Question: (fixed type font picture)
L1 N L2
| | |
|____@@_____|____@@_____|
Outlet 1 Outlet 2

Are these statements correct using the above simple MW circuit?

If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won’t work.

If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

Yes

"Limp Arbor" wrote in message
...
Question: (fixed type font picture)
L1 N L2
| | |
|____@@_____|____@@_____|
Outlet 1 Outlet 2

Are these statements correct using the above simple MW circuit?

If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won’t work.

Correct, just as in most circuits, no neutral = not operating.

If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

Wrong. You will have 240V across BOTH circuits, if both lamps are the same
wattage, that will provide 120V for each. Most power supply is now 120V not
110V.

"Limp Arbor" wrote in message
...
Question: (fixed type font picture)
L1 N L2
| | |
|____@@_____|____@@_____|
Outlet 1 Outlet 2

Are these statements correct using the above simple MW circuit?

If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won’t work.

If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

**You'll get 240 volts at each outlet when equal loads are plugged in, as
EXT explained. Your voltage will vary with dissimilar loads. For this
reason, the neutral conductors of an Edison circuit cannot be dependent upon
a device. In other words, they must be pigtailed

On Mar 29, 5:04*pm, "EXT" wrote:
"Limp Arbor" wrote in message

...

Question: (fixed type font picture)
L1 * * * * *N * * * * *L2
| * * * * * | * * * * * |
|____@@_____|____@@_____|
* Outlet 1 * *Outlet 2

Are these statements correct using the above simple MW circuit?

If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won t work.

Correct, just as in most circuits, no neutral = not operating.



If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

Wrong. You will have 240V across BOTH circuits, if both lamps are the same
wattage, that will provide 120V for each. Most power supply is now 120V not
110V.

Professorpaul is correct. Actually, you're both correct. It's also
important
to define between which two points we're measuring.
With a lamp plugged into outlet one, at outlet 2 with a high impedance
VOM you'd measu

Across the two outlet terminals: 240V
From either terminal to ground: 120V

"Limp Arbor" wrote in message
...
Question: (fixed type font picture)
L1 N L2
| | |
|____@@_____|____@@_____|
Outlet 1 Outlet 2

Are these statements correct using the above simple MW circuit?

If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won’t work.

If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

1. If you loose the neutral and a lamp in outlet 1, it will not work if
nothing is in outlet 2. Neither will one in outlet 2 if nothing is in
outlet 1.

2. If anything is connected to outlet one and two at the same time, the
voltage will split depending on the load. The higher the resistance of the
load at one outlet compaired to outlet 2, the higher resistance will have
the most voltage across it. If the loads are exectally equal, they will
both have half the supply voltage across them. That is in your example, if
you put in a 60 watt lamp in each outlet and they both come up to the same,
they will appear normal with 110 volts across each one. If you just put one
lamp in one of the outlets, you will measuer 220 volts ( more like 219.99999
or so) on the other one that does not have a load. As you put a load on
that outlet , the voltage will start dropping. It may reach almost but not
exectally 0 volts, while the other outlets voltage will rise toward 220
volts.

On Mar 29, 5:04*pm, "EXT" wrote:
"Limp Arbor" wrote in message

...

Question: (fixed type font picture)
L1 * * * * *N * * * * *L2
| * * * * * | * * * * * |
|____@@_____|____@@_____|
* Outlet 1 * *Outlet 2

Are these statements correct using the above simple MW circuit?

If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won t work.

Correct, just as in most circuits, no neutral = not operating.



If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

Wrong. You will have 240V across BOTH circuits, if both lamps are the same
wattage, that will provide 120V for each. Most power supply is now 120V not
110V.

I'm old school
http://www.youtube.com/watch?v=iX3kxAA2L4Q

On Mar 29, 7:07*pm, "Ralph Mowery" wrote:
"Limp Arbor" wrote in message

...
Question: (fixed type font picture)
L1 * * * * *N * * * * *L2
| * * * * * | * * * * * |
|____@@_____|____@@_____|
* *Outlet 1 * *Outlet 2

Are these statements correct using the above simple MW circuit?
If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won’t work.
If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

1. *If you loose the neutral and a lamp in outlet 1, it will not work if
nothing is in outlet 2. *Neither will one in outlet 2 if nothing is in
outlet 1.

