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Finding thevenin equivalent

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    ohms law

    3. 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)
     
  2. jcsd
  3. Feb 28, 2013 #2

    phinds

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    Your equivalent circuit has different output nodes than the original, so you're headed for the wrong answer right there.
     
  4. Feb 28, 2013 #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?
     
  5. Feb 28, 2013 #4

    phinds

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    NO ... I say again: Your equivalent circuit has different output nodes than the original, so you're headed for the wrong answer right there.
     
  6. Feb 28, 2013 #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.)
     
  7. Mar 1, 2013 #6

    phinds

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    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: Mar 1, 2013
  8. Mar 1, 2013 #7

    NascentOxygen

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    Staff: Mentor

    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
     

    Attached Files:

  9. Mar 1, 2013 #8

    NascentOxygen

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    Staff: Mentor

    A 10V ― 1kΩ source converts into a 10mA ― 1kΩ Norton source. You appear to make it 100mA.
     
  10. Mar 2, 2013 #9
    Ok, I'm going to rewrite my original post with pictures, and my detailed efforts at solving this thing.
     
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