Re: [femm] complex inductance???

[dcm3c@xxxxxxx]

In a message dated 10/5/01 3:19:17 AM Eastern Daylight Time, erdal903@xxxxxxxxx writes:


> Thanks Mr. Meeker for the comprehensive explanation. Now I have the following question: We have two methods to calculate inductance:
> 1) using the integral of A.J over the current carrying coils:
> Results: integral of   A.J = [(8.31e-4) -(j7.99e-6)] Henry Amp^2
> current=225 Amper-Turn
> gives an inductance of L=[(3.69 e-6)- (j3.55 e-8)] Henry
> 2) using the integral of magnetic energy. 
> integral of magnetic field energy over the whole area (instead of current carrying components)=2.081 e-4 joule
> current=225 Amper-Turn
> gives an inductance of L=[1.849 e-6] Henry
>  
> Now ignoring the imaginary part of the first solution (so the eddy currents and the hysteresis losses) which one should be taken as the inductance? 3.69 mH  or 1.84 mH? Thanks in advance.


The trick here is that you are evaluating inductance for a harmonic problem.  Now, if the current, i, is a DC current, the stored energy would be (1/2) L i^2.  However, for a harmonic problem, i is the amplitude of a sinusoidally varying current--the average stored energy over a cycle is (1/4) L i^2.  If you use the DC formula for inductance via energy for harmonic problems, you'll come out low by a factor of 2--this appears to be what's going on.  If I take your 1.84 uH value and multiply it by 2, I get 3.68 mH, which closely matches the real component that you got with the A.J integral (3.69 uH).

Dave.
--
David Meeker
http://members.aol.com/_ht_a/dcm3c