2. *If anything is connected to outlet one and two at the same time, the
voltage will split depending on the load. *The higher the resistance of the
load at one outlet compaired to outlet 2, the higher resistance will have
the most voltage across it. *If the loads are exectally equal, they will
both have half the supply voltage across them. * That is in your example, if
you put in a 60 watt lamp in each outlet and they both come up to the same,
they will appear normal with 110 volts across each one. *If you just put one
lamp in one of the outlets, you will measuer 220 volts ( more like 219.99999
or so) *on the other one that does not have a load. *As you put a load on
that outlet , the voltage will start dropping. *It may reach almost but not
exectally 0 volts, while the other outlets voltage will rise toward 220
volts.

And if you put any lamp in outlet one, and nothing in outlet two, the
potential
across the terminals of outlet two will be 240V. I believe that was
the
question.

On Mar 30, 10:30*am, "
wrote:
On Mar 29, 7:07*pm, "Ralph Mowery" wrote:





"Limp Arbor" wrote in message

....
Question: (fixed type font picture)
L1 * * * * *N * * * * *L2
| * * * * * | * * * * * |
|____@@_____|____@@_____|
* *Outlet 1 * *Outlet 2

Are these statements correct using the above simple MW circuit?
If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won’t work.
If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

1. *If you loose the neutral and a lamp in outlet 1, it will not work if
nothing is in outlet 2. *Neither will one in outlet 2 if nothing is in
outlet 1.

2. *If anything is connected to outlet one and two at the same time, the
voltage will split depending on the load. *The higher the resistance of the
load at one outlet compaired to outlet 2, the higher resistance will have
the most voltage across it. *If the loads are exectally equal, they will
both have half the supply voltage across them. * That is in your example, if
you put in a 60 watt lamp in each outlet and they both come up to the same,
they will appear normal with 110 volts across each one. *If you just put one
lamp in one of the outlets, you will measuer 220 volts ( more like 219.99999
or so) *on the other one that does not have a load. *As you put a load on
that outlet , the voltage will start dropping. *It may reach almost but not
exectally 0 volts, while the other outlets voltage will rise toward 220
volts.

And if you put any lamp in outlet one, and nothing in outlet two, the
potential
across the terminals of outlet two will be 240V. *I believe that was
the
question.- Hide quoted text -

- Show quoted text -

And there in lies the fundamental flaw in these circuits. If the
neutral in a regular circuit breaks nothing works. If the neutral in
an edison circuit breaks you have a mess.

On Mar 30, 11:15*am, jamesgangnc wrote:
On Mar 30, 10:30*am, "
wrote:





On Mar 29, 7:07*pm, "Ralph Mowery" wrote:

"Limp Arbor" wrote in message

....
Question: (fixed type font picture)
L1 * * * * *N * * * * *L2
| * * * * * | * * * * * |
|____@@_____|____@@_____|
* *Outlet 1 * *Outlet 2

Are these statements correct using the above simple MW circuit?
If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won’t work.
If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

1. *If you loose the neutral and a lamp in outlet 1, it will not work if
nothing is in outlet 2. *Neither will one in outlet 2 if nothing is in
outlet 1.

2. *If anything is connected to outlet one and two at the same time, the
voltage will split depending on the load. *The higher the resistance of the
load at one outlet compaired to outlet 2, the higher resistance will have
the most voltage across it. *If the loads are exectally equal, they will
both have half the supply voltage across them. * That is in your example, if
you put in a 60 watt lamp in each outlet and they both come up to the same,
they will appear normal with 110 volts across each one. *If you just put one
lamp in one of the outlets, you will measuer 220 volts ( more like 219.99999
or so) *on the other one that does not have a load. *As you put a load on
that outlet , the voltage will start dropping. *It may reach almost but not
exectally 0 volts, while the other outlets voltage will rise toward 220
volts.

