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How is Pmax derived?

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Why is pmax=v^2/4r?

    2. Relevant equations

    I understand that p=v^2/r using the equations:
    p=iv and v=ir


    3. The attempt at a solution

    My problem relates to when I have to find the pmax. Looking over my professors solutions for the homework, he uses pmax=vt^2/4rt.
     
  2. jcsd
  3. Mar 13, 2013 #2

    gneill

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    Staff: Mentor

    Hi xbuoix, Welcome to Physics Forums.

    Can you post the original problem so that we can see the issue in context? It's hard to say anything meaningful about it without knowing where the 't' comes in or how it comes about.
     
  4. Mar 13, 2013 #3
    Thanks, I'm glad to be on this forum!

    Sorry, vt and rt is the Thevenin equation for the voltage and resistance.

    The question goes like this:
    (There is a circuit.)
    Draw the circuit on your solutions and show how you obtain a Thevenin equivalent for everything to the left of the load resistor. Then draw the Thevenin equivalent that you obtain connected to the load resistor. Using the results, what is the maximum power that can be delivered to Rl. What is the value of Wl would result in maximum power?

    Finding the Thevenin equation, I found:
    VT= 30 V
    RT=3kohms
    Imax=10mA
    RLoad=RT
    VMax=VT


    At first, I thought that P=VT^2/RT. After I turned in my homework, I looked at the solution my professor posts online, and I see that he used P=VT^2/4RT. Where did the 1/4 come from?
     
  5. Mar 13, 2013 #4

    gneill

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    Staff: Mentor

    Okay, it's clearer now (By the way, you might try using the x2 and x2 icons in the edit panel to create superscripts and subscripts).

    The 1/4 comes from the mathematical operations involved in determining the conditions for maximum power delivered to the load resistor. Find the value of RL that results in the maximum power delivered to the load, then use that value of RL to determine that power.
     
  6. Mar 14, 2013 #5
    So this case, the RLoad=RT=3kΩ. I'm still confused on how it came about.
     
  7. Mar 14, 2013 #6

    gneill

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    Staff: Mentor

    Have you studied the math behind finding the maximums and minimums of functions?
     
  8. Mar 14, 2013 #7
    As in finding the double derivative of the function? Yes

    Oh.. I get it now..I think.

    The double integral of v2 would give 1/4? Shouldn't we have taken the derivative instead?

    Edit: I meant 1/R. Why are we taking the maximum of the resistance and not the voltage?
     
  9. Mar 14, 2013 #8

    gneill

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    Staff: Mentor

    You're looking for the value of RL which maximizes the power (in RL). Then use that RL to write the expression for the maximum power.
     
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