Re: R = V/I with real lightbulbs

[Doug Craigen <dcc@ESCAPE.CA>]

Leigh Palmer wrote:
> >        I finally have time to put my foot into this mess.
> > William Lichten in his book "Data and Error Analysis" gives as an example
> > the current verses voltage relationship for a tungsten lamp from an applied
> > voltage of 30 mV to 5.3 volts. The log-log graph starts out with a slope of
> > about one (where Ohm's law is approximately correct) and then at higher
> > voltages it shifts over to about a 5/3 power law relationship where
> > radiation cooling becomes important. It is a nice example of how different
> > relationships are at work depending on the physical conditions of the system.
>
> In what way does radiation influence the I vs. V relationship? This
> will surely confuse students, as it confuses me.

At the risk of repeating repeating myselfself, this could be a great
experiment or it could be something which the students do what they're
told with little concept of what's going on.  Furthermore, it might be
that it is producing the right answer for the wrong reasons.  The above
power law tells me that either there is more going on than just T^4 of
the filament even at high temperatures (of course, we already know this
for low temperatures) or the correct relationship between resistivity
and the temperature is in fact linear:

P = IV = k1*T^4
& V/I = k2*T

so IV = k*V^4/I^4
I^5 = k*V^3
hence our 5/3 relationship
however the data you found in the rubber bible seems to exclude this
explanation.

using V/I = k*T^0.5 - the 'expected' relationship for a metal, gives I^9
= k*V^7 for the high end of our I-V plot which doesn't match the claim
in the book.  So, its up to whoever out there has some real data at hand
to provide some insight about what is happening.

Once caution about log graphs with varying slope regions, it is often
possible to prove whatever slope you are expecting.

()-()-()-()-()-()-()-()-()-()-()-()-()-()-()-()

Doug Craigen
          http://www.dctech.com/physics/