Norton's theorem big problem

Bassalisk

Hello,

I encountered a paradox.


When I solve this system using matrix form of electric potentials between circuit knots( i think its called that) I get that the current in R4 is ~86 mA. Using simulations in National Instruments, I got that the current is indeed ~86 mA.

How I did this? Well, I unhooked the branch where R4 is and then calculated what is R Thevenin is 19,9 Ohms. Using further this method, I found that the potential difference between those 2 knots where R is hooked is 5,2 V which in deed is the case.

Inorton=0,26 A. After that calculation of current through resistor R4 is trivial.(86 mA)



But when I use 'classsic' way of solving this circuit and finding Inorton, I get weird results.

When finding R thevenin(for the circut) u unhook the R4 right? and then make all voltage sources short circuited.(I transformed those current sources into voltage sources)

Again I get the right result of ~19 Ohms for R Thevenin. But when I calculate Inorton, I short circuit a branch I am looking current in.

http://physicsforums.com/attachment....8&d=1307125539

I get this case.


By my thoughts I should delete this branch with source and resistor. But that gives me wrong result. I get that the current in that branch is: ~ 760mA waaay off.

Problem here is only this branch with source and resistor, what do I do when I short circuit branch with resistor R4.

Any thoughts here?

I hope I was clear

Attached Thumbnails

 

jegues

Yes. You find the equivalent resistance as seen by the terminals of the load.

I'm not sure what you're asking in the rest of your post...

Are you telling us that you've solved it with Thevenin's theorem, but can't with Nortons?

Post the circuit after you've done all the source transformations and we'll go from there.

Bassalisk

When I use Norton's classic approach, I get bad results. Rthevenin=Rnorton, i just misplaced the therms.

Hmmm I will upload that in about an hour. Very long calculations.

Bassalisk

Here u go, an attachment. Hope you can see my problem

Attached Files

scan0004.pdf (757.7 KB, 15 views)

jegues

In your last circuit on page 3, why did you remove R3 and E3? (I jumped to the last page, assuming all your work up to that point is correct)

How did you get the value for the voltage source on the LHS to be 80V?

Bassalisk

Because in norton's process of finding its current, You have to short circuit that branch. Resulting in shout circuiting a voltage source. So you have to remove that.

80V=40V+Ig2*R1. Look at the first page. I transformed current sources into voltage sources.

jegues

Taking a look at the first few pages now, it seems as though the first source transformation you did to obtain the voltage source with a value of Ig2*R1 is incorrect.

R1 is not connected in parallel with the current source.

Do a source transformation with E1 and R1 followed by combining the parallel current sources/resistors. If you wish do to so you can then convert the resulting current source and parallel resistance to a branch with a voltage source and series resistance.

Bassalisk

I tested it in National instruments, this transformation, and it is correct. This is not the problem. Problem is that short circuit on R4.

jegues

Yes you are correct, sorry I had made a mistake in my calculations. Let me take a look at your work in full.

jegues

EDIT: I am making mistakes all over the place today. Working on it!

...

Bassalisk

... I disagree. I checked, me and my mates, and following strictly nortons procedure, Rth is 19,90 ohms. Checked with matrix forms. Resistance is not the problem.

jegues

This is a good problem!

I can't see what's not clicking... I'll give it a fresh look again later.

Bassalisk

I think there is a problem because u neglect this voltage source, work around that. Studiot where are you :(