From: John Woodgate
Subject: Re: Deriving H for magnetic cores
Date: Thu, 26 Dec 2002 22:29:20 +0000
Organization: JMWA Electronics Consultancy
Reply-To: John Woodgate
NNTP-Posting-Date: Thu, 26 Dec 2002 23:15:11 +0000 (UTC)
X-Newsreader: Turnpike (32) Version 4.01 <5Z8C9wtxbnpWyFnyfFzqmVF739>
I read in sci.electronics.design that Roy McCammon
wrote (in <3E0B66D0.email@example.com>) about 'Deriving H for magnetic
cores', on Thu, 26 Dec 2002:
>John Woodgate wrote:
>> That is quite unfair and you know it. H = I/2[pi]r is not 'qualitative',
>> nor is the argument that you cannot break the closed contours of
>> constant H by deforming the conductor.
>you wound me sir. It is true that I sometimes
>speak incorrectly, but never when I know it.
Well, I had no intention to wound; I simply inferred from your
preference fro a high-level analysis that the simplest end-result would
be familiar to you.
>Lets look at a similar situation in circuit
>theory. The cube made of identical resisters.
>Now, when we teach circuit theory, we teach
>how to use node or mesh analysis to generate
>A system of simultaneous linear equations which
>may be solved to node voltages of mesh currents.
Bit of post-Christmas spirit scrambling the typing there?(;-)
>Then we present the cube. The student may start
>off writing 6 equations, but we chuckle and
>laugh and show how to use symmetry in this case
>arrive at the answer with much less effort.
>But the way we teach magnetics is analogous to
>starting with the cube and symmetry and that
>dropping the subject without ever teaching the
>more general analysis. The student thinks
>he knows all there is too it, but the first
>time he encounters a non symmetric situation,
>he is lost.
Well, I'm sure you know what you mean, but it isn't clear to me. What
non-symmetrical situation' do you have in mind, and how does 'the
student' fail to cope?
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
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