Filament lamp model (derivation)

[rohanlol7]

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

 

[BvU]

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 ?

 

[rohanlol7]

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

 

[rohanlol7]

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

 

[BvU]

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.

 

[rohanlol7]

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

 

[ehild]

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).

 

[rohanlol7]

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

 

[rohanlol7]

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

 

[ehild]

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 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?

 

[lychette]

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'