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KCL / KVL to determine Vo

  1. Jan 27, 2005 #1
    Quick question that is throwing me off....

    If you have a dependant current source in a parallel circuit, how do you represent that in your KCL equation? The one in this problem is Vo/2000.

    I took that as I=V/R so R=2000. Since it is in a parallel circuit on its own 'branch' I used Vs/2000. But when I plug it all into the equations, the answer is wrong.

    The circuit is this....

    Parallel circuit with 4 branches. The following are the branches from left to right.

    1. Dep. Current Source = Vo/2000
    2. 6k ohm resister with + on top. (Marked also with Vs).
    3. 1k ohm resister and a 2k ohm resister. Vo is indicated across the 2k ohm. The + sign is between the two resisters and the - is at the bottom of the 2k ohm.
    4. indep current source at 2mA.

    I took Vs to be the voltage across each branch since they are all in parallel and it is stated that Vs is across the 6k ohm resister. I tried to do the....

    Vs/2000 + Vs/6k + Vs/3k + 2mA = 0

    and then solved for Vs which ended up being 2V. Then used that to find Vo... except it is way off as the answer says Vo is 8V.

    So, I am either screwing up the representation of the Vo/2000 in the equation or I am screwing up the entire KCL for the equation.

    If someone could give a hint, Id appreciate it.
     
  2. jcsd
  3. Jan 27, 2005 #2
    Your mistake was the V/R thing for branch 1.
    Branch one is an dependant current source whose value is Vo/2000. No ohm's law trickery is necessary.
    Therefore KCL at node Vs is:
    Vo/2000-Vs/6000-Vs/3000+2/1000=0 (using your sign convention)
    You can relate Vs to Vo by noting branch 3 is a voltage divider.
    Therefore:
    Vo=2Vs/3
    I am sure you have it from here...
     
  4. Jan 27, 2005 #3
    Yes, that did the trick.

    Thanks!
     
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