Which 5w bulb uses more power?

Which 5w bulb uses more power. Luxeon LED, standard bulb, etc?

  • The 5w Luxeon LED uses more power.

    Votes: 0 0.0%
  • The 5w standard bulb uses more power.

    Votes: 4 25.0%
  • The 5w Xenon bulb uses more power.

    Votes: 2 12.5%
  • They ALL use 5w.....duuuuh!

    Votes: 8 50.0%
  • Almost the same, must calculate resistence and ohms in each case.

    Votes: 2 12.5%

  • Total voters
    16

tron3

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Ok, which bulb uses more power? A 5w Luxeon LED, or any standard 5w dc bulb.

Please explain your answers, I wish to hear from all the scientific brains out there.

Use the force, and choose wisely.
 
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twentysixtwo

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to paraphrase Ms Stein, 5 watts is 5 watts is 5 watts.....


That said, if you have a converter on a LED, then 5 watts at the LED is NOT 5 watts at the battery due to converter losses. In addition, as the batteries run out on a DD LED and an incandescent, you lose light and consume power much faster on the incandescent - Incandescent bulbs drop in resistance (and therfore suck up power faster) and put out less light as the voltage drops.

On the other hand, a DD LED will actually increase in efficiency (light out per unit energy) up to a point as the voltage drops.
 
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twentysixtwo

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Yes.
A hotter filament has higher resistance, which is why incandescent bulbs become more efficient (more light per unit power) when overdriven.

Their resistance drops when underdriven and they are run colder.
 

Flash_Gordon

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AndrewL said:
I thought hotter wires increased resistance

You are correct. An important principal of basic physics. Could be explained with a discussion of ions, metal cations and atoms.

Unnecessary, just think of superconductivity. For example, pure copper attains zero resistance at about -245C. Real tough to obtain/maintain even in a lab experiment. That's why current superconductors are ceramic and not metals.

It is this increase in resistance, that causes the filament to become brighter and brighter up to the point where there is no longer enough available current flow to to overcome the resistance.
 

tron3

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Ok, explaining my vote.

I said, almost the same, but you have to measure...

Why? Well, for lightbulbs, the wattage rating is sometimes a funny thing.
An Incandescent 20 watt bulb uses 20 watts of power. But a 20w florescent outputs the same wattage with less current. Why? Because they blink 60 times a second...not a constant draw.

You have clear bulbs which clearly are the biggest wasters. Some frosted and coated bulbs reflect light energy back in causing the filiment to get hotter and generate more watt output without using more power. 60 watt MISER bulbs from 55 watts of power are a prime example.

Then you have Bulbs with magnified heads, such as many penlights. You think the output is weak now? Look at it when it isn't focused, or reflected.

In short, it's a VERY tricky question to answer clearly in any one fasion. Look at the magnified Striker-VG. 3w of power amplified to 5 or 6 watts. Still love that thing.

That is why it is hard to convert lux to lumens, and vice verser. Bulb design and size has much to do with it. ex: A 5w Xenon bulb for a flashlight, is maybe 3/8" long. The actual gas glows to create light. If you have a 5w Xenon night light, it is about 2" long. The larger size is going to look like much more light. Why? More surface area.

It's the age old argument of comparing apple and oranges, but saying the best apples taste more like pears. Hmm, pear-apple...is there such a thing? I want one now.
 

Flash_Gordon

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Hi tron3-

Your explanation in your last post is generally correct, however it is not an explanation of your original question. Your original question was "Ok, which bulb uses more power? A 5w Luxeon LED, or any standard 5w dc bulb."

There is of course only one correct answer which is neither. A watt is the standard unit of power (energy per unit time), the equivalent of one joule per second. The watt is used to specify the rate at which electrical energy is dissipated, or the rate at which electromagnetic energy is radiated, absorbed, or dissipated.

There is no variance allowed or explanation possible. If there were, Ohm's Law and in fact a large number of other physical principles would not exist, and we would all be in deep sauce.

Had you stated your original question in any number of subtly different ways, such as which uses the most or least current per watt, which produces the most lumens per watt, which produces the most or least heat per watt, your explanation and discussion of the different devices would apply.

Mark
 
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tron3

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Flash_Gordon said:
Hi tron3-

Your explanation in your last post is generally correct, however it is not an explanation of your original question. Your original question was "Ok, which bulb uses more power? A 5w Luxeon LED, or any standard 5w dc bulb."

There is of course only one correct answer which is neither. A watt is the standard unit of power (energy per unit time), the equivalent of one joule per second. The watt is used to specify the rate at which electrical energy is dissipated, or the rate at which electromagnetic energy is radiated, absorbed, or dissipated.

There is no variance allowed or explanation possible. If there were, Ohm's Law and in fact a large number of other physical principals would not exist, and we would all be in deep sauce.

Had you stated your original question in any number of subtly different ways, such as which uses the most or least current per watt, which produces the most lumens per watt, which produces the most or least heat per watt, your explanation and discussion of the different devices would apply.

Mark

Now you know why I posted the question here. :) Thanks for pointing out that watts is a measure of energy dissapation. That explains why watts is a unit of measure for lightbulbs, hair dryers, etc.
 

AndrewL

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So will batteries last longer with a 5watt LED or a 5watt bulb (I know we'll need specific bulbs but I'm not great with all the differences)?
 

jayflash

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If both driven near their specs the LED will be more efficient in producing light - even with reasonable converter (if used) loses. The LED + electronics will consume slightly more current if exactly 5W is flowing through each lamp. With DD both are the same.
 

asdalton

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twentysixtwo said:
A hotter filament has higher resistance, which is why incandescent bulbs become more efficient (more light per unit power) when overdriven.

Resistance only tells you how much voltage is required to push a given current, not how much visible light will be emitted at a given power. Resistance can also be used to convert voltage or current into power consumed (power = I^2*R = V^2/R), but that still tells us nothing about light.

The amount of radiation emitted by a hot filament is directly determined by its temperature. A hotter temperature means more radiation (power ~ T^4), and the filament must consume at least that much electrical power to match. But a hotter filament also emits a greater proportion of its radiation at shorter wavelengths--more visible light rather than infrared, and more yellow and green light rather than red and orange (lambda_max ~ 1/T). This phenomenon results in more lumens per watt, and it is the reason why incandescent lamps are more efficient if you run them very hot.
 
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