100W lamp and a 500mA fuse.

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Following on from a recent rather bold statement I thought I'd do a little experiment for everyone here. I was quite surprised by the result.


Sadly your advice is misleading. Anyone fitting a 500 mA fuse (next size above 0.43 amp) for a 100 watt bulb would be confused when the switch on surge blew it every time.



 
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Cool vid, was that a real 100W lamp?0 A few recent ones I tested were really nearer 60W!
 
Never thought about that! Just done a couple more tests.


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248.2 x 0.42 = 104.244W

I'd say that's pretty close!
 
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It is what I would have expected, a fuse takes time to rupture so within that time the lamp will have reached temperature. I know my clamp on has a max/min reading, but no idea how quick it will react, so cold resistance is a better way to work out the maximum current while it warms up.

Motors are slower reaching speed than a simple bulb so we have fuses designed to allow that in-rush. Semi-conductors are the reverse with overloads of very short duration causing damage, again we have special fuses to protect them. To get the same level of protection with mechanical overloads devices is much harder and with semi conductors near impossible.

I see the simple fuse being used for many years to come. Often the manufacturer publishes charts to show when the fuse will rupture. Some of these are also published in BS7671 even a semi conductor fuse to BS3036 rated 20A for 0.1 seconds will take around 250 amp. I would have thought any real electrician would realise this, and should not need a demo to prove it.

It however does show very well how for a simple bulb a 500 mA fuse will not rupture, they should be built into the bulb so on failure any ionisation will blow fuse in bulb rather than the main fuse. However bitter experience when the main fuse was replaced with a MCB, often when a bulb failed with the typical ionisation flash it would take out the MCB.

With the new LED bulbs watching a video it seems they use a simple resistor as a fuse. I will guess on switch on the cheap LED bulbs also use more power while the capacitor charges and the resistor reduces the rate of charge as well as doubling as a fuse.
 
I had intended to use use slow blow fuses for the 12 strings of 31 lamps in the lighting for my daughter's wedding. The lamps are 12 volt 2 watt with a hot resistance of 72 ohms ( calculated ) and a cold resistance of 10 ohms. When lit a string took approx 4 amps due to voltage on the lamps being adjusted to approx 10 volts to provide a warm glow rather than harsh white light.

Normal 5 amp ( one per string ) fuses were OK when testing the transformer for temperature rise with dummy loads ( resistance wire ).

When supplying a dummy load of 31 lamps a 5 amp "normal" fuse did blow ( peak current of 24 amps was measured ) so the use of slow blow fuses was essential
 
they use a simple resistor as a fuse
It should be a resistor designed to be a fusible resistor, in some circumstances a simple resistor can fail to be just a higher resistance than it was before it "failed". It is not unknown for a overheating resistor to fail and become a lower resistance

But fusible resistors that are garanteed to fail open circuit cost more than simple resistors.
 
It's a 240v bulb as can be seen 15 seconds in on the video.
His mains is 248.2v not 230v as many would like to tell us.
Yes, Winston but his meter does not measure nominal voltage.
It wouldn't be a lot of good if it did, would it?

Do you have a multitude of tables for whatever the voltage is at the time when you are working?



I'm all for what some people may call pedantry but when you are actually wrong and just keeping on, it makes you look silly.
The abbreviation for voltage is a capital V because it is derived from someone's name.
 
And here I frequently measure between 230 and 238 Volts (and typically nearer the 230V end of the spectrum).

I wonder what Winston makes of that?
 

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