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Norton Equivalent Circuit

  1. Sep 20, 2010 #1
    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. jcsd
  3. Sep 20, 2010 #2

    vela

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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.
     
  4. Sep 20, 2010 #3
    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.
     
  5. Sep 20, 2010 #4

    vela

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    It doesn't make sense when you say the "resistor will be 0." What is that supposed to mean?
     
  6. Sep 20, 2010 #5
    Sorry, I meant to say that the voltage across the resistor will be 0.
     
  7. Sep 20, 2010 #6

    vela

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    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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