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Norton circuit

  1. May 21, 2010 #1
    Hello,

    I have the following circuit:

    [PLAIN]http://img15.imageshack.us/img15/6981/nortonm.png [Broken]

    I need to determine the Norton equivalent for this circuit.
    I wonder whether the easiest way to solve this is by zeroing the independent source, connecting a test voltage source and calculate the equivalent resistance.

    The dependent source confuses me somewhat, being positioned in parallel with the resistance.

    Any ideas?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 21, 2010 #2
    It confuses me too. What does the diamond represent? Is it a current source, or something else?
     
    Last edited: May 21, 2010
  4. May 21, 2010 #3
    Oh, forgot to write it out.

    It's a current source.
     
  5. May 21, 2010 #4
    Convert the current source and its parallel 2 ohm resistor into a voltage source in series with a 2 ohm resistor. That should remove the confusion.
     
  6. May 21, 2010 #5

    CEL

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    Where is V in your circuit?
     
  7. May 21, 2010 #6
    V is the voltage across the 3ohm resistor.
     
  8. May 21, 2010 #7

    vela

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    The potential difference from bottom to top or top to bottom?

    Also, the units of K don't make sense. An ampere is equal to a volt per ohm, not a volt-ohm.
     
  9. May 21, 2010 #8
    I think K is the transconductance: amps/volt.
     
  10. May 22, 2010 #9
    The drawing a made seem to have caused a bit of confusion, since i forgot to put all the details in:redface:

    Anyway, here is how it is supposed to look:
    [PLAIN]http://img709.imageshack.us/img709/9333/nortony.png [Broken]
     
    Last edited by a moderator: May 4, 2017
  11. May 22, 2010 #10

    vela

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    I suggest you use a 1-A test current source instead of a 1-V test voltage source.
     
  12. May 22, 2010 #11
    Ok, so zero the independent source, connect a test current across the terminals and find the voltage across the 3 ohm resistor?
     
  13. May 22, 2010 #12

    vela

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    Yeah, that's how you'd start. Then once you know Vx, you can calculate the current through the 2-ohm resistor, and finally calculate the total voltage drop across the test source.
     
  14. Dec 8, 2010 #13
    So,

    If you imagine the voltage source as a short, and a 1A current source applied to the two terminals.

    You'd get

    [tex]\frac{6}{6+3} \cdot 1 = 0.667A[/tex]

    [tex]0.667 \cdot 3 = 2[/tex]

    So [tex]V_x = 20 V [/tex]

    2 + 18(from the Voltage source) ?
     
  15. Dec 8, 2010 #14

    vela

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    You removed the 18-V source and replaced it with a short. It's not in the circuit anymore, so Vx is simply 2 V. (And you wouldn't just arbitrarily add 18 V in anyway, even if the source was still there.)
     
  16. Dec 9, 2010 #15
    Why not?, 18V appear across the 3 ohm resistor ?
     
  17. Dec 9, 2010 #16

    vela

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    1. You removed the 18-V source from the circuit.
    2. Even if it were still in the circult, the entire 18 V wouldn't appear across just that one element.
     
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