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# [femm] Re: Drawing circles & Another AC analysis question

- To: femm@egroups.com
- Subject: [femm] Re: Drawing circles & Another AC analysis question
- From: Dcm3c@aol.com
- Date: Thu, 11 Nov 1999 16:06:00 EST
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    In a message dated 11/11/99 9:54:31 AM Eastern Standard Time, Marctt2@aol.com
    writes:

    > Ah, I see; you're specifying a voltage drive at the coil terminals which
    in
    > turn sets the coil current. This is why my calculation works fine at DC.

    Yes, you could interpret it that way. That is, the J that you specify is
    equal to
    - sigma*grad(Phi), where Phi is voltage.

    > I believe other solvers do this a bit differently.

    Well, some other solvers have more flexibility in the eddy current
    formulation. Some nice additions to femm would be the ability to define
    regions in which eddy currents are either conserved or in which the total
    current is constant. These features would be useful to me for some work I've
    been doing, so I've been working on adding them. Hopefully these features
    will be available soon.

    > So, what's the simplest way to calculate AC resistance of the one-turn
    wire ?
    > I'm trying to generate plots of L vs. frequency and R vs. frequency
    > (basically to find the AC resistance and inductance of my coil).

    Well, the way you should do it is to take the A.J integral over the volume of
    the coil. This integral includes in J both the induced and eddy current.
    Then, divide by the square absolute value of the total current in the coil
    (from the integral "total current", which combines both the source and eddy
    currents) to get inductance.

    However, I tried this on a quick example problem that I cooked up just now,
    and there seems to be a bug in this integral. It doesn't include the
    contribution of the eddy currents in the correct way for harmonic
    axisymmetric problems--bummer. I'll fix this and post a modified version
    soon. Looking further, there is also a similar bug in computing Lorentz
    forces on induced currents in the harmonic axisymmetric case. I'll fix this
    one, too.

    In the mean time, you can just take the "stored energy" integral. This gives
    the right result, as long as you don't apply a "mixed" boundary condition to
    the outer boundary of your domain. Denote the result of the stored energy
    integral as E. To convert E to inductance, get the total current in the coil,
    i, by taking the "total current" integral. Now, inductance is:
    L=4*E/abs(i)^2
    The 4 gets in there (rather than a 2, like in the mstatic case) because we
    are working in terms of amplitude, rather than rms quantities. Likewise,
    resistance is:
    R=2*(Total Losses)/abs(i)^2

    Dave.
    --
    http://members.aol.com/dcm3c

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