Re: Parallel and Series LCR Circuit Qs

[ghub005-at-xtra.co.nz (by way of Terry Fritz <twftesla-at-uswest-dot-net>)]

On 23 Aug 2000, at 18:53, 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
> > 
> > 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.

Exactly my reasoning - I just let Z go to zero and intuitively guessed 
that V would also go to zero (as a zero-impedance has a zero 
voltage drop).

In circuit notation:

+----+         ^
|    |         |
CS   Z (=0)    | V (=0)
|    |         |
+----+         -

It seems that the equivalence between linear voltage and current 
sources only applies to circuits with a non-zero internal impedance.


An interesting (and unrelated) observation; although in the general 
case there is no discernable external difference between the 
Thevenin and Norton equivalent circuits, there is an interesting 
difference in the internal source conditions - because the current 
through Z and the power that Z dissipates are different in each 
circuit! e.g. If no load is attached to a linear source then I = 0 inside 
the Thevenin equivalent so no power is dissipated in Z. But in the 
Norton equivent a current flows through Z, so Z is dissipating some 
power (assuming it has a resistive component).

> > 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

OK

Here is my reasoning for the circuit.

First transform the series combination of the right-hand inductor and 
the 9 ohm resistor into a parallel combination:

Q = 3/9 = 1/3

R = r*(1+Q^2) = 9*(1+1/9) = 10 ohm

Xp = Xs*(1+1/Q^2) = 3*(1+9) =30 ohm

In euivalent circuit notation:

    3j     
o--www-+-----+------+
       |     |      |
       c     w      R
   -3j c     w 30j  R 10
       c     w      R
       |     |      |
o------+-----+------+

Since the resulting reactance is positive, I have shown it in the 
circuit as an inductance. 

Now combine the capacitor with this resulting inductance (since the 
reactances are signed, they can be treated like they are 
resistances):

X1 = (-3j)*(30j) / (30j - 3j) = 90/27j = -j*10/3 ohm

In circuit notation:

    3j     
o--www-+-----+
       |     |
       c     R
-3.33j c     R 10
       c     R
       |     |
o------+-----+

Since the resulting reactance is negative, I have shown it in the 
circuit as a capacitance.

Now transform the parallel combination of this negative reactance 
and the 10 ohm resistance back into a series combination:

Q = R/Xp = 10*(3/10) = 3

ri = R / (1 + Q^2) = 10 / (1 + 9) = 1 ohm

X2 = Xs / (1 + 1 / Q^2) =(-10/3)*(1 / (1 - 1 / 9)) = -3 ohm

In circuit notation:


    3j    -3j
o--www-+--ccc-+
              |
              R
              R 10
              R
              |
o------+------+

Since this is just three impedances in series, the impedance of the 
circuit is:

Z = 1 + 3j - 3j = 1 

So the circuit is resistive.

BTW I got the idea for this transform from a network theorem called 
the two terminal equivalence theorem:

"Any passive linear network can be represented, at any one 
frequency, by either a series combination of a single resistance with 
a single reactance, or a parallel combination of a single 
conductance with a single susceptance"

My TC modelling code just models the secondary as a large network 
(of linear components) and reduces it down to an equivalent circuit 
using the impedance operators. Then it evaluates the response of 
the equivalent circuit across an angular frequency range (in equal 
logarithmic steps). Once I have the numerical data, I just feed it into 
the Matlab graphing program to produce plots of the transfer 
function - including the Bode and Nyquist plots.

I will, of course, have to develop a more sophisticated model for the 
primary circuit. One that accounts for the non-linearity of the SG. 
Then I will need to add a function that models the coupling between 
the primary and secondary networks.

Of course all this can be done in PSpice (and probably more 
accurately too), but I am enjoying working through the various 
circuit elements. It is, of course, a very simple (naive?) way of 
analysing a TC. But it is giving me a feel for what happens when the 
component values are changed etc.

I will post my Matlab code, and TC models, when I have some more 
useful results.

Best regards,

Gavin Hubbard