into
And to add to that, the CFL lamp is a lower wattage than the incandescent. As they are in series, most of the voltage will be across the CFL, so it will light. Also, there are peculiarities of both lamps, i.e. the incandescent lamp has a very low resistance when not lit and the CFL, having electronics in it, can do weird things. But, this seems perfectly normal for the abnormal wiring.
Thanks to all for the replies, both the correct ones and the incorrect ones. :-)
As I suspected, the CF was the culprit. When he described the symptoms, it didn't make sense which is why I asked him if the fixtures were the same. At first he said "Yes...I bought a matching fixture." Then I took it to the next level: OK, what about the bulbs? That's when I learned of the CFL.
As far as using an extension ring...well... it's a finished ceiling which means he's going to have to remove all the existing wires to replace the box/add a ring and then put it all back together. Based on past experiences, I'm afraid to even suggest that to him.
Thanks to all again.
According to Doug Miller :
I'll be expecting it ;-)
Neither "voltage" or "current" are technically quite right in that sentence, but the correct way takes longer to explain ;-)
CF's can't really be considered resistive devices, and IC bulbs when cold have much lower resistance than in continuous normal operation, so, the only real way to know what the voltage across them is to set up the circuit and measure it ... ;-)
I prefer to think of LF1 (when not visibly showing any light) as being close to a dead short (very low resistance) when in series with another device with considerably higher effective resistance. I'd expect the voltage across LF2 to be a lot more than 96V as a simple-minded series "hot" resistance calculation would imply.
You'd be better off measuring the cold resistance of LF1, and using that instead of the hot resistance at 60W output (~240ohms).
According to Terry :
That's correct. What makes you think that I implied otherwise? That's what a switch leg is.
Dat's what I said.
According to DerbyDad03 :
Check the ceiling box depth. It may be deep enough for the 5 cables.
Please save me (and him) the research...what would be the minimum depth for 5 cables? Isn't there also an issue with how many wires can be nutted together in given box? I'm not sure if LF1 has screws or pigtails. Pigtails would, obviously, mean more wires to be nutted.
Sounds cool. When the main light is off, he has a night-light.
Box fill summary: minimum required box volume is a sum of terms, each of which depends on the wire size, in particular 2.00 in^3 for a #14 wire and 2.25 in^3 for a #12 wire. You individually count all current carrying conductors entering the box (hot and neutral), you count all the EGCs as only one wire, you count any internal wire clamps, and you count each device (yoke) in the box as two allowances.
Cheers, Wayne
How many #12s can you fit in a handy box?
What is the record? :)
Several have had it right: Try to clear up the argeument with a diagram:
Present wireing:
~--------S1-----------LF1-----------Return \\---LF2--/
Should have been wired:
~-------S1----------LF1-----------Return \\----LF2---/
Although that CFL will be operating on a lower voltage than usual. Is that OK?
I need some clarification...
re: count all current carrying conductors entering the box (hot and neutral), you count all the EGCs as only one wire
So, is a piece of 14/2 with ground counted as 3 "terms" or 2? i.e. requireing 6 in^3 or 4 in^3?
re: you count any internal wire clamps
Is an internal wire clamp the same as a wire nut or do you mean the clamp that secures the romex in the hole? What is the X in^3 for either item? If it's the wire nut, wouldn't it depend on the size of the wire nut which is dependent on the number of conductors it secures?
re: count each device (yoke) in the box as two allowances
What's a yoke? In the case of this "light fixture ceiling box" how does it (device/yoke) enter into the equation? It's not *in* the box, per se.
Thanks
Depends on how many *other* cables are present. All the equipment grounding conductors together count as only one. So if you have only one 14/2 WG cable, that counts as three conductors. Two such cables counts as five conductors (two black, two white, plus one for the two EGCs). Three cables counts as seven (3 black, 3 white, plus one for the 3 EGCs). And so on.
If the cables have conductors of different sizes, count the all EGCs as one conductor of the largest size. For example, if the box contains two 14/2 WG cables and two 12/2 WG, you count (4) 14ga and (5) 12ga to determine the required capacity.
Note also that conductors that don't leave the box (e.g. pigtails) are not counted at all. Neither are up to 4 fixture wires 14ga and smaller.
No
Yes -- if the clamp is inside the box. The standard Romex connectors that mount through a knockout and are secured with a locknut inside the box have the clamp *outside* the box, and are not counted.
Same as that for the largest conductor present. For example, if the box contains both 12ga and 14ga wires, you count the clamp as a 12ga.
A strap that holds wiring devices such as switches or receptacles -- or fixture nipples.
If the fixture mounts directly to the box, it doesn't. If the fixture mounts to a strap that mounts to the box, that's your yoke.
Jeez, could you have been *any* clearer in your explanation? ;-)
That's a keeper!
Thanks!
Actually, I probably could have -- failed to specify a couple things:
A device yoke counts as *two* of the largest conductor connected to that device. And a support fitting (such as a fixture is attached to) counts as one of the largest conductor present in the box.
You're welcome. Glad I could help.
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