Скачать или смотреть A broken or floating neutral and the effect in three phase electricity systems

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I HAVE ADDED MORE DETAIL AT THE BOTTOM AFTER MANY REQUESTS ABOUT THE THEORY OF WHAT HAPPENS.

A broken or floating neutral in a three phase electricity system is disastrous. The single phase voltages will vary wildly from very low even up to a full phase voltage which will damage electronics and electrical appliances.

This demonstration is to show how the voltage will move from 230 volts or 240 volts anywhere up to 400 volts to 415 volts when the neutral is disconnected.
I've used light bulbs as they are really simple, they show brightness variations easily and they can take the overvoltage but carry on working.


UPDATE: Have a look at the following video. This illustrates a real life situation where thieves stole the neutral wire out in the street.
   • Great Britain: Gas leak - the house explod...  


Please don't work with electricity unless you really know what you are doing. I have been doing this for 26 years and safety is paramount. Phase to phase voltages of 400 volts are not forgiving of mistakes.



ADDITIONAL THEORY:
A few people have asked me to explain what happens more theoretically, so here is an addition.

The current (amps) will force themselves to be equal through all phases when the neutral is open. This is why the voltages change so much.

In a three phase system the sum of the three phases wants to be zero. So the currents want to be equal when the neutral is broken.

With the first set of bulbs, they are all the same wattage, so the currents are equal and voltages are equal.

With the second set of bulbs, the wattages are different, but the systems forces the currents to be equal. That's why the smaller bulbs increase in brightness and the bigger bulb decreases. Based on the formula V=IR, the Resistance won't change, but both V and I will change.

So using P=V²/R or rather R=V2/P bulbs normally are

Red:

R1=240²/60
R1=57600/60
R1=960 Ohms

White:
R2=57600/40
R2=1440 Ohms

Blue:
R3=57600/77
R3= 748,052 Ohms


If we work out current from those and the measured voltages


I1=V1/R1
I1=256/960
I1=0.26667 Amps

I2=292/1440
I2=0.202778 Amps

I3=185/748,052
I3=0.2473 Amps

Ok, so the reason these aren't exactly identical here is probably because the bulbs are not perfect to their rated outputs.

The real result will be that each bulb will carry exactly the same amps. The resistance of each won't change so the 3 phases will just shift voltages to get the current balanced and that is what will determine the voltage each bulb gets. Remember V=IR and P=VI.