Thevenin & Norton Circuit with Current source?

Click For Summary

Discussion Overview

The discussion revolves around finding the Thevenin and Norton equivalents of a circuit containing a current source. Participants explore the implications of a dependent current source and the relationships between voltages and currents in the circuit, particularly focusing on the open-circuit voltage (Voc) and the Thevenin resistance (Rth).

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about finding Voc, questioning how to apply Kirchhoff's laws without knowing the resistance or voltage of the current source.
  • Another participant points out that if the node voltage leads to the equation ix = 3ix, then ix must be zero, raising questions about the implications of this result.
  • Some participants discuss the potential difference across the 2 Ω resistor when ix is zero, suggesting that it must also be zero.
  • There is a debate about whether the Thevenin voltage can be concluded as zero if the current source is zero, with some arguing that a current source maintains its designated current regardless of the voltage across it.
  • One participant suggests using KVL to find the voltage across the current source, while others challenge this approach, indicating that the behavior of controlled sources complicates the analysis.
  • Participants explore the relationship between the Thevenin voltage and resistance, with some concluding that Rth is 11 Ω based on their calculations, while others caution against suppressing controlled sources in this determination.

Areas of Agreement / Disagreement

Participants do not reach consensus on the value of Voc or the implications of a zero current source. There are competing views on how to analyze the circuit and the role of controlled sources in determining Thevenin resistance.

Contextual Notes

There are unresolved assumptions regarding the behavior of the current source and the implications of dependent sources on circuit analysis. The discussion reflects uncertainty about applying KVL in the presence of a current source with unknown resistance.

  • #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
 
Physics news on Phys.org
  • #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.
 
  • Like
Likes   Reactions: Marcin H

Similar threads

Engineering How do I find Vth with the node voltage method?
  • · Replies 10 ·
Replies
10
Views
4K
Engineering Norton/Thevenin Equivalent circuits with a dependent voltage source
  • · Replies 9 ·
Replies
9
Views
6K
How Do Controlled Sources Impact Thevenin Equivalent Calculations?
  • · Replies 5 ·
Replies
5
Views
2K
Engineering Attempt at Thevenin Equivalent Circuit
  • · Replies 8 ·
Replies
8
Views
3K
Engineering How to Calculate Thevenin Equivalent Circuit
  • · Replies 5 ·
Replies
5
Views
3K
Engineering Finding Current in a Thevenin Circuit
  • · Replies 5 ·
Replies
5
Views
2K
Three-part Problem: Thevenin + Dependent current source
  • · Replies 6 ·
Replies
6
Views
2K
Engineering Thevenin/Norton circuit with dependent voltage source
  • · Replies 20 ·
Replies
20
Views
12K
Engineering Finding the Thevenin Equivalent Circuit for a Network with a Load Resistor
  • · Replies 1 ·
Replies
1
Views
2K
Engineering Find the Thevenin equivalent circuit
  • · Replies 42 ·
2
Replies
42
Views
7K