What is the Norton equivalent current for terminals a and b in this circuit?

AI Thread Summary
The discussion focuses on finding the Norton equivalent current for terminals a and b in a given circuit. Initial calculations for currents i1, i2, i3, and i4 were presented, with some confusion regarding the correct values and signs. A key point is that when terminals a and b are shorted, the voltage across the 30k-ohm resistor becomes zero, leading to the conclusion that all current flows through the short. There was a correction regarding the direction of current i1, which should be negative, and clarification on the assumption of zero voltage across the resistor when shorting the terminals. The conversation emphasizes the importance of careful application of KVL and KCL in circuit analysis.
gomezfx
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Homework Statement


Find the norton equivalent with respect to terminals a and b
[PLAIN]http://img821.imageshack.us/img821/7177/nortonequivalent.jpg

Homework Equations


KVL, KCL

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.
 
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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.
 
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.
 
gomezfx said:
How come I can just assume that the 30k ohm resistor will be 0 when I short a and b.
It doesn't make sense when you say the "resistor will be 0." What is that supposed to mean?
 
vela said:
It doesn't make sense when you say the "resistor will be 0." What is that supposed to mean?

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