- #1
[Circuits] Finding the Thevenin and Norton Equivalents #2
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SUMMARY
This discussion focuses on finding Thevenin and Norton equivalents in electrical circuits, specifically using nodal and mesh analysis techniques. The user successfully calculated the Thevenin equivalent values of Vth = 10/21 V and Rth = 10/21 Ω. The conversation emphasizes the importance of applying mesh analysis correctly and suggests using source transformation to derive the Norton equivalent from the Thevenin equivalent. Ultimately, the user resolves their confusion regarding the circuit analysis and confirms their results.
PREREQUISITES- Understanding of Thevenin and Norton theorems
- Proficiency in nodal and mesh analysis techniques
- Familiarity with voltage dividers in circuit analysis
- Knowledge of circuit components such as resistors and dependent sources
- Learn advanced techniques for circuit analysis using Thevenin and Norton equivalents
- Explore source transformation methods in circuit design
- Study the application of mesh analysis in complex circuits
- Investigate the impact of dependent sources on circuit behavior
Electrical engineering students, circuit designers, and anyone involved in circuit analysis and simplification techniques will benefit from this discussion.
- #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.
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- #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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