- #1
[Circuits] Finding the Thevenin and Norton Equivalents #2
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Discussion Overview
The discussion revolves around finding the Thevenin and Norton equivalents of a given circuit. Participants explore various methods for calculating these equivalents, including nodal and mesh analysis, as well as source transformation techniques. The context is primarily homework-related, focusing on circuit analysis in electrical engineering.
Discussion Character
- Homework-related
- Technical explanation
- Exploratory
Main Points Raised
- One participant suggests writing a nodal equation for a single unknown nodal voltage to find the Thevenin equivalent.
- Another participant proposes using a voltage divider approach to derive the Thevenin and Norton equivalents simultaneously through node voltage analysis.
- A participant indicates they have calculated the Thevenin equivalent but are struggling to find the Norton equivalent directly from the circuit, expressing confusion over mesh analysis.
- There is a suggestion to redraw the circuit with a short circuit in place to simplify the analysis, questioning what the voltage Vx would be under that condition.
- One participant confirms that Vx would be 0V when the circuit is shorted, leading to the removal of certain components from the analysis.
- A later reply expresses relief at resolving the problem, indicating a sense of accomplishment.
Areas of Agreement / Disagreement
Participants generally agree on the methods to approach the problem, but there is some uncertainty regarding the application of mesh analysis and the correct identification of Vx. The discussion reflects a mix of exploration and clarification without a definitive resolution on all points.
Contextual Notes
Some participants express confusion over specific calculations and the application of analysis techniques, indicating potential limitations in their understanding or the clarity of the problem statement.
- #2
- #3
gneill
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Here's a hint that might help.
If you consider a general Thevenin model with a resistive load attached, you've really got a simple voltage divider:
If you were to tack a load resistor onto the output of your circuit and use node voltage analysis to solve for the output voltage, then you should be able to rearrange the resulting expression into the form of the voltage divider expression and just pick out Vth and Rth by inspection, thus solving for both at the same time.
This technique can be very handy; as long as the circuit in question has at least one power source (so any controlled sources can be "stimulated"), you don't have to bother with sticking external sources onto the output to find resistances or voltages.
If you consider a general Thevenin model with a resistive load attached, you've really got a simple voltage divider:
If you were to tack a load resistor onto the output of your circuit and use node voltage analysis to solve for the output voltage, then you should be able to rearrange the resulting expression into the form of the voltage divider expression and just pick out Vth and Rth by inspection, thus solving for both at the same time.
This technique can be very handy; as long as the circuit in question has at least one power source (so any controlled sources can be "stimulated"), you don't have to bother with sticking external sources onto the output to find resistances or voltages.
Attachments
- #4
ainster31
- 158
- 1
Sorry, I wasn't clear enough in my original question. I'm not sure how I forgot put these details but here they are now. I've calculated the Thevenin equivalent and I'm currently trying to get the Norton equivalent. I realize that I can just use source transformation to convert from the Thevenin equivalent to the Norton equivalent but I am trying to get the Norton equivalent directly from the circuit.
Since I already did the Thevenin equivalent, I already know that Vx=10/21 V and Rth=10/21 Ω. Now, I'm trying to calculate IN. I've connected nodes a and b but I am having trouble applying mesh analysis. I've already applied mesh analysis to mesh 1 but after looking at it, it seems to be incorrect. Here is the correct mesh analysis:
$$-50+12{ i }_{ 1 }+60({ i }_{ 1 }-{ i }_{ 2 })=0\\ -50+12{ i }_{ 1 }+\frac { 10 }{ 21 } =0$$
Edit: Actually, I don't even know what Vx is. That was incorrect.
I've tried applying nodal analysis at node X:
$$\frac { { v }_{ x }-50 }{ 12 } +\frac { { V }_{ x } }{ 60 } +2{ V }_{ x }+{ I }_{ N }=0$$
but I'm still stuck.
Since I already did the Thevenin equivalent, I already know that Vx=10/21 V and Rth=10/21 Ω. Now, I'm trying to calculate IN. I've connected nodes a and b but I am having trouble applying mesh analysis. I've already applied mesh analysis to mesh 1 but after looking at it, it seems to be incorrect. Here is the correct mesh analysis:
$$-50+12{ i }_{ 1 }+60({ i }_{ 1 }-{ i }_{ 2 })=0\\ -50+12{ i }_{ 1 }+\frac { 10 }{ 21 } =0$$
Edit: Actually, I don't even know what Vx is. That was incorrect.
I've tried applying nodal analysis at node X:
$$\frac { { v }_{ x }-50 }{ 12 } +\frac { { V }_{ x } }{ 60 } +2{ V }_{ x }+{ I }_{ N }=0$$
but I'm still stuck.
Last edited:
- #5
gneill
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Instead of nodal analysis, first redraw your circuit with the short circuit in place. A whole bunch of things get "squashed" by this short circuit. For example, what would Vx be?
- #6
ainster31
- 158
- 1
gneill said:
Hmmm, Vx would be 0V, right? And that means I can remove the 60 ohm resistor and the dependent current source, right?
- #7
gneill
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ainster31 said:
Try it and see
Your results should match your Thevenin values via the simple transformation.
- #8
ainster31
- 158
- 1
Haha, I got it. Wow, I was stressing out over such a simple problem.
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