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Small drop in voltage change in power?

  1. Sep 11, 2006 #1
    In a brownout the power company's voltage drops. Assuming the voltage drop is small, how would I figure out how much of a voltage drop it takes for a 60W light bulb to act like a 50W bulb? The relevant equation I am guessing would be P=V^2/R. I also know the resistance doesn't change. I tried P=(V-h)^2/R for small h, representing the power in a small voltage drop. thus P is approximately V-2Vh/R but I don't think this helps me any since I haven't eliminated h. What do you think?
     
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  3. Sep 11, 2006 #2

    Andrew Mason

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    Your idea is right. Use [itex]P_2 = V_2^2/R[/itex] where [itex]P_2 = 5P_1/6 =\frac{5}{6}V_1^2/R[/itex] and find the ratio of V2 to V1.

    AM
     
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