How can I solve the paradox in Norton's theorem with unhooked branches?

  • Thread starter Thread starter Bassalisk
  • Start date Start date
  • Tags Tags
    Theorem
Click For Summary

Discussion Overview

The discussion revolves around a paradox encountered when applying Norton's theorem to a circuit with unhooked branches, specifically focusing on the calculation of currents and equivalent resistances. Participants are exploring the differences in results obtained through various methods, including Thevenin's and Norton's approaches, and the implications of short-circuiting branches in the circuit analysis.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant reports obtaining a current of ~86 mA through resistor R4 using matrix methods and simulations, while encountering discrepancies when applying Norton's theorem.
  • Another participant confirms that unhooking R4 is necessary to find the equivalent resistance as seen by the load terminals.
  • Some participants express confusion regarding the application of Norton's theorem and the necessity of short-circuiting branches, particularly when dealing with voltage sources.
  • There is a mention of a potential error in source transformations, with one participant suggesting that a transformation involving R1 and a current source may have been incorrectly applied.
  • Disagreement arises over the calculated value of Rth, with one participant asserting it is 19.90 ohms based on their calculations, while another claims it to be 12.75 ohms.
  • Participants are attempting to clarify the process of short-circuiting branches and the implications for the overall circuit analysis.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct application of Norton's theorem or the values of Rth. Multiple competing views remain regarding the calculations and transformations applied in the circuit analysis.

Contextual Notes

Participants express uncertainty about the correct handling of voltage sources during the short-circuiting process and the implications for the calculated currents. There are unresolved mathematical steps and assumptions regarding the circuit configuration that may affect the outcomes.

Bassalisk
Messages
946
Reaction score
2
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.

https://www.physicsforums.com/attachment.php?attachmentid=36138&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
 

Attachments

  • sadfsadfsd.jpg
    sadfsadfsd.jpg
    15.8 KB · Views: 422
Last edited:
Engineering news on Phys.org
Bassalisk said:
When finding R thevenin(for the circut) u unhook the R4 right?

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

Attachments

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

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?
 
Last edited:
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.
 
Bassalisk said:
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.

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.
 
jegues said:
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 current source and parallel resistance to a branch with a voltage source and series resistance.

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

Yes you are correct, sorry I had made a mistake in my calculations. Let me take a look at your work in full.
 
  • #10
EDIT: I am making mistakes all over the place today. Working on it!

...
 
  • #11
jegues said:
After looking over your work, I think the Rth you've calculated may be your problem.

I calculated Rth from the original circuit posted and found that,

[tex]R_{th} = 12.75 \Omega[/tex]

... 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.
 
  • #12
Bassalisk said:
... 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.

This is a good problem!

I can't see what's not clicking... I'll give it a fresh look again later.
 
  • #13
I think there is a problem because u neglect this voltage source, work around that. Studiot where are you :(
 

Similar threads

Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K