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

[theplanck]

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

 

[cjl]

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.

 

[Xerxes1986]

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".

 

[theplanck]

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

 

[sardar]

Thank you for sharing your experience, however, I must caution against conducting experiments with electricity without proper training and safety precautions. The result you observed is due to the electrical properties of the light bulb resistor. A resistor is a component that limits the flow of electricity, and in this case, the 100 watt light bulb resistor is designed to withstand and dissipate the high amount of energy generated by the bulb. When you touched the wires to each end of the resistor, the electricity flowed through it, causing it to heat up and emit light. The smashing of the light bulb may have caused the resistor to break and release more energy, resulting in the brighter light you observed.

Touching the wires to the metal on the back of a light bulb is not the same as connecting them to a resistor. The metal on the back of the light bulb is not designed to handle the high amount of electricity flowing through it, and it may result in a short circuit or damage to the bulb. Additionally, the gas inside the light bulb plays a crucial role in the functioning of the bulb. It helps to regulate the flow of electricity and contributes to the emission of light.

In summary, the difference in results between touching the wires to a resistor and the back of a light bulb is due to the design and function of each component. It is essential to understand the properties and limitations of electrical components before conducting experiments with them.