Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-l] light bulbs



Hi,

I suspect that the human eye is
not a very good sensor for light output.

Another considerations, what were the
actual areas that were glowing. A 50
watt soldering iron does not glow (or
just barely). If the total power is
constant, and the area doubles the
power per unit area would be halved.

Thanks
Roger Haar
U AZ

==================================================================
On 3/8/2011 11:16 AM, brian whatcott wrote:
On 3/8/2011 7:06 AM, Anthony Lapinski wrote:
I have some incandescent bulbs from Europe. In America, the voltage is
half as much, so the current is half as much. Thus, the power should be
one quarter as much (P = IV)

A 60-W European bulb should behave like a 15-W American bulb (4X less
power). However, when I plug it in, it looks more like the glowing
filament of a 7.5-W bulb (8X less power). So I am a bit puzzled with this.
I am wondering if it has something to do with the filament specs. I
believe the electrical frequency in Europe is 50 Hz, but that shouldn't
really matter with 60 Hz in the US.

Anyone have experience with this, or a similar demo?

_
The resistance of incandescent light bulbs varies with filament temperature.
Resistance = Rref*( 1 + alpha (T - Tref))

Rref is the resistance at reference temperature T (Tref)
Alpha the thermal coefficient for tungsten is 0.004403 /K at Tref = 293K
T is the temperature of interest.

A 120volt 60 watt lamp has an R(hot) = 240 ohms
a 240 volt 60 watt lamp has an R(hot) = 960 ohms

Let's suppose that a 240 volt, 60 watt lamp that would burn at 5000K
only heats to 3000K at the lower voltage.
What would its effective resistance be for this condition?
First, its reference resistance = Hot Resistance/(1 + alpha(Thot - Tref))
= 960 ohms/(1 + 0.00444403*(5000K - 293K)) for the 240 volt bulb at 240
volt.
= X ohms/(1 + 0.004403*(3000K - 293K) for the 240 volt bulb at 120 volts.

so X ohms = 960 ohms * (1 + alpha(3000K- 293K) / (1 + alpha(5000K - 293K))
say 550 ohms that's the 240 volt filament resistance when burning dimly
at 3000K

That leads to a power estimate of V^2/R = 120^2 volts*volts/550ohms = 26
watts,
mostly in the infra red....
26 watts to produce what looks like 7.5 watts of light equivalence? How
inefficient!
That's why some economists argued for 1000 hour bulbs - as more efficient
when burning the candle at both ends.

Brian W


_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l