Multi-Wire or Edison Circuit questions

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

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.
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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 measure:
Across the two outlet terminals: 240V From either terminal to ground: 120V
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I'm old school

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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
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Outlet 1 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.
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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.
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wrote:

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

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.
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On Wed, 30 Mar 2011 07:30:59 -0700 (PDT), " snipped-for-privacy@optonline.net"

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.
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On Mar 30, 2:16 pm, snipped-for-privacy@snyder.on.ca wrote:

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

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.
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On 3/30/2011 4:30 PM, snipped-for-privacy@snyder.on.ca wrote: ...

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)
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Are you sure about that?... (I could be missing something.)
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On Wed, 30 Mar 2011 15:10:06 -0700 (PDT), Larry Fishel

You are not missing anything, Larry.
An usual Nonan is.
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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.
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Nope, higher wattage bulb is going to have the lowest resistance.
Jimmie