Find Thevenin Equivalent for Circuit - Basic Principles

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Discussion Overview

The discussion revolves around finding the Thevenin equivalent of a circuit that includes a controlled current source. Participants explore the implications of this configuration on the calculation of Thevenin voltage and resistance, addressing conceptual difficulties and procedural questions related to the presence of controlled sources.

Discussion Character

  • Homework-related
  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant describes the circuit configuration, noting the presence of a controlled current source and expressing conceptual difficulty in calculating the Thevenin equivalent voltage and resistance.
  • Another participant questions the feasibility of the configuration without a load resistor, suggesting that non-zero current may not be possible.
  • There is a discussion about the method for finding Thevenin equivalent resistance, specifically regarding the treatment of controlled versus independent sources when applying a test source.
  • One participant clarifies that only independent sources are turned off during the calculation, while controlled sources remain active in the circuit.
  • An alternative method for calculating equivalent resistance is proposed, involving the short circuit current for the Norton equivalent.

Areas of Agreement / Disagreement

Participants express differing views on how to handle controlled sources in the context of Thevenin equivalent calculations. While some agree on the treatment of independent sources, there remains uncertainty about the implications of controlled sources, indicating that the discussion is not fully resolved.

Contextual Notes

Participants acknowledge limitations in understanding the role of controlled sources, particularly in relation to the circuit's behavior under different conditions. There is also mention of potential confusion arising from the representation of controlled sources in the circuit diagram.

Who May Find This Useful

This discussion may be useful for students and practitioners dealing with circuit analysis, particularly those encountering controlled sources in Thevenin equivalent calculations.

cepheid
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Homework Statement



Find the Thevenin equivalent of the circuit shown below

Homework Equations



Nothing but basic principles

The Attempt at a Solution



Alright, I realize that it will take some time for the attachement to be approved, but it's a really simple circuit. It has just one loop. Starting from ground, you go up through a 10 V voltage source, across a 1k resistor in series with it, and then back down to ground through a "controlled" current source, which means going in the opposite direction of the current indicated by the arrow ('up'). Note that it is a controlled source, even though it doesn't look like it in the diagram. This is because Circuit Maker is mildly retarded and doesn't represent controlled sources using diamonds, but using some other much more complicated circuit symbol. So I just threw an independent current source in there. Anyway, the output voltage is taken to be across this branch (the current source). Anyway, the current across the 2k is ix, and the current supplied by the dependent source is 3ix. I realize I haven't done anything towards a solution yet, but that's because I'm having conceputal difficulty with the controlled source here. So to be fair, for the time being I won't ask for help with how to solve it except to ask, "what do I do with the controlled source when I am trying to calculate the Thevenin equivalent voltage and resistance?"
 

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cepheid said:

Homework Statement



Find the Thevenin equivalent of the circuit shown below

Homework Equations



Nothing but basic principles


The Attempt at a Solution



Alright, I realize that it will take some time for the attachement to be approved, but it's a really simple circuit. It has just one loop. Starting from ground, you go up through a 10 V voltage source, across a 1k resistor in series with it, and then back down to ground through a "controlled" current source, which means going in the opposite direction of the current indicated by the arrow ('up'). Note that it is a controlled source, even though it doesn't look like it in the diagram. This is because Circuit Maker is mildly retarded and doesn't represent controlled sources using diamonds, but using some other much more complicated circuit symbol. So I just threw an independent current source in there. Anyway, the output voltage is taken to be across this branch (the current source). Anyway, the current across the 2k is ix, and the current supplied by the dependent source is 3ix. I realize I haven't done anything towards a solution yet, but that's because I'm having conceputal difficulty with the controlled source here. So to be fair, for the time being I won't ask for help with how to solve it except to ask, "what do I do with the controlled source when I am trying to calculate the Thevenin equivalent voltage and resistance?"

Applying KCL to the circuit you get
[tex]i_x = -3i_x[/tex]. So, [tex]i_x = 0[/tex] and [tex]e_{oc} = V_s = 10V[/tex]
 
Hmm, that's kind of weird. So this configuration is not really possible with non-zero current unless a load resistor is attached?

Also, I know that you're supposed to find the "Thevenin equivalent" resistance of the network by hooking up a "test source" across the terminals and calculating the current drawn from it. When you do so, you are supposed to "turn off" the network's internal sources. Again, how does this apply to the controlled current source? Do I leave it as is, and add its current to that drawn by the test source?
 
cepheid said:
Hmm, that's kind of weird. So this configuration is not really possible with non-zero current unless a load resistor is attached?
Yes.
Also, I know that you're supposed to find the "Thevenin equivalent" resistance of the network by hooking up a "test source" across the terminals and calculating the current drawn from it. When you do so, you are supposed to "turn off" the network's internal sources. Again, how does this apply to the controlled current source? Do I leave it as is, and add its current to that drawn by the test source?
Only the independent sources are cancelled. The controlled ones remain in the circuit.
There is an alternative way to calculate the equivalent resistance. You can calculate the short circuit current for the Norton equivalent. Since the Thevenin and Norton equivalents use the same resistance, you can calculate it as the quocient of the open circuit voltage and the short circuit current.
 
I'll try all that, thanks for the tips.
 

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