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

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Discussion Overview

The discussion revolves around the phenomenon of a light bulb's resistor burning out when directly connected to an electrical outlet. Participants explore the reasons behind this behavior, particularly comparing the effects of connecting the resistor directly versus connecting the metal of an intact light bulb.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant describes an experiment where a resistor from a smashed light bulb was connected to a 120V outlet, resulting in immediate burning and blackening of the resistor.
  • Another participant explains that a light bulb is typically evacuated or filled with a gas mixture, which prevents oxidation of the filament at high temperatures, unlike in open air.
  • A later reply clarifies that the gas inside the bulb (noble gases mixed with nitrogen) prevents combustion, while exposure to air leads to oxidation, increasing resistance until the filament fails.
  • One participant summarizes that the combination of heat and oxygen causes the resistor to burn out quickly when exposed to air.

Areas of Agreement / Disagreement

Participants generally agree on the role of gas in preventing oxidation within the light bulb, but the discussion includes varying levels of detail and understanding regarding the chemical processes involved.

Contextual Notes

There are assumptions about the conditions under which the resistor operates, including the nature of the gas inside the bulb and the effects of temperature and oxidation that remain unresolved.

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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