# discrepancy in watts of a light bulb

Discussion in 'Homework Help' started by RevitRed, Dec 4, 2014.

1. ### RevitRed Thread Starter New Member

Dec 4, 2014
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0
the question is:

A 60W incandescent light-bulb is designed to operate from 120V AC RMS. Its filament resistance is measured with a laboratory Ohm meter. The meter reads 17.5 Ohms however using the formula (VRMS2 / R) = Power, the 17.5Ohm filament should dissipate over 820 Watts. Explain the discrepancy of 60W vs 820W

ok so my "best educated guess" for this was that the filament resistance was being measured at room temperature but at operating temperature the filament resistance is much much greater which causes the decrease in power i am not 100 percent sure about this is this somewhat close?

2. ### MaxHeadRoom Expert

Jul 18, 2013
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3,521
Resistance cold is much lower than illuminated using current through the filament.
Max.

3. ### WBahn Moderator

Mar 31, 2012
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5,648
You've got a working hypothesis. Given the internet, how might you go about finding evidence to support/refute your hypothesis (assuming that you can't just ask someone that can spit the answer out at you since, in many cases in engineering, this will be the case)?

You might consider finding out what the material is that is used in incandescent light bulb filaments. Then you might see if you can find the resistivity of that material at room temperature and how it changes with temperature. You might try to find out what temperature it would have to be to be consistent with the nominal power and might then try to see if this is at all close to the temperature at which these filaments actually operate.

If you track down that information and use it, along with references of where you got it, in your answer you will duly impress you teacher and might even get some extra credit.

MrAl likes this.
4. ### KMoffett AAC Fanatic!

Dec 19, 2007
2,641
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Google Images: resistance vs voltage current of incandescent lamp

Ken

5. ### JoeJester AAC Fanatic!

Apr 26, 2005
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He hasn't measured the current in the 60 W bulb yet and reconfirmed the calculations of R.

P = E^2/R
P = E * I
P = I^2 * R

Do some measurements and see get E, I, and R to produce the 60W, or close to it.

6. ### MrAl Distinguished Member

Jun 17, 2014
3,599
754
Hi,

I measured a standard 100 watt light bulb a long time ago and if i remember right the resistance changed by a factor of 10 to 1. So when hot it had 10 times more resistance or thereabouts.
I also did a curve test on a small 1.5v bulb just for the fun of it. I have the data somewhere that shows the resistance changes quite a bit too as the voltage comes up from 0 to max.

Feb 17, 2009
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8. ### alfacliff Well-Known Member

Dec 13, 2013
2,449
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some years ago, there was a circuit for a magic candle, it had a bulb on the end, and when you held a match to it, it lit up. works by the filament changing resistance when heated, even by an external source. might make a good test of thermal effects on lamps. by the way, the bulb must be a filament bulb, no led or pigtail flourescents.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,167
1,188
But we can use a RBG diode to get the same effect