# Filament lamp model (derivation)

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1. Aug 11, 2016

### rohanlol7

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. Aug 11, 2016

### 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 ?

3. Aug 11, 2016

### 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

4. Aug 11, 2016

### rohanlol7

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. Aug 11, 2016

### 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.

6. Aug 11, 2016

### rohanlol7

OH SORRY! i didnt copy it well. It meant P^5 proportional to V^8 sorry about that

7. Aug 11, 2016

### ehild

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. Aug 11, 2016

### rohanlol7

So okay. ill try using stefan boltzmann law

9. Aug 11, 2016

### rohanlol7

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. Aug 11, 2016

### ehild

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
11. Aug 11, 2016

### lychette

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'