The discussion centers on deriving the relationship that power (P) is proportional to voltage (V) raised to the fifth power (V^5) for a pure tungsten filament bulb. Participants clarify that the relevant equations include Ohm's law (V=IR) and the Stefan-Boltzmann law, which relates power dissipation to temperature. The consensus is that for P to be proportional to V^5, the resistance (R) must vary with voltage in a specific manner, specifically R being proportional to V^-3, which is deemed impractical. The discussion also highlights the significant role of temperature in power dissipation for tungsten filaments.
PREREQUISITESPhysics students, electrical engineers, and anyone interested in the thermal and electrical properties of materials, particularly in the context of incandescent lighting technology.
NEWBvU 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 ?
OH SORRY! i didnt copy it well. It meant P^5 proportional to V^8 sorry about thatBvU 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.
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 said:OH SORRY! i didnt copy it well. It meant P^5 proportional to V^8 sorry about that
So okay. ill try using stefan Boltzmann lawehild 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 constantsehild 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).
No, that is not true.rohanlol7 said:T= temperataure of filament Ti=temperature outside k,m,a are constants
So the temperature is proportional to the Power disspated.
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.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
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'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