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Question

Explain why lightbulbs almost always burn out just as they are turned on and not after they have been on for some time.

Solution

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When a light bulb is first turned on, its temperature will be as room temperature and the filament will have a lower resistance than when it is hot.

R=ρ0LAR=\dfrac{\rho_0L}{A}

Noting that

ρ=ρ0[1+α(TT0)]\rho=\rho_0\left[1+\alpha\left(T-T_0\right)\right]

Therefore,

ρ=ρ0[1+αΔT]\rho=\rho_0\left[1+\alpha\Delta T\right]

This lower resistance means that there will be more current through the lightbulb filament while it is colder since the resistivity ρ\rho is at its minimum value, so do RR.

I=VRI=\dfrac{V}{R}

This suddenly high current flow will make the filament temperature increases very fast and this temperature increase will melt part of the filament immediately.

From all the above, if the temperature change is huge, the filament will be destroyed, and the current will stop.

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