# What is changing in latest light bulb technologies?

I'm confused with the latest home lightning bulbs.

Understanding filament bulbs was easy. For example take 220V, 100W filament bulb:

Power = $V^2/R$ Filament gets heated and emits energy in the form of light & heat. Whose sum is 100J per second. I'm getting a light energy of little less than 100J per sec. Its very clear!

Now lets take latest CFL :

How can it be equivalent to 6 filament bulbs in light and yet consume same amount of energy as a single bulb (or even less in some cases).

Isn't conservation of energy getting violated here? Isn't More light implies, more light energy implies more electrical energy consumption?

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light != visible light –  Captain Giraffe Sep 28 '11 at 17:52

Incandescent bulbs are tremendously inefficient in producing visible light. If you model the filament as a black body at 3000 C you will see that the majority of light emitted is in the infrared.

CFLs, on the other hand operate through the principle of fluorescence rather than incandescence, which means that less of the energy they consume is put out as heat.

Addendum: for some more concrete figures on just how much energy incandescents waste as heat, check these wikipedia page sections:Incandescent light bulb efficiency & Luminous efficacy examples. It would perhaps be more fitting to call them 'heat bulbs'.

[edit: implemented Mark Booth's comment.]

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So, in these technologies, they are just eliminating heat losses & other (invisible) radiation losses? –  claws Sep 28 '11 at 8:23
mainly heat! try touching a CFL and a incandescent lamp (P.S. This comes with a "don't try not this at home" warning!!!) –  Vineet Menon Sep 28 '11 at 8:26
Wow! I never knew incandescent bulbs were so inefficient! I'm still using some of those at home.I'll change these today itself. –  claws Sep 28 '11 at 8:30
Incandescents were recently phased out here in Australia. Just be aware that CFLs contain a (tiny) quantity of mercury and thus should be recycled properly. –  Richard Terrett Sep 28 '11 at 9:02
I think "very little of the energy they consume is put out as heat" is rather misleading "relatively little of the energy they consume is put out as heat" would be more accurate, since a CFL with 12% efficiency is still emitting 88% of it's power as heat rather than say 98% for an equivalent incandescent bulb. –  Mark Booth Sep 28 '11 at 11:43

I'm answering because the prior answer doesn't give a full picture of the energy pie (although it was still a good answer).

The equation in question is $P=V^2/R$, and this can't be changed. This is the power going into a bulb, every bit of that power goes somewhere else because of conservation of energy. For a lightbulb, the breakdown of energy we are interested in is as follows.

$$P=\frac{V^2}{R}=P_{heat}+P_{light}$$

At the common wavelength of 550 nm (green light, middle of the visible spectrum), the relationship between lumens and power is $683 lm=1 W$. Some common attributes of bulbs are:

• CFL bulb: 40 W, 2,600 lm
• Incandescent bulb: 40 W, 450 lm

If we assume all of the light has that wavelength, then we will find values of $P_{light}$ for the bulbs above of $3.8 W$ and $0.6588W$. If you compare two light bulbs of the same power consumption with one more efficient, then you will find that the more efficient only produces less heat, because more of the input electrical power was converted to energy carried by the light itself.

Of course, this can't continue on forever. A light bulb with 100% efficiency that consumed 40 W would produce 27,320 lm, and this could not be increased regardless of the technology.

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