Thevenin & Norton Circuit with Current source?

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SUMMARY

The discussion focuses on solving a Thevenin and Norton equivalent circuit involving a current source and resistors. The key conclusion is that the only value for the current i_x that satisfies the equation i_x = 3 i_x is zero, leading to a Thevenin voltage V_th of 8V and a Norton current I_n of 8/11 A. The participants emphasize the importance of applying Kirchhoff's Voltage Law (KVL) and understanding the behavior of dependent sources in circuit analysis.

PREREQUISITES
  • Understanding of Thevenin and Norton equivalents
  • Familiarity with Kirchhoff's Voltage Law (KVL)
  • Knowledge of dependent and independent sources in circuits
  • Ability to analyze circuits using node voltage and mesh current methods
NEXT STEPS
  • Study the application of Kirchhoff's Voltage Law (KVL) in circuits with dependent sources
  • Learn how to derive Thevenin and Norton equivalents from complex circuits
  • Explore the implications of controlled sources on circuit behavior
  • Practice solving circuit problems using node voltage and mesh current analysis techniques
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing and simplifying electrical circuits, particularly those dealing with Thevenin and Norton equivalents.

  • #31
gneill said:
Check your node equation. You've got terms for currents flowing out of the node on both sides of the equation. If you're going to do it that way, one side or the other should describe a currents flowing into the node.

The best way to write a node equation consistently is to write a sum that equals zero. Choose either all currents flowing into the node or all currents flowing out. Then write all the terms on one side of the equation and set it equal to zero. So in this case if we assume that all currents are flowing out of the node:
$$\frac{V_1 -8}{2} + \frac{V_1}{9} + 3 \left(\frac{8 - V_1}{2} \right) = 0$$
Did I draw my arrows at V1 incorrectly? I just look at current in = current out and then write the equation. I'm not too sure why your equation is like that, but using it I get V1= -9V and that would give me IN= -1A
 
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  • #32
Marcin H said:
Did I draw my arrows at V1 incorrectly? I just look at current in = current out and then write the equation. I'm not too sure why your equation is like that, but using it I get V1= -9V and that would give me IN= -1A

In your equation (post #29) your terms all describe currents leaving the node. That means, for example, if you moved the term on the left side over to the right it would be negated and describe a current entering the node. Then your equation would say that some current entering the node plus all the currents leaving is zero. That's not KCL.

You seem to have made a sign error in solving the equation from post #30. Can you show your work?
 
  • #33
gneill said:
In your equation (post #29) your terms all describe currents leaving the node. That means, for example, if you moved the term on the left side over to the right it would be negated and describe a current entering the node. Then your equation would say that some current entering the node plus all the currents leaving is zero. That's not KCL.

You seem to have made a sign error in solving the equation from post #30. Can you show your work?
I just double checked it. I got + 1A now.

New Doc 27.jpg
 
  • #34
Good! So now you have both the open circuit voltage and the short circuit current. What then is the Thevenin resistance?
 
  • #35
gneill said:
Good! So now you have both the open circuit voltage and the short circuit current. What then is the Thevenin resistance?
RT = Vt/Isc = 8V/1A = 8ohms
 
  • #36
Right.
 
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