On Fri, Jun 27, 2003 at 12:31:27PM +1000, Russell Chapman wrote: > Erik Reuter wrote: > > >By the way, that shunt is interesting. How do you suppose burning a > >coating off of a wire actually DECREASES the resistance? The only > >explanation I can come up with is that the coating insulates the wire > >and keeps it hot, and a hot wire has a higher resistance. > > I was assuming that the shunt wraps around the filament terminals in > some way, so that when the coating burns off you end up with a better > contact between the shunt and the filament...
That would require that the shunt actually move by at least the thickness of the coating, right? If so, what provides the force for it to move? Gravity? Springy wire (I wonder what material can retain its springiness for so long, esp. since I guess it would be quite warm)? Also, from the picture it seems like the coating covers a lot more area than would be necessary for such an operation. Hmmm. One curiosity begets another... > I'm curious how we get 50 light strands on 240volts. The plug is too > small to contain any serious componentry, and doesn't get hot at all, > and from there it is just simple 2 strand wires out to the bulbs... Another interesting question. I can't think of any quick way to check if your 240V bulbs are the same as our 120V ones. I assume yours look like the picture of the 2.5V mini-bulbs in the link? I guess a bulb could be made that took 5V instead of 2.5V but emitted the same light (it would use less current and consume about the same power). If you double the voltage and quadruple the resistance, you get the same power dissipation. -- "Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/ _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l
