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Power Dissipated by a Lightbulb

  1. Jan 15, 2012 #1
    1. The electric power P is dissipated by a light bulb of resistance R is P=v^2/R, where V represents the line voltage. During a brownout, the line voltage is 10.0% less than its normal value. How much power is drawn by a light bulb during the brownout if it normally draws 100.0 W (watts)? Assume that the resistance does not change.



    2. P = V^2/R



    3. Since resistance does not change, I disregarded it completely. I then took 100 in place for P and solved for V in the equation. Then I took 10 (my solution to the first problem) in place of V and subtracted 10.0% to find the power that is drawn by the lightbulb during the brownout. But my answer does not match up to the books.
     
  2. jcsd
  3. Jan 15, 2012 #2

    jedishrfu

    Staff: Mentor

    I think you use the 100 watts as P and 120v as v to get the R. Then use the v x.9 and R to get the brownout wattage.
     
  4. Jan 15, 2012 #3
    you do not need to know R because it does not change.
    The power is proportional to V^2
    If V is only 90% 0f the original then the power will be proportional to (0.9)^2 x original V
     
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