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Thevenin equivalent, (superposition?)

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data

    http://imageshack.us/a/img14/2387/homeworktest2prob2.jpg [Broken]

    a. Find the Thevenin equivalent circuit with respect to terminals a, B for the circuit.

    b. Find also the Norton equivalent circuit with respect to the terminals A, B, for the circuit



    2. Relevant equations

    V = IR

    current division, voltage division,

    KCL, KVL


    Thevenin procedure, Norton procedure.

    3. The attempt at a solution


    With the load cut out, 4Ω disappears and

    25V/5Ω + 3A = Vb/20Ω

    100V + 60V = Vb

    Vb = 160V, VTh = 160V


    I need help mainly with getting RTh


    I don't know if I can just simplify the resistors from A to B with that current source (do I cut it out so it becomes an open circuit? I know you put a wire to short the 25V while doing it though).

    Trying RTh = VTh/Ino instead, then a wire goes from a to b, then once I do that I try to use superposition (so might need help here with that too:)

    With the 3A cut out the equivalent R is 8.3Ω with 25V so I' = 3A

    then with the 25V silenced the current is just I'' = 3A, right? Because that's the total current in?

    So 3A + 3A = 6A

    Ino = 6A


    So for RTh I get 160V / 26.6A

    = 6Ω


    But I don't think it's correct.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 25, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Your node equation is not correct. The potential difference across the 5Ω resistor is not 25V. What potentials are at either end of this resistor?

    I find that the simplest way to write a node equation is to assume that either all currents flow into the node or all currents flow out. Then all the terms can be written on one side of the equation and their sum must equal zero.

    Since there are no dependent supplies, you can suppress all the supplies and calculate the net resistance of the network as seen from terminals a-b.
     
  4. Feb 25, 2013 #3
    Ahh thanks, my bad, the nodal equation should be:

    (25V - Vb)/5 + 3A = Vb/20Ω

    So Vb = VTh = 32V then.


    For the second part do you mean that by suppressing everything that I can do that with the current source too (but only to find RTh)? Would it then just be 4Ω?
     
  5. Feb 25, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    Yes :approve:
    Yes. Suppress all sources. Remember that current sources aren't suppressed in the same way that voltage sources are...
     
  6. Feb 25, 2013 #5
    Ah ok, so it's 8Ω then.


    Thanks again.
     
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