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Re: brightness vrs. power



I've always assumed that ohm's law was stated as R is independent of I.
Now I see that Ohm's law is really that R is independent of I at constant
T.

Now I realize that I've always assumed R was defined as V/I. Is this
correct?

Ohm's law is not a law of Nature. It is not even unique, as we have
seen in this discussion. It is not applicable to light bulbs except
in a perverted form of very limited applicability.

Could you describe your series-parallel configuration? I'm assuming
you are comparing a single light bulb...

---------------X-------------

with the following series-parallel configuration...

/-----X------X-----\
----- -------
\-----X------X-----/

If R was the same in each case, then the total current would be the same
and the current through each light bulb would be halved, correct? Due to
the dependence of R on temperature, there is actually less resistance in
the second circuit and thus more total current. Is this your point? It
seems you are implying something about how the brightness of the bulb
might be (naively) expected to be brighter, the same as, or dimmer than
the single bulb alone, but I didn't catch it. Perhaps I am still lacking
your insight. Please illuminate.

That's right. If the power dissipated is greater, the beginning
student should expect the four lights to be brighter. They aren't.

You understand everything. My intent was to show that light bulbs are
not a suitable example for use in teaching Ohm's "law" to children.
There are the complicatiing effects of large nonlinearity and of
confounding radiometry with photometry. An honest observer will wind
up thoroughly confused, having learned nothing from a teacher who is
poorly informed.

Light bulbs are, on the other hand, a great tool for teaching physics
(and critical physical thinking) to students who already know Ohm's
law and can learn Wien's law and Planck's law.

Leigh