How Do You Determine the Thevenin Equivalent Circuit for Complex Networks?

In summary, the conversation discusses different methods for finding the Thevenin equivalent circuit at terminals (a,b) for a given circuit. These methods include loop analysis, delta-Y transforms, and voltage divider expressions. It is also mentioned that components only need to share two nodes to be in parallel, and that there are multiple ways to approach finding the Thevenin equivalent.
  • #1
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Homework Statement


Find the Thevenin equivalent circuit at terminals (a,b) for the circuit below.


Homework Equations





The Attempt at a Solution


I am having a conceptual issue for this circuit. I can't combine any of the resistors, and my gut is telling me that something about the fact that its between terminals (a,b) means that one of these resistors can be neglected and essentially erased. However, I don't understand how that works as far as what signs I should look for and what should be neglected.

I removed all the sources in an attempt to find the thévenin equivalent resistance.
 

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  • #2
Nothing can be neglected (your gut is wrong) and since you (as you correctly stated) cannot combine, as serial or parallel, any of the resistors, how might you go about finding the equivalent circuit ?
 
  • #3
If I can't combine them, then I don't know how to find an equivalent resistance.
 
  • #4
Do you know how to do loop analysis ? Do you know Kirchhoff's laws?
 
  • #5
Yes I do know them. I don't see how I can use that to find the equivalent resistance when I set all the sources to zero
 
  • #6
Ah, well how about delta-Y transforms, do you know them?

An alternative is to create a Thevenin equivalent for a subset of the circuit and go from there. Do it more than once if that's what it takes.
 
  • #7
I was told that the 2 ohm resistor that is vertical (between d and e) has no current going through it, thus can be essentially ''erased'' from the circuit. It gives the correct answer, but I can't figure out why it is the case
 
  • #8
Maylis said:
I was told that the 2 ohm resistor that is vertical (between d and e) has no current going through it, thus can be essentially ''erased'' from the circuit. It gives the correct answer, but I can't figure out why it is the case

That certainly isn't obvious from looking at the circuit but it might be true ... only an analysis will tell, unless I'm missing something.
 
  • #9
Maylis said:
Yes I do know them. I don't see how I can use that to find the equivalent resistance when I set all the sources to zero

If you suppress all the sources in the given circuit then you're left with just a network of resistors. If you then apply a known voltage across the output terminals (a - b) and determine the current that source drives into the port, you can determine the net resistance.

attachment.php?attachmentid=66735&stc=1&d=1392725701.gif


If you intend to use loop analysis on the resulting circuit, I might recommend removing the 4 Ω resistor across the applied source first. It's a known resistance in parallel with the rest and can be "added" back later. The benefit of removing it is that it reduces the number of loops by one.

Another method for finding the Thevenin equivalent of a circuit is to tack a load resistor RL onto the output and solve for the voltage Vout across RL (leaving RL as a variable, of course).

attachment.php?attachmentid=66736&stc=1&d=1392727207.gif


Use any method you like (loop, mesh, nodal). Take the expression you get for Vout and hammer it into the form of a voltage divider expression, since for a voltage divider formed from a Thevenin equivalent circuit with a load attached,
$$V_{out} = V_{th} \frac{RL}{RL +R_{th}}$$
Then simply pick out the terms for Vth and Rth from the result. Easy.
 

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  • #10
If I don't have a value for RL, how can I find Rth?
 
Last edited:
  • #11
Components only need share exactly two nodes in order to be in parallel. It makes no difference if other components also connect to those nodes. In the proposed circuit RL and the 4 Ω resistor in question are certainly in parallel.
 
  • #12
Maylis said:
If I don't have a value for RL, how can I find Rth?

Simple. Look at the resulting expression for the voltage across RL. This is a derivation leaving RL as a symbol. The important bits (Rth and Vth) are in the rest of the expression.
 
  • #13
What I did was solve for the currents. Using your labeling, I get

I1 = -1/10 A
I2 = -19/10 A
I3 = -3/2 A

Then I multiplied the 4 ohm resistor by I2 and got the Vab = -7.6 V, which is the correct answer for Vth. Now I am struggling to solve for Rth. I thought I could add a short circuit to the terminal (a,b), solve for the short circuit current, then divide Vth by Isc to get Rth
 
  • #14
Maylis said:
What I did was solve for the currents. Using your labeling, I get

I1 = -1/10 A
I2 = -19/10 A
I3 = -3/2 A

Then I multiplied the 4 ohm resistor by I2 and got the Vab = -7.6 V, which is the correct answer for Vth. Now I am struggling to solve for Rth. I thought I could add a short circuit to the terminal (a,b), solve for the short circuit current, then divide Vth by Isc to get Rth

Yes, dividing Vth by the short circuit current will also give you Rth.
 
