Scott Lurndal wrote:
IanM writes:

Please explain *CLEARLY* how increasing the thickness of any uniform
substance can *increase* the resistance if everything else remains
unchanged.


The resistance varies inversely to the cross-section of the conductor.

AWG 12 wire resistance/foot = 1.619Ohms.
AWG 10 wire resistance/foot = 1.018ohms.
AWG 8 0.6405

http://www.interfacebus.com/Copper_Wire_AWG_SIze.html

Actually those number are ohms per 1000 feet.


Ergo, heavier wire, less resistance.

Yes. assuming the same length, material, etc.


So assume that a 100watt blub rated at 130V filament consumes
0.769231 amperes of current. From ohms law, one can then derive
the resistance of the conductor as (R=V/I) 156 ohms.

No. The resistance should be 169 ohms at 130 volts. You seem to have
gotten 156 ohms by dividing 120 volts by 0.768231 amps. There is no
justification for this. The resistance of the filament will vary with
its temperature. It will be lower than 169 ohms at 120 volts. but
there is no reason to assume that it will be 156 ohms.


Now run that same bulb at 120volts, the current in the filament
(per again ohms law) will be (I=V/R) 0.769231 (i.e. the same current).

See previous comment. You seem to have derived a result from your
assumptions (you assumed the current at 120 volts is 0.769231 amps
in your previous calculation).


However, the power consumed (P=IV) will only be 92.3 watts, thus
reducing the lumen output of the bulb.

This result is based upon an incorrect assumption that the current is
0.7629231 when the voltage is 120 volts.

The power at 120 volts will be between the 85 watts (the value that would
be calculated if the resistance is constant with temperature) and the 100
watts at 130 volts. Thus 92.3 watts is a reasonable guess for the power
at 120 volts but you have not presented anything to prove it.


Dan