- #1
How Do You Solve Problems Using Thevenin and Norton Equivalents?
- Thread starter euphoria172
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Discussion Overview
The discussion revolves around solving problems using Thevenin and Norton equivalents in electrical circuits. Participants are seeking guidance on methods and approaches for applying these concepts, particularly in the context of circuit analysis involving voltage sources and resistors.
Discussion Character
- Homework-related
- Technical explanation
- Exploratory
Main Points Raised
- One participant expresses uncertainty about their answers and confusion regarding the parallel part of the circuit.
- Another participant suggests that the first method used to find the Thevenin equivalent appears correct and encourages the use of Thevenin or Norton transformations to simplify the circuit step by step.
- A participant questions the clarity of the second method employed, noting that it seems to misinterpret the contributions of the voltage source.
- There is acknowledgment of a mistake made in a similar question, leading to doubt about the current problem's answer.
Areas of Agreement / Disagreement
Participants do not reach a consensus, as there are differing views on the methods used and their effectiveness. Uncertainty remains about the correct application of techniques in the context of the problem.
Contextual Notes
Participants mention specific methods and steps taken, but there are unresolved aspects regarding the application of superposition and the interpretation of circuit components.
- #2
tiny-tim
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welcome to pf!
hi euphoria172! welcome to pf!
show us what you've tried, and where you're stuck, and then we'll know how to help!
hi euphoria172! welcome to pf!
show us what you've tried, and where you're stuck, and then we'll know how to help!
- #3
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gneill
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Why do feel that your Method 1 is incorrect? The results look fine It appears that you first found the Thevenin equivalent of the voltage source and first two resistors and then applied superposition to find the final Voc. Nothing wrong with that.
In your Method 2, it's not entirely clear to me what your steps were trying to accomplish. It appears that you were again trying to use superposition to find the contribution from each source, but for the voltage source you found the total current that it produces, not the portion flowing through the shorted output.
The method that you employed in the first attempt, namely working section by section through the circuit making Thevenin (or Norton) transformations can be very useful. It can allow you to combine sources easily as you go. Try it for the Norton equivalent: first convert the 25V source and 5 Ohm resistor to its Norton equivalent. What's then in parallel and can be combined? Rinse, repeat
It appears that you first found the Thevenin equivalent of the voltage source and first two resistors and then applied superposition to find the final Voc. Nothing wrong with that.In your Method 2, it's not entirely clear to me what your steps were trying to accomplish. It appears that you were again trying to use superposition to find the contribution from each source, but for the voltage source you found the total current that it produces, not the portion flowing through the shorted output.
The method that you employed in the first attempt, namely working section by section through the circuit making Thevenin (or Norton) transformations can be very useful. It can allow you to combine sources easily as you go. Try it for the Norton equivalent: first convert the 25V source and 5 Ohm resistor to its Norton equivalent. What's then in parallel and can be combined? Rinse, repeat
- #5
euphoria172
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Thx gneill. I realize i make a mistake on a similar question that cause me to doubt this answer.
Thx again. I appreciate your help.
Thx again. I appreciate your help.
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