# Two Coupled Inductors Connected in Parallel

Discussion in 'General Electronics Chat' started by TheoStark, Jul 23, 2014.

1. ### TheoStark Thread Starter New Member

Jul 23, 2014
2
0
Hello all,

I'm trying to make sense of two tutorials on another website, the first concerning mutual inductance and the second, two coupled inductors connected in parallel-aiding

and parallel-opposing

configurations. Individually the tutorials are fine but, together, they make no sense at all - at least not to me. I have written the author for clarification but he has not responded.

Firstly we have two mutually-coupled inductors, L1 and L2, whose coupling is 100%. In this case the mutual inductance is calculated as

M = sqrt(L1*L2) .

Next we have two coupled inductors connected in parallel-aiding configuration. The net inductance for parallel-aiding is calculated as

L = (L1*L2 - M^2) / (L1 + L2 - 2M)

and for parallel-opposing as

L = (L1*L2 - M^2) / (L1 + L2 + 2M) ,

but notice that if the two inductors in both configurations are perfectly coupled, the numerator reduces to zero in every case. Yet, in the tutorial text the author claims (without explaining how he arrived at his conclusion) that if the inductances are equal then the net inductance is L = L1 = L2 = M in the first case and, in the second, L = 0.

In the second tutorial's examples, perfect coupling is never assumed and the author never addresses the case where the coupling is 100%. In these examples the mutual inductance is simply a 'given' and always less than 100%.

I am hoping someone here can shed some light on this apparent conundrum.

Thanks!

Last edited: Jul 23, 2014
2. ### Maxfooo New Member

Aug 24, 2013
17
2
Hey TheoStark,

I can set you up in the right direction, but it has been a while since I have dealt with inductors.

The issue your facing has to do with the fact that M != sqrt(L1*L2), but rather M = k*sqrt(L1*L2), where k represents the coupling coefficient and will be a value equal to or less than 1.

I wish I new the conditions you were given so I could help more (are you given number of turns, just inductance, anything else), but you can read this:

http://web.mit.edu/viz/EM/visualizations/notes/modules/guide11.pdf

Go to page 4, there it will give you a relationship between epsilon(voltage in circuits terms) and the number of turns for your inductor as well as flux from one on the other. Ignore the latter term, but replace it with the relationship it provides between that term and M*dI/dT, which you should know dI/dT and possibly the voltage.

Some knowledge about k, it is easily found generally found through experiment and then put through the equations on page 4 of the document, but then again this is pretty much the kindergarten version of E/M, so it may not be the most accurate.

Hope this helps, don't let it confuse you.

3. ### MrAl Distinguished Member

Jun 17, 2014
3,613
756
Hello,

If you have two inductors with the same inductance and they are coupled perfectly, then what you really have here is a transformer with a ratio of 1:1.

If you connect the two windings in parallel all you have is a single winding with twice the copper wire area. The total inductance should take on the value of either winding.

If you connect them in anti parallel and apply a voltage, the current should go up very high because that would produce a short circuit, in theory. That would be equivalent to L=0.

If there was any leakage inductance it would be a different story because that would mean the coupling is not perfect.

4. ### TheoStark Thread Starter New Member

Jul 23, 2014
2
0
Hello Maxfooo,

Yes, in the general case M = k * sqrt(L1*L2) for 0 <= k <= 1.

It is when k == 1 that I could wring these equations' necks! The numerators in both the parallel-aiding and parallel-opposing equations reduce to zero under conditions of perfect coupling, ie, k = 1.

For the sake of discussion, let's assume we have two perfectly coupled inductors whose independent, uncoupled values we know. How we accomplish their perfect coupling is immaterial as is any particular choice of values for L1 and L2. Because the equations should work with any values we choose, let us keep things general and treat L1 and L2 as being their actual values. Toward that end, let us further assume that we measured them separately and uncoupled, and learned that their values are not equal, that is, L1 != L2.

Now couple them. Perfectly. k = 1 and so M = 1 * sqrt(L1*L2) = sqrt(L1*L2). So far so good.

Next, connect them in parallel-aiding configuration:

Now plug M into the equation for two coupled inductors connected in parallel-aiding configuration and see what happens:

L = (L1*L2 – M^2) / (L1+L2 – 2M)
L = (L1*L2 – (sqrt(L1*L2))^2) / (…)
L = (L1*L2 – L1*L2) / (…)
L = 0 / (…)
L = 0

Zero inductance!

But not only; because the numerators in both equations are identical, the same thing happens when the inductors are in parallel-opposing configuration. Either way, it makes no sense. In real life, were L1 == L2 == L, for example, we would measure L as the net inductance overall, yes? Evidently, then, these equations do not reflect reality under conditions of tight coupling. Oops.

Btw, looking at this "L1 == L2, k = 1" scenario from a somewhat different angle, instead of winding two separate, perfectly-coupled inductors connected as parallel-aiding, we could have instead constructed a single inductor wound with same number of turns using wire having twice the cross-sectional area and gotten the same results, yes? It would handle more current but (ideally) the inductance remains the same regardless. Yet, our equation tells us that our net inductance is zero. How can this be?

That
, my friend, is the crux of the problem. If these equations do not reflect the reality, then which equations do, what are they and how are they derived?

The tutorials I've seen thus far are happy to explain what mutual inductance is and how to compute it, but none of them correctly address the situation described here. If they address it at all, they assume in their examples that the coupling coefficient is much less than 1, side-stepping the issue altogether.