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A typical way of an incandescent bulb to burn out is with a bright blueish flash and often an audible sound and (very rarely) an explosion of the bulb glass.

Wikipedia says the following:

Since a filament breaking in a gas-filled bulb can form an electric arc, which may spread between the terminals and draw very heavy current, intentionally thin lead-in wires or more elaborate protection devices are therefore often used as fuses built into the light bulb.

But it's unclear why an arc would spread and cause very high currents.

Why would it not burn out silently instead?

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TL;DR The electric arc may conduct much higher current than the filament, this is not normal mode of bulb operation so you can see a flash and maybe even have the bulb exploded.

Typical incandescent bulb failure develops while current is flowing through the filament. The filament is a piece of wire about half a meter long coiled such it forms a filament which is about 20 millimeters long. The filament wire is therefore very thin and long and its resistance is relatively high. The filament is attached to the upright supply wires which are not coiled - they are pieces of wire much thicker than that of the filament, their resistance is relatively low.

While the bulb is working current flows through the supply wires and the filament. Filament resistance is what limits the maximum current.

If the filament breaks while current is flowing through it then this situation is identical to starting an arc with electric arc welder. The filament breaks while current is flowing though it, its ends move from one another and chances are an electric arc starts.

While the filament was intact the maximum current was initially limited by the filament resistance and there was no way for the current to flow other than through the entire length of the filament. If an electric arc starts then this limitation is no longer true - current can flow through the ionized gas and the length the arc can span is limited by the power supply voltage. This allows the arc ends to travel from the broken filament ends to the supply wires and no longer limit its current by the filament resistance. Once the arc starts it ionizes the surrounding gas and so that gas can conduct current too which allows for much higher currents than the one initially flowing through the thin filament. The current is now limited by the supply wires resistance.

This causes rather intensive arc which produces a bright flash and sometimes makes the bulb explode.

This is why you need a fuse inside the light bulb (near the socket). If arc develops then current through the fuse gets much higher and the fuse is melted and then the circuit is broken and the arc stops. If the fuse fails to melt fast enough you may have a circuit breaker tripped.

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    \$\begingroup\$ We really don't need the extreme-worst-case scare mongering here. I have never seen a light bulb develop any kind of sustained arc, ever. 99% of the time, the bulb burns out when it is switched on, because of the mechanical stress caused by the rapid heating. There's a quick flash of arc and then nothing. And a fuse that could withstand the normal startup surge of the bulb would not be much help in stopping an arc. We certainly don't have a requirement for such fuses here in the US. \$\endgroup\$ – Dave Tweed Aug 16 at 13:53
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    \$\begingroup\$ Interesting. At 220 volts about every second bulb would produce a flash when burning out and certain "lucky" people witness more than one bulb exploding. Of source it's no sustained arc - either the fuse melts or the circuit breaker gets tripped or a bulb explodes and the arc stops promptly. \$\endgroup\$ – sharptooth Aug 16 at 14:02
  • \$\begingroup\$ TL;DR TL;DR There is an open circuit. \$\endgroup\$ – KingDuken Aug 16 at 15:27
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    \$\begingroup\$ Internal lamp fusing: i.imgur.com/272zAWT.png from the Sylvania Catalog An intense arc can certainly develop over ~15-20mm from a high fault current 240V source. I have exploded 5x20mm fuses. It would be interesting to see the CSA standard C22.2 NO. 84-05 for incandescent light bulbs. Doesn't seem to address fault current limiting from the index, so maybe it's a Euro requirement and/or manufacturer's initiative. \$\endgroup\$ – Spehro Pefhany Aug 16 at 15:34
  • \$\begingroup\$ None of the small box of incandescent bulbs I have has a UL listing, CSA only, unlike LED and CFL bulbs, so maybe there were/are effectively no standards at all in the USA for incandescent bulbs? I'm in Canada but we usually get very similar products. Even the CSA standard excludes non-consumer types. \$\endgroup\$ – Spehro Pefhany Aug 16 at 15:35
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When the tungsten filament burns out (or this one which is also cool), the gap is very small, and widens as the tungsten filament gets welded away by the arc. (in the videos you can see tungsten sparks fly away). The flash is from the arc and is the same blue color as arcs from welders.

As the gap widens the amount of voltage needed to sustain the arc increases, until the arc extinguishes itself.

If there is a current increase, it is probably because the tungsten filament has crossed over and shorted. I did have a light bulb that the filiment broke and still lit up for a few minutes, but was several times brighter because the filiment crossed over and made the total path shorter. If the arc shorted any part of the tungsten filament then this would also create a current increase due to the shorter pathway.

Below one can see a graph of what it takes to sustain an arc. As the distance increases the voltage increases also. (its actually a pressure distance product, but the pressure in the bulb will be fixed for the most part).

enter image description here Source: https://www.researchgate.net/publication/288833649_Arc_Breakdown_over_Very_Small_Gap_Distances/figures

Why would it not burn out silently instead?

Sometimes they do (I've had a few just die, unexcitedly) , but if there is a gap, initially there will be arcing.

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