I'm having a little brain-block, apparently, because I can't understand why there is no short when I connect the switch-leg (hot) to the neutral at a light fixture. Why is this different from shorting the hot and neutral at a receptacle? (I'm embarrassed to even be asking this question, and I am sure I will be humbled by the replies. Be gentle with me.)
Thanks, Leland. Here's the sit: A 12-2 NM brings power into a s-p switch. Another 12-2 NM leaves the switchbox as a switchleg to the light fixture. (The two neutrals in the switchbox are spliced together.)
At the 4-0 box for the light fixture (but without the fixture yet installed), the black and the white both from the switchleg are wirenutted together. Why no short? Wouldn't the neutral provide a path back to the breaker?
(Is this gonna be one of those answers where I slap my head and say "OMG, what a dufus I am!"?
Leland - I wish I could say 'Oh yeah, how silly of me!' But the truth is that the breaker is indeed on, as the black conductor at the fixture tests hot; and furthermore the other lights on the same switch (but via a separate switch-leg) come on with the switch. Which partially answers your second suggestion -- that the neutral is not connected.
The only other thing I can think of is that the neutral conductor specifically going to this fixture is not connected in the switchbox. Never looked in that box; and never actually used my Wiggy to test for actual 120v. Why they would leave the black connected in the switchbox but not the neutral is puzzling to me.
Any other possibilities? Cuz I sure felt stupid when the situation had me dumfounded, while the carpenter said, "Of course it doesn't short -- splicing the hot and neutral [of the swsitchleg] is just like what happens when they are both connected to a lightbulb."
When you "connect" the hot and neutral through the light bulb, you're not shorting them out, but connecting them through a resistance.
Volts = Amps x Ohms Amps = Volts/Ohms
A typical 60 Watt light-bulb might have a resistance of 240 Ohms. 120V / 240 Ohms = 0.5A.
A length of wire is a resistor, too, but the resistance is so small that we usually ignore it. (Table 9 in the NEC lists this. It becomes quite important for voltage drop & arc flash calculations.) You're only talking 2 Ohms for 1000' of #12 wire. Or 0.2 Ohms for 100'. So, if you actually DID wire-nutted two 50' lengths of #12 (100' total) and flipped on the light switch, you'd get 120V / 0.2 Ohms = 600A, and would trip the breaker and maybe burn something up.
Since you didn't blow anything up, something's disconnected. Maybe whoever nutted the two together did so while doing continuity tests to try to identify where the neutral was open? It's the first thing I would do if I suspected there might be a break in the cable at some point between the two boxes. (Well, 2nd thing; I'd disconnect the neutral in the light switch box first!)