And if you put any lamp in outlet one, and nothing in outlet two, the
potential
across the terminals of outlet two will be 240V. *I believe that was
the
question.- Hide quoted text -

- Show quoted text -

And there in lies the fundamental flaw in these circuits. *If the
neutral in a regular circuit breaks nothing works. *If the neutral in
an edison circuit breaks you have a mess.

Exactly!

Is the minimal savings in wire really worth it?

Yeah I know there is also additional labor also but why do this? To
me the potential problems outweigh the savings.

On Wed, 30 Mar 2011 07:30:59 -0700 (PDT), "
wrote:

On Mar 29, 7:07Â*pm, "Ralph Mowery" wrote:
"Limp Arbor" wrote in message

...
Question: (fixed type font picture)
L1 Â* Â* Â* Â* Â*N Â* Â* Â* Â* Â*L2
| Â* Â* Â* Â* Â* | Â* Â* Â* Â* Â* |
|____@@_____|____@@_____|
Â* Â*Outlet 1 Â* Â*Outlet 2

Are these statements correct using the above simple MW circuit?
If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp wont work.
If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

1. Â*If you loose the neutral and a lamp in outlet 1, it will not work if
nothing is in outlet 2. Â*Neither will one in outlet 2 if nothing is in
outlet 1.

2. Â*If anything is connected to outlet one and two at the same time, the
voltage will split depending on the load. Â*The higher the resistance of the
load at one outlet compaired to outlet 2, the higher resistance will have
the most voltage across it. Â*If the loads are exectally equal, they will
both have half the supply voltage across them. Â* That is in your example, if
you put in a 60 watt lamp in each outlet and they both come up to the same,
they will appear normal with 110 volts across each one. Â*If you just put one
lamp in one of the outlets, you will measuer 220 volts ( more like 219.99999
or so) Â*on the other one that does not have a load. Â*As you put a load on
that outlet , the voltage will start dropping. Â*It may reach almost but not
exectally 0 volts, while the other outlets voltage will rise toward 220
volts.

And if you put any lamp in outlet one, and nothing in outlet two, the
potential
across the terminals of outlet two will be 240V. I believe that was
the
question.
And the answer is a "qualified yes".

Open circuit there will be 240.
Now, if there is a 50 watt bulb connected to outlet one, and you plug
a 200 watt bulb into outlet 2, the bulb in outlet one will blow
because there well be a higher voltage drop across it than across the
200 watt bulb.The 200 watt bulb will just glow untill the 50 pops.

On Mar 30, 2:16*pm, wrote:
On Wed, 30 Mar 2011 07:30:59 -0700 (PDT), "





wrote:
On Mar 29, 7:07*pm, "Ralph Mowery" wrote:
"Limp Arbor" wrote in message

....
Question: (fixed type font picture)
L1 * * * * *N * * * * *L2
| * * * * * | * * * * * |
|____@@_____|____@@_____|
* *Outlet 1 * *Outlet 2

Are these statements correct using the above simple MW circuit?
If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp won’t work.
If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

1. *If you loose the neutral and a lamp in outlet 1, it will not work if
nothing is in outlet 2. *Neither will one in outlet 2 if nothing is in
outlet 1.

2. *If anything is connected to outlet one and two at the same time, the
voltage will split depending on the load. *The higher the resistance of the
load at one outlet compaired to outlet 2, the higher resistance will have
the most voltage across it. *If the loads are exectally equal, they will
both have half the supply voltage across them. * That is in your example, if
you put in a 60 watt lamp in each outlet and they both come up to the same,
they will appear normal with 110 volts across each one. *If you just put one
lamp in one of the outlets, you will measuer 220 volts ( more like 219..99999
or so) *on the other one that does not have a load. *As you put a load on
that outlet , the voltage will start dropping. *It may reach almost but not
exectally 0 volts, while the other outlets voltage will rise toward 220
volts.

And if you put any lamp in outlet one, and nothing in outlet two, the
potential
across the terminals of outlet two will be 240V. *I believe that was
the
question.

*And the answer is a "qualified yes".

