wattage

[dlbarkl]

Is there any way to tell the wattage of a lightbulb if the glass has been broken.

 

[Big_Spark]

Is there any way to tell the wattage of a lightbulb if the glass has been broken.

If the coil is unbroken you could measure the resistance of the coil and do a simple calculation, however if the glass is broken, it is highly likely the coil is too..so the answer would be a resounding NO..(Unless the manufacturer also wrote on the base of the lamp)

 

[ban-all-sheds]

If the coil is unbroken you could measure the resistance of the coil and do a simple calculation

What is the simple calculation to derive the consumption of a lamp given the resistance of the coil at room temperature?

 

[Big_Spark]

If the coil is unbroken you could measure the resistance of the coil and do a simple calculation

What is the simple calculation to derive the consumption of a lamp given the resistance of the coil at room temperature?

Use Ohms Law fella...

I = V/R so once you know the resistance of the coil and the voltage it works at you then know the current it draws, once you know that it is simply P=VI..

But then I KNOW you KNOW all this BAS...

Example...

The Coil of a lamp has a resistance of 400 Ohms..
The supply Voltage 120Vac

So I= V/R = 120 / 400 = 0.3A

Now we know the current drawn, so P = VI = 120 x 0.3 = 36W

 

[Cork]

Hi Big Spark,
If you think Ohms law will work that easily consider this experiment.
Put about 2 volts across a 60 watt 240volt lamp and measure the current.
I would expect you will get about 25mA.
From Ohms law that means a resistance of 80 ohm.
Try it again with 240 volts power supply and you will find the lamp glows white and about 250mA drawn.
Use ohms law the resistance is about 960 ohm.
eureka: Resistance changes with temperature. You need to consider the "Temperature cooeficient of resistance" if you want to work the math. You will also need to be able to guess at the temperature the filiment will work at and you need to know the the material used in the filiment.

If you want to try the experiment please wire it carefully. Mains voltage bites. We would hate to see a big spark. For US the mains voltage drops but the experiment will still work.

 

[Big_Spark]

Cork, your talking ****e..

The resistance of a fixed wattage Tungsten filament lamp is FIXED..

Here...

240V Lamp...60W

60 / 240 = 0.25A

R = V / I = 240 / 0.25 = 60 Ohms

Now if you want the power at your mythical 2V

I =V /R = 2 / 60 = 0.0333A

P = VI = 2 x 0.033 = 0.066W

Now if you are able to rewrite the laws of physics, be my guest. The temperature coeficient of the TUNGSTEN filament will have been accounted for by the manufacturer and they are always calcualted at room temperature or 20C. This is standard practice according to Osram and Philips.

 

[ban-all-sheds]

B_S - that is complete rubbish.

Tungsten has a positive temperature coefficient of resistance, so it's resistance increases with temperature.

The coefficient of tungsten is 0.004403

Since tungsten filaments operate at 1000's of degrees C, the hot resistance is 10-15x the cold.

And you talk about others re-writing the laws of physics....

 

[dlbarkl]

My original question was about finding out the wattage of a bulb with the glass missing. Is there anything different about the thickness of filaments, mounts or distance between them?

 

[Big_Spark]

B_S - that is complete rubbish.

BAS..as usual you have not read my entire post. now there is a surprise. I have not said the Temperature coeficient does not count, I have simply said that ACCORDING TO PHILIPS AND OSRAM they have taken this into account and that calculating the resistance of the coil at room temperature will give you the resistance of the element and thus the wattage.

The resistance of the coiled coil may increase as it's temperature rises, as many, if not the majority of materials do, however as I said, the resistance of the coiled coil is FIXED..it's resistance at a given temperature is the same at all times, unless the element has started to break down which means the resistance at a given temperature will change ever so slightly. The initial and final resistance of a 60W lamp and a 100W lamp are different, and both are fixed for a given temperature..or do you dispute that too??

Now if you wish to contact these rather large companies and tell them their talking rubbish, be my guest, but as they are the two largest lamp manufacturers and researchers on the planet, I think they know more than you and I and everyone else on this forum.

I do wonder if you even understand what you have posted or simply cut and paste it in an attempt to try to make me look wrong and foolish. That was the intention of your original question..and you went quiet as you got the correct answer.

Perhaps if you wanted to be helpful to the original question poser, you would also have posted this from the same source as your GIF above..

Since the cold resistance of tungsten filaments is about 1/12 to 1/16 the hot resistance, it might be expected that the initial inrush current when rated voltage is applied to a cold filament would be 12 to 16 times the rated current which flows when the filament heats up to rated temperature. The actual initial inrush current is generally limited to some smaller value by circuit reactance and is a function of the position on the ac wave at which the voltage is applied.

So based on this the original example would expect a cold resistance of about 3.75 to 6 Ohms if the lamp was a 60W rated at 240V.

