2007-06-02 03:38:42 · 8 answers · asked by lykh55 1 in Science & Mathematics ➔ Physics
The heat is a result of a resistance to current flow. The filament is designed to resist a current, and the heat is constant to that resistance and current value.
Increase any one of those values and that will change the amount of heat produced. The resistance is inherent in the filament, so the only actual value that can be manipulated is the current. If you put too much current through a filament, the heat is increased to the point that the filament will fail.
To conserve energy, a filament "loses" it's ability to resist current over time. Therefore, the bulb eventually "burns out." Increase the current, and the time of resistance is decreased.
2007-06-02 03:57:14 · answer #1 · answered by Anonymous · 0⤊ 3⤋
The reason is because the rate of heat produced is equal to the rate of heat lost from the filament to the surrounding.
Rate of heat loss is propotional to difference in temperature between filament and surrounding. As the filament gets hotter , rate of heat loss will also increase because of the temperature difference increases. There would come a time when the rate exactly cancels out.
If somehow, rate of heat loss is more than rate of heat produced, temperature will decrease, rate of heat loss will correspondingly decreases. Both ways, a steady state will be achieved.
doug_donaghue answered the question well, but not ... accurately enough. But it's worth reading.
Even, if the coefficient of resistance is negative, the bulb will still reach a constant temperature for the reason stated above.
2007-06-02 11:05:10 · answer #2 · answered by Daniel T 2 · 1⤊ 1⤋
The temperature levels off when an equilibrium (equality) is reached between Ohmic heating power P (which decreases with temperature T due to the fact that the resistance R of the filament increases with temperature P= V^2/R) and radiation loss (which increases in proportion T^4, the *fourth power* of T). There is also thermal conduction loss through the base of the bulb and any surrounding gas (which is proportional to T), but that is usually minimal.
2007-06-02 11:58:50 · answer #3 · answered by Dr. R 7 · 0⤊ 1⤋
Enough heat is lost through the body of the bulb to stabilize the temperature of the filament.
2007-06-09 16:45:29 · answer #4 · answered by johnandeileen2000 7 · 0⤊ 0⤋
A finite amount of power is dissipated. The amount of power dissipated is proportional to temperature of the filament. Pout = f (Temp). So there is some temperature that balances the equation: Pout(Temp) = Pin. The output power is comprised of two parts: Light (photons) and Heat (infrared radiation).
2007-06-09 00:55:22 · answer #5 · answered by Robert T 4 · 0⤊ 0⤋
The material(s) used to make the filament have what is known as a 'positive temperature coefficient of resistance'. That is, as their temperature increases, so does their resistance. This means that, if the current tries to increase, the heating goes up, the filament becomes warmer, the circuit resistance increases, and the current is forced to stay 'relatively' constant.
Doug
2007-06-02 10:45:28 · answer #6 · answered by doug_donaghue 7 · 2⤊ 2⤋
Because heat is lost continuously by radiation through the walls of the bulb to the much cooler surroundings.
2007-06-02 10:44:12 · answer #7 · answered by Mike1942f 7 · 2⤊ 0⤋
Heat is continuously dissipated
2007-06-07 15:11:58 · answer #8 · answered by Brave 3 · 0⤊ 0⤋