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Voltage change to go from 60-W to 50-W?

  1. Dec 14, 2014 #1
    In a "brownout" situation, the voltage supplied by the electric company falls. Assuming the percent drop is small, show that the power output of a given appliance falls by approximately twice that percent (assuming resistance does not change).
    How much of a voltage drop does it take for a 60-W lightbulb to begin acting like a 50-W lightbulb?

    Relevant equations:
    (Power output/Power input)*100%
    P=(v^2/R)

    The attempt at a solution:
    (50/60)*100%=83%
    V[1]=sq.root(60R) v[2]=sq.root(50R)
    sq.root(60R) * 0.83 = sq.root(50R)
    60R * (0.83)^2 = 50R
    60 * ((0.83)^2 )/R = 50
    ((0.83^2)/(60R) = 50/60
    0.83^2=50R
    R= ((0.83)^2)/50
    R= 0.014 Ω
    V[1]=sq.root(60(0.014)) v[2]=sq.root(50(0.014))
    V[1]= 0.91 V v[2]= 0.83 V

    (0.91 V - 0.83 V) *100%= (approximately 8%)

    Answer key told me it was =8.4%, as i was a bit off, was wondering if i had made a mistake or not.
     
  2. jcsd
  3. Dec 14, 2014 #2

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    These are both good.
    This is NOT.
     
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