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Resistance and Power Probelm (again)

  1. Jan 26, 2009 #1
    A light bulb is wired in series with a 132 resistor, and they are connected across a 120 V source. The power delivered to the light bulb is 23.4 W. What are the two possible resistances of the light bulb?

    a. lower value
    b. higher value

    relevant equations V=I/R and P=(I^2)(R)

    attempt at soultion:

    23.4=I^2(132)
    =.4210

    Wrong!


    V=I/R
    120=I/132
    =15840

    Wrong!
     
  2. jcsd
  3. Jan 26, 2009 #2

    mgb_phys

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    Science Advisor
    Homework Helper

    You have two resistors in series (the 'resistor' and the bulb )
    What do you know about the current through these two resistors?
    What equations do you know for the total resistance ?
     
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