Could someone refresh me on the theory, that when a main neutral is lost on a 120/240 volt circuit, it throws 240 volts on one leg?? Thanks, I've run across this several times in the past.. Steve Has something to do with electricity seeking ground, right??

Opening the neutral connection effectively places the 2 halves of the panel in series with each other, across the 240V line, and the voltages across each half divide according to Ohm's law. If the loads on the 2 halves were perfectly balanced, each side would see 120V, and there would be no detectable problem. If one phase is more heavily loaded, it will drop LESS voltage, with the difference appearing across the opposite side.

Worst case would be a small electronic load (VCR?) on one side of the panel, and a space heater on the other side. The VCR is across 240V, with only the low resistance of the heater in series with it. Unless designed to take that voltage (some are), the VCR is toast VERY quickly.

A sure sign of an open neutral somewhere are lamps that get brighter when larger loads cycle on.

Hopefully someone can offer a more technical answer here, but basically the voltage on each 'leg' will depend on the resistance/impedance of the connected load on that 'leg' at that moment.

Both Voltages could be equal if both loads were perfectly balanced, but Murphy's Law says that one will be high and one low.

Bill

NJ,

I've got to learn to type faster.
Your post wasn't there when I started typing.
Good response.


Bill

[This message has been edited by Bill Addiss (edited 12-16-2003).]

Steve, first of all it will not put 240v on one leg, it will be E = I x R on each load in the circuit, which is now a 240v series circuit.


Now, electricity doesn't seek ground, (earth) it seeks it's source.

It can however use ground (earth) as a conductor.

Roger

sometimes a pix helps....

Any circuit that shares a "NOODLE" with the other will be a series circuit.
The circuits would be just fine if they were equally balanced, but they aren't usually, therefore the voltages "float", according to the load.
I'm crappy at explaining this...
Help from Scott T??

y'know, I had this happen...not a week ago...I just got finished making my splices in a j-box in the attic...(of course,where else would I be...?)and I had brought up a 14-3 nm from the basement, to the attic for some receptacles and lighting...the 2 phases were then energized and on finishing, I realized I didn't like the way the noodles were spliced so I untwisted them,and to my surprise...the temporary pig-tail lampholders I had strung up had SUPERNOVA'd...I put the "tidy whities" back together and the lamps went back to normal again...I really got scared...I thought somethin' was really awry here, and I called a buddy o'mine who told me to "calm down" and that "that phenomenon is a normal occurance"...phew, I thought somethin' was wrong with my wiring and the sheet-rocker just got done w/ his 1st coat of spackle...I had split the neutral between the receptacles and the lighting....I'd never run into this until recently....boy did I feel stooopid!!!...I also came up on the reverse of this....after losing the noodle on a 2-wire circuit,that I was troubleshooting, I put my DMM across the hot and neutral and read about 78 volts...I guess thats because the current was reading through the lamp filaments and the resistance thru it lowered the actual 110-120volts supplied to it....I eventually found the bad neutral splice...

[This message has been edited by Attic Rat (edited 12-16-2003).]

[This message has been edited by Attic Rat (edited 12-16-2003).]

I know this is primitive and simple but here goes. Don't pay attention to the arrows.



The circuit has 5 amps flowing.
If we look at each load and use E = I x R

The 36 ohm load would see 180 volts

5 x 36 = 180 volts

The 12 ohm load would see 60 volts

5 x 12 = 60 volts

180 + 60 = 240 volts.

I will practice before I post any more drawings.

Roger

Just to add that the situation will become a little more complex if a parallel neutral path exists.

If you lost the incoming service neutral, there may still be a circuit back to the transformer neutral via the grounding electrode, water pipes, etc.

The resistance of this path will be higher than that of the service neutral. The greater this resistance, the more the imbalance between the two sides of the circuit will be.

So, it stands to reason then, that the jumper wire between the inlet and outlet sides of the water meter is there not just for electrical continuity in the absence of the meter, but to keep the meter guy from getting the "shock-of-a-lifetime" when changing out a meter.

Gene,

In my area there's a heavy metal yoke that goes across the 2 sides that provides continuity without the jumper. Our local AHJ considers the yoke sufficient to bond both sides so only one connection is really required.



Bill

P.S. Folks: I know this shows 2 connections and that they are in bad shape .... picture posted to show the yoke.


Bill

If there were two meters within the yoke, would we need a handle (valve) tie.

Sorry about that, I just couldn't resist.

Roger

LOL!!!!!!!!!...but not if it were split yolks..{rimshot}...ok,ok,..I'll stick to my attic work!!:0:0

OK,

ENOUGH with the Yolks Folks! ...

... What was the question?


Bill

I like my yokes over-easy please!

