Neutral lost on house

[bradleymattlin]

On a 120/240 volt system single phase, if the neutral is lost the most the voltage can spike on one phase is 240v right.

 

[gar]

130715-2116 EDT

Under normal circumstances, yes.

But realize the peak voltage for a sine wave is sq-root of 2 times the RMS value. At 120 V that is 169.71 V.

If you feed a half-wave rectifier with 120 V and the load is a capacitor, then the peak inverse voltage on the rectifier is 170 V.

.

 

[K8MHZ]

130715-2116 EDT

Under normal circumstances, yes.

But realize the peak voltage for a sine wave is sq-root of 2 times the RMS value. At 120 V that is 169.71 V.

If you feed a half-wave rectifier with 120 V and the load is a capacitor, then the peak inverse voltage on the rectifier is 170 V.

.

. .

 

[K8MHZ]

130715-2116 EDT

Under normal circumstances, yes.

But realize the peak voltage for a sine wave is sq-root of 2 times the RMS value. At 120 V that is 169.71 V.

If you feed a half-wave rectifier with 120 V and the load is a capacitor, then the peak inverse voltage on the rectifier is 170 V.

.

I'm not sure you answered his question. He asked if the most voltage would be 240 with no neutral. So, are you saying that if the RMS value is 240, then the peak to peak would be 340?

That's what we were taught.

 

[hbiss]

Geeze, just answer his question. A simple YES is sufficient for his purposes.

-Hal

 

[iwire]

Geeze, just answer his question. A simple YES is sufficient for his purposes.

-Hal

I whole heartily agree.

 

[iwire]

On a 120/240 volt system single phase, if the neutral is lost the most the voltage can spike on one phase is 240v right.

It would take some odd conditions to get the full 240 but you can get close.

 

[goldstar]

On a 120/240 volt system single phase, if the neutral is lost the most the voltage can spike on one phase is 240v right.

As Iwire stated you'll can get close to that 240V point but the sad part is that whatever you have connected to the breaker panel that is electronic or has a digital display will be toast. So, aside from the curiosity of what the exact voltage might be the exact voltage becomes meaningless compared to the cost of the equipment and appliances you'll have to replace. Of course if you're one of those people who really wants to know the exact voltage you can always remove the neutral at your own house and take a measurement.

 

[don_resqcapt19]

... Of course if you're one of those people who really wants to know the exact voltage you can always remove the neutral at your own house and take a measurement.

You will have to be quick because as the electronic parts give up their magic smoke the voltage will be changing

 

[ptonsparky]

As Iwire stated you'll can get close to that 240V point but the sad part is that whatever you have connected to the breaker panel that is electronic or has a digital display will be toast. So, aside from the curiosity of what the exact voltage might be the exact voltage becomes meaningless compared to the cost of the equipment and appliances you'll have to replace. Of course if you're one of those people who really wants to know the exact voltage you can always remove the neutral at your own house and take a measurement.

Yea, you could but it is only good for your house at the time of what items are connected and on. This method may require several do overs. 10 tries should be close enough. Fluke makes some pretty good meters that can catch and record the peak that Gar was talking about. $350 for the ones I have, but that was several years ago. Let us know what you record and I am sure you will get more help or suggestions. Some may even be helpful. OT, but would this type of research be deductible?

 

[ggunn]

On a 120/240 volt system single phase, if the neutral is lost the most the voltage can spike on one phase is 240v right.

Yes, as a theoretical limit. If you turn off all the loads and then turn on one 120V load, it will not operate because the open neutral is not a return path. If you turn on another 120V load that is on the opposite leg of the 240V supply, then they form a voltage divider across the 240V supply and the voltages they see will be proportional to their resistance. If one is very high and one is very low, then the higher one sees a proportionately higher voltage approaching but not exceeding 240V. If one of the loads is a dead short, then the other will see 240V.

 

[templdl]

Yes, as a theoretical limit. If you turn off all the loads and then turn on one 120V load, it will not operate because the open neutral is not a return path. If you turn on another 120V load that is on the opposite leg of the 240V supply, then they form a voltage divider across the 240V supply and the voltages they see will be proportional to their resistance. If one is very high and one is very low, then the higher one sees a proportionately higher voltage approaching but not exceeding 240V. If one of the loads is a dead short, then the other will see 240V.

This is a very good explanation.

 

[roger]

Here's a visual

Roger

 

[kwired]

Here's a visual

Roger

Depending on the load, the overvoltage may not be the damaging factor, but consider that the 30 ohm load in the example is carrying 4 amps in the "normal" condition and 1.5 times that (6 amps) in the open neutral condition. This means more heating than usual even if it can withstand the higher voltage.

A true resistive load maybe survives this condition if not for too long of duration. Place inductive or capacitive loads in there and impedance of loads are going to change as voltage changes when neutral is lost.

 

[gar]

130716-1706 EDT

Put a 1 ufd capacitor from 120 V hot to neutral, and a 7.023 H inductor with an internal resistance of 120 ohms between neutral the opposite 120 V phase.

With the neutral intact what are the voltages and currents?

Same question with the neutral opened?

.

 

[GoldDigger]

130716-1706 EDT

Put a 1 ufd capacitor from 120 V hot to neutral, and a 7.023 H inductor with an internal resistance of 120 ohms between neutral the opposite 120 V phase.

With the neutral intact what are the voltages and currents?

Same question with the neutral opened?

.

One of my favorite though experiments.
(It gets even more interesting if the inductor has a really low internal resistance. May I assume that by internal resistance you are talking about equivalent series resistance rather than equivalent parallel resistance? The results change mightily between those two cases.)

 

[gar]

130716-1837 EDT

GoldDigger:

Yes equivalent series resistance. I chose to put all the loss in the inductor because very good capacitors are available, and wire wound inductors will always have a substantial internal series resistance. Also note the arbitrary selection of 120 ohms for the resistance. That should provide a hint to some readers. I don't know if I can easily make such an inductor.

.

 

[cowboyjwc]

Geeze, just answer his question. A simple YES is sufficient for his purposes.

-Hal

No, that's not how it works here, we can't simply answer a question with out first showing how smart we is. We'll post 120 times discussing time warps and variable speed drives, and the OP will then give up and just Google the answer.

Now what do you get if you "find" a neutral.:lol:

 

[roger]

No, that's not how it works here, we can't simply answer a question with out first showing how smart we is. We'll post 120 times discussing time warps and variable speed drives, and the OP will then give up and just Google the answer.

Couldn't have said it better.

Now what do you get if you "find" a neutral.:lol:

Try to find a gear?

Roger

 

[iwire]

No, that's not how it works here, we can't simply answer a question with out first showing how smart we is. We'll post 120 times discussing time warps and variable speed drives, and the OP will then give up and just Google the answer.

Couldn't have said it better.

I also agree.

At least start out by just answering the question and then if the OP requests detail we can provide it.