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Homework Help: Power Dissipated - Kirchhoff's Rule

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Using symbols, write PRL, the power dissipated in RL as a function of RL (your other symbols are R0 and the power supply voltage epsilon both of which are constant. 10V and 100ohms) and then take the derivative of this function with respect to RL for which PRL has an extreme value, in this case a maximum.


    2. Relevant equations
    V = IR


    3. The attempt at a solution
    So here's what I was thinking...
    PRL = VI where I = V/R so replace that with Epsilon2/R0
    Then plug in the constant values 10V and 100ohms to get PRL = 1W dRL

    Not sure where to go from here though as the derivative gives zero. What am I missing? Did I do it all incorrectly? Thanks in advance!
     
  2. jcsd
  3. Mar 28, 2010 #2

    tiny-tim

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    Hi Pete_01! :smile:

    You're confusing R0 with RL.

    I haven't worked out what R0 is (what is it? and what is the context?), but clearly P will depend on RL, either instead of or as well as R0. :wink:
     
  4. Mar 28, 2010 #3
    Hi Tim,

    R0 is a constant resistor of value 100ohms.
     
  5. Mar 28, 2010 #4

    tiny-tim

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    But where is it, and what's RL ? :confused:
     
  6. Mar 28, 2010 #5
    Oops, sorry. RL is a variable resistor that was changed in value from 10-1280ohms. I attached the circuit picture as well.

    Thanks
     

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  7. Mar 28, 2010 #6

    tiny-tim

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    ohh! they're two resistors in series! :rolleyes:

    ok, then your formula I = V/R0 should be changed to … ? :wink:
     
  8. Mar 28, 2010 #7
    So it would be I = V/(R0+RL)?

    Then your left with PRL = 102/(R0+RL)d(RL) Correct?
     
  9. Mar 28, 2010 #8

    tiny-tim

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    Yes. :smile:
    what's d ? :confused:

    and what formula are you using for PRL ?
     
  10. Mar 28, 2010 #9
    So I used P=VI where I is substituted in from above. 10 is the voltage.

    PRL = VI = V*(V/R) = V2/(R0+RL)

    It said to find RL where it is maximum I need to take the derivative of the function with respect to RL and set the derivative equal to 0, then solve for RL.
     
  11. Mar 28, 2010 #10

    tiny-tim

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    But 10 isn't the voltage across RL, is it? :wink:

    (and that's not the quickest formula, anyway)
     
  12. Mar 28, 2010 #11
    Ah yes, it is not. What is the best way to solve this?
     
  13. Mar 28, 2010 #12

    tiny-tim

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    erm :redface:

    … one of Kirchhoff's rules ? :smile:
     
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