Mutual Inductance

Discussion in 'Homework Help' started by Hitman6267, May 6, 2010.

Apr 6, 2010
82
0

I have no idea how to even start this. All I know about mutual inductances is how to calculate the V of each inductance (taking into consideration the mutual inductance)

V= L di/dt ± M di'/dt

But I don't see how that helps me.

2. The Electrician AAC Fanatic!

Oct 9, 2007
2,308
343
Have you studied phasors yet? Do you know how to represent the impedance of an inductor as jωL, and a capacitor as 1/jωC?

Apr 6, 2010
82
0
No, all what I've been taught about inductances is how to calculate their equivalent inductance if they're in series/parallel and the formula in my first post.

The solution of this problem is 11.33 and it's done using a formula that only has L1 L2 L3 but I don't know it and it's of the form of a fraction. (I've seen people solving it with the formula but I don't know from where they derived or what it is.)

To help you determine what the proper method is I feel telling you that I'm a civil major and that this is not an advanced electric circuits class might help you.

4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Since the inductors L1 & L2 have mutually aiding flux (according to the dot convention per your diagram) their individual effective inductances are

L1_mutual=L1+M and L2_mutual= L2+M

Inductors in parallel are treated the same as resistors in parallel.

So the effective circuit inductance Lab would be

Lab=L3+(L1_mutual||L2_mutual) where || indicates the parallel combination.

This gives the correct answer of 11.337H (to be exact)

5. The Electrician AAC Fanatic!

Oct 9, 2007
2,308
343
This is a good approximation, but is not perfectly accurate. If you use this method to calculate the effective inductance of two such coupled inductors in parallel, you get "formula 1" for the equivalent inductance of two coupled inductors in parallel:

f1(L1,L2,M)=(L1+M) || (L2+M) = $\frac{(L1+M)*(L2+M)}{(L1+M)+(L2+M)}=\frac{L1L2+M(L1+L2)+M^2}{L1+L2+2M}$

If you set up loop equations for the given circuit, but set the 6.6 H inductor to zero so that when looking into the circuit from the left end we will now just see the effective value of L1 and L2 in parallel (with a mutual inductance of M included), we will get "formula 2" for the equivalent inductance of two coupled inductors in parallel::

f2(L1,L2,M)=$\frac{L1L2-M^2}{L1+L2-2M}$

Now if we plug in the numbers from the problem we get:

f1(9,8,1) = 4.73684210526

f2(9,8,1) = 4.73333333333

If we add the 6.6H for L3, we get the equivalent inductance seen at terminals a and b. The two values are:

Leq1 = 11.3368421053

Leq2 = 11.3333333333

Both would seem to be in agreement with the given answer. How can we determine which is correct? Perform an experiment! It's what Faraday would have done.

I was sure that the difference in the two formulas would be apparent in the case of two inductors of different value, closely coupled. The inductors in the problem aren't closely coupled, so the two formulas give almost the same result. And, examination of the two formulas shows that if M=0, they both reduce to the formula L1 || L2, so we would expect that if M is small compared to SQRT(L1*L2), both formulas would give nearly the same result.

To make a couple of closely coupled inductors, I took a 3 inch OD toroid made of permalloy which I had lying around, and which already had 18 turns of wire wound on it (for who knows what reason!), and I added another winding of 9 turns interleaved with the existing winding. This should give a couple of closely coupled inductors whose inductances are in the ratio 4:1. I then carefully measured the inductance of the two inductors, with the following result:

L1 = 12.003 mH
L2 = 3.006 mH

I then connected the two inductors in series-aiding configuration and measured the inductance, and then made another measurement in series-opposing configuration. The results were:

Laiding = 26.75 mH
Lopposing = 2.992 mH

The mutual inductance is given by (Laiding-Lopposing)/4, which in this case was 5.9395 mH. This method of measurement is notorious for not being very accurate, but it's a start.

If we plug these numbers into the formulas, we get:

f1(12.003mH,3.006mH,5.9395mH) = 5.96938 mH
f2(12.003mH,3.006mH,5.9395mH) = .25666 mH

These results are sufficiently different that it should lead us to suspect that only one of them is correct. So the next step is to parallel the two inductors and actually measure the equivalent inductance.

Think what you would expect to happen if you wound a couple of windings of substantially different number of turns on a core and then connected them in parallel. If the core were excited by a third winding, the first two windings would output voltages in the ratio 2:1, and connecting them in parallel would be equivalent to shorting a winding on the core. The effective inductance of the two in parallel should be very small.

I measured the inductance of the two inductors connected in parallel, and the result was an inductance of 2.48 μH. This is very much less than the approximately 6 mH that formula 1 predicts, but it is also much less than the 257 μH formula 2 predicts. Why is it so much less? It's because the coupling is much better than the (Laiding-Lopposing)/4 formula gives us.

If the two inductors were perfectly coupled, M would be SQRT(L1*L2) = 6.00674770571. So, apparently if M is a little bit larger than 5.9395 mH, the effective paralleled inductance will be smaller. It turns out that if M is about 6.0061, the effective inductance of the paralleled inductors given by formula 2 is about 2.5 μH. This value of M when used with formula 2 gives the actual measured inductance of the two paralleled inductors, and this is a more accurate way to determine the actual M between the two inductors than the series-aiding, series-opposing formula.

This value of M still gives about 6 mH when used with formula 1, so we see that formula 1 is not the correct expression for the equivalent inductance of paralleled inductors which are coupled. If two inductors are only loosely coupled then formula 1 may be good enough.

Last edited: May 7, 2010
6. The Electrician AAC Fanatic!

Oct 9, 2007
2,308
343
Are you sure the formula didn't also involve M? Without M you can't get the right answer.

7. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Yes Electrician - rather sloppy of me I'm afraid.

Your correct equation for Leq of parallel mutually coupled coils should be applied.

8. The Electrician AAC Fanatic!

Oct 9, 2007
2,308
343
I'm wondering how Hitman6267, who is apparently a civil engineer (?), could be expected to solve this? He says:

In the loosely coupled case, simply paralleling the inductances is reasonably accurate, but then why would the problem give the mutual inductance if they didn't want the student to use it in the solution?

Apr 6, 2010
82
0
I think it was the formula t_n_k posted. It did involve M.

10. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Hi Hitman,

As the venerable (esteemed) Electrician points out, the two results are quite close - an artefact of the given question parameters [M in particular]. The answer Electrician gives as Leq2 is correct under all circumstances and probably better approximates the answer you were given as 11.33H.

I agree with Electrician that this problem is quite tough to expect a civil eng. student to solve.

Apr 6, 2010
82
0
I asked my professor this is how the solution goes:
Write V1= L1 di1/dt + M di2/dt
V2= L2 di2/dt + M di1/dt
V2=V1 because they are in parallel.
and use the fact that i1= I-i2 and i2= I-i1 to get di2/dt and di1/dt in terms of L and M and dI/dt.
In the end you will have an equation of V and the Leq would be the term before dI/dt

The answer is Leq= L3 + (L1*L2 - M^2)/ (L1+L2-2M)

12. mbleslie New Member

Feb 6, 2012
1
0
Thank you for your post #11 hitman. The derivation of 'formula 2' is actually quite difficult to find on the internet. It might be in a circuits text but I have misplaced mine at the moment

13. gatecrazy New Member

Jan 22, 2013
1
0
When is L(equivalent)= L + L' + M
and when is L(equivalent)= L + L' - M
?

Oct 9, 2007
2,308
343