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Bogged down on Norton equivalent problem

  1. Oct 4, 2011 #1
    Picture an 8ma ideal current source pointing downward in parallel to a 20kohm resistor on the left of this picture. Wouldn't these elements contribute to the voltage drop across A and B.

    Rth makes sense because the ideal voltage source will be shorted eliminating the last resistor but the voltage is confusing me.

    Thanks for the help

    Screenshot2011-10-04at50201PM.png
     
  2. jcsd
  3. Oct 5, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    You mean like this:

    attachment.php?attachmentid=39635&stc=1&d=1317828530.gif

    The 30V voltage source is not going to change its voltage no matter what. So any KVL equations you care to write concerning what's to the right of that source are going to have 30V there. Nothing (physically realizable) that happens to the left of the voltage source can affect that.

    So the voltage source acts as an effective "barrier" to what goes on to the left hand and right hand sides of it.
     

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  4. Oct 5, 2011 #3
    Ya same picture except the 8mA is pointing downward..

    Allright ya so since it has to be 30V on the upper and lower nodes nothing can change that. And you can find the current though the 20kohm resistor to be 1.5mA.

    Doing KCL on the lower node 6.5mA is moving to the right in the lower node. That 6.5 A will move through the voltage source since they are modeled as having 0 Resistance.

    So all the current will stay on the left to satisfy KCL and there is no way 30V is going to change therefore you can throw the stuff on the left out.
     
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