Understanding Thevenin's Theorem

In summary, the conversation discusses finding the Thevenin equivalent resistance and voltage for a given circuit. There are multiple ways to approach this problem, such as using circuit theory or analyzing the circuit from AB with voltage sources short-circuited. The final step is to combine the information gathered to form the Thevenin equivalent representation.
  • #1
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Homework Statement


You have a circuit (attached pic) and want to find the Thevenin equivalent resistance and voltage.

Homework Equations



See below

The Attempt at a Solution


OK< here's what confuses me. If I simply look at the total resistance of the circuit from A and B, it seems to me that I should be adding up R1 with the parallel sum of R2 and R3. From there I would get the total resistance of the circuit and if I hook up A and B to a voltmeter, I should take the current in the circuit and multiply that by R1, giving me a voltage drop. I take that drop, subtracting from V1, and run that across the parallel sum of R2 and R3. Current remains the same, so the total voltage there (call it V3) should be V1 - I*R1 - I*R2.

But the weird thing to me is that the actual answer is treating R2 and R1 as though they are in parallel. I get that with the A and B connections that's how it would look. OK then, but if that's the Thevenin resistance then should not Vth = I * Rth?

And if that's the case, total I (current) is going to be V(source) / total resistance, which is (R1 + (R2R3/ R2 + R3)). Bt that isn't what I am told it is, Vth = R2*V / R1+ R2. This is not making sense to me.

So any help in parsing this would be appreciated.
 

Attachments

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  • #2
There are a couple of ways to go about determining the Thévenin equivalent.

Choose any two of the following tasks:

-▸ place yourself at AB and look into the network once all voltage sources have been short-circuited; what you see is the Thévenin resistance. Here, you'll see the 3 resistors are all in parallel.

-▸ use circuit theory, and calculate how much current will flow into short-circuited terminals AB.

-▸ use circuit theory to determine open-circuit voltage, VAB.

Finally, put any two pieces of this data together and form the Thévenin equivalent representation.
Good luck! http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg [Broken]
 
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  • #3
NascentOxygen said:
There are a couple of ways to go about determining the Thévenin equivalent.

Choose any two of the following tasks:

-▸ place yourself at AB and look into the network once all voltage sources have been short-circuited; what you see is the Thévenin resistance. Here, you'll see the 3 resistors are all in parallel.

-▸ use circuit theory, and calculate how much current will flow into short-circuited terminals AB.

-▸ use circuit theory to determine open-circuit voltage, VAB.

Finally, put any two pieces of this data together and form the Thévenin equivalent representation.
Good luck! http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg [Broken]
OK, when you say "ue circuit theory" which one on particular do you mean? KVL? KCL?
 
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  • #4
Emspak said:
OK, when you say "ue circuit theory" which one on particular do you mean? KVL? KCL?
KVL, KCL, nodal analysis, mesh analysis, source replacement (successive Thevenin/Norton applications with network reduction), parallel/series reductions,... anything you want, really, it's your choice. There are many ways to tackle any given circuit.
 
  • #5
OK, when you say "ue circuit theory" which one on particular do you mean? KVL? KCL?
Anything that works.

If I simply look at the total resistance of the circuit from A and B, it seems to me that I should be adding up R1 with the parallel sum of R2 and R3. From there I would get the total resistance of the circuit and if I hook up A and B to a voltmeter, I should take the current in the circuit and multiply that by R1, giving me a voltage drop. I take that drop, subtracting from V1, and run that across the parallel sum of R2 and R3. Current remains the same, so the total voltage there (call it V3) should be V1 - I*R1 - I*R2.
Total R = R1 + R2.R3/(R2+R3)
∴ current in R1 = V/(Total R)

Open circuit VAB = current x (R2.R3/(R2+R3)

= ## {V}\over {1 + \frac{R_1(R_2 + R_3)}{(R_2 \cdot R_3)}}##

You'll get this result whatever analytical approach you use.
 
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1. What is Thevenin's Theorem?

Thevenin's Theorem is a fundamental concept in circuit analysis that allows us to simplify complex circuits into an equivalent circuit with a single voltage source and a single resistance. This simplified circuit is called the Thevenin equivalent circuit.

2. How is Thevenin's Theorem used in circuit analysis?

Thevenin's Theorem is used to analyze complex circuits and determine their behavior and performance. It allows us to calculate the voltage and current at any point in the circuit, without having to go through the lengthy process of solving multiple equations and using complex circuit analysis techniques.

3. What are the conditions for Thevenin's Theorem to be applicable?

Thevenin's Theorem is applicable to any linear circuit with two terminals, as long as the circuit contains only independent sources (voltage and current sources) and resistors. The theorem is also valid for circuits with dependent sources, as long as the dependency is linear.

4. How do you find the Thevenin equivalent circuit?

To find the Thevenin equivalent circuit, we need to follow these steps:
1. Disconnect all the loads from the circuit.
2. Calculate the open circuit voltage (Voc) across the two terminals.
3. Calculate the equivalent resistance (Req) seen from the two terminals.
4. Draw the Thevenin equivalent circuit with a voltage source of value Voc and a resistor of value Req in series.

5. What are the advantages of using Thevenin's Theorem in circuit analysis?

Thevenin's Theorem allows us to greatly simplify complex circuits, making them easier to analyze and understand. It also helps in predicting the behavior of a circuit under different conditions. Additionally, it reduces the number of calculations required and saves time in circuit analysis.

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