On Nov 6, 10:26 am, DerbyDad03 wrote:
A co-worker, who should *not* be doing his own wiring attempted to add
a 2nd light fixture in his mudroom this weekend. Here is the situation
he presented to me this morning.

Light Fixture 1 (LF1) is an existing ceiling fixture in the mud room.
An existing Switch (S1) controls LF1.

He added LF2, a second ceiling fixture in the mud room, and wants to
control it with S1.

LF1 holds an incandescent bulb and LF2 holds a compact fluorescent.

He took the cover off of S1 and found a black and white from length of
Romex attached to S1. That's all that was in the box. So far, nothing
out of the ordinary, as far as I can tell. I'm assuming this length of
Romex is the switched hot from LF1.

His simple mind told him that if S1 controls LF1 via the black and
white from LF1, then all he had to do was run a length of romex from
LF2 and attach black to black and white to white at S1.

Here's the reported result of his ignorance: He says that if S1 is
on, LF1 is on and LF2 is off. That makes sense to me. However, he
says that if he turns S1 off, LF1 goes off and LF2 comes on.

If this is wired the way I think it is, then shouldn't *both* LF1 and
LF2 come on when S1 is off? LF2 is in parallel with an open S1 and
LF1 is now in series with LF2? Wouldn't that create a complete circuit
through both fixtures?

Does the fact that LF2 holds a compact fluorescent enter into this?

This is a stretch, but a possibility. If he put white wire 2 (w2) and
b2 onto different screws in the switch (because they were sitting
there not doing anything) instead of onto the same screw as the
existing wires, AND that the switch happened to be a double-throw
switch; then you would get the situation that you described.