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Power in circuits

  1. Jan 2, 2016 #1
    1. The problem statement, all variables and given/known data

    Why in a circuit in parallel I can't use P= i^2 R where R is the equivalent resistance? Why do I have to use P= V^2/R ?

    2. Relevant equations P= Vi , P= V^2/R , P= i^2R


    3. The attempt at a solution
     
  2. jcsd
  3. Jan 2, 2016 #2

    CWatters

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    Please give an example circuit. Power dissipated in what?
     
  4. Jan 2, 2016 #3
    Dissipated in resistance,
    an usual circuit with parallel resistances and a V created by a battery
     
  5. Jan 2, 2016 #4

    berkeman

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    I dunno. Why do you think you cannot? Can you post a diagram of the circuits?
     
  6. Jan 2, 2016 #5

    berkeman

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    Well, beaten to the punch by CWatters as usual, but we still want more details to help you out... :confused:
     
  7. Jan 2, 2016 #6
    Q30, it asks to find the circuit which dissipate more power, all the resistance are R and the difference of potential is ε, why can't I find in each circuit the total resistance and say that the power is P= i^2 R and say that i is the same in each circuit? (So I can find out which dissipate more power by looking at the total resistance)

    Isn't the current i the same in each case?

    Why do I have to use P= ε^2 /R ? Isn't the same thing as using P= i^2 R ?
     

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    Last edited: Jan 2, 2016
  8. Jan 2, 2016 #7

    berkeman

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    I'm not able to see your very dim attachment.

    Are you familiar with how current divides into parallel circuits? Maybe that's the disconnect here. For parallel circuits, the voltage is the same but the current flows through both parallel branches...
     
  9. Jan 2, 2016 #8
    Yes I knew that, but I thought that calculating the total resistance is was able to "transform" each circuit into one with one resistance with a current i like that: (photo)

    Then other approach that I had thinking at the problem was thinking that maybe i would have been different in each circuit

    (Sorry for bad english )

    Just thought that the current was the same in each case
     

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  10. Jan 3, 2016 #9

    haruspex

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    Quite so. Reducing the parallel resistors to an equivalent single resistance allows you to work out the total current, but it does not tell you how much of that flows in each. The two individual currents will be in inverse proportion to the resistances.
     
  11. Jan 3, 2016 #10
    So for this problem -I'm just assuming so don't take this to be an answer OP - but isn't it basically finding information from simplifying the current to have an equivalent resistance then applying that newfound info to the individual branches? Please let me know if that's the case, as I would like to solve a similar problem like this myself but cannot properly see this from the OP's posts. Thank you!
     
  12. Jan 3, 2016 #11

    haruspex

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    yes, you can do it that way, but it is simpler just to apply V2/R to each in the first place.
     
  13. Jan 3, 2016 #12
    Ah okay, I'll look into that formula. Do you know what this equation is known as/called? I'd like to look into how this is derived because it's not immediately apparent to me at the moment.
     
  14. Jan 3, 2016 #13

    haruspex

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    V2/R? You can derive it from V=IR and P=IV.
     
  15. Jan 3, 2016 #14
    Thanks! I'll do that now and work it through.
     
  16. Jan 3, 2016 #15

    cnh1995

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    P=I2R. Put I=V/R. Voltage is same across parallel resistors. So, power will be inversely proportional to the resistance value.
     
  17. Jan 3, 2016 #16
    Thanks, thought there was more to it than there actually is haha.
     
  18. Jan 3, 2016 #17
    Does P= V^2/R works also for series circuits? In series i is constant and V vary right?
     
  19. Jan 3, 2016 #18

    gneill

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    It works fine if V is the voltage across the entire branch and R the total resistance of the branch.

    Also, if you can determine how the total voltage divides across each of the series resistors then you can apply V2/R to each of them individually and sum the results.
     
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