Hello all;

Everyone already contributed perfect explanations and examples of the "What-For's" and 'Why's" regarding this topic (the original topic of open noodle problems - that is!)

To add more of the common Hyper Nerd inclusion here as usual, consider the following:
(Use the 1Ø 3W center tapped Transformer Schematic - attached by Sparky - as working example)

Scenario #1: NJ Wirenut' example situation;

As posted by 'NJ, here is an example of two different types of loads, one with a very LOW Resistance to the current flow (the portable heater), and the other with a very HIGH Impedance to the current flow (the VCR's Power Supply).
Just a side note: the different terms used "Resistance" and "Impedance" are due to the way each type of load "works" against the current flow.
The Heater (without a Fan), is a simple Resistor, however the VCR's Power Supply will be connected to the AC line via the Primary of a Transformer - which will "Impede" the current flow through it; hence the term "Impedance" is used. Impedance still results in an opposition to current flow, so for sake of "Easier Understood Terms", I will refer to all opposition as "Resistance".

Using the Schematic quoted earlier, if the two separate loads are placed individually across A-center tap (or B-center tap), each will see only a potential of 120 Volts.
If one load is placed between A and center tap, and the other placed between B and center tap, the results are still the same: 120 Volts is the potential.
Now, if we cut the wire going to the center tap - right outside of the winding, the two loads are connected in a Series fashion.
In this setup, the load with the HIGHEST Resistance to the flow of current will have the HIGHEST potential (Voltage) impressed across it.
The opposite can be said about the other load - the one with the LOWEST Resistance to current flow will have the LOWEST potential impressed across it.

The VCR will have the HIGHEST Resistance to current flow, since it is designed to draw only minimal amount of True Power (Wattage) from the AC Power Source (the PoCo's Transformer, and eventually the Generator its self). This would be in the range of maybe 35 Watts. Under normal operation, the Transformer will "allow" maybe upto ½ ampere @ 120 VAC. This translates into 240 Ohms Impedance (in this case, Z, or Impedance is more accurate than Resistance - but for simplicity, think "Resistance).
The Heater will have the LOWEST Resistance to current flow, since it is designed to draw a large amount of True Power from the AC Power Source. For this example, we will say the Heater is designed to draw 1200 Watts @ 120 Volts - which results in 10 amps flowing at 120 volts, or a Resistance of 12 Ohms. In this case, the proper term will be "Resistance".

With these two in series, the overall Resistance will be 252 Ohms. In a Series setup, the total potential (Voltage) will now be 240 Volts (since we cut the center tap conductor outside the Transformer).

This is where a wierd thing happens! 252 Ohms Resistance connected to a power supply with a potential of 240 Volts, results in a current flow of 0.95 amperes. This will have odd results in each element.
The VCR's Transformer has an Impedance of 240 Ohms. According to Ohms Law, if the element has 0.95 amps flowing through it and the Resistance is 240 Ohms, the Voltage needed to push that 0.95 amps through it will be 228 Volts.
The Heater has a Resistance of 12 Ohms. With 0.95 amps flowing through it, the Voltage required to push that "almost an amp" of current will be 11.4 Volts - or for sake of simplicity, 12 Volts. Notice that this equals the "Missing" Voltage from the VCR (240 V - 228 V).

Here's the crazy part! With 228 Volts pushing 0.95 amps across the 240 Ohm Impedance of the VCR's Power Supply, this will result in 216.6 Watts of True Power being drawn from the AC Power Source and "Dumped" into the Transformer - resulting in smoke loss within an hour.
The Heater will hardly even get hot! Only 11.4 Watts of True Power will be dumped into the Heater! The total True Power (Wattage) drawn by both loads will be 228 Watts.

Scenario #2: Roger's Drawing;

Roger has included a great drawing, which will physically show the results of the stuff described above! The resulting failure will not be as much smoke as it will be bright flash of light, then no light at all!
In Roger's example, the values are 12 Ohms and 36 Ohms.

Let's see if you can conclude the results of this example below:

120 Ohm load + 180 Ohm load across 240 Volt circuit. Each load is designed to be connected to 120 Volt system only.

Good luck, and have fun!!!

Scott35

P.S. There was a thread from a few weeks / months back, which asked:
"What Happens if a 40 Watt 120 Volt Lamp and a 60 Watt 120 Volt Lamp are connected in Series, then connected to a 120 Volt power source - which Lamp will be the brightest". This is a great brain picker!

S.E.T.

Scott, thank you for the compliment.

Don't you want to say something about Bill's yoke?

Roger

Wow! And I thought just a simple question would get a simple answer

Bill,
Thanks for the picture. I've never seen one of those before. They must not be required around here.

Scott35,
Did you ever consider writing books?

[This message has been edited by Mean Gene (edited 12-22-2003).]