How Is This Switch Wired?

[DerbyDad03]

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?

[trader4]

If there is only one romex coming into the box, that is out of the ordinary. You should have at least 2 romex, with one being the feed, other going to LF1. Could also be other romex that tap off feed at that point, etc.

That is exactly correct. IF wired that way, it would work. The other choice would have been to run a romex from the switcht to then new fixture, if it were easier. Either way they wind up in parallel and have to go on and off together.

No, they should both come on when S1 is on.

LF2 is in parallel with an open S1 and

It can't be wired the way he's telling you, or both would go on and off together, simple as that.

No.

[RayV]

Sounds strange to me as well. Possibly LF2 is completing the circuit back through the ground and bypassing LF1 by following the path of least resistance.

[Pat]

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.

[Doug Miller]

No, that's perfectly normal -- it's called a switch leg. Power feed comes into the box at the light fixture, and is tied to the white wire of the switch leg, which goes to one side of the switch. The black wire of the switch leg goes to the other side of the switch, and to the black fixture wire.

[terry]

Without getting into black connected to black and white to white etc. it sounds as though his new wiring to LF2 is connected 'across' the switch? So when the switch is 'off' electricity then is flowing through LF2 and then on its way to LF1. In other words the two lights are 'in series' and electricity is flowing through both of them? In other words electricity is now never off. It's on all the time? The reason for LF2 lighting is probably because with LF2 and LF 1 in series the compact fluorescent which takes much less electricity gets enough voltage to light. One way to check this assumption is to exchange the two lamps (the regular bulb and the CFL) and it should be found that then LF1 (the CFL) will light with the switch off? Gee it takes longer to write this up than to do it! But should that individual really be doing any wiring at all????? Has the proper wire been used? Items grounded as required? Is the LF2 lamp socket wired with correct polarity? Wires clamped properly? Etc. etc.

[DonC]

That's actually quite common. Power for LF1 is fed directly to LF1. The black wire in the LF1 box connects to one of the wires heading to S1. The other wire from S1 connect to the hot side of LF1. This circuit is controlled by the position of S1.

S1 ---------------------Hot->--LF1--

[DerbyDad03]

Agreed...I have no problem with the way the LF1 is switched.

But, Doug...of all people ;-) you didn't comment on why only LF2 is on with the switch off.

Any thoughts?

[Chris Lewis]

According to DerbyDad03 :

It's the hot and switched hot. _No_ neutral. This is a switch leg, you can't get "proper" power from the switch end of this circuit.

When S1 is on, LF1 will be on, but the leads to LF2 are shorted by the switch. When S1 is off, LF1 and LF2 are in series. Since LF2 is a CF, it's current draw is very low, and it lights up, but the current is insufficient to visibly light LF1. If LF1 and LF2 were identical incandescent bulbs, they'd both light dimly. If they were both CFs, can't predict precisely what they do without knowing what the CFs do with 60V. Might not do anything.

He needs to take the wire from LF2 off S1, and move it to be across the leads on LF1 (inside the LF1 junction box). Eg: LF2's black wire wirenutted to the wire going to the center pin on LF1's bulb base, and white wire wirenutted to the wire going to LF2's base shell. Then the voltage seen on both bulb bases will be identical.

[DerbyDad03]

re: But should that individual really be doing any wiring at all?

That's why I started my post with: "A co-worker, who should *not* be doing his own wiring..."

This is the second time he's come to me with questions *after* what he's attempted has not worked, always adding "electrical wiring is not my strong suit."

The scary part is that he's never questioned anything that's he's done that *did* work.

[Chris Lewis]

According to DerbyDad03 :

See my other posting. When you put a 100W incandescent in series with a, say, 13W CF, the CF is passing so little current that the IC bulb may not visibly light up at all, or only have a faint red glow. If they were both ICs of the same wattage, they'd both light up (dimly).

[Doug Miller]

That's because I'm still trying to figure it out!!

My first thought is that something there is *not* as it was described to you.

[DerbyDad03]

I knew that ;-) but he sure didn't. He always comes to me *after* he's screwed something up.

Yep ,that's exactly how I told him it should be wired, but I wasn't sure why only the CF lit. It makes sense that's it's a current issue.

Now he has another problem...He opened LF1 and found that it is being used as a junction box (not his doing) to supply power to an outlet in the mudroom and an outside light with it's own switch. The key point being there are already at least 4 pieces of Romex coming into LF1 (source in, outlet out, S1 in/out, S2 in/out) therefore, I'm sure the box is way too crowded for another length of Romex from LF2.

[Terry]

With the swtich open the two lights are in series. With the switch closed L2 is shorted.

[Doug Miller]

We have a winner, ladies and gentlemen! Give that man a cigar!

Just one quibble: you meant the *voltage* is insufficient to light LF1. In a series circuit, the current is the same everywhere. Assuming a single 60W lamp in LF1, and a 15W CF lamp in LF2, at 120V the voltage drop across the CF lamp in LF2 is 96V, and incandescent lamp in LF1 is seeing only 24V.

Exactly.

[Doug Miller]

That's what they make extension rings for: you double the capacity of the box at the cost of two or three dollars and five minutes of work.

[Mark Lloyd]

If could be that LF2 is drawing insufficient current to light LF1. In this case, unscrewing the bulb in LF1 will make LF2 go out.

I don't remember ever knowing as little about electricity as some people.

BTW, this should be fixable by wiring LF2 across LF1 rather than the switch.

[Terry]

The op said there was only two wires at the switch.

The way to make the light burn is to take it out of the switch and wire it in parallel to the existing light.

=== If you are going to do something stupid, always get it on film.

[Mark Lloyd]

Apparently the 2 Romex were at the light fixture. Most of mine are wired that way. Only 2 wires are needed for a switch.

It wouldn't. LF2 needs to be wired across LF1. not across the switch.

No, they shouldn't. The switch has the effect of creating a short across LF2. There's no way it could light in this condition.

Not necessarily with different bulbs in each fixture.

Yes.

[Phisherman]

Sounds like S1 is a 3-way switch, then the above makes sense. Typically 3-way switch will not have "ON" nor "OFF" printed on the switch. It doesn't matter if LF2 is fluorescent or incandescent.