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?

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.

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.




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.

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.



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?

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



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?

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




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

No.

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?

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.

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.

In article . com, wrote:

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.

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.

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

It's time to throw all their damned tea in the harbor again.

On Nov 6, 12:54 pm, wrote:
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.

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.



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.

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



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?

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

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?

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



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

No.- Hide quoted text -

- Show quoted text -

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.

wrote in message
ups.com...
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.

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'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---Ground -----------------------------
Power Source W
\--------------------------------------------------------------------------
Power Source B

On Nov 6, 11:09 am, (Doug Miller) wrote:
In article . com, wrote:
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.

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.

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

It's time to throw all their damned tea in the harbor again.

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?

According to 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.

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.
--
Chris Lewis,

Age and Treachery will Triumph over Youth and Skill
It's not just anyone who gets a Starship Cruiser class named after them.

On Nov 6, 11:10 am, terry wrote:
On Nov 6, 12:54 pm, wrote:





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.

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.

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.

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

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?

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

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?

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

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

No.- Hide quoted text -

- Show quoted text -

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.- Hide quoted text -

- Show quoted text -

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.

According to DerbyDad03 :

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

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).
--
Chris Lewis,

Age and Treachery will Triumph over Youth and Skill
It's not just anyone who gets a Starship Cruiser class named after them.

In article om, DerbyDad03 wrote:
On Nov 6, 11:09 am, (Doug Miller) wrote:
In article . com,
wrote:
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.

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.

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.

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

Any thoughts?

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

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

It's time to throw all their damned tea in the harbor again.

On Nov 6, 11:33 am, (Chris Lewis) wrote:
According to 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.

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.

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

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.


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.

On Tue, 06 Nov 2007 16:38:35 GMT, (Doug Miller)
wrote:

In article om, DerbyDad03 wrote:
On Nov 6, 11:09 am, (Doug Miller) wrote:
In article . com,
wrote:
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.

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.

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.

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

Any thoughts?

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

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

In article , (Chris Lewis) wrote:
According to 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.

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.

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.

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.

Exactly.

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

It's time to throw all their damned tea in the harbor again.

In article . com, DerbyDad03 wrote:

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.

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.

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

It's time to throw all their damned tea in the harbor again.

On Tue, 06 Nov 2007 07:26:23 -0800, 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?


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.

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

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
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On Tue, 06 Nov 2007 16:33:53 -0000, (Chris
Lewis) wrote:

According to 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.

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.


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.

On Tue, 06 Nov 2007 07:54:12 -0800, wrote:

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.

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.


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




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.

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.


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



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?

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


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.



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?

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


Not necessarily with different bulbs in each fixture.




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

No.


Yes.



--
49 days until the winter solstice celebration

Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On Tue, 06 Nov 2007 07:26:23 -0800, 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?


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.

Terry wrote:
On Tue, 06 Nov 2007 16:38:35 GMT, (Doug Miller)
wrote:

In article om, DerbyDad03 wrote:
On Nov 6, 11:09 am, (Doug Miller) wrote:
In article . com,
wrote:
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.
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.
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.
That's because I'm still trying to figure it out!!
Any thoughts?

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

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

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.

On Nov 6, 11:51 am, (Doug Miller) wrote:
In article . com, DerbyDad03 wrote:

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.

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.

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

It's time to throw all their damned tea in the harbor again.

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

I'll be expecting it ;-)

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.

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
It's not just anyone who gets a Starship Cruiser class named after them.

According to Terry :

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

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

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

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

Age and Treachery will Triumph over Youth and Skill
It's not just anyone who gets a Starship Cruiser class named after them.

According to DerbyDad03 :

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.

Check the ceiling box depth. It may be deep enough for the 5
cables.
--
Chris Lewis,

Age and Treachery will Triumph over Youth and Skill
It's not just anyone who gets a Starship Cruiser class named after them.

On Nov 6, 1:16 pm, (Chris Lewis) wrote:
According to DerbyDad03 :

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.

Check the ceiling box depth. It may be deep enough for the 5
cables.
--
Chris Lewis,

Age and Treachery will Triumph over Youth and Skill
It's not just anyone who gets a Starship Cruiser class named after them.

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.

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.

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

On 2007-11-06, DerbyDad03 wrote:

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?

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

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

On 2007-11-06, DerbyDad03 wrote:

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?

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?

On Nov 6, 11:16 am, (Chris Lewis) wrote:
According to DerbyDad03 :

Age and Treachery will Triumph over Youth and Skill
It's not just anyone who gets a Starship Cruiser class named after them.


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---/

On Tue, 6 Nov 2007 15:54:00 -0600, "HeyBub"
wrote:

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.

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


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

Mark Lloyd
http://notstupid.laughingsquid.com

"Unlike biological evolution. 'intelligent design' is
not a genuine scientific theory and, therefore, has
no place in the curriculum of our nation's public
school classes." -- Ted Kennedy

On Nov 6, 7:13 pm, Wayne Whitney wrote:
On 2007-11-06, DerbyDad03 wrote:

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?

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

I need some clarification...

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?

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?

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

In article . com, DerbyDad03 wrote:

I need some clarification...

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?

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.

you count any internal wire clamps

Is an internal wire clamp the same as a wire nut

No

or do you mean the
clamp that secures the romex in the hole?

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.

What is the X in^3 for either item?

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.

count each device (yoke) in the box as two allowances

What's a yoke?

A strap that holds wiring devices such as switches or receptacles -- or
fixture nipples.

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.

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)

It's time to throw all their damned tea in the harbor again.

On Nov 7, 11:40 am, (Doug Miller) wrote:
In article . com, DerbyDad03 wrote:

I need some clarification...

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?

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.



you count any internal wire clamps

Is an internal wire clamp the same as a wire nut

No

or do you mean the
clamp that secures the romex in the hole?

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.

What is the X in^3 for either item?

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.

count each device (yoke) in the box as two allowances

What's a yoke?

A strap that holds wiring devices such as switches or receptacles -- or
fixture nipples.

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.

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)

It's time to throw all their damned tea in the harbor again.

Jeez, could you have been *any* clearer in your explanation? ;-)

That's a keeper!

Thanks!

In article om, DerbyDad03 wrote:

Jeez, could you have been *any* clearer in your explanation? ;-)

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.

That's a keeper!

Thanks!

You're welcome. Glad I could help.

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

It's time to throw all their damned tea in the harbor again.