How Is This Switch Wired?

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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
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On Wed, 07 Nov 2007 00:13:54 GMT, Wayne Whitney

How many #12s can you fit in a handy box?
What is the record? :)
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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
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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.
--
Regards,
Doug Miller (alphageek at milmac dot com)
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On Nov 7, 11:40 am, snipped-for-privacy@milmac.com (Doug Miller) wrote:

Jeez, could you have been *any* clearer in your explanation? ;-)
That's a keeper!
Thanks!
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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.
--
Regards,
Doug Miller (alphageek at milmac dot com)
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On Nov 6, 11:16 am, snipped-for-privacy@nortelnetworks.com (Chris Lewis) wrote:

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---/
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(Chris Lewis) wrote:

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.
--
Regards,
Doug Miller (alphageek at milmac dot com)
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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).
--
Chris Lewis,

Age and Treachery will Triumph over Youth and Skill
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On Tue, 06 Nov 2007 16:33:53 -0000, snipped-for-privacy@nortelnetworks.com (Chris Lewis) wrote:

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.
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That's correct. What makes you think that I implied otherwise? That's what a switch leg is.

Dat's what I said.
--
Chris Lewis,

Age and Treachery will Triumph over Youth and Skill
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wrote:

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.
--
49 days until the winter solstice celebration

Mark Lloyd
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wrote:

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.
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DerbyDad03 wrote:

Sounds cool. When the main light is off, he has a night-light.
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wrote:

Although that CFL will be operating on a lower voltage than usual. Is that OK?
--
48 days until the winter solstice celebration

Mark Lloyd
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