Talk:Norton's theorem

This picture is so wrong
http://en.wikipedia.org/wiki/Image:Thevenin_to_Norton.PNG

And if you have some weird explanation about this picture referencing current flowing towards higher voltages, the thevenin page has the right picture.

http://en.wikipedia.org/wiki/Image:Thevenin_to_Norton2.PNG

They should both say the same thing. I'm of the opinion that some real engineer made that 2nd picture but forgot to fix the "norton equivalent" wiki page. I fixed it, don't put it back. —Preceding unsigned comment added by 171.66.32.204 (talk) 23:33, 17 October 2007 (UTC)

Bringing reactive devices into the mix
What about capacitors and inductors? the should also be allowed to be used. Thanks, --Abdull 12:51, 24 Apr 2005 (UTC)


 * You would have to perform the analysis at a specific frequency, substituting the resulting phasor that each reactance represents at the analyzed frequency. But unlike pure resistances (where the Norton Equivalent holds for all frequencies), the Norton Equivalent of a reactive circuit would only hold true for the one frequency at which you analyzed (because other frequencies might have resonances, etc.).


 * Atlant 15:39, 24 Apr 2005 (UTC)


 * The answer to the last comment is not entirely correct. The reason the impedance has to be re-calculated at each frequency is not necassarily as a result of resonances. Although it is true that resonances will occur at the resonant frequency (if one exists for a particular circuit).


 * The reason imedance has to be re-calculated at each frequency is simply because the impedance of each reactive element is a direct function of frequency. ie.  $${\it Xc}=1/2\,{\frac {1}{f\pi\,L}}$$ and $${\it Xl}=2\,f\pi\,L$$.


 * Nozog 15:44, 5 April 2006 (UTC)


 * I thought that was implied (by my earlier comments about "the ... phasor that each reactance represents at the analyzed frequency"), but if you prefer, I'll just claim that that point was hiding in the "etc." ;-). Thanks for your clarification!


 * Atlant 16:25, 5 April 2006 (UTC)


 * Sorry, I guess it just wasn't quite clear to me. Partly the reason I decided to write my comment above is becuase it actually took me bit of thought to consider how phasor analysis handles resonance frequency, which I realized is that the reactive components will simply dissapear.  I think it's an very interesting aspect, and although the solution is trivial, envisioning it is anything but.


 * Nozog 17:54, 5 April 2006 (UTC)


 * Long live Fourier! lol --euyyn 20:53, 14 March 2007 (UTC)

Co-discovery
There were a few comments I wanted to make:

1. This theorem was actually founded by two people in the same year (Norton of course being one) and Hans Ferdinan Mayer being the other and in fact Mayer was the only one of the two who actually published the result. Norton did write a report on the subject, but it was only circulated within Bell Labs. I thought this was rather interesting, and might be a nice addition to the article. Here is a link to an article citing this:


 * http://www.ece.ubc.ca/~jrms/eece560/Helmholtz-Thevenin-Equivalents.pdf

2. The other thing I wante to ask was, what is meant by "prototype circuit" in this article. It seems like the writer is saying Norton equivalents are used primarily to prototype power supplies and devices. If that's the case it is also not entirely accurate. Norton (and Thevenin) equivalents are parhaps most prevalent in modeling communications and power networks, in efforts to find solutions for the propagation of waves and such through them and this is actually the type of work that Norton was doing when he discovered the theorem.

3. In devising a Norton (or Thevenin) equivalent circuit, the more general way of determining the impendance of the system is to connect a fictitious current source at the output terminals, and then determining the voltage at the those terminals. The reason for this is that dependant sources within the circuit must be taken into consideration when determining effective impedance of a circuit.

Nozog 16:06, 5 April 2006 (UTC)


 * Please feel free to be bold and add this information to the article (including citations as necessary); improvements are always welcome!


 * Atlant 16:27, 5 April 2006 (UTC)


 * I have made the changes, I hope nobody minds. I thought the example was quite nice and I haven't touched it, I can see it took someone quite a bit of time.


 * A copy of the section I adjusted:
 * Norton's theorem for electrical networks states that any collection of voltage sources and resistors with two terminals is electrically equivalent to an ideal current source I in parallel with a single resistor R. For single frequency AC systems the theorem can also be applied to general impedances, not just resistors. The Norton Equivalent is a prototype circuit used to represent a power supply or battery. The circuit consists of an ideal current source in parallel with an ideal resistor.


 * The theorem was published in 1926 by Bell Labs engineer Edward Lawry Norton (1898-1983).

To calculate the equivalent circuit:


 * 1) Replace the load circuit with a short.
 * 2) Calculate the current through that short, I, from the original sources.
 * 3) Now replace voltage sources with shorts and current sources with open circuits.
 * 4) Replace the load circuit with an imaginary ohm meter and measure the total resistance, R, with the sources removed.
 * 5) The equivalent circuit is a current source with current I in parallel with a resistance R in parallel with the load.


 * Nozog 17:54, 5 April 2006 (UTC)

conversion to thevanin
source not that its hard to calculate on your own. Comments on my pretty large changes? Fresheneesz 05:09, 14 April 2006 (UTC)

Popular culture?
Seems saying too much for me. At least as we define popular culture. I'll reword it as in thévenin if nobody objects. --euyyn 20:57, 14 March 2007 (UTC)
 * It's really irrelevant. Just got rid of that section in this article. Alexius08 (talk) 11:27, 10 March 2009 (UTC)

not clear / correct
Connect a constant current source at the output terminals of the circuit with a value of 1 Ampere and calculate the voltage at its terminals. This voltage divided by the 1 A current is the Norton impedance RNo.

"calculate the voltage at the terminals" ??? Measure to terminal voltage. Apply 1 amp. measure the voltage. The delta is the resistance if using 1A. delta V / I otherwise. — Preceding unsigned comment added by 23.233.28.77 (talk) 04:10, 22 April 2015 (UTC)