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Power Dissipated on light bulb

  1. Mar 10, 2013 #1
    1. A 25 ohm resistor is connected in series with a 100 W light bulb. The standard 120V/60Hz AC outlet voltage is applied to the series combination. Determine the real power dissipated on the light bulb. Assume the internal resistance of the light bulb is independent of the power dissipation. Round off your answer to two decimal places.



    2. P=Vrms x Irms



    3. I used 120 V as Vrms. To find Irms I did 120/25. Is power lost over the resistor? Thanks in advance
     
  2. jcsd
  3. Mar 10, 2013 #2

    rude man

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    What is the resistance of the light bulb? Then, what is the current? And finally, what is the power dissipated by the light bulb?
     
  4. Mar 11, 2013 #3

    CWatters

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    estsleep - A 100W light bulb only dissipates 100W when connected directly to 120VAC.
     
  5. Mar 11, 2013 #4

    rude man

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    120/25 would be the current in the resistor if the resistor were connected directly across the 120V. But it's not - there is a light bulb in series with it dropping some of the voltage ...
     
  6. Mar 11, 2013 #5
    can i use the formula P = V^2/R to find the resistance of the bulb? so, 100 = 120^2/R. R=1440 ohms?
     
  7. Mar 11, 2013 #6

    rude man

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    Almost. I compute 144 ohms.
     
  8. Mar 11, 2013 #7
    ok, to find the current can I just Vrms/R? 120/144 = .8333. Then power is I^2 x R. Doing that gives me 99.99
     
  9. Mar 11, 2013 #8

    rude man

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    No, because the 120V is not all across the light bulb, is it?
     
  10. Mar 11, 2013 #9
    Will doing voltage division to find the Vrms over the bulb work? Then finding the Irms from Vrms/Rbulb?
     
  11. Mar 11, 2013 #10

    rude man

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    Yes it would.
     
  12. Mar 11, 2013 #11
    got it! thanks for the help
     
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