Bogged down on Norton equivalent problem

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SUMMARY

The discussion centers on analyzing a circuit with an 8mA ideal current source and a 20kΩ resistor in parallel. The participants clarify that the 30V voltage source remains constant and acts as a barrier, ensuring that the voltage across nodes A and B is unaffected by elements to the left of the voltage source. The current through the 20kΩ resistor is calculated to be 1.5mA, and KCL confirms that the total current remains on the left side of the circuit, validating the voltage source's role in maintaining a stable voltage.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Familiarity with Kirchhoff's Current Law (KCL)
  • Basic knowledge of circuit components such as current sources and resistors
  • Concept of Thevenin's Theorem and equivalent resistance
NEXT STEPS
  • Study Thevenin's Theorem applications in circuit analysis
  • Learn about voltage and current source behavior in parallel circuits
  • Explore advanced KVL and KCL problem-solving techniques
  • Investigate the impact of ideal versus real voltage sources on circuit performance
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or designing electrical circuits will benefit from this discussion.

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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