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...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'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...
Cheers Russell C.
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