Final Temperature of a Lightbulb filament

AI Thread Summary
The discussion focuses on calculating the final operating temperature of a tungsten filament in an incandescent light bulb during its warm-up period. When the bulb is first turned on, the steady-state current is only 1/6 of the initial current, and the initial filament temperature is assumed to be 20 °C. The resistivity of tungsten increases linearly with temperature, and the relevant equation used is R = Ro[1+α(T-To)]. The calculated final temperature is 1131 °C, although there are some notational errors and a need to specify the temperature coefficient of resistance for tungsten. The solution appears correct, provided the appropriate values and symbols are used.
Augustine Duran
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Homework Statement


When an incandescent light bulb is switched on, it can take a few moments for the filament to fully heat up and reach its equilibrium temperature. Assume the filament of a given light bulb is made of Tungsten, and also
assume the potential difference across the filament is maintained at a constant value during this short warm up
period. If the ‘steady state’ current is measured to be only 1/6 of the current drawn when the lamp is first turned on, then what is the final operating temperature of the filament? Assume the initial temperature of the filament is 20 °C, and assume the resistivity increases linearly with increasing temperature.

Homework Equations


R = Ro[1+∝(T-To)]

The Attempt at a Solution



So I am not given the resistance here but i am told the voltage is constant throughout this warm up. So i can use
R = V/I

Plug into R = Ro[1+∝(T-To)]

V/I = V/Io [1+∝(T-To)]

Since voltage is constant i can divide both sides to cancel them out, also multiply both sides by I and Io leaving me with

Io = I [1+∝(T-To)]

Im told the steady state current is only 1/6 of the initial current when the bulb is first turned on so i can say
I= 1/6 Io and plug it into the equation.

Io = 1/6 Io [1+∝(T-To)]

From here i cancel out Io from both sides and solve for T. The answer is get is
1131 oC, is this correct?
 
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What did you use for the constant of proportionality ∝ ?
 
Augustine Duran said:

Homework Statement


When an incandescent light bulb is switched on, it can take a few moments for the filament to fully heat up and reach its equilibrium temperature. Assume the filament of a given light bulb is made of Tungsten, and also
assume the potential difference across the filament is maintained at a constant value during this short warm up
period. If the ‘steady state’ current is measured to be only 1/6 of the current drawn when the lamp is first turned on, then what is the final operating temperature of the filament? Assume the initial temperature of the filament is 20 °C, and assume the resistivity increases linearly with increasing temperature.

Homework Equations


R = Ro[1+∝(T-To)]

The Attempt at a Solution



So I am not given the resistance here but i am told the voltage is constant throughout this warm up. So i can use
R = V/I

Plug into R = Ro[1+∝(T-To)]

V/I = V/Io [1+∝(T-To)]

Since voltage is constant i can divide both sides to cancel them out, also multiply both sides by I and Io leaving me with

Io = I [1+∝(T-To)]

Im told the steady state current is only 1/6 of the initial current when the bulb is first turned on so i can say
I= 1/6 Io and plug it into the equation.

Io = 1/6 Io [1+∝(T-To)]

From here i cancel out Io from both sides and solve for T. The answer is get is
1131 oC, is this correct?
That looks correct to me! :smile:

'Just a heads-up though,

1) The symbol you used "∝" is, I think, the "proportional to" symbol. I believe you meant to use "α" alpha.

2) In your final equation, Io = 1/6 Io [1+α(T-To)], you've used two "Io"s. I'm pretty sure one of them should be an "I".

3) You haven't specified your value for Tungsten's temperature coefficient of resistance. In the future, you should specify such things in either the problem statement section or the relevant equations section.

Otherwise it looks good to me. :smile:
 
I used the temperature coefficient for tungsten which is 4.5E-3 according to my book
 
collinsmark said:
That looks correct to me! :smile:

1) The symbol you used "∝" is, I think, the "proportional to" symbol. I believe you meant to use "α" alpha.

2) In your final equation, Io = 1/6 Io [1+α(T-To)], you've used two "Io"s. I'm pretty sure one of them should be an "I".

1) oops, they look similar ;p

2) well since I is going to be 1/6 of Io can't i just substitute 1/6 Io for I?
 
Augustine Duran said:
I used the temperature coefficient for tungsten which is 4.5E-3 according to my book
OK, good. Then the correct equatiion to solve as collinsmark suggested is ##I_0 =I( 1+\alpha(T-T_0))##
 
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