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Homework Help: Maximum power

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine, for the network shown in fig 4 the value of the load
    Impedance that will dissipate the maximum power and the value of
    this power.


    2. Relevant equations
    Using Thevenin theorem to simplify the circuit then,I believe maximum power transfer theorems apply.

    P=I2R
    R=√(R2+x2)
    I=E/Zt
    Zt=z+R

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    But E is in Hz not in Volts?Do you think it could be a printing error?


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Nov 3, 2013 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    hi shaltera! welcome to pf!! :smile:
    no

    the impedance of a circuit (or of a component) does not depend on the voltage (the emf)

    but it does depend on the frequency, ω, doesn't it? :wink:
     
  4. Nov 3, 2013 #3
    Thevenin Impedance

    Thank you for your reply.I calculated Thevenin impedance.Calculation below:

    R1=220Ω
    R2=1KΩ
    R3=150Ω

    (R1xR2)/(R1+R2)=(220ΩX1000Ω)/(220Ω+1000Ω)=180.32Ω

    Resistance of L1
    Zl=2πfL=2xπ50x700x10-3=220Ω (round)

    R3+RL1=150Ω+220Ω=370Ω

    ZTH=180Ω+370Ω=550Ω

    And then?
     
  5. Nov 3, 2013 #4

    gneill

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    Staff: Mentor

    Careful with the inductor. The inductor has impedance, an imaginary value, not a real values resistance. ZL = jωL.
     
  6. Nov 3, 2013 #5
    How ω can be calculated?There is no ω?
     
  7. Nov 3, 2013 #6

    gneill

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    Staff: Mentor

    ω is the angular frequency. ##\omega = 2\pi f##.
     
  8. Nov 3, 2013 #7
    Thank you
     
  9. Nov 3, 2013 #8
    ZL = jωL=j2πfL=j220
    R3+RL1=150+j220
    ZTH=180+150+j220=330+j220
     
  10. Nov 3, 2013 #9
    OK.I calculated Zth what should I do next?Thank you
     
  11. Nov 3, 2013 #10

    gneill

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    Staff: Mentor

    Looks good. Don't forget that the units of the impedance is still Ohms, just like resistance.
     
  12. Nov 3, 2013 #11
    Sorry I forgot to add Ω at the end

    330+j220Ω
     
  13. Nov 3, 2013 #12

    gneill

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    Staff: Mentor

    Okay! So now you either apply the Maximum Power Transfer Theorem, or you try to derive it from scratch. It's simple for real-valued resistances, but it gets a bit "mathy" for complex impedance values. The result is very simple, but getting there is tedious! You might want to do a little investigation on the web to see how its done.
     
  14. Nov 3, 2013 #13
    You mean:

    ZL=ZTH
    Pmax=|VTH|2/8RTH

    Thanks
     
  15. Nov 3, 2013 #14

    gneill

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    Staff: Mentor

    Nope ZL is not Zth for maximum power transfer! That's the tricky bit. It's very closely (amazingly, really) related to Zth though...

    Do a web search on "Maximum Power Transfer Theorem".
     
  16. Nov 3, 2013 #15
    Thanks again.I keep looking and looking.All examples have a voltage source.
     
  17. Nov 3, 2013 #16

    gneill

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    Staff: Mentor

    A voltage source and a series resistance or impedance. In other words, a Thevenin model...

    Check out the Wikipedia entry on the MPTT. It has a section devoted to impedance.
     
  18. Nov 3, 2013 #17
    Searched in Wiki on the MPTT,weird results.I'm fed up now,spent too much time on this silly problem.Thanks anyway
     
  19. Nov 3, 2013 #18

    gneill

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    Staff: Mentor

  20. Nov 3, 2013 #19
    Yes I did, but examples are with a voltage source.I'm sure you know what you are talking about,its probably me.I really wanted to do it myself but its getting ridiculous and I have to move to Plan B.
     
    Last edited: Nov 3, 2013
  21. Nov 3, 2013 #20
    I started college 4 weeks ago, and they want me to solve this problem,ridiculous. That's why people noways do not remember simple things.Colleges,Uni give so much information for short time.Electrical and Electronic Theory in some countries is divided in 3 part (12 months), in UK in 7 weeks.
     
    Last edited: Nov 3, 2013
  22. Nov 3, 2013 #21

    gneill

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    Staff: Mentor

    Your circuit also has a voltage source. It's an AC voltage source, only they haven't specified a particular voltage; they've just supplied the frequency of the AC source. The frequency is the important piece of information for this problem since it determines the impedance of any reactive components (inductors or capacitors). You can assume a 1V AC source if you wish.

    While the total power depends on the voltage, the fraction of the total that ends up in the load depends upon the source impedance (your Thevenin equivalent) and the load impedance. The source voltage is just a scaling factor. The frequency determines the impedances, and so determines how the power will be split between the source impedance and load impedance.
     
  23. Nov 4, 2013 #22
    what a headache?

    I can't find a solution
     
    Last edited: Nov 4, 2013
  24. Nov 4, 2013 #23

    gneill

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    Staff: Mentor

    If the problem is just looking for the required load impedance then you are free to use the result of the Maximum Power Transfer Theorem and be done with it. Just state that that is what you used; it's a well known theorem. It's like using KVL or KCL or Ohm's law, which you don't derive every time you use them.

    If you are required to derive the MPTT from scratch, that is another matter.
     
  25. Nov 4, 2013 #24
    The maximum power transfer theorem states: A load will receive maximum power from a network when its resistance is exactly equal to the Thevenin resistance of the network applied to the load.

    image.jpg
     

    Attached Files:

  26. Nov 4, 2013 #25

    gneill

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    Staff: Mentor

    Yes, that's true for resistances. There's a bit of a twist when impedances are involved since there are separate real and imaginary quantities that interact through the mathematical operations. The result is still very simple though. Look at the last line of the Wikipedia article...
     
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