Loose Neutral


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Old 12-19-03, 09:34 AM
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Loose Neutral

I've seen several posts over time saying that a loose neutral would cause lights to become brighter. Can anyone explain why this happens. Seems to me that if a neutral becomes loose or disconnected current to that circuit would become erratic and the supply current may go up or down resulting in the lights on that circuit either dimming or brightening. Can't understand why they should become brighter only.

Also seems to me that a loose connection to the hot end of the breaker should have the same effect.
 
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Old 12-19-03, 11:54 AM
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The loose or lost neutral / bright lights phenomenon occurs when the main grounded conductor (coming from the transformer) loses it's path back to the transformer. What happens is the current travels throught the hot wire to the light bulb, through the filament, and then back to the panel via the white neutral wire. If it's path back to the transformer is impeded, it will travel through the neutral wires on other circuits (backwards if we were dealing with direct current as opposed to alternating current) and through those light bulbs which are being fed with a different hot circuit. If the bulb other bulb is being fed with a circuit that is 180 degrees out of phase with the first bulb, then both bulbs will measure 240 volts across their filaments, and will burn incredibly bright before blowing out.
 
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Old 12-19-03, 12:42 PM
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Not too sure I understand this. As I understand it each circuit is powered starting with a breaker. The breaker is connected to one of two phases on the hot buses. The current passes through the breaker and then goes to all the loads in the circuit and then returns via the neutral to the breaker box. If the neutral becomes disconnected would this not break the circuit?

Seems to me that you are saying that current from one circuit is going to leak into another. However, I thought that the common point is the breaker box only. Also don't understand the bit about the transformer - which one are you refering to.
 
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Old 12-19-03, 01:06 PM
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In this case the loose neutral being discussed is the one that goes back to the power company transformer. The power company supplies three lines. Two are hot and one is a neutral.

The power that the power company delivers has to get back to the transformer one way or another. if the neutral is loose or disconnected, then it will get back via another hot wire, as desceibed.
 
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Old 12-19-03, 01:24 PM
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OK now it makes sense. Thanks for both your inputs.
 
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Old 12-19-03, 02:43 PM
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It doesn't work exactly as described. You won't get 240 volts across the each of bulbs. That would blow the bulb instantly.
What happens is the 240 volts splits between the bulbs. If the bulbs are the exact same wattage then they share the voltage equal and you don't notice anything. If one bulb is higher wattage then the voltage does not split equal and one bulb gets for example 90 volts and the other gets 150 volts. That's what you see when light dim and brighten.
 
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Old 12-19-03, 05:56 PM
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Now that I've understood where the neutral being referred to is, let me explain the situation to see if I have it right.

If the neutral becomes disconected, the total load in the house is now connected across the two phases. So as I see it if the loads on each of the two phases are equal then no effect should be noticed. However, if one phase is fully loaded and the other marginally, the loads on the first phase will have a potential diference of almost 240V and the second phase will have a very small voltage drop (ie they form a voltage divider). I don't think that you can consider each light individually - you need to look at the total load on each phase and this will determine what will happen. By this reasoning I agree that the lights may dim or brighten.

Does this make sense or is it completely wrong.
 
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Old 12-19-03, 09:23 PM
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You're right, you do have to consider all the loads on each leg of the power. Essentially, all the loads on each leg are in parallel with the other loads on the same leg, and then the two legs are in series with each other. It would take quite a bit of math to figure out precise what voltage everything would get.
 
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Old 12-20-03, 01:38 AM
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Actually John the calculation are pretty straightforward. Iíve calculated this and my model for doing it is explained below.

Letís say that we have 3 loads of 10W, 50W and 90W on one phase and 40W and 60W on the other. Letís also say that the neutral is no longer in place. So now according to what we discussed the voltage at the neutral point will be at some potential Ė letís call it Vx.

