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Incandescent Light Bulb, Resistance and Brightness

  1. Jan 13, 2009 #1
    Hi all, there are a few questions regarding the incandescent light bulb, which I believe are rather elementary, but has nevertheless been bothering me for quite some time. I don't know if my chain of thought is correct, please point out my mistakes if there's any. Thanks! Here goes (they are not separate questions, but rather, a continuous narrative):

    1) Filament light bulbs (incandescent) are non Ohmic conductors, and we know that on the I-V graph, their resistance increases with voltage, because the filament gets hot and increases the frequency of collisions between the particles in the filament and the flowing electrons.

    2) Thus, if I were to plot a graph of resistance against temperature for the filament, I'll get one where resistance increases with temperature.

    3) It is a known fact that as an incandescent light bulb gets hotter, it gets brighter as well. This can well be explained by the processes which give rise to incandescence - the faster the charges in the filament vibrate, the greater the intensity of the EM light waves they produce.

    4) If I were to follow through statements 2 - 3, I can come to the conclusion that the greater the resistance, the brighter the incandescent light bulb

    5) Herin lies the possible contradiction: While I have just reasoned that the greater the resistance, the brighter the incandescent light bulb (statement 4), the equation P =V2/R gives us the contradictory result that power output would in fact DECREASE if resistance of the filament increases.

    6) Ive been thinking of this for quite some time, and the only explanation I could find, which falls in line with both sides of the contradiction, is this:

    7) Going by P =V2/R , it is true that power output decreases if resistance of the filament increases (for the purpose of this argument, the temperature increase from prolonged use leads to increased resistance) and the voltage across the light bulb remains the same (single light bulb in a simple circuit with a single power cell).

    8) Put simply, there is high power output of the bulb initially, and low power output eventually.

    9) However, the efficiency of the light bulb in producing light energy is low initially, but high eventually. Points 10 and 11 will explain why:

    10) This is because when the light bulb is first switched on and the filament is cool, a large amount of power output goes into heating the filament to high temperatures, and only a small amount of energy becomes light energy (cuz at low temperatures, incandescence effects are minimal).

    11) At high temperatures, however, the only amount of power output that is converted to heat energy of the filament is that required to replace the heat lost to the surroundings. This amount is significantly less than that in point 10. So even though there is now a lower power output, more energy is being converted to light energy than before. While this mathematically explains why it is possible for more light energy to be emitted than when the filament was cool, the argument that incandescence effects are more substantial at high temperatures provides a theoretical justification for increased luminance in this situation.

    12) Summarizing points 10 and 11, the filament of the incandescent light bulb is more efficient in producing light energy at high temperatures, compared to low temperatures.

    13) Summarizing the entire post, the equation P =V2/R cannot be used single mindedly in explaining the relationship between temperature, resistance and luminance of an incandescent light bulb. Efficiency of light production must be considered as well.

    So after reading all of this (thanks for your time!), can anyone point out mistakes if there are any? I would certaintly like to correct them. Thanks again!
  2. jcsd
  3. Jan 13, 2009 #2


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    Good analysis, monstersaurou. Your points look fine.
  4. Jan 13, 2009 #3
    Well, R is indeed increasing, but V^2/R defines the total power delivered from the source to the bulb, which is more than 90% heat. The fraction of the total power being converted to optical power changes however. At higher temp, the radiated power is greater. An incandescent bulb is more efficient at hotter filament temp.

    Yes, I agree that one cannot single mindedly pull out one equation and run with it. There is more than one thing going on here.

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