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What will be the power output

  1. Jan 6, 2014 #1
    A load has a power rating of 160 w when the current is 6.0 A. What will be the power output if the current increases to 15 A?

    I know how to do it : use the formula v=I^2R
    In this case 15/6 = 2.5^2 = 6.25
    160 * 6.25 = 1000W.
    My question is why can't you use p = VI ? Because in this case , it would be 160* 2.5?
     
  2. jcsd
  3. Jan 6, 2014 #2

    gneill

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    Staff: Mentor

    Coco12, please be sure to use the posting template provided when a new thread is started here in the homework sections of Physics Forums. Otherwise you're setting yourself up to incur infraction points....

    You cannot assume that V remains unchanged when I changes.
     
  4. Jan 6, 2014 #3

    Thank you very much. I posted this from an ipad, and the template did not come up. I wonder why?
     
  5. Jan 6, 2014 #4

    gneill

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    Staff: Mentor

    Ah. It's a known deficiency in the app, and is being looked into. I suggest manually formatting your initial posts along the lines of the approved manner until the deficiency is corrected.

    1) Problem Statement
    2) Relevant Equations
    3) Attempt at Solution
     
  6. Jan 6, 2014 #5
    Thank you.
     
  7. Jan 7, 2014 #6

    CWatters

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    I believe you will also need to provide more details of the load. Is it safe to assume it's a resistor? The answer would be different for a motor.
     
  8. Jan 16, 2014 #7

    Another question: so you can assume that resistance doesn't change?
     
  9. Jan 16, 2014 #8

    CWatters

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    You tell us. Does the problem statement describe the load?

    If the load is a resistor then yes it's probably safe to assume that the resistance doesn't change. In which case the if the current changes that implies a change in the voltage.

    However if the load is a motor then the above isn't necessarily true. A DC motor operating on a fixed 12V could draw a wide range of currents (and hence power) depending on the load on the motor shaft.

    If the problem statement doesn't describe the load in any more detail then you should probably state that you are assuming the load is resistive, and not something like a motor, then proceed accordingly.
     
  10. Jan 16, 2014 #9

    OK thank you.
     
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