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

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In summary, the current in R4 is ~86 mA using simulations and ~760 mA using the "classics" method. However, when short circuiting a branch with a resistor, the "classics" method gives a result that is off by a factor of two.
  • #1
Bassalisk
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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
 

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  • #2
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.
 
  • #3
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.
 
  • #4
Here u go, an attachment. Hope you can see my problem
 

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  • #5
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?
 
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  • #6
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.
 
  • #7
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.
 
  • #8
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.
 
  • #9
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 :(
 

1. What is Norton's theorem and why is it important in science?

Norton's theorem is a fundamental principle in electrical engineering and circuit analysis. It states that any linear, bilateral network of multiple sources and resistors can be simplified into a single current source and a single parallel resistor. This theorem is important because it allows for easier analysis and understanding of complex circuits, and is the basis for many circuit analysis techniques.

2. What is the difference between Norton's theorem and Thevenin's theorem?

Both Norton's theorem and Thevenin's theorem are methods for simplifying complex circuits. However, while Norton's theorem represents a circuit as a single current source and resistor in parallel, Thevenin's theorem represents a circuit as a single voltage source and resistor in series. They are essentially two different approaches to achieve the same goal of simplifying circuit analysis.

3. What are the limitations of Norton's theorem?

Norton's theorem can only be applied to linear, bilateral networks with multiple sources and resistors. It also assumes that the circuit is in steady-state and does not take into account any non-ideal components such as capacitance or inductance. Additionally, Norton's theorem is only valid for DC circuits and does not apply to AC circuits.

4. How do you calculate the Norton equivalent current and resistance?

To calculate the Norton equivalent current, you need to short circuit all voltage sources in the circuit and calculate the resulting short circuit current. This value will be the Norton current. To calculate the Norton equivalent resistance, you need to find the equivalent resistance of the circuit between the two terminals where the current source will be placed. This can be calculated using Ohm's law or by simplifying the circuit using series and parallel resistance rules.

5. Can Norton's theorem be applied to circuits with dependent sources?

Yes, Norton's theorem can be applied to circuits with dependent sources, as long as they are linear and bilateral. In this case, the dependent source should be treated as a regular source for the purpose of calculating the Norton equivalent current and resistance.

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