Open circuit there will be 240.
Now, if there is a 50 watt bulb connected to outlet one, and you plug
a 200 watt bulb into outlet 2, the bulb in outlet one will blow
because there well be a higher voltage drop across it than across the
200 watt bulb.The 200 watt bulb will just glow untill the 50 pops.- Hide quoted text -

- Show quoted text -

You have to be careful with that a bit. Light bulbs have a variable
resistance that changes based on the temp. I'm not saying it wouldn't
do what you say just that occasionally they behave in unexpected
manners.

On Wed, 30 Mar 2011 08:22:51 -0700 (PDT), Limp Arbor
wrote:

On Mar 30, 11:15Â*am, jamesgangnc wrote:
On Mar 30, 10:30Â*am, "
wrote:





On Mar 29, 7:07Â*pm, "Ralph Mowery" wrote:

"Limp Arbor" wrote in message

...
Question: (fixed type font picture)
L1 Â* Â* Â* Â* Â*N Â* Â* Â* Â* Â*L2
| Â* Â* Â* Â* Â* | Â* Â* Â* Â* Â* |
|____@@_____|____@@_____|
Â* Â*Outlet 1 Â* Â*Outlet 2

Are these statements correct using the above simple MW circuit?
If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp wont work.
If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

1. Â*If you loose the neutral and a lamp in outlet 1, it will not work if
nothing is in outlet 2. Â*Neither will one in outlet 2 if nothing is in
outlet 1.

2. Â*If anything is connected to outlet one and two at the same time, the
voltage will split depending on the load. Â*The higher the resistance of the
load at one outlet compaired to outlet 2, the higher resistance will have
the most voltage across it. Â*If the loads are exectally equal, they will
both have half the supply voltage across them. Â* That is in your example, if
you put in a 60 watt lamp in each outlet and they both come up to the same,
they will appear normal with 110 volts across each one. Â*If you just put one
lamp in one of the outlets, you will measuer 220 volts ( more like 219.99999
or so) Â*on the other one that does not have a load. Â*As you put a load on
that outlet , the voltage will start dropping. Â*It may reach almost but not
exectally 0 volts, while the other outlets voltage will rise toward 220
volts.

And if you put any lamp in outlet one, and nothing in outlet two, the
potential
across the terminals of outlet two will be 240V. Â*I believe that was
the
question.- Hide quoted text -

- Show quoted text -

And there in lies the fundamental flaw in these circuits. Â*If the
neutral in a regular circuit breaks nothing works. Â*If the neutral in
an edison circuit breaks you have a mess.

Exactly!

Is the minimal savings in wire really worth it?

Yeah I know there is also additional labor also but why do this? To
me the potential problems outweigh the savings.
In my opinion, and the opinion of my dad, an electrician, the ONLY
time an "edison circuit" makes sense is for a "split receptacle", as
required on countertops in Canada. You have 2 separate 15 or 20 amp
circuits that are not shared anywhere else - and the "tied" breakers
make sure there is NEVER power on only one side.
Each half of the duplex can handle it's full rated current, without
dependency on any other outlet. It's code in Canada.
NO legitimate load will EVER trip the breaker, so the breaker is there
to do it's primary job - protect against "fault" currents.

On Wed, 30 Mar 2011 12:08:29 -0700 (PDT), jamesgangnc
wrote:

On Mar 30, 2:16Â*pm, wrote:
On Wed, 30 Mar 2011 07:30:59 -0700 (PDT), "





wrote:
On Mar 29, 7:07Â*pm, "Ralph Mowery" wrote:
"Limp Arbor" wrote in message

...
Question: (fixed type font picture)
L1 Â* Â* Â* Â* Â*N Â* Â* Â* Â* Â*L2
| Â* Â* Â* Â* Â* | Â* Â* Â* Â* Â* |
|____@@_____|____@@_____|
Â* Â*Outlet 1 Â* Â*Outlet 2

Are these statements correct using the above simple MW circuit?
If you lose the neutral and have a lamp plugged into Outlet 1 and
nothing in Outlet 2 the lamp wont work.
If you lose the neutral and have a lamp plugged into Outlet 1 you will
have the potential of 220V at Outlet 2.

