Norton Equ. homework help.

Thread Starter

thatsmessedup

Joined Feb 15, 2018
59
I am having a hard time figuring out who it correct, my teacher or myself. For problem (a) I get opposite polarities on my norton equ. but as far as I can tell mine is still equivelent to the teachers. On problem (b) I get a different answer completely. Can anyone shed a little light on who went wrong and where?

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This is the Word document that the teacher provided:
http://courses.egr.uh.edu/ECE/ECE2201/Dr_Dave_Project/Module04/PEQWS_Mod04_Prob03_v04.doc
 

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WBahn

Joined Mar 31, 2012
30,071
Take your solution and your teacher's solution and see which one of them, if either, is actually a solution to the problem given. Your Norton equivalent gives an open circuit Vab of 125.3 V.

Is that the open circuit voltage in the original problem?

Now check the instructor's solution.

Keep in mind, too, that with a dependent source inside the equivalent, it is possible to have a negative equivalent resistance.
 

Thread Starter

thatsmessedup

Joined Feb 15, 2018
59
When I plug in the numbers the answer is -6.03mA which is the instructors answer. Is my Norton equivalent circuit the same as my instructors even though it has current in the opposite direction but with a negative resistance? Because the only difference I can see is that when I switch my signs I get the right answer.
 

Thread Starter

thatsmessedup

Joined Feb 15, 2018
59
I think I discovered the problem. If im taking the short circuit current from point b to point a then I suppose the norton current should be pointing down and not up. If I take that line of reasoning my numbers work out.

20180415_120906.jpg
 

WBahn

Joined Mar 31, 2012
30,071
You can orient you Norton circuit either way, but doing so only flips the sign on the current source, not on the equivalent resistance. The negative sign on the equivalent resistance comes as a result of the dependent current source in the circuit and it will be negative regardless of how you orient the Norton source. This can be seen from the fact that Rnorton can be determined by shutting off all of the independent sources and applying a test source between 'a' and 'b'. As you apply a positive voltage, you will get a negative current, regardless of how you connect it.
 
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