2
\$\begingroup\$

Why electric light bulb more often burns out when we switch from OFF to ON our wall switch, but not during regular work? What happens to it so it burns?

\$\endgroup\$
  • \$\begingroup\$ Why, just why is this close-voted as off-topic? It's a perfectly good question about how electrical device failure develops. \$\endgroup\$ – sharptooth Oct 5 '15 at 13:38
  • \$\begingroup\$ I agree with @sharptooth; this seems reasonably on-topic. \$\endgroup\$ – Adam Haun Oct 5 '15 at 21:30
5
\$\begingroup\$

The sudden change in temperature from cold to hot is quite a violent thermal-mechanical shock to a light bulb. Of course you get the same when you turn it off too but then you'll only notice it has failed when you come to switch it back on.

Also, the cold-resistance of the filament is quite low so, when power is initially applied there is a very high current surge that settles down in a few tens of milliseconds to normal running current. Wiki says this: -

The actual resistance of the filament is temperature dependent. The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating. For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold resistance is much lower (about 9.5 ohms).

\$\endgroup\$
  • \$\begingroup\$ Wow, the resistance fact is so obvious, but I never thought of it, until you pointed out. \$\endgroup\$ – Nazar Oct 5 '15 at 12:09
  • \$\begingroup\$ So it acts as PTC thermistor? Why people didn't made it from some material whose resistance doesn't change with temperature? \$\endgroup\$ – TicTacToe Oct 5 '15 at 12:10
  • 1
    \$\begingroup\$ @Zola one of the least temperature dependent materials would be Constantan or Manganin, but they have melting points at around 900-1200°C, at that temperatures they would glow red but nowhere near any usable light emission. Another thing is that the temperature coefficient changes with temperature and is mostly constant in a region of far less degree than 900°C or even 2700K (basically somewhere where a resistor is used as a resistor, so up to 200°C maybe). \$\endgroup\$ – Arsenal Oct 5 '15 at 13:15
  • 1
    \$\begingroup\$ Cooling down of the filament is much less violent than heating it up. \$\endgroup\$ – PlasmaHH Oct 5 '15 at 13:38
  • 1
    \$\begingroup\$ It's worth noting that some incandescent light applications. Eg theatre lights have a standby mode that keeps them warm but not emitting visible light. \$\endgroup\$ – BenG Oct 6 '15 at 1:46
1
\$\begingroup\$

The failure mechanism for incandescent bulbs needs to be looked at a microscopic scale. Over time, tungsten evaporates from the filament and condenses on the glass bulb. This creates the dark discoloration you see on well-used bulbs. As this occurs, you get local erosion of the filament, which ultimately creates a positive feedback cycle. Where a filament gets thinner, its local resistance goes up, so it gets hotter than the other parts of the filament. This increased resistance causes a local hotspot to occur, which increases the evaporation rate, which increases the temperature, etc. At some point the filament fails. If the failure occurs simply because the filament gets too weak, the bulb fails during normal operation. More often the shock of turn-on is too much for the local thin point, since it is (having gotten thinner) physically weaker than the rest of the filament, and the bulb fails on turn-on.

\$\endgroup\$
  • \$\begingroup\$ Never bothered to find out why bulbs get black spot. I thought it was a consequence of the flash pop off the bulb dying, not a slow progressive failure \$\endgroup\$ – Passerby Oct 5 '15 at 21:40
  • \$\begingroup\$ @Passerby - Nah. It's just that you don't look closely at the bulb until after it burns out. \$\endgroup\$ – WhatRoughBeast Oct 5 '15 at 21:41
  • \$\begingroup\$ In halogen bulbs the evaporated tungsten will redeposit back onto the filament, and therefore they don't get black spots. Source: en.wikipedia.org/wiki/Halogen_lamp#Halogen_cycle \$\endgroup\$ – Mads Skjern Sep 14 at 18:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.