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Regarding modeling of light bulb circuit

  1. Dec 9, 2013 #1
    Hello, I'm trying to apply conservation of energy to a lightbulb circuit, but I can't think of what term should be included for heat loss / convection / conduction


    Here's what I've got
    Pin = Pradiated + i2Rwire, connections, etc + Power lost to heating surroundings? (what can I use for that?) c(T-T0)?

    Are there any other terms I should include?
    thanks
     
  2. jcsd
  3. Dec 9, 2013 #2

    SteamKing

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    The power lost to heating the surroundings is a component of the power radiated from the bulb, i.e., not all of the energy coming out of the bulb is visible light. The amount of power radiated is proportional to the difference in the fourth power of the absolute temperatures (Stefan-Boltzmann Law) of the body and its surroundings:

    http://en.wikipedia.org/wiki/Stefan–Boltzmann_law
     
  4. Dec 9, 2013 #3
    I intended for Pradiated = Stefan-Boltzmann law.
    But there is power lost to heating the air which is not from radiation, right? what could I use for that term?
     
  5. Dec 9, 2013 #4

    jtbell

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    The power radiated as light from the bulb, lost to heating air, etc. are part of the i2R term, not in addition to it.

    In an alternating-current circuit you also have some energy radiated in the form of electromagnetic waves at the AC frequency. That is, the circuit acts like a radio antenna although at much lower frequency. I don't have any equations handy, but I'm sure this effect is very small unless the circuit is specifically designed to work as an antenna.
     
  6. Dec 9, 2013 #5

    sophiecentaur

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    I think he's referring to the R losses in the supply.

    Different filament bulbs of the same nominal power will radiate different powers of optical light. It will depend upon the shape and length of the filament (coil or coiled coil) and it would need actual measurements to characterise a particular bulb. Heat lost to the surroundings will depend upon the size of the glass envelope, the gas it's filled with and the cables conduct heat away.
    This is not hard to do in an arm waving sort of a way but pretty difficult if you want to make an accurate prediction.
     
  7. Dec 9, 2013 #6
    just in case if your wondering , well if the bulb is incandescent one then it emitts only 5 to 10% in visible light , the rest or up to 90% and more (depending on the bulb type) goes away as heat or infrared.
    This is not a precise calculation but those are some approximate numbers that are true.
     
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