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Why is it light bulbs always tend to break the moment when you switch them on.
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Light bulb
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- #2
marcus
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Originally posted by username
Why is it light bulbs always tend to break the moment when you switch them on.
I don't know the answer but I do know that the tungsten
filament is subject to various different kinds of stress.
One is the sudden change in temp from room temperature to half the temperature of the surface of the sun.
The sun is around 6000 K and a 100W bulb filament is about 3000K. The temperature shock must be significant.
Another thing is that the filament may actually have broken when you turned the bulb off and the filament cooled (which would also involve some stress) so when you turn it on you get the effect of a break that happened earlier.
And if, while the bulb has been sitting, oxygen has leaked in from the outside, then when you turn it on is the time it will burn the filament.
Finally, and this is something I am not at all sure of, I have always suspected is that the filament
acts on itself with magnetic forces at the moment you run
current thru it. The mechanical stress of the surge of current might be just enough to break the filament if it is ready to go.
Anyway, I can't say for sure but I have noticed the same thing. If it is going to burn out it is very likely to happen when you turn it on.
- #3
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Originally posted by username
Why is it light bulbs always tend to break the moment when you switch them on.
As you repeatedly switch the bulb on and off, the filament repeatedly heats and cools. This thermal cycling causes the stress in the filament to also cycle, which in turn causes the tiny cracks that exist in all materials to grow bigger. Eventually, there is insufficient cross section of material to take the current and it breaks.
The effect is called thermal fatigue - it occurs in all sorts of electrical connections - including the pins on microchips. That's why it's sometimes recommended that you don't switch off the power to a computer.
Cheers,
ron.
- #4
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Here's why!
When the filament is at room temperature it only has about 15% the resistance it has when hot. So nearly seven times as much current flows in the circuit until it warms up. As the filament ages some parts of it become thinner than the rest, this subject has been studied very carefully by the light bulb people, and these thinner parts overheat during the surge.
When the filament is at room temperature it only has about 15% the resistance it has when hot. So nearly seven times as much current flows in the circuit until it warms up. As the filament ages some parts of it become thinner than the rest, this subject has been studied very carefully by the light bulb people, and these thinner parts overheat during the surge.
- #5
BoulderHead
Hahaha, the way mine only last a couple of months I'd agreeOriginally posted by Tyger
...this subject has been studied very carefully by the light bulb people...
- #6
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Oh.. oh... oh...
I actually know this one.
Light bulbs are designed for a certain operating temperature. It is intended that the majority of the life time of the filament is spent at this temperature, and the filament is made so that it has the correct resistence at exactly this temperature.
As we all know, increased temperature increases resistance of metals. Hence, a filament that is off has a significantly lower resistance than a filament that is on, and hence heated to the operating temperature.
Now, formula to convert electrical power...
P = IV
Now, substitute in ohm's law...
I = V/R
Therefore: P = V2/R
As R is smaller, the power output of the filament increases exponentially. Hence, the energy coming out of the filament is largest when it is being switched on, so it has the largest chance of overheating (some energy going into heat) and melting at this time.
Simple, isn't it?
Damn... Tyger got there first... Ah well...
I actually know this one.
Light bulbs are designed for a certain operating temperature. It is intended that the majority of the life time of the filament is spent at this temperature, and the filament is made so that it has the correct resistence at exactly this temperature.
As we all know, increased temperature increases resistance of metals. Hence, a filament that is off has a significantly lower resistance than a filament that is on, and hence heated to the operating temperature.
Now, formula to convert electrical power...
P = IV
Now, substitute in ohm's law...
I = V/R
Therefore: P = V2/R
As R is smaller, the power output of the filament increases exponentially. Hence, the energy coming out of the filament is largest when it is being switched on, so it has the largest chance of overheating (some energy going into heat) and melting at this time.
Simple, isn't it?
Damn... Tyger got there first... Ah well...
- #7
Integral
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The net result is, are you really saveing money or energy by turning off a light every time you leave a room. It seems to me, that rather then rating a bulb by the number of hours it will run continuously, they ought to be rated in On/off cycles. Wonder if anyone has done a study on the optimal time to leave a bulb on, rather then switching it off. e.g. If I am going to be out of a room 5min, should I leave the light on or turn it OFF? How about an hour?
Costs to consider:
CP=Cycle price= Price of Bulb/Rated number of on/off cycles
RR=Run Rate = local cost of electriciy/Rated run life of bulb
Well, I really do not have time to complete such an analysis, given those numbers as starting point perhaps others can continue. I will put pensil to paper and be back with more later.
Costs to consider:
CP=Cycle price= Price of Bulb/Rated number of on/off cycles
RR=Run Rate = local cost of electriciy/Rated run life of bulb
Well, I really do not have time to complete such an analysis, given those numbers as starting point perhaps others can continue. I will put pensil to paper and be back with more later.
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