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  • #15
Okay, well I redid the circuit with a short circuit, and here are my new equations

-2 -2 8 0 = -8
8 -2 -2 0 = 6
-2 8 -2 0 = -12
0 -4 0 4 = 0

then I get i4 (Isc) is -1.9 A, and divide 7.6/1.9 and it gives me 4 ohms, not the correct answer
 
  • #16
Maylis said:
Okay, well I redid the circuit with a short circuit, and here are my new equations

-2 -2 8 0 = -8
8 -2 -2 0 = 6
-2 8 -2 -4 = -12
0 -4 0 4 = 0

then I get i4 (Isc) is -1.9 A, and divide 7.6/1.9 and it gives me 4 ohms, not the correct answer

You forgot the item shown in red above.
 
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  • #17
Okay, I see.

I am so confused because there are too many ways to find the thévenin equivalent circuits. I kind of want to master one method at a time instead of flip flop around with different methods for every problem. Are all methods applicable for all circuits, or is it just convenient to use some methods over others in particular cases?
 
  • #18
You usually can use any network analysis method. One method may be easier for a given circuit than another.

Finding Thevenin equivalents consist of finding Vth, which is just finding the voltage across a couple of nodes, and finding Rth. Finding Rth can be done by finding the current in a short circuit across those same two nodes and dividing Vth/Isc. Or you can apply an external test current or voltage with the internal sources set to zero (except the dependent sources).

For instance, in post #9, gneill's first image shows the method of applying a test voltage. If you make the test voltage 1 volt, then your equations would become:

-2 -2 8 0 = 0
8 -2 -2 0 = 0
-2 8 -2 -4 = 0
0 -4 0 4 = 1

If you solve that system, the current in your added loop is the reciprocal of Rth; couldn't be easier.
 
  • #19
Maylis said:
Okay, I see.

I am so confused because there are too many ways to find the thévenin equivalent circuits. I kind of want to master one method at a time instead of flip flop around with different methods for every problem. Are all methods applicable for all circuits, or is it just convenient to use some methods over others in particular cases?

In general the idea is to find the open source voltage and output resistance of the network in question. How one goes about doing that for any particular circuit is a matter of applying favorite techniques. If you're really comfortable with loop analysis, use it. Same goes or nodal analysis. All the tools one has learned to use (parallel/series simplification, Ohm's Law, KVL, KCL, loop, nodal, mesh,...) can come in handy. Choosing the best technique for a given circuit is largely a matter of experience gained through practice.

For circuits comprised of resistors and fixed sources only, the open circuit voltage and network reduction method always works. Finding the short circuit current can replace the network reduction (essentially finding the Norton equivalent and using the Norton <---> Thevenin equivalency).

When controlled sources are present it can be trickier. One cannot simply suppress them as you do fixed sources to find the circuit equivalent resistance; There must be some power source active in the circuit in order to "stimulate" the controlled source behavior. Here the short circuit current method is useful if the controlled sources in the circuit are activated by fixed supplies internal to the circuit. If not you need to apply an external source to the output to get things working.

My own favorite method involves placing a load on the circuit and solving (symbolically) for the voltage across the load. Since a Thevenin source with a load is just a voltage divider scenario, one can pick out the Thevenin voltage and resistance directly from the resulting expression for Vout, thus finding both values at the same time.
 

Related to How Do You Determine the Thevenin Equivalent Circuit for Complex Networks?

What is a Thévenin equivalent circuit?

A Thévenin equivalent circuit is a simplified representation of a complex circuit that contains a voltage source and a resistor in series. It is used to analyze the behavior of a circuit at a specific point by replacing the original circuit with a simpler equivalent circuit.

How is a Thévenin equivalent circuit calculated?

To calculate the Thévenin equivalent circuit, you need to find the open-circuit voltage and the equivalent resistance. The open-circuit voltage is the voltage measured across the terminals when the circuit is disconnected, and the equivalent resistance is the resistance between the terminals when all the voltage and current sources are removed.

What is the purpose of using a Thévenin equivalent circuit?

The main purpose of using a Thévenin equivalent circuit is to simplify a complex circuit and make it easier to analyze. It also helps in calculating the behavior of a circuit at a specific point without having to go through the entire circuit.

What are the limitations of a Thévenin equivalent circuit?

A Thévenin equivalent circuit is only accurate at a single frequency and cannot account for the frequency response of a circuit. It also assumes that the circuit is linear, and all the elements are independent. Additionally, it only works for circuits that have a finite number of elements.

How is a Thévenin equivalent circuit used in practical applications?

A Thévenin equivalent circuit is commonly used in circuit analysis and design. It can help in predicting the behavior of a circuit and optimizing its performance. It is also used in electronic device modeling and simulation, as well as in troubleshooting and fault finding in circuits.

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