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Re: Parallel and Series LCR Circuit Qs
Sorry I wasn't more clear. What I meant was that in the Norton
equivalent, a zero output-impedance (in parallel with the ideal
current source) intuitivly results in a zero-output voltage.
Speaking of equivalent circuits (Thevenin, Norton etc). I have been
doing some simple circuit analysis for fun (I'm making some simple
numerical models for a classical TC in Matlab). Here is a useful
result that I derived about a week ago - it is a simple transformation
for a two terminal, two element linear network that lets me
progressively collapse and absorb the elements in a network
(reducing the number of numerical parameters by 1 each time).
This transformation probably has a name - has someone on the list
has come across it before?
fixed width (courier) font etc:
o---WWW---+
r |
E
E Xs
E
|
o---------+
o---+------+
| |
W E
W R E Wp
W E
| |
o---+------+
To transform in either direction use
R = r(1 + Q^2)
Xp = Ws (1 + INV[Q^2])
where Q = Xs/r = R/Xp as usual
Note that Q will be negative for capacitive reactances, but since the
equations use Q^2 this isn't a problem.
Example:
Consider the following circuit:
3j 3j
o--www-+-www-+
| |
c R
-3j c R 10
c R
| |
o------+-----+
In this example, applying the transformation one element at a time
results in a purely resistive circuit (1 ohm) - which is not
immediately apparent at first sight.
Best regards,
Gavin Hubbard
P.S. I realise that my second paragraph in that previous post was
incorrect also - of course a zero-impedance output impedance in
parallel with a ideal current source won't give an infinite loop current!
I have to stop writing these emails so late at night :-o
On 22 Aug 2000, at 19:24, Tesla list wrote:
> Original poster: "Antonio Carlos M. de Queiroz"
> <acmq-at-compuland-dot-com.br>
>
> Tesla list wrote:
> >
> > Original poster: ghub005-at-xtra.co.nz
>
> > I almost always instinctively think of a power source as being an
> > ideal current source in parallel with an internal impedance (the
> > Norton equivalent?). With that model, a zero-impedance intuitively
> > results in a zero-output voltage.
>
> The Norton equivalent has a -current- source (V/Z) in parallel
> with the output impedance Z. The short-circuit current is just
> V/Z.
>
> Antonio Carlos M. de Queiroz