Inductance (mutual and self)

Thread Starter


Joined Jan 5, 2008
So i have a very basic circuit, as shown in the diagram attached.

My question is, is how do i work out the total inductance.

i know Leq=L1+L2+L3+2M1+2M2+2M3

BUT, as one of the dots is in the other direction, i'm getting confused on how to work out the total impedance.

Any help would be greatly appreciated.



Joined Apr 20, 2004
I could be mistaken, but I think the significance of the dots on inductors is a red herring. In a transformer, they show phasing for series and parallel connections of multiple windings, but "loose" inductors can only have the polarity imposed by the current through them. Ignore the dots and solve for three inductors in series.


Joined Apr 2, 2007
i havent touched his subject for a long time myself..i remember that for the dot encountered while current entering has opposite sign to that when encountered while current exiting.

The Electrician

Joined Oct 9, 2007
In the figure are some arcing lines with values like 2H in the middle of them. They are clearly intended to indicate the inductance that would be measured between the nodes spanned by the arcing lines.

The longest arc indicates that the inductance measured between the far left and far right nodes is 2 henries, the Leq for the whole thing.

The inductances for the individual inductors are also given. The only thing not given is the mutual inductances, so apparently that is the problem to determine.

You will need 3 equations to determine the 3 unknown mutual inductances. The first would be an expression for the total inductance from left to right:

L1 + L2 + L3 + 2*M12 - 2*M13 - 2*M23 = 2

The second will be for the inductance across L2 and L3 in series, given as 2H:

L2 + L3 - 2*M23 = 2

The minus signs for some of the mutual inductances are because of the reversed dot on L3.

But now we have a problem. The smallest arc only goes across L1, and tells us that the inductance measured across L1 is 2 henries. But we already knew that, and this gives us no new information. We don't have enough equations to solve for the 3 mutual inductances. We can solve for 1 of them directly with the second equation above.

If the smallest arc were extended to go across L1 and L2, then we could solve for all 3 mutual inductances.