# Final Temperature of a Lightbulb filament

## 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 im 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?

kuruman
Homework Helper
Gold Member
What did you use for the constant of proportionality ∝ ?

collinsmark
Homework Helper
Gold Member

## 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 im 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!

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.

I used the temperature coefficient for tungsten which is 4.5E-3 according to my book

That looks correct to me!

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 cant i just substitute 1/6 Io for I?

kuruman