What is the Thevenin equivalent for this circuit?

In summary, the Thevenin equivalent for a circuit is a simplified version that models the behavior of the original circuit at a specific node. It consists of a voltage source in series with a resistance and is used to analyze and design complex circuits by reducing them to a simpler circuit with only a few components. The Thevenin equivalent is useful for calculating voltage drops and current flow in a circuit, and can be used to determine the maximum power that can be transferred to a load.
  • #1
Kevin2341
57
0

Homework Statement



https://www.circuitlab.com/editor/#?id=wt59bd

Homework Equations



ohms law

The Attempt at a Solution



Ok, from what I understand about this circuit, I can:
eliminate R1 (resistors in parallel with a voltage source don't effect voltage), and eliminate R3 (resistors in series with a current source don't effect current).

From there I can combine R4, R5, and R6 to get a equivalent resistance of 666.66 Ohms. So there I am left with:

https://www.circuitlab.com/circuit/77x7da/thq2question5phase2/

Well, here I could combine my voltage source with the resistor in parallel, OR I can combine the current source with the parallel resistor.

Doing the voltage gives:

0.1mA current source in parallel with a 1k ohm resistor

Doing the current gives:
1.234V voltage source in series with a 666.66 Ohm resistor

Which now I have the question, does it matter the orientation in which I place the voltage source:

https://www.circuitlab.com/circuit/axaacm/thq2question5phase3/?
Because if i keep it as the current sources and combine it, I end up with 0 amps right? (0.1A - 0.1A = 0)
 
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  • #2
Your equivalent circuit has different output nodes than the original, so you're headed for the wrong answer right there.
 
  • #3
oops, yeah I forgot those, just pretend they are (or I'll just update the schematic). I think I also swapped polarities with the bottom right circuit. The second voltage source should be negative on top, positive on bottom.

Anyhow, am I on the right track with my bottom right circuit?
 
  • #4
Kevin2341 said:
oops, yeah I forgot those, just pretend they are (or I'll just update the schematic). I think I also swapped polarities with the bottom right circuit. The second voltage source should be negative on top, positive on bottom.

Anyhow, am I on the right track with my bottom right circuit?

NO ... I say again: Your equivalent circuit has different output nodes than the original, so you're headed for the wrong answer right there.
 
  • #5
I guess I don't understand what you mean. Are you talking about my node a and node b on the original circuit? I was just using those originally because I thought I had to assign nodes like that on Circuit Lab in order to test the circuit (I updated the circuit to exactly how it is on my assignment.)
 
  • #6
You said you are trying to find the Thevinin equivalent for a specific circuit. A Thevinin equivalent is based on a set of nodes. If you choose a different set of nodes, you get a different equivalent circuit. This is what you have done.

You can certainly GET a Thevinin equivalent for your redrawn, simplified, circuit but because of the way you have simplified it, and changed the effective output nodes, it will not be the equivalent circuit for the original circuit's nodes A and B. If these are NOT the nodes for which you are supposed to be finding the equivalent circuit, then you need to show what ARE those nodes before I or anyone else can tell you whether or not you are getting the equivalent circuit correctly.
 
Last edited:
  • #7
Kevin2341 said:
In future could you post schematics as a picture, e.g., a jpeg. It should be possible for you to take a screenshot of the circuitlab screen and post that jpeg. A picture loads much faster and this will overcome some readers not being able to see your schematic:

attachment.php?attachmentid=56232.jpg
 

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  • #8
A 10V ― 1kΩ source converts into a 10mA ― 1kΩ Norton source. You appear to make it 100mA.
 
  • #9
Ok, I'm going to rewrite my original post with pictures, and my detailed efforts at solving this thing.
 

What is Thevenin equivalent?

Thevenin equivalent is a concept in electrical circuit analysis that allows complex circuits to be simplified into a single voltage source and series resistor. This simplified circuit is equivalent to the original circuit in terms of voltage and current at a specific load or terminal.

Why is Thevenin equivalent important?

Thevenin equivalent is important because it allows engineers and scientists to analyze complex circuits more easily. By simplifying the circuit into a single voltage source and resistor, calculations and analysis become simpler and more manageable.

How do you find the Thevenin equivalent?

To find the Thevenin equivalent, you need to follow these steps:

  1. Remove the load or load resistor from the circuit.
  2. Determine the open-circuit voltage (Voc) by using a voltmeter across the load terminals.
  3. Determine the short-circuit current (Isc) by connecting a ammeter across the load terminals.
  4. Calculate the equivalent resistance (Req) by removing all independent sources and calculating the resistance between the load terminals.
  5. The Thevenin equivalent circuit is then Voc in series with Req.

What are the applications of Thevenin equivalent?

Thevenin equivalent has various applications in circuit analysis and design. Some common applications include:

  • Predicting the behavior of a circuit at different loads.
  • Designing and optimizing complex circuits.
  • Simulating a complex circuit with a simpler one.
  • Calculating maximum power transfer in a circuit.

Are there any limitations of Thevenin equivalent?

While Thevenin equivalent is a useful tool, it does have some limitations. These include:

  • Thevenin equivalent can only be used for linear circuits.
  • It assumes that the circuit parameters are constant and do not change with load.
  • It does not take into account non-linear elements such as diodes and transistors.
  • The accuracy of the Thevenin equivalent depends on the accuracy of the measurements taken.

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