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Voltage Thevenin and norton

  1. Mar 17, 2014 #1
    1. The problem statement, all variables and given/known data

    find the voltage by using norton and thevenin equation
    232323232fp83232_uqcshlukaxroqdfv4938=ot_3368=428=;25=XROQDF_2646882988255ot1lsi.jpg
    is this method correct?
     
    Last edited: Mar 18, 2014
  2. jcsd
  3. Mar 18, 2014 #2
    if I change the pararelled resistor and send the most right resistor to the left, the voltage changed
    is it right if im doing it like this?
    thx for reply in advance
    sory for bad english
     
  4. Mar 18, 2014 #3
    Hey there!

    You might notice that the voltage over your last resistor (on the right) will be the Voltage across your two nodes. So, if you were to find the voltage over that resistor, that will be your Thevenin voltage.

    For your Thevenin resistance, try to deactivate all of your independent devices and starting at point A on your node, calculate the equivalent resistance from that point. That will be your Thevenin resistance.

    From there you know the Norton circuit is a source transformation. Good luck!
     
  5. Mar 18, 2014 #4
    thx for the advice
    after I deactivate, I found the resistor from paralleled 10k, 6k and 8k
    then the ampere source have the resistor (4k) below it paralleled with volt source and the resistor ( from paralleled 10k, 6k and 8k).
    after that I'm stuck.
    any idea how to go on?
     
  6. Mar 18, 2014 #5

    gneill

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    Staff: Mentor

    Hi darwinharianto, Welcome to Physics Forums.

    Some of the component values in your attached figure are hard to make out. Can you clarify their values?
     
  7. Mar 18, 2014 #6
    It looks like you might benefit from using source transformations to make both of your sources the same, if you want to use this method instead. I see a voltage source in series with a resistor right off the bat. Also, if you have a resistor in series with a current source, what can you say about the current in that section of wire?

    From there, it seems like it would only be current division and Ohm's law to find the voltage over your resistor on the right.
     
  8. Mar 18, 2014 #7

    NascentOxygen

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    Staff: Mentor

    Hi darwinharianto! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    You converted the voltage source to its Norton equivalent. Why not also convert the left-most source and resistors to its Norton equivalent? You'll then have two current sources in parallel .....

    .... the rest is easy! :smile:
     
    Last edited by a moderator: May 6, 2017
  9. Mar 18, 2014 #8
    actually the left one is not a volt source, its an ampere source (1mA)
    which is trouble me
    hope this pic will be giving a better view
    Picture 002.jpg
     
    Last edited by a moderator: May 6, 2017
  10. Mar 18, 2014 #9

    NascentOxygen

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    Staff: Mentor

    I know it's a current source. You can still represent it by an ideal current source in parallel with a resistance.
     
  11. Mar 18, 2014 #10
    how is that?
    so the current source is multiplied? I got confused
     
  12. Mar 18, 2014 #11

    NascentOxygen

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    Staff: Mentor

    What is the procedure for representing any source by a Norton (or Thévenin) equivalent? Follow that.
     
  13. Mar 18, 2014 #12
    okiee
    ill figure something out
     
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