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Bulb Brightness

  1. Feb 26, 2014 #1
    1. The problem statement, all variables and given/known data

    Normally, household lightbulbs are connected in parallel to a power supply. Suppose a 40 W and a 60 W lightbulb are, instead, connected in series. (see here) Which bulb is brighter?

    2. Relevant equations

    [tex]P_{resistor}=I\Delta V=I^2R =\frac{(\Delta V)^2}{R}[/tex]

    3. The attempt at a solution

    I have found out the answer is the 40 W light bulb but am not sure why. Since both bulbs are getting the same current it means the 60 W has a higher resistance according to the formula [tex]P=I^2R[/tex] Doesn't a higher resistance mean more energy dissipated means more light given off? Or am I thinking about it wrong?

    Thanks
     
  2. jcsd
  3. Feb 26, 2014 #2

    haruspex

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    How do you deduce that?
     
  4. Feb 26, 2014 #3
    If I is the same for both resistors then according to P=I^2(R), R for the 60 W bulb must be higher.

    But I am confused because according to P = V^2/R the 60 W has a higher R...how do I know which equation to use?
     
  5. Feb 26, 2014 #4

    NascentOxygen

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    Why don't you work out the resistance of a normally-operating 60w bulb in your house? How many ohms would it be?
     
  6. Feb 26, 2014 #5

    haruspex

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    The "60W" rating assumes you are connecting it to the standard mains supply. There is nothing inherent to the bulb that guarantees it consumes 60W. The fundamental constant of the bulb is its resistance.
     
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