fools who run by you making signs of encouragement. Persist some months, and you'll have trouble remembering what was so hard about running 10km a day. Math has the bonus you don't get injured, you just become eccentric.

Master this course and one day you'll look back and wonder what the trouble was.