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Superposition, Moving Around CKT Elements

  1. May 31, 2013 #1
    Uxse28D.png



    2. V=IR, Voltage division, KCL, KVL



    3. When I short out the dependent current source by replacing it with an open, can I switch the location of the independent source and the 20 ohm resistor so I can do a voltage division to find the voltage across the 20ohm resistor which is then equal to Vo because they are in parallel?
     
    Last edited by a moderator: May 31, 2013
  2. jcsd
  3. May 31, 2013 #2

    The Electrician

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    Gold Member

    The conventional wisdom is that you can't deactivate dependent sources when doing superposition. See this for more info:

    http://users.ece.gatech.edu/mleach/papers/superpos.pdf

    But, if you did short the dependent current source in your circuit, the 20 ohm resistor would also be shorted, wouldn't it? So your voltage divider wouldn't have any output since one leg of it would be zero ohms.
     
  4. May 31, 2013 #3

    NascentOxygen

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    "short" ≠ "open"

    I'd retain the dependent current source where it is. Find the contribution of the voltage source to Vo. Then find the contribution of the current source to Vo.
     
  5. Jun 1, 2013 #4

    rude man

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    Solve the equations twice, once with the 3A source removed and once again with the 15V source set to zero volts (a wire in other words).

    In each case the dependent current source 2Ix must remain in place.
     
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