# Homework Help: Equivalent Mutual Inductor Circuits

1. May 3, 2013

### anol1258

1. The problem statement, all variables and given/known data

Consider the circuit which consists of two ideal inductors $L_{S}$ ,$L_{P}$. These inductors are magnetically coupled with mutual inductance, M. It is possible to replace these magnetically coupled coils by three coils which do not share magnetic flux linkages. The two topological possibilities include the “Y” and the “Δ”. Calculate all inductance values:$L_{1}$,$L_{2}$,$L_{3}$,$L_{A}$$L_{B}$,$L_{C}$

circuit:

2. Relevant equations

V=L*di/dt

3. The attempt at a solution

Once I get either a "Δ" or "Y" circuit I would be able to convert between them.

Last edited: May 3, 2013
2. May 3, 2013

### SammyS

Staff Emeritus
I don't see an image posted.

I got a error message when I tried to follow the link to that image.

Added in Edit:

OK. It's visible now.

Last edited by a moderator: May 6, 2017
3. May 3, 2013

### anol1258

Ok I think that should work

4. May 3, 2013

### Staff: Mentor

Interesting problem. Is there anything else in the problem description? Does it say anything about the turns ratio of Ls and Lp? That can be derived from the two inductances, I suppose. And does it mention any loading of the secondary? The problem seems to be pretty straightforward if the secondary is not loaded...

For the "T" circuit, consider the unloaded case where there is no secondary load resistance. If M=1, the secondary voltage equals the primary excitation voltage. So what does that imply about the bottom inductor value?

And when M=0.5, you get half of the primary voltage making it to the secondary, so what does that imply about the ratio of the upper inductors to the lower inductor?

I haven't gotten very far yet on the case where there are source and load impedances, and including the turns ratio part. But I think approaching it from the ideal cases is helpful to start to figure out the method to use...

5. May 3, 2013

### anol1258

It doesn't mention any loading of the secondary. He does give us the values of L1,L2, and M though. Yeah my professor said this was a pretty straightforward problem.

Thanks for your contribution. I will work on what you have provided :)

6. May 3, 2013

### Staff: Mentor

If you write the portion of the KCL loop equations which include the inductances in Laplace form for the original and "Y" (or "T") cases, you can equate the coefficients. The result is quite nifty.

7. May 4, 2013

### rude man

g, have you done this with arbitrary secondary load?

I have my doubts about this. How can you get a voltage gain > 1 from either of those circuits, or can one or more inductances be negative?

8. May 4, 2013

### Staff: Mentor

No, but I didn't think it would matter. I essentially assumed a current source driving each side. Whether that's a real current source or just what current obtains when a load is present shouldn't matter; if you replace a load that draws a given current with a current source of that same value, the result should be indistinguishable.
Negative inductance not required. The mutual inductance is always less that or equal to both individual inductors. Or, in terms of the coupling coefficient,
$$K = \frac{M}{\sqrt{L_1 L_2}}$$
and 0 ≤ K ≤ 1.

9. May 4, 2013

### rude man

There is no mutual inductance among L1, L2 and L3, nor LA, LB and LC.

Take the L1-L2-L3 network, ascribe any values you like to those inductors, and show me how the output can exceed the input?

10. May 4, 2013

### anol1258

whoa whoa whoa guys I'm still stumped on how I can do KCL on this. I'm a mechanical engineer not EE.

11. May 4, 2013

### rude man

What's preventing you? Not that I especially recommend it, especially with no output load element. Then it's just ac voltage dividers ...

Anyway, these "equivalent" circuits can't handle the case where k*sqrt(Ls/Lp) > 1, k = M/sqrt(Lp*Ls), 0< k< 1.
In other words, turns ratio > 1.

12. May 4, 2013

### SammyS

Staff Emeritus
To whom are you addressing this remark?

We have this nifty "Quote" feature to resolve this issue --- especially when more than two people are involved in the discussion.

After all, it is anol1258's thread.

13. May 4, 2013

### Staff: Mentor

Good point. I hadn't considered the step-up case in my earlier post...

14. May 4, 2013

### rude man

That's to whom it was addressed, as the sequence might suggest ... but yes you're right & I apologize for the faux pas.

Last edited: May 4, 2013
15. May 4, 2013

### rude man

's OK, neither did anyone else ...

These ckts are really, really pointless. They don't include winding resistance, not to mention step-up capability. They do include leakage inductance, ya gotta give 'em credit for that i suppose.

16. May 4, 2013

### Staff: Mentor

It turns out, if one allows negative inductance then step-up will work just fine too Winding resistance can be pushed into the external circuit as a lumped value.

A negative inductance is really a capacitance. -jωL looks just like 1/(jωC) for a suitable choice of C. So the circuit becomes second order and resonance effects obtain. For a sinusoidal stimulation, this can appear as amplification (step-up).

The relationship that I found using term matching does yield a negative value for one of the "T" inductors when the mutual inductance exceeds the value of either winding's inductance.

17. May 5, 2013

### rude man

Looking worse & worse. Resonance effects? Aren't supposed to be any resonance effects - not at any frequency.

If the equiv ckt is only good at one frequency then I wouldn't want to put a step function into it ... use it as a pulse transformer ...

18. May 5, 2013

### rude man

Hmmm ... really! :uhh:

19. May 5, 2013

### Staff: Mentor

Well, I assume that the idea is to use the transformed circuit for analysis purposes, not to build a substitute for a transformer (losing isolation amongst other things). The resulting equations do appear to match the original though.

20. May 5, 2013

### rude man

Well, if anyone can make sense of it I imagine it would be you. To me this is the equiv ckt from hell!

BTW sorry about my comment on M vs. L1 and L2. I believe you corrected that later & I forgot.

21. May 5, 2013

### Staff: Mentor

Sorry 'bout that. I was thinking about the single-loop 'proof' of that. Doesn't apply to multiloop coils. There M ≤ √(LpLs).

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