Filament lamp model (derivation)

  • #1
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2

Homework Statement


State and use simple assumptions to show that ' Power is proportional to V^5' is the expected relationship for a pure tungsten filament bulb.

Homework Equations


V=IR
I=dq/dt
Q=mcT

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
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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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 ?
 
  • #3
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2
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
 
  • #4
67
2
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 ?


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
 
  • #5
BvU
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Homework Helper
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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.
 
  • #6
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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.
OH SORRY! i didnt copy it well. It meant P^5 proportional to V^8 sorry about that
 
  • #7
ehild
Homework Helper
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OH SORRY! i didnt copy it well. It meant P^5 proportional to V^8 sorry about that
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).
 
  • #8
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2
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).
So okay. ill try using stefan boltzmann law
 
  • #9
67
2
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).
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
 
  • #10
ehild
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T= temperataure of filament Ti=temperature outside k,m,a are constants
So the temperature is proportional to the Power disspated.
No, that is not true.
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
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:
  • #11
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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
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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