 

[Big_Spark]

My original question was about finding out the wattage of a bulb with the glass missing. Is there anything different about the thickness of filaments, mounts or distance between them?

I think the only way you would find that information out is from the manufacturer. I suspect that the pillars of the lamp will be equal distances apart for lamps of the same bulb diameter. The resistance of the filament is the only surefire way to go, but if this is damaged then your knackered.

 

[ban-all-sheds]

I will look into this (I haven't "gone quiet" - I've been busy), and I will measure some lamps.

One observation I will make, though - if you're right a "100W" lamp actually consumes 6-10W when running. Which makes me wonder why there is such an interest in making, selling and buying 20W CFLs as "energy saving" equivalents...

 

[Big_Spark]

I will look into this (I haven't "gone quiet" - I've been busy), and I will measure some lamps.

One observation I will make, though - if you're right a "100W" lamp actually consumes 6-10W when running. Which makes me wonder why there is such an interest in making, selling and buying 20W CFLs as "energy saving" equivalents...

BAS..I think it is simply the manner in which the companies give out the information, they quote the cold figures for a given wattage..but in fact it is the resistance at operating temperature. I had never given this any consideration as I have never had a reason to measure the resistance of a lamp element...this thread made me look into it a bit deeper..

As for the 100W lamp, standard Tungsten Filament lamps are only about 20% to 30% efficient, depending on how they are manufactured. The vast bulk of the consumed emergy is converted into Infra-Red (Heat) radiation not visible light.

For comparison, a 20W Compact Flourescent lamp is equivelent in light output to a filament lamp of about 100-120W, yet will only consume about 35W of energy..

 

[ban-all-sheds]

B_S - that is complete rubbish.

BAS..as usual you have not read my entire post. now there is a surprise.

I did read your entire post, which is why I was able to say it was complete rubbish.

I have not said the Temperature coeficient does not count, I have simply said that ACCORDING TO PHILIPS AND OSRAM they have taken this into account and that calculating the resistance of the coil at room temperature will give you the resistance of the element and thus the wattage.

Firstly, surely even a brief moment of reflection would have at least raised doubts about whether, and why, a lamp manufacturer would quote the power that the item would consume if somehow the filament could be persuaded to remain at room temperature. Ignoring, of course, the fact that if it remained at room temperature it wouldn't give out any light. Why would they say that a lamp that when it was at operating temperature was a 7-10W device, was a "100W lamp"?
Secondly, what on earth would such a practice do to the validity of calculating loads on circuits? Why have I never before seen any hint of the idea that an incandescent lamp actually consumes one tenth to one fifteenth of its rated power when running? Could it be because it doesn't?
Thirdly, as I realised, such a scenario makes a complete nonsense of all the arguments in favour of more efficient "energy saving" lamps. Why didn't you spot that?
Fourthly, no matter whether you think a lamp consumes 60W at room temperature, or 60W at operating temperature, and whether or not you think that the resistance value you have been given for a filament is its resistance at room temperature or its resistance at operating temperature, your example assumed that the resistance of the filament would be the same at operating temperature as it would at whatever temperature the filament reached when supplied with 2V:

240V Lamp...60W

60 / 240 = 0.25A

R = V / I = 240 / 0.25 = 60 Ohms

Now if you want the power at your mythical 2V

I =V /R = 2 / 60 = 0.0333A

P = VI = 2 x 0.033 = 0.066W

It's also got a dreadful arithmetic error - 240 ÷ 0.25 is 960, not 60.

And you had the gall to preface that with "Cork, your talking shite.."

The resistance of the coiled coil may increase as it's temperature rises,

"May" increase? Implying it may not?

as many, if not the majority of materials do,

Name me a metal that does not have a positive temperature coefficient of resistance.

however as I said, the resistance of the coiled coil is FIXED..it's resistance at a given temperature is the same at all times, unless the element has started to break down which means the resistance at a given temperature will change ever so slightly. The initial and final resistance of a 60W lamp and a 100W lamp are different, and both are fixed for a given temperature..or do you dispute that too??

No - I do not dispute that at a constant temperature the resistance of a lamp filament is constant, nor do I dispute that at the same temperature the resistances of 60W and 100W filaments are different.

What I disputed was the ridiculous assertion that the resistance of any particular filament is the same at room temperature as it is at its operating temperature of thousands of degrees. And you now realise that it was tosh.

Now if you wish to contact these rather large companies and tell them their talking rubbish, be my guest, but as they are the two largest lamp manufacturers and researchers on the planet, I think they know more than you and I and everyone else on this forum.

I'm sure they know a great deal more about their field of expertise than you or I. I'm equally sure that you either did not understand, or completely failed to think about what you had read from them.

I do wonder if you even understand what you have posted

Of course I do - why should I not? It's incredibly basic stuff.

or simply cut and paste it in an attempt to try to make me look wrong and foolish. That was the intention of your original question

No it wasn't. That may be your intention when you post in disagreement with others, but not everyone is like you. And anyway, you do a good enough job of that yourself.

and you went quiet as you got the correct answer.

