Help please with the Thevenin Equivalent of this circuit

In summary: The resistors connected to point "A" are 9 Ohm resistors.The resistors connected to point "B" are also 9 Ohm resistors.The voltage at point "A" is 9 volts.The voltage at point "B" is also 9 volts.
  • #1
asd852
12
2
Homework Statement
1.whats the equivalant circuit for circiut shown in figure 3???????
2.how can we determine the Thevenin parameter for question 2?????
Relevant Equations
mesh/nodal analysis
LC{YV]S6@{HJ@[2]Y`ZQ)9U.png
 

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  • #2
Hi asd852, Welcome to Physics Forums.

We can offer help only after you've shown your own efforts so far. What have you tried?
 
  • #3
I am a bit clueless
 
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  • #4
asd852 said:
I am a bit clueless
Then you have taken on a problem that is too advanced for you. Go back and study the basics.
 
  • #5
asd852 said:
Homework Statement:: 1.whats the equivalent circuit for circiut shown in figure 3??
2.how can we determine the Thevenin parameter for question 2?
Relevant Equations:: mesh/nodal analysis

View attachment 273298
Seeing that you're new to PF, a bit of help.

For the problem you show (Question 2 with Figure 3):
Consider the circuit with the 9 Ω resistors removed from terminals A and B.

For Question 2: Make a new thread, showing the attachment as full size. Above all, show an attempt and/or be specific about what is confusing to you and what you do know.

asd852 Prob2.png
 
  • #6
There are some good explanations on the web of Thevenin's theorem as well as YouTube videos. I suggest you look into those, try to apply those methods to your problem, and come back here with questions about the confusing bits if you need to. Sometimes you need a different explanation than the one you got in class, or you just need to hear it a few times.
 
  • #7
I suspect that part of your problem is that you find this:
1606593655108.png

part of the circuit confusing. This is a VERY standard exercise in redrawing the circuit and ends up being totally trivial.
 
  • #8
SammyS said:
Seeing that you're new to PF, a bit of help.

For the problem you show (Question 2 with Figure 3):
Consider the circuit with the 9 Ω resistors removed from terminals A and B.

For Question 2: Make a new thread, showing the attachment as full size. Above all, show an attempt and/or be specific about what is confusing to you and what you do know.

View attachment 273308
Could you please draw an equivalent circiut diagram and calculate the results,after which I assume I would understand my confusing points
 
  • #9
phinds said:
I suspect that part of your problem is that you find this:
View attachment 273318
part of the circuit confusing. This is a VERY standard exercise in redrawing the circuit and ends up being totally trivial.
Sorry,I find the website you presented irrelevant to the above question
 
  • #10
One more try with clues to assist you.

You have probably been told that any points connected by a wire are at the same potential (voltage), at least I hope that was part of your curriculum!

Keeping that in mind, look at the junction between the upper two 9 Ohm resistors.

Can you say what the voltage is there, and why?

If that helped any, pretend the drawing is on a sheet of rubber that you can stretch out of shape. Now mentally drag the common point of the two upper 9 Ohm resistors to a position that makes more visual sense for you.

If you have gotten this far, please post a sketch of the re-arranged circuit.

Cheers,
Tom

(Hang in there asd582, the exercise you were given is designed to be somewhat confusing, and to get that "AH HA" moment in your head when you realize the solution.)
 
  • #11
asd852 said:
Sorry,I find the website you presented irrelevant to the above question
Huh? I did not put a link in my answer. Perhaps you are talking about the link in my signature? That has nothing to do with you. AND you are avoiding my point. DO you find that part of the circuit confusing?
 
  • #12
You are not listening:

... show an attempt and/or be specific about what is confusing to you and what you do know.

If you have gotten this far, please post a sketch of the re-arranged circuit.
Could you please draw an equivalent circiut diagram and calculate the results,after which I assume I would understand my confusing points
We're here to HELP you, not do it for you. Drawing an equivalent circuit is the heart of the problem and exactly what YOU need to figure out.
 
  • #13
Tom.G said:
One more try with clues to assist you.

You have probably been told that any points connected by a wire are at the same potential (voltage), at least I hope that was part of your curriculum!

Keeping that in mind, look at the junction between the upper two 9 Ohm resistors.

Can you say what the voltage is there, and why?

If that helped any, pretend the drawing is on a sheet of rubber that you can stretch out of shape. Now mentally drag the common point of the two upper 9 Ohm resistors to a position that makes more visual sense for you.

If you have gotten this far, please post a sketch of the re-arranged circuit.

Cheers,
Tom

(Hang in there asd582, the exercise you were given is designed to be somewhat confusing, and to get that "AH HA" moment in your head when you realize the solution.)
A picture may help.(two points connected with a viire are the same point in the circuit. )
1606662653601.png

Which resistors are connected to point A ? And to point B?
 
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  • #14
You get this sort of problem when drawing schematics from PCB traces. My technique was to colour each track, so that wherever it reached, I could see from the colour what it was connected to.
This one is trivially easy, because you use only two colours.
In this case I'd put the A track (red) across the top and B track (blue) along the bottom. Then just put the resistors wherever they'll fit neatly, with their ends connected to the right colour.
curcuit.png
 
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  • #15
Merlin, that's a good technique but I think we've given hints aplenty with no response from the OP to them. So now, let's wait until he shows some effort in using what we've already provided.
 
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What is the Thevenin Equivalent of a circuit?

The Thevenin Equivalent is a simplified representation of a complex circuit that consists of a single voltage source and a single resistor. It is used to analyze the behavior of a circuit and can be used to find the voltage and current at any point in the circuit.

How do you calculate the Thevenin Equivalent?

To calculate the Thevenin Equivalent, you need to find the open circuit voltage and the equivalent resistance. The open circuit voltage is the voltage between the two terminals of the circuit when there is no current flowing. The equivalent resistance is the resistance seen by the load when all the voltage sources are removed from the circuit.

Why is the Thevenin Equivalent useful?

The Thevenin Equivalent is useful because it simplifies a complex circuit into a simpler one, making it easier to analyze. It also allows us to find the voltage and current at any point in the circuit without having to go through complicated calculations.

What are the limitations of the Thevenin Equivalent?

The Thevenin Equivalent assumes that the circuit is linear, meaning that the relationship between voltage and current is constant. It also only works for circuits with two terminals and cannot be used for circuits with active components such as transistors.

How is the Thevenin Equivalent used in practical applications?

The Thevenin Equivalent is used in practical applications to design and analyze circuits, especially in electronic devices. It is also used in circuit simulation software to model and test circuits before they are physically built.

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