Why does a light bulb burn out?

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Homework Help Overview

The discussion revolves around understanding why a light bulb filament burns out in a specific location. Participants are exploring the relationship between resistance, current, and temperature in the context of electrical components, particularly focusing on the behavior of tungsten filaments in light bulbs.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to connect the concepts of resistance, temperature, and the physical properties of the filament material. Questions are raised about the mechanisms of heat dissipation in resistors and the molecular interactions that lead to changes in the filament's structure.

Discussion Status

Some participants are providing insights into the process of filament burnout, suggesting that uneven thinning leads to localized heating and eventual failure. Others are seeking deeper understanding of the molecular processes involved in energy conversion and heat generation within the filament.

Contextual Notes

There is an acknowledgment of varying levels of knowledge among participants, with some expressing uncertainty about the underlying physics of resistors and thermal energy generation.

peteza
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Homework Statement



The problem is that I cannot figure out why a filament will burn out in a particular spot. Higher resistance = lower current why does the filament heat up more?

Homework Equations




R2= R1[1+a(T2-T1)]
Resistance = (resistivity * L)/ A
R= V/I

The Attempt at a Solution



I know that tungsten oxide evaporates form the filament when a critical temperature is reached. This causes the cross sectional area of the filament to decrease and increases resistance. Increased resistance results in increased temperature. The filament gives off more heat as the cross sectional area decreases. This could be because the current is reduced so this excess energy has to go somewhere so it is converted to heat energy and radiated from the filament. I just don't understand how a resistor dissipates heat and why the light bulb 'choses' to burn out in a certain spot.
 
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peteza said:

Homework Statement



The problem is that I cannot figure out why a filament will burn out in a particular spot. Higher resistance = lower current why does the filament heat up more?

Homework Equations




R2= R1[1+a(T2-T1)]
Resistance = (resistivity * L)/ A
R= V/I

The Attempt at a Solution



I know that tungsten oxide evaporates form the filament when a critical temperature is reached. This causes the cross sectional area of the filament to decrease and increases resistance. Increased resistance results in increased temperature. The filament gives off more heat as the cross sectional area decreases. This could be because the current is reduced so this excess energy has to go somewhere so it is converted to heat energy and radiated from the filament. I just don't understand how a resistor dissipates heat and why the light bulb 'choses' to burn out in a certain spot.

I'm not very knowledgeable about such things but it seems to me you have answered your own question.

First:
"...tungsten oxide evaporates form the filament when a critical temperature is reached..."
then:
"...Increased resistance results in increased temperature..."
and finally:
"...tungsten oxide evaporates form the filament when a critical temperature is reached..."

i.e. it's a runaway process.

Whatever spot starts thinning first is the spot that will burn out first. A theoretically perfectly manufactured filament will evaporate evenly until the whole filament burns out in a puff.
 
so how exactly does a resistor decrease current and convert energy to heat energy? what is happening on a molecular level?
 
Thermal energy is generated via electronic collisions with the scattering center (fixed lattice atoms). At one position in the lattice you may have an energetic electron collide with the fixed lattice, this can change the structure of the lattice at that point which will interrupt the flow of electrons in that region. With this "defect" now in place, other electrons will collide with the defect further altering the current path, etc.
 

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