Solving Filament Failure Mode with Maths

[CWatters]

Homework Statement

Not homework but this seems the best place to post this. In a private conversation someone posted this explanation from a book on how a lamp filament fails..

when it comes to the filament in a lamp the book said the thinner part of the filament is hotter than other parts and evaporation occurs faster there because of smaller cross sectional area and higher resistance. This makes the thinner part even thinner until it breaks.

I'm having some trouble doing the maths to show that the thinner higher resistance part gets hotter...

Homework Equations

Ohms law

The Attempt at a Solution

[/B]
Imagine the filament is made up of two resistors in series R₁ and R₂ (See diagram). Let R₂ be the part that's getting thinner for some reason...

The voltage on R₂ is given by the potential divider rule..

V_R2 = V * R₂/(R₁+R₂)

The power dissipated in R₂ is

W_R2 = (V * R₂/(R₁+R₂)²/R₂

If you expand and simplify you get..

W_R2 = (V² * R₂) / (R₁² + 2R₁R₂ + R₂²)

So if R₂ increases the power in R₂ decreases due to the R₂². That should reduce the temperature in R2 which is the opposite of what the book suggests. Have I made an error?

 

[CWatters]

I suspect the problem is to do with the physics of the filament. eg the thinner part has less surface area so although the power dissipated in that bit is lower the surface area is also lower so it's harder to dissipate the power.

 

[tech99]

Homework Statement

Not homework but this seems the best place to post this. In a private conversation someone posted this explanation from a book on how a lamp filament fails..
I'm having some trouble doing the maths to show that the thinner higher resistance part gets hotter...

Homework Equations

Ohms law

The Attempt at a Solution

[/B]
Imagine the filament is made up of two resistors in series R₁ and R₂ (See diagram). Let R₂ be the part that's getting thinner for some reason...

View attachment 85636

The voltage on R₂ is given by the potential divider rule..

V_R2 = V * R₂/(R₁+R₂)

The power dissipated in R₂ is

W_R2 = (V * R₂/(R₁+R₂)²/R₂

If you expand and simplify you get..

W_R2 = (V² * R₂) / (R₁² + 2R₁R₂ + R₂²)

So if R₂ increases the power in R₂ decreases due to the R₂². That should reduce the temperature in R2 which is the opposite of what the book suggests. Have I made an error?

If R2 is small compared with the total resistance, the current through the combination will be roughly constant. The dissipation in in R2 will be I^2 x R2, so it is proportional to the resistance of R2. In your formula, R2 appears in both top and bottom, so your statement is not quite accurate.

 

[CWatters]

Ah yes I'm forgetting that the square of a number less than 1 is smaller than the original number.

 

[rude man]

If R2 is the part getting thinner then, assuming R2 is same length as the rest of the filament R1, R2 will be bigger than R1, and
power in R2 = i*R2 = [V/(R1 + R2)]R2 which is > [V/(R1 + R2)]R1.

Question for you: why is R2 bigger than R1?