qude said:
After thinking it over. I got the main idea. So the load is
related to it all. But you said lower resistant load requires
thicker material.. What's the physical basis? Is it because when
the material is thicker, there are more spaces for the electrons to
pass through that's why resistance is less. While in high
resistance load, it needs thinner material so the electrons
would have to struggle to pass thru it and hence the entire
electron line is adjusted to be lesser (lower current)??
That is pretty much what is happening. Instead of "more spaces",
people usually think in term of more area (cross-section), but
you have the right idea.
If right. Well. How do you select material where thicker
means more electrons can pass thru and less resistance
yet have same lighting power to the thinner wire with
more resistance? Can you use same material to build
thicker or thinner parts or does it depends on material
properties for each (higher or lower resistance) application?
The material selected is the one that can withstand the highest
temperature and conduct electricity. Of all known materials,
tungsten is the one that fits this description. As far as I
know, all light bulb filaments made these days are made from
tungsten, regardless of power or voltage or other requirements.
Then the wire's length and thickness must be chosen.
This can be done by knowing the necessary
resistance and operating temperature of the wire. The resistance
can be calculated from the available line voltage and the power.
Operating temperature is usually chosen to be around 2200 to
2800 C for tungsten. Hotter temperatures result in unreasonably
short burn-out lifetime. Lower temperatures result in less
efficient operation (less light produced per Watt of electricity).
Even though many length & thickness combinations will give the
necessary wire resistance, only one combination of length and
thickness will give BOTH the desired resistance AND operating
temperature for the wire. A wire of that length and thickness will
only "work" for a given supply voltage. At lower voltages, the
wire will run less efficiently. At higher voltages, the
light will be more efficient BUT will burn out faster.
I want to be able to visualize it all and not just memorizing
formulas and equations. Thanks.
For resistance, visualize that a longer path length OR smaller
cross-section area will increase the resistance:
Resistance is proportional to ( Length / diameter^2 )
For temperature, a smaller outside surface area will increase the
temperature, since the power leaves the wire mainly by radiating
from this surface.
Temperature depends (approximately) on ( Diameter / Length^2 )
Mark