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Thevenin Equivalent Problem

  1. Dec 11, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-12-11_15-51-25.png

    2. Relevant equations


    3. The attempt at a solution
    First I found the equivalent resistance since the 3 ohm resistor and 6 ohm resistor are in series. 9ohm. Then I did chose the node where the arrow points to as Voc, which is also va.

    KVL:
    (va+24V)/9ohm=1.33va
    I get va to be 2.18

    Then looking for Isc, I close the output terminals and then do kcl again on the node on top of the voltage controlled current source

    1.33va-isc-24V/9ohm=0

    and get isc=0.23

    so Rt=2.18/0.23=9.47

    The book says

    isc=.24 and Rt=-3

    I don't know what I'm doing wrong.
     
  2. jcsd
  3. Dec 11, 2017 #2

    Drakkith

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    That doesn't look right to me. From your picture, it looks like ##V_a## is between the resistors, not at the node above the dependent source.
     
  4. Dec 11, 2017 #3
    Yikes. ok so if i apply kcl on va i between the resistors i'd have:

    (va+24)/3ohm + (voc-va)/6ohm=1.33va

    since i have 2 variables i need 2 eq and doing kcl on voc:

    (voc-va)/6ohm -24V/9ohm= 1.33va

    I get that va is 2.29 and voc is 36.58
     
  5. Dec 11, 2017 #4

    NascentOxygen

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    Staff: Mentor

    It helps if you sound out what your equations are saying. For instance, your first eqn says that the current through the 3Ω plus the current through the 6Ω is equal to the current from the dependent source. Is that what you intend?

    direction is consistently clockwise
     
  6. Dec 12, 2017 #5

    scottdave

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    Like @NascentOxygen said, look at how many different currents you should have in that one loop.
     
  7. Dec 15, 2017 #6

    gneill

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    Staff: Mentor

    Beware! When you have controlled sources like the ##1.33 v_a## source in the network, the effective Thevenin resistance will very likely not be simply a function of the network resistors! The controlled source will actively alter the V vs I characteristics that are "seen" from the output terminals. You cannot simply suppress controlled sources to find the Thevenin resistance. You need to take its effects into account.

    A typical way to approach such problems is to attach a "test" source to the output and then analyze the circuit to find the current and voltage at the output, and hence by Ohm's law, the resistance. There are other methods, too. One that I favor involves assuming that there is some load resistance ##R_L## connected to the output, then using nodal analysis to find the voltage across that load resistor. This will yield an expression involving ##R_L##. Since a Thevenin source driving a load resistance is a classic voltage divider circuit, the expression that you find for the output voltage can be manipulated to the form of the standard voltage divider equation, and the Thevenin voltage and Thevenin resistance can be picked out of that equation by inspection :wink:
     
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