# Norton Equivalent Circuit

1. Sep 20, 2010

### gomezfx

1. The problem statement, all variables and given/known data
Find the norton equivalent with respect to terminals a and b
[PLAIN]http://img821.imageshack.us/img821/7177/nortonequivalent.jpg [Broken]

2. Relevant equations
KVL, KCL

3. The attempt at a solution
I set i1 = .008 A
Then I found the rest of the currents.

KVL@ loop i2:
2000(i2-i1)+30=0
i2 = .0065 A

KVL@ loop i3:
-30+15000i3=0
i3=.002 A

KCL@ node D
i3-i4-.010=0
(.002)-i4-.010=0
i4= -.008 A

KVL@ loop iab:
30000(iab-i4)=0
iab= -.008A

I got the same current for i4 and iab, unless I did my math wrong.
Can someone check my equations and set up to make sure I'm even finding the right current for the norton equivalent circuit?

EDIT. I redid i4 by doing (i4-i3)=.010 A
I got i4 = .012 A so I'm not sure which current to o use.

Last edited by a moderator: May 4, 2017
2. Sep 20, 2010

### vela

Staff Emeritus
There's nothing wrong with having iab = i4. The current will choose the path of least resistance, so if you short a and b, all the current flows through the short. Equivalently, you can think of it this way. If you short a and b, the voltage across the 30k-ohm resistor will be 0, so no current flows through it. Applying KCL to E then gives you that iab = i4.

Your initial attempt to find i4 is correct. In the second try, you introduced a sign error. At node D, i3 enters and i4 and 10 mA exit, so you have i3 = i4 + 0.010 A.

I didn't check everything in the rest of the problem, but I'll note that given how you drew the direction of i1, you should have used i1=-0.008 A.

3. Sep 20, 2010

### gomezfx

How come I can just assume that the 30k ohm resistor will be 0 when I short a and b. My professor made did this same thing in class with another problem and a lot of things I've been reading do the same.

I'll redo the problem with the -0.008 A as well.

4. Sep 20, 2010

### vela

Staff Emeritus
It doesn't make sense when you say the "resistor will be 0." What is that supposed to mean?

5. Sep 20, 2010

### gomezfx

Sorry, I meant to say that the voltage across the resistor will be 0.

6. Sep 20, 2010

### vela

Staff Emeritus
Oh, OK. When you short a and b, the two ends of the resistor are connected to the same node in the circuit, so the voltage across the resistor must be 0.