# Homework Help: Inductors in series

1. Sep 28, 2011

### geft

Referring to the attached image, find the equivalent inductance.

L1 = 2 + 2 - 2 = 2H
L2 = 1 + 2 - 2 = 1H
L3 = 3 - 2 - 2 = -1H
Leq = 2 + 1 - 1 = 2H

Just checking to see if I got the concept right.

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2. Sep 28, 2011

### Staff: Mentor

Are the dots meant to imply that the coils are all wound on the same core? If so, I get a different answer from you.

3. Sep 28, 2011

### Staff: Mentor

Hmm. I'd like to know how the coupling was arranged for these fancy coils, and where can I buy some! In general, for a pair of inductors L1 and L2 the mutual inductance M must satisfy

$M \leq L1$

$M \leq L2$

$M^2 \leq L1 L2$

So how were the 2H mutual inductances with the 1H coil managed?

4. Sep 28, 2011

### rude man

While the first two inequalities of gneill do not necessarily hold, the third one certainly does.

I would throw this question back at whoever gave it to you and call him something less than complimentary.

5. Sep 28, 2011

### Staff: Mentor

I'm looking at it more from the perspective of measuring each inductance separately, with the other coils open circuited (like you would measure Lm for each coil on a transformer), and then the question is asking for the overall inductance if you measure them end-to-end connected. In that case, there is a physical answer to the question.

6. Sep 28, 2011

### geft

The dots indicate the 'polarity' of the inductors (i.e. the way they are wound). Could you please elaborate?

Last edited: Sep 28, 2011
7. Sep 28, 2011

### rude man

I know the problem has an answer assuming the M's given are possible. My view is that the instructor (if so) was an idiot for giving students a scenario violating basic physical laws. I mean other than the usual (necessary) assumptions of zero weight of a cable, no friction, infinite conductivity, etc. etc.

I might even boycott this problem until feasible M's were given, just out of spite!

<< proposed solution deleted by berkeman >>

Last edited by a moderator: Sep 28, 2011
8. Sep 28, 2011

### Staff: Mentor

So far the problem is under-defined. The problem needs a better definition before it can be answered.

Are these coils wound on the same core? How were the inductances that are shown measured? If the coils are wound on the same core, what was done with the other coils as each coil's inductance was measured?

When you can define the problem better, we can perhaps give you some hints to help you figure out the question.

9. Sep 28, 2011

### geft

L1 + L2 + L3 + M12 + M23 + M13
= 2 + 1 + 3 + 2 - 2 - 2
= 4H

10. Sep 28, 2011

### geft

Here's the complete question.

11. Sep 28, 2011

### Staff: Mentor

How rude! LOL

The problem posted here is under-defined, but that may not be the instructor's fault if the OP didn't post the full problem information and definition. And even if the OP did, there may have been some context provided by the instructor in class that the OP did not provide to us.

In any case, in the real world we deal with transformers with multiple windings all the time, and the dot conventions and Lm values are things we use. I agree with you that the OP did not make it clear that these are Lm values for windings on a common core.

12. Sep 28, 2011

### geft

The problem only states that there are 3 coils in series. There is no mention of cores. Anyway, this is not a transformers topic (it's the next chapter), but a mutual inductance one.

13. Sep 29, 2011

### rude man

The problem is not really underdefined. OK, here's the solution, assuming the M values could be realized, which they can't:

L = L1 + L2 + L3 + 2M12 + 2M23 + 2M13. This could also e.g. be written
L = L1 + L2 + L3 + 2M21 + 2M32 + 2M31 since Mij = Mji.

M is positive if the currents thru the two coils augment each other, and negative if they oppose each other.

The dot convention is such that if current enters the dotted ends of both coils, they augment each other's fields. (Also true if the current enters both undotted ends). But if the current enters one dotted end and one undotted end, the fields oppose.

Consequently, M12 > 0 while M23 and M13 < 0.

It occurs to me that whoever came up with this problem didn't mean mutual inductance, which is M, but actually 2M, in which case the numbers given are OK. But 2M is not the mutual inductance between two coils. M12 is rigorously defined by the voltage induced across coil 1 by a changing current in coil 2: V1 = M12*di2/dt. di1/dt is assumed = 0. The reason the effective inductance is L1 + L2 +/- 2M instead of just L1 + L2 +/- M is that both coils carry current so the effect is doubled, each coil inducing voltage in the other, and M12 = M21.

Here is the best link I've found on the subject:
http://www.electronics-tutorials.ws/inductor/series-inductors.html

Last edited: Sep 29, 2011