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Equivalent Mutual Inductor Circuits

by anol1258
Tags: analysis, circuit, coupled, inductors, mutual
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anol1258
#1
May3-13, 04:28 PM
P: 26
1. The problem statement, all variables and given/known data

Consider the circuit which consists of two ideal inductors [itex]L_{S}[/itex] ,[itex]L_{P}[/itex]. 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:[itex]L_{1}[/itex],[itex]L_{2}[/itex],[itex]L_{3}[/itex],[itex]L_{A}[/itex][itex]L_{B}[/itex],[itex]L_{C}[/itex]

circuit:

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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.
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SammyS
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May3-13, 05:19 PM
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Quote Quote by anol1258 View Post
1. The problem statement, all variables and given/known data

Consider the circuit which consists of two ideal inductors [itex]L_{S}[/itex] ,[itex]L_{P}[/itex]. 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:[itex]L_{1}[/itex],[itex]L_{2}[/itex],[itex]L_{3}[/itex],[itex]L_{A}[/itex][itex]L_{B}[/itex],[itex]L_{C}[/itex]

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.
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.

anol1258
#3
May3-13, 06:34 PM
P: 26
Ok I think that should work

berkeman
#4
May3-13, 07:45 PM
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Equivalent Mutual Inductor Circuits

Quote Quote by anol1258 View Post
1. The problem statement, all variables and given/known data

Consider the circuit which consists of two ideal inductors [itex]L_{S}[/itex] ,[itex]L_{P}[/itex]. 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:[itex]L_{1}[/itex],[itex]L_{2}[/itex],[itex]L_{3}[/itex],[itex]L_{A}[/itex][itex]L_{B}[/itex],[itex]L_{C}[/itex]

circuit:

Attachment 58453

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.
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...
anol1258
#5
May3-13, 07:55 PM
P: 26
Quote Quote by berkeman View Post
does it mention any loading of the secondary? The problem seems to be pretty straightforward if the secondary is not loaded...
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 :)
gneill
#6
May3-13, 10:21 PM
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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.
rude man
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May4-13, 11:42 AM
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Quote Quote by gneill View Post
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.
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?
gneill
#8
May4-13, 01:40 PM
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Quote Quote by rude man View Post
g, have you done this with arbitrary secondary load?
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.
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?
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.
rude man
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May4-13, 02:25 PM
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Quote Quote by gneill View Post
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.
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?
anol1258
#10
May4-13, 02:28 PM
P: 26
whoa whoa whoa guys I'm still stumped on how I can do KCL on this. I'm a mechanical engineer not EE.
rude man
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May4-13, 02:50 PM
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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.
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May4-13, 02:55 PM
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Quote Quote by rude man View Post
What's preventing you? Not that I especially recommend it, especially with no output load element. Then it's just ac voltage dividers ...
...
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.
berkeman
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May4-13, 04:07 PM
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Quote Quote by rude man View Post
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.
Good point. I hadn't considered the step-up case in my earlier post...
rude man
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May4-13, 05:35 PM
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Quote Quote by SammyS View Post
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.
That's to whom it was addressed, as the sequence might suggest ... but yes you're right & I apologize for the faux pas.
rude man
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May4-13, 05:42 PM
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Quote Quote by berkeman View Post
Good point. I hadn't considered the step-up case in my earlier post...
'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.
gneill
#16
May4-13, 10:34 PM
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Quote Quote by rude man View Post
'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.
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.
rude man
#17
May5-13, 11:16 AM
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Quote Quote by gneill View Post
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.
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 ...
rude man
#18
May5-13, 11:28 AM
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Quote Quote by gneill View Post
The mutual inductance is always less that or equal to both individual inductors.
Hmmm ... really!


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