Why Does a Light Bulb's Resistor Burn Out When Directly Connected to an Outlet?

AI Thread Summary
When a light bulb's resistor is connected directly to an outlet, it burns out due to oxidation, which occurs when the filament is exposed to air. The filament heats up rapidly in the presence of oxygen, leading to combustion and a quick failure. In contrast, when the filament is enclosed in a gas mixture like argon or nitrogen, combustion is prevented, allowing it to operate without burning out. The resistor's failure is a result of the chemical reactions that happen when it is exposed to air at high temperatures. Thus, the gas inside the bulb plays a crucial role in preventing the resistor from frying.
theplanck
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I was with my electrician friend the other day, and we stuck two copper wires into each of the holes of a regular 120V outlet then smashed a 100W light bulb and took the resistor (small coiled looking thing) out of it and set it on a piece of wood and touched the copper wires to each end of the resistor, and it immediately lit up and was friend black.

My question is why does this not happen when you do the virtually the same thing without smashing the light bulb, that is, you touch the ends of the wires to the metal on the back of a light bulb? Does it have to do with the gas that is inside the light bulb? Because I would think that the resistor should just get hot and radiate light according to blackbody radiation and not fry up either way? Hope I explained my problem well enough
 
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A lightbulb is essentially evacuated normally - when you heat a filament to that kind of temperature normally, it will instantly oxidize and burn out. Basically, it fries because of the chemical reactions between the filament and air, exacerbated by the extreme temperature.
 
cjl said:
A lightbulb is essentially evacuated normally - when you heat a filament to that kind of temperature normally, it will instantly oxidize and burn out. Basically, it fries because of the chemical reactions between the filament and air, exacerbated by the extreme temperature.


Almost correct. The inside of a light bulb isn't a vacuum, or else the glass would have to be extremely thick for the bulb not to implode! Normally the gas on the inside of the bulb is a mixture of some noble gas (Usually Argon, Krypton, or Xenon) and nitrogen. When the filament is in this gas mixture, combustion is not possible (combustion requires oxygen). In air the filament would heat up just as much as it does in the bulb, but now there is oxygen. Heat + Oxygen = oxidation. An oxidized metal does not carry current as well as metal itself, so the resistance gets higher and higher until the filament "fries".
 
hey thanks for the answer, so basically the mixture of oxygen and heat together causes combustion and it lights up extremely bright for less than a second and then it;s a black fried resistor
 
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