filament lamp?

Question

There is a significant current surge when you first switch on a filament lamp .
what's the theoretical model of this phenomenon ?
I assume that there is a huge different between ac & dc in this case .
some one told me that i should get help from faradic , or this formula : L( di / dt )
Please help me out !
I'm so confused

while we say about ohm's law ( V = RI ) it's about the baytry & ...
but i want this case to be solved in alternative current , and as I told it has something to do with L ( di / dt)

while we say about ohm's law ( V = RI ) it's about the batry & ...
but i want this case to be solved in alternative current , and as I told it has something to do with L ( di / dt)

Answer 1

Interesting question. My opinion is that it's predominantly an increase in the resistance of the filament with temperature that causes the current to decrease from a cold start. The inductance L at low frequencies would be negligible, so I wouldn't suspect that the inductive reactance has much effect even during the transient turn on stage. There must also be some temperature expansion and contraction effects that stress the filament and make it burn out sooner the more you switch it on and off. Of course vibration would be another factor on the service life of the filament. Aren't those new flourescent bulbs great?

Answer 2

The resistance of a tungsten filament will boost with temperature. in the previous switching the lamp on the filament is chilly and has a low resistance. The preliminary modern-day will subsequently be particularly severe. (for this reason a kin easy bulb frequently fails at change-on.) The filament heats up without postpone (some tens of milliseconds), the resistance will boost and the present diminishes. It makes no distinction no remember if the filament is provided with d.c. or a.c.

Answer 3

The main reason for the current surge when first switched on is because the Tungsten filament is cold.

When tungsten is cold it has a low resistance, and thus the current is high - as the filament quickly warms up, the resistance increases rapidly, lowering the current level.

If I = V / R ----- you can see that if R is small, V doesn't change and therefore I is large. As R increses I becomes smaller.

This is the main reason a bulb will blow when you first switch it on.....