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Thevenin/Norton Problem

  1. Jun 25, 2013 #1
    1. The problem statement, all variables and given/known data
    2013-06-25215856_zps5b4eb0e5.jpg


    2. Relevant equations
    V = IR
    I1 + I2 .... + In = 0 for a node (KCL)
    V1 + V2 .... + Vn = 0 for a loop (KVL)

    3. The attempt at a solution
    I've really been struggling with this problem and Thevenin/Norton problems in general. I just can't seem to perform the proper circuit analysis. I know that I can set IA = 0 and VAB = VVoc in order to find Thevenin values, but I just keep getting stuck. I tried to do KCL on the top node, which got me -kvx - (vx/50Ω) + ((vs - vx)/200Ω), but that appears to be getting me nowhere. I've been trying to find a relation between vx and vAB so I can substitute in, but to no avail. I'd really appreciate a nudge in the right direction here as well as general advice, as I'm pretty lost.
     
  2. jcsd
  3. Jun 25, 2013 #2

    gneill

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    Staff: Mentor

    Are you familiar with nodal analysis? If so, use it to find the potential at the top of the controlled current source when the output is open-circuited. That'll be your Thevenin voltage. Show your work so we can see how you're doing.
     
  4. Jun 25, 2013 #3
    -vx/50Ω + vAB/200Ω -kvx = 0

    vAB/200Ω = vx/50Ω + kvx

    vAB = 4*vx + 4*kvx

    additionally,

    vAB - vs + vx = 0

    vx = vs - vAB

    so

    vAB = 4*(vs - vAB) + 4*k*(vs - vAB)

    vAB = 4*vs - 4*vAB + 4*0.025*vs - 4*0.025*vAB

    5.1*vAB = 4.1*vs

    vAB = 16.078... V

    I don't think that's right. Things aren't adding up when I plug it back in. Where'd I go wrong?
     
  5. Jun 25, 2013 #4

    gneill

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    Staff: Mentor

    Okay, for starters don't treat Vx as "known" value for your node equation, except where it is referenced on controlled sources. Vx is a variable that you'll find an expression for based upon fixed circuit parameters. Write your node equation for the 50 Ω branch as though Vx wasn't labeled.

    What's the expression for the current in the 50 Ω branch supposing that the node voltage is VAB? Given that current flowing through the 50Ω resistor, what's an expression to replace Vx?
     
  6. Jun 25, 2013 #5
    Isn't that what I did? Can't vx be written as -(vAB - vs) or vs - vAB if we're treating it as the voltage difference between the nodal voltage and vs?
     
  7. Jun 25, 2013 #6

    gneill

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    Staff: Mentor

    Okay, you're right, you can do it that way. My own preference is to write the node equations ignoring the "sampled" values except for their use in the controlled sources, then later replace those instances with expressions derived from the circuit.

    Looking at your node equation manipulations:

    -vx/50Ω + vAB/200Ω -kvx = 0

    vAB/200Ω = vx/50Ω + kvx

    vAB = 4*vx + 4*kvx

    check the coefficient for the last term on the right hand side.
     
  8. Jun 25, 2013 #7
    Well that was stupid of me. Figures that I thought to check my circuit before I thought to check my actual math, lol. VAB = Voc = 18 V.

    So now I need either Rth or Isc. If I do the same nodal analysis on the top node but this time include a -IA, does that make my VAB = 0? It would be a short circuit, correct?
     
  9. Jun 26, 2013 #8

    gneill

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    Staff: Mentor

    Your open-circuit voltage looks fine.

    Yes, you can repeat the analysis with the output shorted to determine the Norton current. Just find the node voltage and divide by the 40Ω "load" to find the current in that branch.

    Note that for part (b) you're going to need a version of the solution for the Thevenin voltage that leaves k as a variable...
     
  10. Jun 26, 2013 #9
    Got it. Thanks so much for your help- it's much appreciated.
     
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