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Maximum Power Transfer problem

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data
    For the following circuit, find the values of RL and CL that maximize the power extracted from the generator.

    https://dl.dropboxusercontent.com/u/15390227/circuit.jpg [Broken]


    2. Relevant equations
    Z = R + jX
    XL = jωL
    XC = 1/jωC


    3. The attempt at a solution
    Immediately, the impedance of the source can be easily determined, as the source resistor and inductor are in series:
    ZS = 50 + j47.879 Ω

    Normally, if the load's resistor and capacitor were in series, this could easily be solved by setting the load impedance equal to the complex conjugate of the source impedance:
    ZL = 50 - j47.879 Ω

    However, I can't do this since the load's resistor and capacitor are in parallel, not in series. I could determine the equivalent impedance of the load with this formula:
    ZL = XCRL / XC + RL

    Once I have these two equations for the load impedance, I'm kind of stumped on what to do next. I think I'm on the right track, but I've hit a brick wall. Any help would be greatly appreciated!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 25, 2014 #2

    gneill

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    Staff: Mentor

    Have you considered the maximum power transfer theorem from the point of view of admittance rather than impedance?
     
  4. Mar 25, 2014 #3
    Thanks for replying. Well, I hadn't considered trying that before. Mainly because I'm more familiar with impedance.

    So going down that route, the admittance formula is:

    Y = (R - jX) / (R2 + X2)

    After plugging in the given resistance and inductance values, I can determine the admittance of the source:

    YS = 0.0104332 - j0.00999

    The complex conjugate gives us the admittance of the load:

    YL = 0.0104332 + j0.00999

    Since the admittance is the inverse of the impedance, I also have this formula:

    YL = XC + RL / XCRL

    There must be something else I'm not understanding, because this just leaves me in the same position I was when I tried to solve using the impedance. I can't use the complex conjugate method because the capacitor and load resistor are in parallel instead of series, and I'm not sure how I can use the second formula to solve the problem.
     
  5. Mar 25, 2014 #4

    gneill

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    Staff: Mentor

    Conductances or admittances in parallel simply add. What are the two expressions for the admittances of the load components?
     
  6. Mar 25, 2014 #5
    For the capacitor: Ycapacitor = jωC (since Y = 1/Z)
    For the resistor: Yresistor = 1/R

    Is this what you were looking for?
     
  7. Mar 25, 2014 #6

    gneill

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    Staff: Mentor

    Yup. So what's the net admittance of the load?
     
  8. Mar 25, 2014 #7
    The net admittance of the load would be:
    YL = 1/RL + jωC

    Or, since we know ω:
    YL = 1/RL + j377C
     
  9. Mar 25, 2014 #8

    gneill

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    Staff: Mentor

    Okay, now what might you do with that? (Max power transfer theorem....)
     
  10. Mar 25, 2014 #9
    I'm not quite sure I'm following. I know that to achieve maximum power transfer, we need to set the impedance of the load to the complex conjugate of the source (or vice versa). I have the admittance for the load:

    YL = 1/RL + j377C
    ...whose complex conjugate is YS = 1/RL - j377C

    I also have the impedance for the source:
    ZS = 50 + j47.879
    ...whose complex conjugate is: ZL = 50 - j47.879

    I could set the load admittance equal to the inverse of the load impedance, but I'm not sure what I could solve with that, as I would still have multiple unknowns (the capacitance and the load resistance).
     
  11. Mar 25, 2014 #10

    gneill

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    Staff: Mentor

    You have a numerical value for the impedance of the source. The corresponding admittance is trivially obtained...
     
  12. Mar 25, 2014 #11
    YS = 0.01 - j0.01
    or: YL = 0.01 + j0.01

    So I have two equations for the admittance of the load:
    YL = 0.01 + j0.01
    YL = 1/RL + j377C

    If I set these equal to each other, though, I still have two unknowns.

    Wait: could I set 0.01 = 1/RL and 0.01 = 377C?
     
  13. Mar 25, 2014 #12

    gneill

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    Staff: Mentor

    Yup. One term is real and the other imaginary. They do not mix, so they represent separate "equations". Simply equate the like terms as you did.

    By the way, you might want to hang onto some extra digits in intermediate steps so that truncation error doesn't bite you along the way :smile:
     
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