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[Circuits] Finding the Thevenin and Norton Equivalents #2
In summary, the conversation discusses finding the Thevenin and Norton equivalents of a circuit, and using nodal and mesh analysis techniques to solve for the unknown variables. The method of using a voltage divider and short circuit to find the Thevenin equivalent is also mentioned. Ultimately, the solution is found by redrawing the circuit and removing certain components.
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- #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
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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
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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
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Haha, I got it. Wow, I was stressing out over such a simple problem.
FAQ: [Circuits] Finding the Thevenin and Norton Equivalents #2
What is the Thevenin equivalent circuit?
The Thevenin equivalent circuit is a simplified equivalent circuit that represents a more complex circuit with a single voltage source and a single resistance.
How is the Thevenin equivalent circuit calculated?
The Thevenin equivalent circuit is calculated by determining the open-circuit voltage (Voc) and the equivalent resistance (Req) of the original circuit. The equivalent voltage source is then placed in series with the equivalent resistance.
What is the purpose of finding the Thevenin equivalent circuit?
The purpose of finding the Thevenin equivalent circuit is to simplify a complex circuit into a simpler one, making it easier to analyze and design. It is also useful for determining the maximum power transfer in a circuit.
What is the difference between Thevenin and Norton equivalent circuits?
The main difference between Thevenin and Norton equivalent circuits is that Thevenin uses a voltage source with an equivalent resistance in series, while Norton uses a current source with an equivalent resistance in parallel. They are mathematically equivalent and can be converted from one form to another.
What are the limitations of using Thevenin and Norton equivalents?
Thevenin and Norton equivalents are only applicable for linear circuits and do not account for non-linear components. They also assume that the circuit is in steady state and do not take into account transient effects. Additionally, they are only accurate for a specific load, and the equivalent circuit may change if the load is changed.
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