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Electric circuits

  1. Mar 11, 2008 #1
    1. The problem statement, all variables and given/known data
    The power rating of a 1000-W heater specifies the power consumed when the heater is connected dto an AC voltage of 120 V. Explain why the power consumed by two of these heaters connected in series with a voltage of 120 V is not 2000-W.



    2. Relevant equations
    P= (V)^2/R

    (?) P = IVsin^2(2 pi ft)

    f= frequency
    t= time



    3. The attempt at a solution
    Is it because there are 2x the resistance? Therefore lowering P? In my text it starts to explain something about sinusoidal fluctuation but I don't really get how it pertains to this question, however I feel it may...
     
  2. jcsd
  3. Mar 11, 2008 #2

    Doc Al

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    Staff: Mentor

    What's the voltage across each if they are in series?
     
  4. Mar 11, 2008 #3
    I'm not completely sure, but is it 60 V each?
     
  5. Mar 11, 2008 #4

    Doc Al

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    Right. What does that tell you?
     
  6. Mar 11, 2008 #5
    That the watts used doesnt change.
     
  7. Mar 11, 2008 #6

    Doc Al

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    :bugeye:

    So you think that if you have a device that uses 1000W when it is hooked up to 120V it will still use up 1000W if you turned the voltage down by half? :wink:
     
  8. Mar 11, 2008 #7
    (blush) Well, I thought that each unit was turned down by half. There are 2 heaters connected to the in the series.
     
  9. Mar 11, 2008 #8
    Still stumped here :(
     
  10. Mar 11, 2008 #9

    Doc Al

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    Yes.

    The only way that each heater can produce 1000W (for a total of 2000W) is if each heater is given the full 120V. But we just showed that each heater gets only half the voltage, thus the total power must be less than 2000W.
     
  11. Mar 11, 2008 #10
    Ok :) So is the total power between the 2 heaters 1000-W?
     
  12. Mar 11, 2008 #11

    Doc Al

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    Not necessarily. You can answer that by applying one of the equations from your first post. What happens to the total resistance?
     
    Last edited: Mar 11, 2008
  13. Mar 11, 2008 #12
    What?! I am so lost here! Ok, so if the total watts for one heater connected to 120 V is 1000 watts, and if 2 heaters are in a series with 120 V and each heater has 60 V, then wouldnt the total watts add back up to 120 and leave us where we started?
     
  14. Mar 11, 2008 #13
    Ok, I think I got it. So P = V^2/R that gives me R = 14.4 When the voltage is cut in half then it becomes P = 60^2/14.4 so P = 250......is this correct?!
     
  15. Mar 11, 2008 #14

    Doc Al

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    The total voltage (not wattage) would add up to 120 V. But that voltage is now spread out over both heaters. Use your power equation: P = V^2/R. If the total volts remains the same, but the resistance doubles, what happens to the total power consumed?
     
  16. Mar 11, 2008 #15
    So R = 14.4 when voltage is 120 for ONE heater. When there are 2 R doubles. So, the total watts for the 2 heaters is 500 W?
     
  17. Mar 11, 2008 #16

    Doc Al

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    Right. Since the total resistance doubles while the voltage remains the same, the power drops to half.
     
  18. Mar 11, 2008 #17
    Thank you so much Doc Al!
     
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