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  1. Feb 7, 2008 #1
    1. The problem statement, all variables and given/known data
    The corroded contacts in a lightbulb socket have 8 Ω total resistance.How much actual power is dissipated by a 126 W (115 V) lightbulb screwed into the socket?

    2. Relevant equations
    P = I*V
    I = E/R

    3. The attempt at a solution
    I = 115/8 = 14.375
    I don't know how to calculate power...
  2. jcsd
  3. Feb 7, 2008 #2
    Calculating Power

    Electric Power in Watts is defined as Volts x Current

    for example A device connected to a 12Volt DC source is measured to Draw 6 Amperes of current what is the Power Consumption of this Device Ans P = V x I = 12V x 6 A = 72 Watts

    of Course algebraic substiutions from Ohms Lawcan lead to differnt forms such as

    P = V x I I = V/R Therefore P = VxV/R

    Or V = I x R so P = V x I = IxRxI = IxIxR
  4. Feb 7, 2008 #3
    Yea I saw that theres like 3 different equations you can use but I had tried them all and none of them are getting the right answer...
    P = V/I = V^2/R = I^2 * R
    I =126/115 = 1.096 A
    R = 8 Ohms
    V = 115 V

    V/I = 104.96
    V^2/R = 1653.125
    I^2 * R = 9.6

    Why am I getting a different for all of them? I thought they were all equal??
  5. Feb 7, 2008 #4
    You had to add the 8 Ohms Resistance as another load. Think about it, if they tell you the bulb is rated at 126 W at 115 V then I would = 126/115 = 1.1 Amps. This implies that Internal resistance of Bulb is 115V/1.1A = 105 Ohms. So Total resistanceis way more than 8ohms it is actually 8 + 105 = 113 ohms. the 8ohms is the resistance of the corrosion only. SO.... Total Power = VxV/R = 115x115/113= 117 Watts. The resistance due to corrosion ACTUALLY causes the light bulb to use less power OF COURSE this means the Bulb will provide less light. This is exactly how a dimmer switch works
  6. Feb 7, 2008 #5
    Oh I see...so I have to subtract the power lost due to the corrosion right? And I would use the I^2 * R = 9.6 W. 117 W - 9.6 W =107.5 W
    Is that right?
  7. Feb 7, 2008 #6
    Yes, That is a another very good way to think of it you attached another load in series (the corrosion) so there is leass power available to the light bulb both ways should yield same answer good luck
  8. Feb 7, 2008 #7
    except you have to subtract the 9.6 watts ( this is the power consumed by the 8 ohms of corrosion) from the original 126 watts 126 - 9.6 ~ 117 Watts.
  9. Feb 7, 2008 #8
    BTW Please tell me how you are able to type symbols (such as ohm sign) into your posts?
  10. Feb 7, 2008 #9
    Actually I don't know how to I copied and pasted the problem
  11. Feb 7, 2008 #10
    Why do I subtract it from 126 and not the 117??
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