# How To Calculate Fluorescent To Incandescent Wattage? [closed]

This is simple mathematical question but I cannot figure it out:

Compact fluorescent lamps consume 75% less wattage and still deliver equal power of incandescent lamp. If 15W compact fluorescent lamp which consumes 75% less wattage, delivers same result as 60W incandescent lamp then how much wattage of incandescent lamp delivers 9W compact fluorescent lamp?

Please tell me calculation. Thanks.

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## closed as off topic by Brian Carlton, endolith, Leon Heller, Kevin VermeerOct 17 '11 at 20:18

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Homework? I don't see the electronics question either. – Brian Carlton Oct 17 '11 at 18:43
Would this be a better fit for mathematics stackexchange maybe? – AndrejaKo Oct 17 '11 at 19:09
@Brian: Not homework for anyone old enough to be allowed here. – Olin Lathrop Oct 17 '11 at 19:11
@AndrejaKo - Have you looked at the front page of Mathematics or Mathoverflow recently? This would not be a good homework question there. – Kevin Vermeer Oct 17 '11 at 20:18
@Kevin Vermeer Actually, yes, but what caused my suggestion was the official stance of Math that they accept questions for any educational level. On the other hand Mathoverflow is a no go since they are for math research. – AndrejaKo Oct 17 '11 at 20:22

## 2 Answers

This is off topic, but it's right out of 7th grade arthmetic and easy to answer, so I'll just do it:

If johnny leaves his apples on the table and Suzy always takes 3/4 of them at noon, then how many apples did Johnny start with if Suzy left 9 of them?

F = I - (75% I)
F = 25% I
F = .25 I

I = F / .25
I = 4 F

Therefore according to your 75% figure, a 9W fluorescent would put out as much light as a 36W incandescent.

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Thanks. What 75% in brackets does with I? Multiplies or divides? – Boris_yo Oct 19 '11 at 15:33
@Boris: You mean the "75% I" in the parenthesis? In general when a mathematical operator is left out, it is understood to be multiplication. Therefore (75% I) = 75% * I = .75 * I – Olin Lathrop Oct 19 '11 at 16:44

I'll not go into detailed description why this is not a good way to compare the light sources. Basically the only real way to be sure is to read the specification of the 9 W CFL or to do measurements yourself.

Anyway, to answer the question here we have a basic elementary-school lever proportion.

If we assume that both CFLs are of same efficiency and so on (and we can't safely do that), we can say that:

$P_{1I}= 60 \mbox{ } W$
$P_{1CFL}=15 \mbox{ } W$
$P_{2I}=X\mbox{ }W$
$P_{2CFL}= 9 \mbox{ }W$

We may also dare to claim that:
$P_{1I}:P_{1CFL}=P_{2I}:P_{2CFL}$
From that, we can calculate that:
$P_{2I}=\frac{P_{1I}P_{2CFL}}{P_{1CFL}}$
The final result would be that $P_{2I}= 36 \mbox{ }W$

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Ha, beat you to the answer by 2 seconds! – Olin Lathrop Oct 17 '11 at 19:12
@Olin Lathrop But MINE is better! :) – AndrejaKo Oct 17 '11 at 19:13