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Power dissipation

  1. Mar 18, 2008 #1
    1. The problem statement, all variables and given/known data
    The corroded contacts in a lightbulb socket have 5.0 Ohms of resistance. How much actual power is dissipated by a 100 W (120 V) lightbulb screwed into this socket?

    2. Relevant equations
    P = V^2/R
    V = IR
    P = RI^2

    3. The attempt at a solution
    I found the resistance of the lightbulb to be 120^2/100 = 144 Ohms, then added 5 Ohms to it to get the resistance of the entire circuit. Then I found the current in the circuit by using I = V/R = 0.8054 A, and then I found the power dissipated by the lightbulb by using P = R(of the lightbulb)I^2 = 144 x 0.8054 = 93.4 W as my final answer, but I don't know if this is right and electric circuit analysis is NOT my specialty in physics.
  2. jcsd
  3. Mar 19, 2008 #2
    This is completely right, and clearly explained!
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