Filament lamp model (derivation)

In summary, in order to show that power is proportional to V^5 for a pure tungsten filament bulb, one must consider the heating effect of the filament and its dissipation through radiation. Using the Stefan-Boltzmann law and considering the filament's temperature and resistance, it can be shown that the power dissipated in the filament is proportional to V^8. Additionally, it is important to note that the resistance of the filament increases with temperature, making the relationship between power and voltage non-linear.
  • #1
rohanlol7
67
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
 
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  • #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 ?
 
  • #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
 
  • #4
BvU said:
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
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
BvU said:
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
rohanlol7 said:
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
ehild said:
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
ehild said:
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 can't quite continue
 
  • #10
rohanlol7 said:
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.
rohanlol7 said:
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 can't 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
rohanlol7 said:
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 can't 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'
 

1. What is the filament lamp model?

The filament lamp model is a theoretical model that explains the relationship between temperature and resistance in a light bulb. It states that as the temperature of the filament increases, the resistance also increases, leading to a decrease in current and ultimately the production of light.

2. How is the filament lamp model derived?

The filament lamp model is derived using the Stefan-Boltzmann law, which relates the temperature of a blackbody (such as the filament in a light bulb) to the rate at which it emits thermal radiation. This is combined with Ohm's law, which describes the relationship between current, voltage, and resistance, to create a mathematical model for the behavior of the filament in a light bulb.

3. What are the assumptions made in the filament lamp model derivation?

The filament lamp model makes several simplifying assumptions, including that the filament behaves as a blackbody, that the resistance of the filament is proportional to its temperature, and that the temperature of the filament is uniform along its length.

4. How accurate is the filament lamp model?

The filament lamp model is a simplified theoretical model and may not accurately reflect the behavior of a real light bulb. Factors such as the composition of the filament and the surrounding gas can affect the accuracy of the model. However, it provides a good approximation for understanding the general behavior of a light bulb.

5. What practical applications does the filament lamp model have?

The filament lamp model is used in the design and optimization of light bulbs, as well as in the study of thermodynamics and electrical circuits. It also has applications in areas such as materials science and engineering, where the behavior of high-temperature materials is of interest.

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