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Find maximum W for a generator

  1. Nov 23, 2015 #1
    1. The problem statement, all variables and given/known data
    So a generation with an internal resistance r delivers a voltage E. The generator is connected to a electrical circuit with a resistance R. The work W done each second in sending a current through a circuit with a resistance R is given by W =E^2*R/(R+r)^2.

    For constant r and E, show that W is a maximum when R=r. Sketch a graph of W as a function of R

    2. Relevant equations
    v=ir

    3. The attempt at a solution
    So, I am confused. I have solved for R, which is

    R= (E^2/(2W))-r

    If someone can explain the question, I will try to answer to question.

    P.S it may look like a question from some electrical engineering book but its from a calculus book -.- (no answer)
     
    Last edited by a moderator: Nov 23, 2015
  2. jcsd
  3. Nov 23, 2015 #2

    Mark44

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    Solving for R won't help you. To find the maximum value of a function of a single variable (R here, since r and E are assumed to be constants), find the derivative W'(R) and set it to zero.
     
  4. Nov 23, 2015 #3

    Mark44

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    BTW, in the future, please try to choose a thread title that is more informative. I have changed the title to "Find maximum W for a generator".
     
  5. Nov 25, 2015 #4
    Okay, I have done some work. I found the derivative of W'(R) and set it to 0. Here is what I got
    upload_2015-11-25_17-24-52.png
     
  6. Nov 25, 2015 #5

    SteamKing

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    You're not interested in dW/dx, whatever x is. You want to calculate dW/dR and set that equal to zero, not set R = 0. You also want to show that W is a maximum when r = R.

    BTW, if you express W as the product of two expressions, rather than the quotient, you can save yourself some calculation when finding the derivative.
     
    Last edited: Nov 25, 2015
  7. Nov 29, 2015 #6
    I find product of two expression harder to do.

    So I derived for dW/DR and this is what I got
    upload_2015-11-29_17-20-47.png

    If I set R = 0
    I would just get 0.
     

    Attached Files:

  8. Nov 29, 2015 #7

    Mark44

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    Here's what I said earlier, in post #2.
    I didn't say anything about setting R to zero -- I said you should set W'(R) to zero.
     
  9. Nov 29, 2015 #8

    SteamKing

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    I find that astounding.

    If f(x) = g(x) * h(x), then f'(x) = g(x) * h'(x) + g'(x)*h(x)

    whereas

    if f(x) = g(x) / h(x), then f'(x) = [g'(x)* h(x) - g(x)*h'(x)] / h2(x)
    Which is incorrect, BTW.
    And Mark44 is right. You don't set R = 0, you set dW/dR = 0, which is different. But you must calculate the correct expression for dW/dR.
     
  10. Dec 3, 2015 #9
    So, I was wondering, When you find the expression of dW/dR, do you need to find the second derivation of it to graph it or just find the fist derivation of dW/dR and set it to 0.
     
  11. Dec 3, 2015 #10

    SteamKing

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    The value of R which makes dW/dR = 0 is the value which either maximizes or minimizes W.

    It should be easy to determine which is which without calculating d2W/dR2
     
  12. Dec 3, 2015 #11
    So I found the expression for dW/dR
    upload_2015-12-3_18-51-35.png

    After the deravation, if I set dW/dR to 0, and set r and E to 1, I get a L shape graph. The x and y basically start from origin and y is + and x is positive
     

    Attached Files:

    Last edited: Dec 3, 2015
  13. Dec 3, 2015 #12

    Mark44

    Staff: Mentor

    Please post your work as text rather than as an image. Posting an image makes it impossible for us to insert a comment at the point of a mistake. Instead, we have to give the location where you are in error, which makes more work for us.

    There is no need to set r and E to any particular values.

    "Keep your eye on the prize." -- the goal here is to show that W is at its largest when R = r.
    The variables are R and W. Neither x nor y appear in this problem.

    What you ended with is correct, but it can be written in a slightly more simplified form:
    ##\frac{dW}{dR} = \frac{E^2(r - R)}{(R + r)^3}##
    What does it take for a fraction to be zero? There is absolutely nothing to gain by expanding ##(R + r)^3##.
     
  14. Dec 3, 2015 #13

    SteamKing

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    The first line of your work contains an error.

    W = E2*R/(R+r)2 not dW/dR = E2*R/(R+r)2

    You would run into less algebra if you wrote

    W = E2*R * (R+r)-2

    and then used the Product Rule to find dW/dR
     
  15. Dec 5, 2015 #14
    if somebody has full solution to this problem pls post an image solving or a word file.
     
  16. Dec 5, 2015 #15

    cnh1995

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    Its actually the 'Maximum power transfer theorem' in electrical engineering.
    W=E2*R/(R+r)2.
    Find dW/dR and equate it to 0 as already explained in #13 by SteamKing.
     
  17. Dec 5, 2015 #16
    I have found the solution to the answer. Thanks for helping everyone
     
  18. Dec 5, 2015 #17

    Mark44

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    We don't do that here. From the rules for this forum:
    From the rules (https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/):
     
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