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Finding Equivalent Resistance in a Circuit

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
    The problem wants me to find the equivalent resistance at the terminals. Here is the circuit plus some of my work. The Answer according to the teacher is C.
    http://i.imgur.com/JaW7f.jpg


    2. Relevant equations



    3. The attempt at a solution
    As seen from my work, I added a test source. By doing this I'm trying to solve for I norton so i can divide the test source by I norton for the resistance. I start doing node voltage analysis at the node closest to the test voltage. From the calculations I find the voltage of the node farthest of the voltage source to be 2 volts. I then use node voltage at that node to solve for i0. I found i0 to be 3/4A however after that i'm completely lost.
     
  2. jcsd
  3. Oct 12, 2011 #2

    gneill

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    Staff: Mentor

    Surely if you put a voltage supply v across ab then the current io will be v/1Ω.
     
  4. Oct 12, 2011 #3
    But isn't there a voltage at the bottom too? so it i0=v1/2i0?
     
  5. Oct 12, 2011 #4

    gneill

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    Staff: Mentor

    If you place a voltage source across a-b, that voltage and no other will appear across the 1 Ohm resistor at the input terminals.
     
  6. Oct 12, 2011 #5
    if it was that simple then why isn't the answer just 1? I'm really confused on why you disregard the voltage at the bottom. Underneath the 1ohm resister is not ground but another voltage which is 2i0.
     
  7. Oct 12, 2011 #6

    gneill

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    Staff: Mentor

    io is the current through that "input" resistor, and is so defined by the circuit diagram. Call it R. You place a voltage source V across that resistor R and the resulting current through that resistor is V/R. It doesn't matter what the potential at b is because node a is V volts greater than that due to that voltage source. If V is an ideal voltage source, NOTHING can alter that.

    Note that io is not the total current being "pushed" into the circuit by V. There are still calculations to do to find that!
     
  8. Oct 12, 2011 #7
    what would be your suggestion to go from here then? Do i not use a test source? because i really don't know any other way to solve it then.
     
  9. Oct 12, 2011 #8

    gneill

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    Staff: Mentor

    Yes, use a test source (I thought that was what the recent exchange concerned!).

    If you place a test voltage source Vin across the input a-b, then you immediately know the current io. This gives you the controlled source values. From there look at writing KCL node equations. Start with the node that the current source flows into.
     
  10. Oct 13, 2011 #9
    WOW. i'm SOOO STUPID. I did that exactly and i got it. Thanks that problem even stumped my professor in class. Thank you soooo much
     
  11. Oct 13, 2011 #10

    gneill

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    Staff: Mentor

    We aim to please :biggrin:
     
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