Find the Thevenin equivalent circuit:

In summary, the conversation discusses finding the Thevenin equivalent circuit with respect to the terminals A and B. The homework equations and methods used include V=IR, KVL, KCL, voltage division, current division, and the Thevenin formula/procedure. The expert suggests using nodal analysis with respect to B as the reference node to find the potential at A with respect to B. The conversation also clarifies that the 5 and 20 ohm resistors are not in parallel with respect to the 72V source and that the voltage goes "across" rather than "through". It is also mentioned that the potential at A is to start from the battery across the resistors and to the point, then finding out
  • #1
Color_of_Cyan
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Homework Statement




http://imageshack.us/a/img809/9307/homeworkprobsg210omgthi.jpg [Broken]


Find the Thevenin equivalent circuit with respect to the terminals.


Homework Equations



V = IR,

KVL, KCL

voltage division, current division

Thevenin formula / procedure


The Attempt at a Solution



I think the Thevenin resistance is from A to B so with the 72V source surpressed, then 5 and 20 are in parallel which is 4, and then that would be in series with 8, which is then 12. Then 12 and the original 12 are in parallel, so that would mean

R(Th, A to B) = 6Ω

But I'm having trouble finding V(Th) from a to B (if that's what I'm supposed to find)


I forgot this, but from the diagram, are the 5 ohm and 20 ohm in parallel resistors in parallel with respect to the 72V source? I think the potential from A to B would be the voltage "through" the 8 and 20 ohm resistors, but not sure.
 
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  • #2
Color_of_Cyan said:

Homework Statement




http://imageshack.us/a/img809/9307/homeworkprobsg210omgthi.jpg [Broken]


Find the Thevenin equivalent circuit with respect to the terminals.


Homework Equations



V = IR,

KVL, KCL

voltage division, current division

Thevenin formula / procedure


The Attempt at a Solution



I think the Thevenin resistance is from A to B so with the 72V source surpressed, then 5 and 20 are in parallel which is 4, and then that would be in series with 8, which is then 12. Then 12 and the original 12 are in parallel, so that would mean

R(Th, A to B) = 6Ω
Yes, that looks good for Rth.
But I'm having trouble finding V(Th) from a to B (if that's what I'm supposed to find)


I forgot this, but from the diagram, are the 5 ohm and 20 ohm in parallel resistors in parallel with respect to the 72V source? I think the potential from A to B would be the voltage "through" the 8 and 20 ohm resistors, but not sure.
No, they're not in parallel. And voltage doesn't go "through". Current goes "through"; voltage is always "across" or between two points.

For this problem I would suggest applying a method such as nodal analysis to find the potential at a with respect to b.
 
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  • #3
Okay, so using nodal analysis with respect to b do you mean using b (the node by it) as a reference node? I mean, since the a and b points were part of the load and constituted the load, aren't they cut off while doing this? So I would find the voltages at the points in the circuit where they would be connected to then instead, right (ie. just at the "node" that would connect point a)?Because if a and b are cut off from the rest of the circuit then the current doesn't go anywhere and then I would be able to put the 12 and 8 resistors in series no matter what.
 
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  • #4
Color_of_Cyan said:
Okay, so using nodal analysis with respect to b do you mean using b (the node by it) as a reference node?
Sure. Since you are looking for the potential at a with respect to b, it would make sense to choose b as the reference node.
I mean, since the a and b points were part of the load and constituted the load, aren't they cut off while doing this? So I would find the voltages at the points in the circuit where they would be connected to then instead right (ie just for point a)?

Terminals a and b are where a load would be connected. You are looking for the Thevenin equivalent of the network from the point of view of terminals a and b. In other words, imagine that whatever load was previously connected between terminals a and b has been removed for analysis, and the diagram as given is the result.
 
  • #5
So potential at point (node) A is to start from the battery across the resistors and to the point then, and then find out how MUCH more there is to still cross before getting back to the battery then.It's seems there's 2 points though... would V(Th) be different if points A and B were switched? Or does it just matter where the potential from the battery goes across to the first point?
Thanks yet again, by the way.
 
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  • #6
Color_of_Cyan said:
So potential at point (node) A is to start from the battery across the resistors and to the point then, and then find out how MUCH more there is to cross before getting back to the battery then, right?
:confused: If that's a way of saying you want to find the potential at a with respect to b, then yes. Otherwise, I guess I don't understand your paragraph :redface:
Thanks yet again, by the way.
You're welcome :smile:
 

What is a Thevenin equivalent circuit?

A Thevenin equivalent circuit is a simplified representation of a complex circuit, which can be used to analyze and understand the behavior of the original circuit. It consists of a voltage source in series with a resistor, and can accurately predict the voltage and current at any point in the original circuit.

Why is it important to find the Thevenin equivalent circuit?

Finding the Thevenin equivalent circuit allows us to simplify complex circuits and make calculations easier. It can also help us understand the behavior of the original circuit, and make it easier to troubleshoot and make modifications.

How do you find the Thevenin equivalent circuit?

To find the Thevenin equivalent circuit, we need to remove all the load components from the original circuit and calculate the open circuit voltage (Vth) and the equivalent resistance (Rth) at the load terminals. Vth is the voltage between the load terminals, and Rth is the resistance that would be seen looking into the load terminals. These values can then be used to create the Thevenin equivalent circuit.

What are some applications of the Thevenin equivalent circuit?

The Thevenin equivalent circuit is commonly used in circuit analysis and design. It can also be used in power systems, electronic devices, and telecommunications to simplify complex circuits and make calculations and predictions easier.

Can the Thevenin equivalent circuit be used for all types of circuits?

No, the Thevenin equivalent circuit is only applicable to linear circuits, which follow Ohm's law and have a constant voltage-current relationship. Non-linear circuits, such as those containing diodes or transistors, cannot be simplified using the Thevenin equivalent circuit.

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