Re: Parallel and Series LCR Circuit Qs

["Antonio Carlos M. de Queiroz" <acmq@compuland.com.br> (by way of Terry Fritz <twftesla@uswest.net>)]

Tesla list wrote:
> 
> Original poster: ghub005@xtra.co.nz
> 
> 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.

But it doesn't... The output voltage would be (V/Z)*Z=V, for any Z.
This is why it is an "equivalent" to the Thévenin form.
 
> 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 ...

Interesting relation :-). I didn't know about it.

> 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.

I obtain 9/10 Ohms.

Antonio Carlos M. de Queiroz