Lost neutral explaination

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JasonGSparky

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So this may be a bit of a goofy question, but I was wondering if anyone knew of an easy way to explain how a lost service neutral can result in 240v on 120v circuits, some lights dimming while others get brighter, etc.

I've had a few service calls in last couple months where the home owner complains of the typical lost neutral problems - some devices started smoking, light bulbs burned out, other things aren't working; and aside from one mwbc that was knicked, its almost always the neutral from the power company was broken or corroded, resulting in these problems. (neat tip I learned, if the overhead service lines are hanging really low, its a good indication the neutral is busted off at the power line) I know how to troubleshoot it, I can usually find where the break is, and sometimes I can repair it myself if its on the house side.

The best explanation I can give to the customer is that, 120v circuits require the neutral wire to take the unbalanced load back to the transformer, and when that neutral is gone, the unbalanced voltage backfeeds the circuit. I'm not totally comfortable with this explanation, and I would like to know how this problem actually works. I've seen some diagrams and read other posts, but I just can't wrap my mind around it. I'm stuck on the fact that if both hots come in contact with each other or the neutral, something will trip or blow up, so how can you get 240v on a 120v outlet?:?

*My best thought is that when the service neutral is gone, all the voltages come back to the neutral bar and feed BACK through the other circuit neutrals. Still not sure how it doesn't trip a main breaker, unless the ground wire isn't adequate enough to cause a huge surge in current?
 
JasonGSparky said:
So this may be a bit of a goofy question, but I was wondering if anyone knew of an easy way to explain how a lost service neutral can result in 240v on 120v circuits, some lights dimming while others get brighter, etc.

I've had a few service calls in last couple months where the home owner complains of the typical lost neutral problems - some devices started smoking, light bulbs burned out, other things aren't working; and aside from one mwbc that was knicked, its almost always the neutral from the power company was broken or corroded, resulting in these problems. (neat tip I learned, if the overhead service lines are hanging really low, its a good indication the neutral is busted off at the power line) I know how to troubleshoot it, I can usually find where the break is, and sometimes I can repair it myself if its on the house side.

The best explanation I can give to the customer is that, 120v circuits require the neutral wire to take the unbalanced load back to the transformer, and when that neutral is gone, the unbalanced voltage backfeeds the circuit. I'm not totally comfortable with this explanation, and I would like to know how this problem actually works. I've seen some diagrams and read other posts, but I just can't wrap my mind around it. I'm stuck on the fact that if both hots come in contact with each other or the neutral, something will trip or blow up, so how can you get 240v on a 120v outlet?:?

*My best thought is that when the service neutral is gone, all the voltages come back to the neutral bar and feed BACK through the other circuit neutrals. Still not sure how it doesn't trip a main breaker, unless the ground wire isn't adequate enough to cause a huge surge in current?
Click to expand...

It's actually a lot simpler than that. Picture two 120V (L1 and L2) lines to neutral with different sized loads (resistances) on each line to neutral. The voltage across each load is 120V irrespective of the its resistance. Now disconnect the neutral from the service. It's now a voltage divider. Now you have two loads in series across L1 and L2; the voltage across the series pair is 240 V, but the voltage across each of them is determined by its resistance. Ohm's Law. The voltage to ground at the point between them (what was the neutral before you disconnected it) could be anything between +120V and -120V depending on the size of the loads.
 
And because of the voltage divider effect, the lower power, higher resistance, loads will be the ones that see the higher voltage and burn out.
The higher wattage loads on the other leg will see a reduced voltage and will be dimmer, cooler, slower, but will probably not be damaged in the short term.
 
I explain this using a tug-of-war analogy. Tell them this.

Your house service is like a tug-of-war rope before the guy lets go of the middle of the rope. The two lengths are balanced with 120V on each side making a total of 240V between the ends of the rope. The neutral is the anchor that keeps the tape centered in the middle. When the neutral goes, the rope gets tugged to various sides depending on how my load (pulling) is happening at any given moment. Hence the voltage fluctuations.

Mark
 
an image I have posted here before, hopefully explains things for you or helps narrow down any further questions

 
When the neutral is lost in a MWC the circuit becomes a series circuit and you can simply use Ohms law to figure it out or show it.

3wire3.gif


Roger
 
Thank you guys, every one of you. These were perfect explanations. The tug-of-war is a good analogy. I often find myself trying to keep explanations simple for customers, because a lot of times they don't care, they just want it to work.

Thanks again! :thumbsup:
 
Additionally to the damaging overvoltage, when the overvoltage item fails, the line serving the items receiving the lower voltage will then get 240 volts, most likely damaging them.
 
JDawn said:
Additionally to the damaging overvoltage, when the overvoltage item fails, the line serving the items receiving the lower voltage will then get 240 volts, most likely damaging them.
Click to expand...
In most cases when the load getting overvoltage fails the circuit is opened and current flow ceases.

Roger
 
JDawn said:
Additionally to the damaging overvoltage, when the overvoltage item fails, the line serving the items receiving the lower voltage will then get 240 volts, most likely damaging them.
Click to expand...
Only if the damaged item develops into a low resistance after damaged. If it burns to an open circuit condition, then you just have open circuit with 240 volts across the open point.
 
I agree, generally on a 120volt circuit, residential, you would be dealing with light bulbs or plug connected items. Failure of these type items would just eliminate resistance and the circuit would still be completed. Usually.
 
JDawn said:
I agree, generally on a 120volt circuit, residential, you would be dealing with light bulbs or plug connected items. Failure of these type items would just eliminate resistance and the circuit would still be completed. Usually.
Click to expand...
I would think the opposite, burn out a light bulb that is in series with other loads - the burned out light bulb acts like an open switch to whatever is in series with it.
 
kwired said:
I would think the opposite, burn out a light bulb that is in series with other loads - the burned out light bulb acts like an open switch to whatever is in series with it.
Click to expand...
I agree.

Roger
 
JDawn said:
Failure of these type items would just eliminate resistance.
Click to expand...
Which is the reason the circuit would not see any current flow and the other loads would be safe.

Roger
 
kwired said:
Resistance isn't "eliminated", when an item burns out, it is raised to/near "infinity"
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Since "infinity" is unachievable, I think for the sake of conversation and looking at a lowly incandescent lamp, "eliminated" works for this conversation.

Roger
 
roger said:
Since "infinity" is unachievable, I think for the sake of conversation and looking at a lowly incandescent lamp, "eliminated" works for this conversation.

Roger
Click to expand...
Well "zero resistance" is pretty much as unachieveable but at the other extreme. Couldn't that be as easily seen as "eliminating resistance"?
 
kwired said:
Couldn't that be as easily seen as "eliminating resistance"?
Click to expand...
Now we're getting somewhere but zero resistance would be a complete circuit so it wouldn't work for this conversation.

Roger
 
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