Solving Filament Failure Mode with Maths

Click For Summary

Discussion Overview

The discussion revolves around the failure mode of lamp filaments, specifically focusing on the mathematical modeling of temperature differences in filament sections of varying thickness. Participants explore the relationship between resistance, power dissipation, and temperature in the context of filament failure, incorporating concepts from electrical engineering and physics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes a book's explanation that the thinner part of a filament gets hotter due to higher resistance and smaller cross-sectional area, leading to faster evaporation and eventual breakage.
  • Another participant suggests that the thinner part's reduced surface area makes it harder to dissipate heat, potentially complicating the relationship between power dissipation and temperature.
  • A participant attempts to derive the power dissipated in the thinner section (R2) using Ohm's law and the potential divider rule, concluding that an increase in R2 leads to a decrease in power, which contradicts the book's claim.
  • Another participant points out that if R2 is small compared to the total resistance, the current remains roughly constant, and power dissipation in R2 is proportional to its resistance, challenging the initial conclusion drawn from the mathematical derivation.
  • One participant acknowledges a misunderstanding regarding the mathematical implications of squaring a number less than one, indicating a potential error in reasoning.
  • A further contribution questions the assumption that R2, the thinner part, is larger than R1, prompting a discussion on the relationship between the two resistances.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between resistance, power dissipation, and temperature in filament sections. There is no consensus on the correctness of the mathematical approach or the underlying physical principles, indicating ongoing debate and exploration of the topic.

Contextual Notes

Participants note limitations in their assumptions regarding filament geometry and resistance values, as well as the dependence on definitions of power dissipation and temperature relationships. The discussion remains open to further clarification and refinement of these concepts.

CWatters
Science Advisor
Homework Helper
Gold Member
Messages
10,546
Reaction score
2,324

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...

Filament.jpg


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

VR2 = V * R2/(R1+R2)

The power dissipated in R2 is

WR2 = (V * R2/(R1+R2)2/R2

If you expand and simplify you get..

WR2 = (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?
 
Physics news on Phys.org
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.
 
CWatters said:

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..

VR2 = V * R2/(R1+R2)

The power dissipated in R2 is

WR2 = (V * R2/(R1+R2)2/R2

If you expand and simplify you get..

WR2 = (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.
 
Ah yes I'm forgetting that the square of a number less than 1 is smaller than the original number.
 
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?
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 26 ·
Replies
26
Views
4K