And that correct answer was?

Perhaps if you wanted to be helpful to the original question poser, you would also have posted this from the same source as your GIF above..

Why?

How would any of those formulae answered the question "Is there any way to tell the wattage of a lightbulb if the glass has been broken"?

And how would any of them have helped show that your claim that the resistance of a filament does not change with temperature was false?

So based on this the original example would expect a cold resistance of about 3.75 to 6 Ohms if the lamp was a 60W rated at 240V.

No - see above for details of your order-of-magnitude arithmetic error.

If the filament at its operating temperature has a resistance of 960 ohms, then we'd expect it to be 60 - 80 ohms, but I'm not sure I'd want to trust (potentially) my life to picking a safe lamp power for a luminaire on the basis of a measurement and calculation like that....

BAS..I think it is simply the manner in which the companies give out the information, they quote the cold figures for a given wattage..but in fact it is the resistance at operating temperature. I had never given this any consideration as I have never had a reason to measure the resistance of a lamp element...this thread made me look into it a bit deeper..

But we still had to go through the cycle (mercifully compressed this time, I will admit) of

- You post tosh.
- People say "that is tosh, and this is why"
- You ignore their arguments and launch into sarcastic and insulting replies
- Eventually you realise that what you posted was tosh.

Why do you keep doing this?

 

[Big_Spark]

BAS..Typographical error not noticed on ther 60 instead of 960..I apologise..

As for the rest you post..*******s fella..as usual your attempting to be the big I am when you don't have a clue what your talking about AGAIN..

Yopu simply go through my posts attempting to pick holes...In this case a typographical error has given you an IN..

You question whether I should question information from manufactirers about a blessed lamp...Unlike you fella I have a life..I stated I had never given it any thought..

This is boring boring boring BAS...

Nothing I have posted in incorrect (save the typo) and yet you waffle waffle waffle as if you know...

Stop making accusations you are not able to support.

 

[ban-all-sheds]

BAS..Typographical error not noticed on ther 60 instead of 960..I apologise..

As for the rest you post..**** fella..as usual your attempting to be the big I am when you don't have a clue what your talking about AGAIN..

Oh look - B_S is challenged on something, so he resorts to insults.

What have I said that is incorrect? Please point to a specific error, and please show how that proves that I don't know what I'm talking about.

And please justify "AGAIN".

Yopu simply go through my posts attempting to pick holes...In this case a typographical error has given you an IN..

I go through your posts trying to get you to see where you've gone wrong, because as I said, this is not atypical behaviour from you. As for the "typo":
Firstly, I just said that it was "also" in terms of an error - your example calculation was hopelessly flawed anyway, as we shall see.

Secondly, it's not really a typo - you didn't misspell something, or accidentally hit an adjacent key on your keyboard, or accidentally misplace the decimal point - you wrote the formula "240 / 0.25" you knew from Ohm's Law that you should be doing the calculation "V / I", and then you performed "240 x 0.25". That is more than just a typographical error - it's an egregious mathematical mistake.

Thirdly, I didn't eventually spot it after ages poring over your post looking for an "in" - at the same time as I read it I instantly saw the mistake. Why didn't you spot it?

You question whether I should question information from manufactirers about a blessed lamp...Unlike you fella I have a life..I stated I had never given it any thought..

Firstly, it doesn't require much thought. You don't have to "not have a life", and it doesn't have to be a welcome diversion for it to be immediately so questionable that you either find the time to check it out and think it through properly or you shelve it as information which is so questionable that you don't use it.

Secondly, if you end up giving out duff information because you are too busy to spot that it's wrong, perhaps you shouldn't be posting in the first place.

This is boring boring boring BAS...

What is boring is the way you sometimes refuse to consider whether you could be wrong until the repeated attempts to get you to do so have dragged on and on...

Nothing I have posted in incorrect (save the typo) and yet you waffle waffle waffle as if you know...

OK - let's correct the typo, and see what you meant to say.

240V Lamp...60W

60 / 240 = 0.25A

R = V / I = 240 / 0.25 = 960 Ohms

Now if you want the power at your mythical 2V

I =V /R = 2 / 960 = 0.0020833A

P = VI = 2 x 0.0020833 = 0.004166W

In other words, you're trying to say that the resistance of the filament is the same when 250mA is passing through it as it is when 2.0833mA is passing through it.

In other words you're trying to say that the filament would be just as hot, and therefore just as bright when supplied with 2V as it would be when supplied with 240V.

Stop making accusations you are not able to support.

What accusation have I made that I cannot support?

That you cannot sensibly determine the rating of a lamp by measuring the cold filament resistance?

That your example calculation which ignores the fact that the resistance of a filament changes with temperature is wrong?

That you failed to spot your egregious "typo"?

Please advise.


Edited to make calculation more clearly a quote