We also know that the total current passing through all the loads on one phase must pass through the other phase. So we solve the problem by equating currents as follows:

Using the relationship that power = voltage * current.

For Phase 1 total current = 10/(120-Vx) + 50/(120-Vx) + 90/(120-Vx) = 150/(120-Vx).

For phase 2 total current = 40/(Vx+120) + 60/(Vx+120) = 100/(Vx+120).

Equating current s we get

150/(120-Vx) = 100/(Vx+120).

Solving for Vx gives us Vx = -24V. So all loads on phase 1 will experience a voltage difference of 144V and on phase 2 96V. Iíve assumed purely resistive loads for the calculations and you can see that all you need to do is add the power requirements of all the elements in each phase and substitute in the numerator in the equations Iíve written down.

I just did this by thinking this out Ė it does not come from any book so Iíd welcome comments on any possible things I may have forgotten about when doing the calcs.
 
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Old 12-20-03, 10:19 AM
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Your answer is kinda correct (but oddly reversed), but your math is flawed. The equation "power=voltage*current" led you to write equations that assumed that as voltage increases, current decreases, with power holding constant. But power is not a constant (i.e., you cannot put a constant "10" in for the 10-watt load). The only constant is resistance (well ... another story). So the calculation cannot be made without also considering Ohm's law: voltage=current*resistance.

Put another way, a 60-watt light bulb is no longer a 60-watt light bulb when the voltage is other than 120 volts (or whatever the bulb is rated for).

Bottom line is that your calculation came up with the right numbers for the voltages (i.e., 144 and 96), but the 96 volts is on the first leg and the 144 volts is on the second leg (just the opposite of what you came up with). Furthermore, your calculation did not correctly predict the current. Your formula predicts the current at 1.04 amps, but the correct calculation predicts it at 1 amp.

Finally, it is indeed a rare home with only five loads running. Try it again with a few hundred loads running.
 

Last edited by John Nelson; 12-20-03 at 11:12 AM.
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Old 12-20-03, 12:25 PM
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Yes, youíre quite right Ė what an oops!! It was rather late last night when I decided to do it. Yes I should have used constant resistance which as you say is not constant anyway but is more so of one than power. Also there will be reactive elements present which I have not taken into account.

My calcs agree with yours now Ė Iím putting the steps down briefly so that anyone interested can see how it was done.

Resistance of the loads calculated in order using R = V**2/P

1440,288,160,360,240

Total resistance in phase 1 is R1

1/R1 = 1/1440+1/288+1/160

R1 = 96 ohms

Total resistance is phase 2

1/R2 = 1/360+1/240

R2 = 144 ohms

So Vx will be 144/(144+96) * 240 = 144 Ė strangely the same answer as before. Iíve assumed that phase 1 is +120V and phase 2 -120V as before in my previous calc so Vx will actually be -120 + 144 = +24V and so the order is revered as you say. I think the voltage drops matching is complete coincidence.

I agree that in a real situation this is too simplified but I just did it to show the principle. I think what would make it complex are not just the static loads but the combination of the loads on each phase as things are switched on and off.

Based on this discussion I guess loose neutral would not affect appliances running off 240V circuits. Also for a bunch of lights that get bright another set should dim.

Is this situation very common. It sounds extremely dangerous to me. In general I wonder why this topology is used in the US and other places. The single phase 220V solution sounds safer to me in this respect. Also with 220V you need should not need such think wires and current carrying capacity for the breakers. Anyone have any idea about the history and reasons for the US style supply system vs the single phase 220V approach.
 
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Old 12-20-03, 02:03 PM
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Judging by the posts in this forum and others, a loose connection from the power company (either the neutral or one of the hots) is pretty common indeed.

And no, it's not just coincidence that the voltages were the same but reversed in your original method of calculation. It'll happen that way every time. Your original calculation held power constant let resistance vary. The correct calculation holds resistance constant and lets power vary. The difference is equal and opposite.

I agree with your new math.
 
 

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