rcd question?

[Spark123]

Depends how much current was flowing, the resistance/impedance of the circuit conductors and versus the resistance/impedance of the meter tails. If the meter tails etc are pretty short you may need to have quite a few amps flowing to get enough PD between the neutral and earth rails in the CU to cause enough current to flow through the N-E fault to cause the imballance (somewhere between 16 - 30mA).
As long as all your tests came out fine you are home dry

 

[laughingjester]

[quote="Spark123";p="1160594] If the meter tails etc are pretty short [/quote] they are 25mm tails 16mm main earth and they are only 50cm long think you might have something there buddy thanks

 

[ColJack]

no he's not ok there..

with a load on the remaining circuits on that side of the RCD, it should have tripped with a N-E fault..

the PD and size of meter tails etc have nothing to do with it for the reasons I stated above..

the neutral current should have been split between the neutral conductor and the earth conductor..

RCD's don't work on PD.. they work on current alone.

if the outgoing current doesn't equal the returning current then it should trip ( a difference of 30mA is enough to operate it.. )

so unless you only had a load of 40mA to begin with ( in which case you'd get about 15 mA running down the earth wire ) then it should operate..

 

[Spark123]

Yes, he is OK there.

RCDs operate on current. In order for current to flow you must have a potential difference - remember CGSE physics.
Where there are two paths present for the current to return i.e. two links between neutral and earth - one being the one in the service head and the other one being the fault, the current will be shared between them.
The cables up until the point of point of the two links will have a resistance/impedance. If the tails etc have a very much lower resistance/impedance than the RFC cables then a bulk of the current will flow back via the meter tails - think resistors in parallel.
Now consider the amount of current which needs to flow in the meter tails in order to push up the potential of the neutral rail in order for enough current to flow - lets say 25mA through a cable of more than likely quite a higher resistance/impedance.

OKay - lets if we introduce some figures we know he has 0.5m long tails and the size.
We know it is a RFC. Don't know how long it is or where the fault it, but for arguament sake lets say 40m long and the fault at the mid point so equates to about the same as a 10m long radial.
The resistance of a 10m long radial (R1+R2) from Appx 9 of the OSG is 19.51mΩ/m so 0.1951 Ω in total.
The resistance of the meter tails is 1.877mΩ/m, so from OSG 0.5m equates to 0.0009385 Ω in total.
In order for say 0.025A to flow down the twin and earth we need a potential of V = IR = 0.025 x 0.1951 = 0.0048775 V
The combined resistance of the tails back to the N-E link and resistance of the path back to the fault will be 0.000934 Ω
In order for a potential of 4.8775mV to exist between the N & E rail in the CU a current of I = V/R = 0.0048775 / 0.000934 = 5.2A will need to be flowing through the CU.

 

[laughingjester]

thats alot to think about spark 123. but im getting through it slowly.. although the other cicuits were on only 1 was drawing a load, wich was garage lights , so i see your point. because the load was so small on the cicuits protected by that rcd it didnt generate enough mA down the nuetral to trip the rcd....sounds about right i think, you both know your stuff though wish i was as technically minded as you both im just a humble sparky with enough testing knowlegde to get by on at the moment. but gaining more knowledge all the time cheers guys. will keep checking out your other threads..