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Circuits Question - Thevein & Norton Equivalents

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data

    http://img69.imageshack.us/img69/1973/captureyxs.png [Broken]

    Find the Thevenin and Norton equivalent of the circuit above between A and B using the node voltage method.

    2. Relevant equations



    3. The attempt at a solution

    I place the ground reference at the node with where the two 10 ohm resistors, 30 volt supply, 40 ohm resistor, and 1 amp source meet in the upper right hand corner.

    I'm confused by this problem. If I put a wire between A and B to find the short circuit current than the equivalent resistance between the 100 ohm resistor and the wire is zero ohms so I can just ignore this particular resistor. However I don't understand how any current is traveling though this part of the wire since it's just one giant node with one voltage so how can any current travel through it? Thanks for any help.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 8, 2013 #2

    ehild

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    Homework Helper
    Gold Member

    Currents flow through wires. The resistance of an ideal wire is zero, and multiplied by any current it gives zero voltage across. And you can make that wire short, even shrink it to a node, the same current flows through. But it is better to keep that wire between A and B and find the current through it.

    The internal resistance both for Norton-equivalent or Thevenin-equivalent sources can be got also as the resistance of the circuit between A and B by replacing all voltage and current sources by their internal resistance (zero for ideal voltage source and infinite for an ideal current source).

    ehild
     
  4. Feb 8, 2013 #3
    Ok so I built it in Circuit Lab to confirm that the node voltages I found where correct.

    http://img59.imageshack.us/img59/9217/capturegvro.png [Broken]

    Node 9 = -61 Volts
    Node 11 = 5 Volts
    Node 12 = 128.75 Volts
    Node 13 = -76.25 Volts

    So agrees with what I calculated.

    [itex]V_{Th} = V_{13} - V_{12} = -76.25 V - (-128.75 V) = 52.5 Volts[/itex]
    [itex]V_{Th}[/itex] has an orientation of
    A + - B
    Where node 13 or A is more positive than node 12 or B

    I than calculated [itex]R_{Th}[/itex] by the method below
    I think that the answer I get in this post is wrong for the Thevenin equivalent resistance because I forgot to disable a current source.
    http://img14.imageshack.us/img14/5950/captureldq.png [Broken]
    http://img15.imageshack.us/img15/2237/captureitw.png [Broken]
    http://img703.imageshack.us/img703/4967/captureuc.png [Broken]
    http://img820.imageshack.us/img820/7518/capturepvv.png [Broken]
    http://img651.imageshack.us/img651/774/capturezpmw.png [Broken]
    http://img401.imageshack.us/img401/3530/capturekva.png [Broken]

    As you can see I get [itex]R_{Th} = 76.19 Ω[/itex]. Below is what I get for the Thevenin and Norton equivalents of the circuit.

    http://img853.imageshack.us/img853/7579/capturedi.png [Broken]
    http://img266.imageshack.us/img266/6669/captureyus.png [Broken]

    As you can see I get [itex]I_{Th} = 689.1 mA[/itex]. I think however that I'm doing something wrong but I'm not sure what.
     
    Last edited by a moderator: May 6, 2017
  5. Feb 8, 2013 #4
    Oh I think I forgot to take out one current source.
    So I think it should be
    [itex]R_{Th} = 100 Ω[/itex]
    [itex]I_{N} = .525 A[/itex]
    [itex]V_{Th} = 52.5 V[/itex]

    I still think I'm doing something wrong though.
     
    Last edited: Feb 8, 2013
  6. Feb 8, 2013 #5

    gneill

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    Staff: Mentor

    Your Vth looks okay, but Rth is a problem. Check to make sure that you suppressed all the sources correctly.
     
  7. Feb 8, 2013 #6
    Shouldn't it just be the equivalent of this circuit?

    http://img845.imageshack.us/img845/6814/capturessle.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. Feb 8, 2013 #7

    gneill

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    Staff: Mentor

    Yup. That looks better. What's Rth then?
     
    Last edited by a moderator: May 6, 2017
  9. Feb 8, 2013 #8
    I still get 100. Am I supposed to include the resistor between nodes A and B as part of the equivalent resistance?
     
  10. Feb 8, 2013 #9

    gneill

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    Staff: Mentor

    Absolutely.
     
  11. Feb 8, 2013 #10
    So the answers I get for using the node voltage method is as follows:

    http://imageshack.us/photo/my-images/259/capturexdf.png/

    The picture won't load but it's the same voltage already found with a 200 ohm resistor.

    Now when I try to use source conversions I get a different answer. I'm really confused by source conversions and hate using this method. Any help would greatly appreciate me.

    http://img715.imageshack.us/img715/1055/capturehav.png [Broken]

    Here's my attempt

    http://img255.imageshack.us/img255/3316/capturexhh.png [Broken]

    http://img842.imageshack.us/img842/3083/captureobp.png [Broken]

    http://img19.imageshack.us/img19/4566/capturezu.png [Broken]

    [PLAIN]http://img208.imageshack.us/img208/3541/capturezop.png [Broken] [Broken]

    [PLAIN]http://img208.imageshack.us/img208/3541/capturezop.png [Broken] [Broken]

    http://img593.imageshack.us/img593/9273/captureezc.png [Broken]

    As you can see this answer is wrong as it dosen't match up.

    Thanks for any help!
     
    Last edited by a moderator: May 6, 2017
  12. Feb 8, 2013 #11

    gneill

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    I don't know how you got from the second diagram in your attempt to the third. Looks like some rules were badly bent out of shape... like adding current sources that are not in parallel.
     
  13. Feb 8, 2013 #12
    Well like what can i do from the second one? I know that I can add that one set of parallel resistors. on the left but I don't see what else I can go from there.
     
  14. Feb 8, 2013 #13

    gneill

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    Staff: Mentor

    Yes, definitely put those two resistors in parallel.

    Something helpful to know: A resistor in series with a current source has no effect -- an ideal current source will produce its specified current no matter what its load is. What does that mean in practical terms? Since it doesn't matter what values those resistors have they might as well be zero! You can eliminate those resistors entirely, replacing them with wires. That should get you going.
     
  15. Feb 8, 2013 #14
    http://img252.imageshack.us/img252/4772/captureopx.png [Broken]

    When I get down to here and covert the to a voltage source and a resistor in series I don't get the same answer =( is the above circuit above correct?
     
    Last edited by a moderator: May 6, 2017
  16. Feb 8, 2013 #15

    gneill

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    Staff: Mentor

    How did the 26.19Ω resistance evolve? It doesn't look right to me.
     
  17. Feb 8, 2013 #16
    Ok so it should be 50, this would give me the correct resistance, but that still dosen't give me equivalent voltage, it would give me 105 volts.
     
  18. Feb 8, 2013 #17

    gneill

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    Yes, but 105 Volts spread across two 100Ω resistances....
     
  19. Feb 8, 2013 #18
    Ya but isn't Rth 200 ohms and it's just two 100 ohm resistors in series with a 105 voltage source so I don't see how it gives me the correct answer it's still 105 volts across 200 ohms?

    I guess I'm just seeing this wrong. Each 100 ohm resistor experiences .525 amperes which isn't I Norton either.
     
  20. Feb 8, 2013 #19

    gneill

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    Staff: Mentor

    The two 100 Ω resistances are in parallel -- remember that you're to "look" at the circuit from the viewpoint of the A-B terminals. The Thevenin voltage is the voltage across that 100 Ω resistor sitting between the two terminals. The other 100 Ω resistance is in parallel with it from that perspective.
     
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