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Filament lamp model (derivation)

  1. Aug 11, 2016 #1
    1. The problem statement, all variables and given/known data
    State and use simple assumptions to show that ' Power is proportional to V^5' is the expected relationship for a pure tungsten filament bulb.

    2. Relevant equations
    V=IR
    I=dq/dt
    Q=mcT

    3. The attempt at a solution
    I tried to use a simple model where the rate of heat loss is proportional to the difference in temperature.
    From there things got complex and i can't seem to see how to derive this
     
  2. jcsd
  3. Aug 11, 2016 #2

    BvU

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    I don't see P (power) in your relevant equations ?
    And what is R ? IF it's the resistance, how could that vary with V ?
    You sure the exercise says ##V^5## and means applied voltage ?

    I=dq/dt and Q=mcT are equations alright, and I can guess what the symbols stand for, but are they relevant equations ?
     
  4. Aug 11, 2016 #3
    The question is asked just as i posed it. I can't figure out how to relate the power to the voltage easily.
    I'm not sure what equations to use actually. I figured maybe it had something to do with the heating effect of the filament.
    Yeah V^5 is the applied voltage
     
  5. Aug 11, 2016 #4

    NEW
    The question is asked just as i posed it. I can't figure out how to relate the power to the voltage easily.
    I'm not sure what equations to use actually. I figured maybe it had something to do with the heating effect of the filament.
    Yeah V^5 is the applied voltage
     
  6. Aug 11, 2016 #5

    BvU

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    Link says ##V^{1.6}##.

    And this is assuming changes in ##V## at normal operating conditions -- something the exercise should have mentioned too !

    Usually we state ##P = V I = V^2/R##. So if ##R## is constant, we have ##P\propto V^2##. To get ##\propto V^5## you would need ##R\propto V^{-3}## which to me seems mission impossible.
     
  7. Aug 11, 2016 #6
    OH SORRY! i didnt copy it well. It meant P^5 proportional to V^8 sorry about that
     
  8. Aug 11, 2016 #7

    ehild

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    The power on the tungsten filament in a light bulb is mainly dissipated by radiation. How does the radiated power depend on temperature (think of black-body radiation).
     
  9. Aug 11, 2016 #8
    So okay. ill try using stefan boltzmann law
     
  10. Aug 11, 2016 #9
    T= temperataure of filament Ti=temperature outside k,m,a are constants
    So the temperature is proportional to the Power disspated. T=kP. P=m(T^4-Ti^4)=mT^4 ( approximately)
    Resistance of the filament lamp =R=Ro(1-a(T-Ti))
    Considering the lamp with current I through it V=I*Ro(1-a(T-Ti))=IRo(1-aT) (approximately)
    From there I cant quite continue
     
  11. Aug 11, 2016 #10

    ehild

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    No, that is not true.
    The filament is much hotter than the surroundings, (about 2500 °C) so you can ignore Ti. And change the minus to plus. Approximately, the resistance is proportional to T.
    How do you get the power dissipated in a resistor R if the voltage across the resistor is U?
     
    Last edited: Aug 11, 2016
  12. Aug 11, 2016 #11
    your equation for resistance variation with temperature for a metal is wrong. The resistance increases with temperature and the variation with temperature is only linear for 'low' temperatures. The temperature of a tungsten filament bulb would not be considered as 'low'
     
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