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Norton's Theorem

  1. Jun 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Find Io using Norton's Theorem

    2. Relevant equations
    Screenshot - 06112015 - 11:09:31 AM.png

    3. The attempt at a solution
    I broke it into the following circuit:
    Snapshot.jpg
    I would like to break it down some more but if I open the node between 3k and 6k (ohms), and 6k and 6k (ohms), I run into problems....

    What should I do to make sure I stick to Norton's Theor
     

    Attached Files:

  2. jcsd
  3. Jun 11, 2015 #2

    gneill

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    Staff: Mentor

    I don't think you're restricted to using Norton's Theorem exclusively. It should be sufficient to determine the Norton equivalent of the network driving the current through the 6 k "load" using whatever means you wish. You might begin by suppressing all the sources and determining the Norton resistance.

    By the way, 2 k + 3 k is not 6 k :wink:
     
  4. Jun 11, 2015 #3

    FOIWATER

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    Gold Member

    I just took a while and solved the problem by hand using mesh analysis with a supermesh and then put the circuit into QUCS and verified my answer.

    I am having a hard time seeing how it would be easily done with norton's theorem or thevenin's theorem. I got a value for the equivalent resistance seen by the load resistor but breaking down the circuit to find the voltage across the terminals of the load resistor would lead me back to using mesh.

    supermesh.png
     
    Last edited: Jun 11, 2015
  5. Jun 11, 2015 #4

    gneill

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    Staff: Mentor

    I'd go for a Thevenin approach and Nodal Analysis. Finding the Thevenin resistance (same as Norton resistance) is pretty easy with the sources suppressed. Removing the 6k load to open-circuit the output (For Thevenin you want the open-circuit voltage) leaves a single essential node. The rest is just finding particular potentials along branches.
     
  6. Jun 11, 2015 #5

    FOIWATER

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    Gold Member

    Yeah I agree that would be another way. converting the thevenin equivalent into a norton equivalent. The equivalent resistance is easy to find

    supermesh.png

    The difference in voltages Vx and Vy is the thevenin equivalent voltage
     
  7. Jun 12, 2015 #6
    At first I didn't know what you meant by "driving the current through the 6 k "load" using whatever means you wish." but with FOIWATER's picture:

    supermesh-png.84725.png

    I am able to understand now. Thank you! I will try my best to solve this using the techniques you suggested!
     
  8. Jun 12, 2015 #7
    I was able to find the Voltage between the open circuit using Thevenin and Node analysis: I got Vth = ~10.8 V (making Vy facing positive and Vx near negative), and Rth = 33/5. So the short circuit current is 10.8/6.6=1.6464 and current division between the load and 6.6 KOhms makes the load current 0.857. This hsould be negative because of the way the current is facing and the way I had my voltages set up. Thanks everyone!

    (Btw FOIWATER, your R6 resister should be 3k)
     
  9. Jun 12, 2015 #8

    FOIWATER

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    Gold Member

    Glad you figured it out friend.
     
  10. Jun 12, 2015 #9
    Follow up question on using QUCS:
    I'm trying to make a current controlled voltage source matching this diagram...:


    Screenshot - 06122015 - 01:09:17 PM.png
    Im not sure how exactly the current controlled voltage source is used in qucs.. Examples help, but I didnt know the right name for it in the examples given in qucs

    Screenshot - 06122015 - 01:10:27 PM.png
    Code (Text):

    <Qucs Schematic 0.0.18>
    <Properties>
      <View=-50,-814,893,-265,1,287,0>
      <Grid=10,10,1>
      <DataSet=5.11.dat>
      <DataDisplay=5.11.dpl>
      <OpenDisplay=1>
      <Script=5.11.m>
      <RunScript=0>
      <showFrame=0>
      <FrameText0=Title>
      <FrameText1=Drawn By:>
      <FrameText2=Date:>
      <FrameText3=Revision:>
    </Properties>
    <Symbol>
      <.ID -20 -16 SUB>
      <Line -20 20 40 0 #000080 2 1>
      <Line 20 20 0 -40 #000080 2 1>
      <Line -20 -20 40 0 #000080 2 1>
      <Line -20 20 0 -40 #000080 2 1>
    </Symbol>
    <Components>
      <R R1 1 230 -450 15 -26 0 1 "3000 Ohm" 1 "26.85" 0 "0.0" 0 "0.0" 0 "26.85" 0 "US" 0>
      <R R2 1 440 -450 15 -26 0 1 "6000 Ohm" 1 "26.85" 0 "0.0" 0 "0.0" 0 "26.85" 0 "US" 0>
      <R R4 1 530 -630 -26 15 0 0 "2000 Ohm" 1 "26.85" 0 "0.0" 0 "0.0" 0 "26.85" 0 "US" 0>
      <Idc I1 1 80 -450 18 -26 0 1 "8 mA" 1>
      <IProbe Pr1 1 600 -430 16 -26 1 3>
      <GND * 1 430 -310 0 0 0 0>
      <CCVS SRC1 1 340 -620 34 -26 1 3 "4000" 1 "4000" 0>
      <.DC DC1 1 40 -540 0 43 0 0 "26.85" 0 "0.001" 0 "1 pA" 0 "1 uV" 0 "no" 0 "150" 0 "no" 0 "none" 0 "CroutLU" 0>
      <Idc I2 1 420 -780 -26 18 0 0 "2 mA" 1>
    </Components>
    <Wires>
      <440 -580 500 -580 "" 0 0 0 "">
      <230 -310 430 -310 "" 0 0 0 "">
      <230 -590 230 -480 "" 0 0 0 "">
      <560 -620 600 -620 "" 0 0 0 "">
      <230 -420 230 -310 "" 0 0 0 "">
      <440 -580 440 -480 "" 0 0 0 "">
      <440 -420 440 -310 "" 0 0 0 "">
      <500 -630 500 -580 "" 0 0 0 "">
      <560 -630 560 -620 "" 0 0 0 "">
      <80 -310 230 -310 "" 0 0 0 "">
      <80 -420 80 -310 "" 0 0 0 "">
      <80 -480 230 -480 "" 0 0 0 "">
      <230 -590 310 -590 "" 0 0 0 "">
      <440 -310 600 -310 "" 0 0 0 "">
      <600 -400 600 -310 "" 0 0 0 "">
      <600 -620 600 -460 "" 0 0 0 "">
      <430 -310 440 -310 "" 0 0 0 "">
      <370 -590 370 -580 "" 0 0 0 "">
      <370 -580 440 -580 "" 0 0 0 "">
      <370 -670 370 -650 "" 0 0 0 "">
      <370 -670 560 -670 "" 0 0 0 "">
      <560 -670 560 -630 "" 0 0 0 "">
      <310 -660 310 -650 "" 0 0 0 "">
      <310 -660 500 -660 "" 0 0 0 "">
      <500 -660 500 -630 "" 0 0 0 "">
      <230 -780 230 -590 "" 0 0 0 "">
      <230 -780 390 -780 "" 0 0 0 "">
      <600 -780 600 -620 "" 0 0 0 "">
      <450 -780 600 -780 "" 0 0 0 "">
    </Wires>
    <Diagrams>
    </Diagrams>
    <Paintings>
    </Paintings>

     
     
  11. Jun 12, 2015 #10

    gneill

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    Staff: Mentor

    R4 is shorted by your wiring of the controlled current source "sense line". Note that there is a path from the top of R5 to the top of R2 that does not pass through R4!
    Fig1.gif
     
  12. Jun 14, 2015 #11
    Thanks again!
     
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