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Electrical Circuit Question - Maximum Power Dispated in Resistor

  1. Feb 21, 2013 #1
    1. The problem statement, all variables and given/known data

    http://img855.imageshack.us/img855/6951/captureaww.png [Broken]

    2. Relevant equations



    3. The attempt at a solution

    Is one allowed to find [itex]R_{Th}[/itex] by simply short circuiting all voltage source and opening all current sources and finding the equivalent resistance while keeping [itex]R_{L}[/itex] in place and treating it's ends as stationary nodes? I know you can do this with independent sources but I'm not really sure about dependent sources. My professor said that he wasn't really sure.

    When I use this method to find [itex]R_{Th}[/itex] I get two ohms.

    However I don't know if this is accurate as it's not what I get when I use other methods to find [itex]R_{Th}[/itex].

    So is this a valid solution to this problem? If it is and [itex]R_{Th}[/itex] is really two than I'm greatly concerned because I have tried solving this problem two other ways and don't get this for my answer. The two other methods I used to solve the problem give me the same answer as well. If it is than I'll post my other solutions if it is necessary.

    Thanks for any help.
     

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    Last edited by a moderator: May 6, 2017
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  3. Feb 22, 2013 #2

    vela

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    No, you have to divide the open-circuit voltage by the short-circuit current when there's a dependent source.
     
  4. Feb 22, 2013 #3
    I get half a ohm for R th when I don't use this method, does this sound correct?
     
    Last edited: Feb 22, 2013
  5. Feb 22, 2013 #4

    ehild

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    I got the same for the resistance with the highest power, which have to be the same as the Th resistance. What method did you use?


    ehild
     
  6. Feb 22, 2013 #5
    You mean you got .5 ohm?

    First Method
    I found open circuit voltage of .25 V
    short circuit current of 2 A
    R th .5 ohm
    Max power in Rl to be 1/32 W

    Second Method
    Considered Rl to be a variable. Found voltage across Rl as a function of Rl. Used power equation as a function of Rl, V/Rl. Took derivative set equal to zero and got
    RL = .5
    I plugged back into my equation for power in terms of Rl and got
    1/32 W

    I have spent several hours on this problem because I was under the assumption I could just short voltage sources and open the dependent current source and solve for Rth
    I got 2 ohm when I did this
    I then solved for the voltage across this resistor (I forget what I got I erased this)
    and then plugged into the power equation
    I was getting a different answer from the other two methods so I questioned if I could actually do this method or not.

    Does the answers I got for the first two sound right?

    Thanks for all your help!
     
  7. Feb 22, 2013 #6

    ehild

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    Yes, RL=0.5 ohm. The second method is correct for sure.
    I do not see yet how you got the open circuit voltage and short circuit current.
     
  8. Feb 22, 2013 #7

    vela

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    I think you meant you got 1/2 A for the short-circuit current, not 2 A. I got the same results you did then.
     
  9. Feb 22, 2013 #8
    Oh ya thanks a lot for all the help guys. The short circuit current should be .5 A and for the open circuit voltage I still get .25

    .25/.5 = .5 ohms

    I'm really glad I struggled through this and learned that I can't open circuit dependent current sources or short dependent voltage sources.

    Thanks again guys.
     
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