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Re: Alternating Current



Hi,
The instantaneous power would be P(t) = V(t) I(t) =R V(t)^2. This
effectively doubles the period because the bulb "flashes" once one each half
cycle. The light intensity is not simply proportional to the power. The
power is just the rate at which energy is used to heat the filament.

The total light output is related to the temperature of the
filament, T^4, I recall.
But the temperature of the filament is related to the energy gained from
heating and lost by emitting light ( all wavelengths) An additional
complication is that the temperature dependence is more complicated for a
given wavelength and thus the range of wavelengths for which your detector
works.

Because surface of the filament cools radiatively and the interior
cools conductivity to the surface there are still more complications.

I think the end effect is that the detected light output is a
constant plus a time varying piece. The time varying part is at 120 Hz and
is blurred to look like a sign wave.

Thanks
Roger haar

*****************************************************
David Abineri wrote:

I just did a lab today with my Pre Calculus class. Since we have
introduced sinusoidal variation I had them use a CBL to look at the
light intensity from an incandescent bulb sampling every .0001 seconds
for 100 data points.

When one does this, the calculator shows a beautiful sinusoidal graph
and the regression analysis shows an excellent fit with supporting
residuals.

However, I would have expected to see a graph more like |sint| which
should not look like a sin but should have the frequency of double the
line voltage frequency since each half of the voltage cycle produces a
full cycle of light intensity. The frequency came out consistently at
124Hz by the way and I wonder if the power company is typically this far
off their standard value of 2*60Hz?

Can anyone help me with this? Am I really seeing |sint| and it just
looks like sint? Has anyone else tried this?

Thanks for the help.
--
David Abineri dabineri@choice.net