Why does a light bulb burn out?

AI Thread Summary
A light bulb filament burns out primarily due to the evaporation of tungsten oxide at high temperatures, which decreases its cross-sectional area and increases resistance. This increased resistance leads to higher temperatures, creating a feedback loop where the filament heats up more in certain spots, ultimately causing it to fail. The process is described as a "runaway" situation, where the weakest point in the filament thins first and burns out. Additionally, molecular interactions, such as electron collisions with lattice atoms, can create defects that further disrupt current flow. Understanding these mechanisms explains why a light bulb filament fails at specific locations.
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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