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Homework Help: 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.

    Attached Files:

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


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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


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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?

  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


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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


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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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