1. Â*If you loose the neutral and a lamp in outlet 1, it will not work if
nothing is in outlet 2. Â*Neither will one in outlet 2 if nothing is in
outlet 1.

2. Â*If anything is connected to outlet one and two at the same time, the
voltage will split depending on the load. Â*The higher the resistance of the
load at one outlet compaired to outlet 2, the higher resistance will have
the most voltage across it. Â*If the loads are exectally equal, they will
both have half the supply voltage across them. Â* That is in your example, if
you put in a 60 watt lamp in each outlet and they both come up to the same,
they will appear normal with 110 volts across each one. Â*If you just put one
lamp in one of the outlets, you will measuer 220 volts ( more like 219.99999
or so) Â*on the other one that does not have a load. Â*As you put a load on
that outlet , the voltage will start dropping. Â*It may reach almost but not
exectally 0 volts, while the other outlets voltage will rise toward 220
volts.

And if you put any lamp in outlet one, and nothing in outlet two, the
potential
across the terminals of outlet two will be 240V. Â*I believe that was
the
question.

Â*And the answer is a "qualified yes".

Open circuit there will be 240.
Now, if there is a 50 watt bulb connected to outlet one, and you plug
a 200 watt bulb into outlet 2, the bulb in outlet one will blow
because there well be a higher voltage drop across it than across the
200 watt bulb.The 200 watt bulb will just glow untill the 50 pops.- Hide quoted text -

- Show quoted text -

You have to be careful with that a bit. Light bulbs have a variable
resistance that changes based on the temp. I'm not saying it wouldn't
do what you say just that occasionally they behave in unexpected
manners.
You saying there is a possibility it would pop the 200? Or that the
50 could survive?
Particularly if both are tungsten bulbs, not a chance of either. Both
bulbs will be cold when the power comes on. Both will be at minimum
resistance (resistance increases and current drops as the filament
heats up) - and the 50 will heat up first, making it's resistance
drop to the point there is not enough current to heat the 200 - which
is already only 1/4 the resistance of the COLD 50.

Only one POSSIBLE result.

On 3/30/2011 4:30 PM, wrote:
....

Only one POSSIBLE result.

Well, if I think of it as a voltage divider, the 200W-er is roughly 4X
the R of the 50W and thus the voltage drop will be about 80% across
it...R1/(R1+R2)

--

On Mar 30, 5:41*pm, dpb wrote:
....the 200W-er is roughly 4X
the R of the 50W...

Are you sure about that?... (I could be missing something.)

On Wed, 30 Mar 2011 16:41:19 -0500, dpb wrote:

On 3/30/2011 4:30 PM, wrote:
...

Only one POSSIBLE result.

Well, if I think of it as a voltage divider, the 200W-er is roughly 4X
the R of the 50W and thus the voltage drop will be about 80% across
it...R1/(R1+R2)
You have it totally backwards. The 200 watt bulb is roughly 1/4 the
resistance of the 50, so will drop roughly 20% of the voltage across
ir, assuming equal temperatures. This means the 50 watt bulb gets 192
volts across it when cold - and as the temperature goes up and the
resistance increases, the current drops, causing the 50 watt bulb to
see closer to 210 volts - for a split second.

On Wed, 30 Mar 2011 15:10:06 -0700 (PDT), Larry Fishel
wrote:

On Mar 30, 5:41Â*pm, dpb wrote:
....the 200W-er is roughly 4X
the R of the 50W...

Are you sure about that?... (I could be missing something.)
You are not missing anything, Larry.

An usual Nonan is.

On Mar 30, 5:41*pm, dpb wrote:
On 3/30/2011 4:30 PM, wrote:
...

Only one POSSIBLE result.

Well, if I think of it as a voltage divider, the 200W-er is roughly 4X
the R of the 50W and thus the voltage drop will be about 80% across
it...R1/(R1+R2)

--

Nope, higher wattage bulb is going to have the lowest resistance.

Jimmie