Bogged down on Norton equivalent problem

AI Thread Summary
The discussion centers on analyzing a circuit with an 8mA ideal current source and a 20kΩ resistor, questioning their contribution to the voltage drop across points A and B. It is clarified that the 30V voltage source remains constant and acts as a barrier, meaning that changes on the left side of the source do not affect the voltage on the right. The current through the 20kΩ resistor is calculated to be 1.5mA, while KCL indicates that 6.5mA flows to the right through the voltage source, which has zero resistance. Ultimately, the conclusion is that the components on the left can be disregarded since the 30V remains unchanged. Understanding these principles helps clarify the voltage behavior in the circuit.
MotoPayton
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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
 
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MotoPayton said:
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

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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Ya same picture except the 8mA is pointing downward..

Allright you 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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