Why doesn't RCD trip?

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6mm Cooker Circuit.

R1 + R2 = 0.06 ohms.

L - E IR > 999 M ohms.

L - N IR > 999 M ohms.

E - N IR = 0 M ohms. (between conductors only - cable removed from earth & neutral bars)

E - N resistance 2.74K ohms.

RCD ramp test 25mA

RCD tests ok at 1/2, 1X & 5X

Why even under load does this circuit not trip RCD?

 
Draw it out as a simple circuit....    (Ignore the Live side).

N coming back from the load through, MCB & RCD (or RCBO) then through main switch then making its way back to supply to re-join Earth at the Star point..  

with a parallel path somewhere part way down this N wire, via a 2.7k fault, back to the installation main earth. (And if this is TN-C-S this is back onto the N wire incoming side of supply)..

So as ProDave said...  the electrical resistance straight down the normal neutral path will be so low that hardly any current will flow down your 2.7k fault path...

or think of it as a voltage divider...

where your neutral current reaches a fault with one half down a copper conductor the other half down a 2.7k fault....

for the sake of arguments imagine the two parallel paths are   2700ohms and 0.35ohms..

calculate the voltage split and thus current you would expect down your neutral fault...

Whereas a fault on the live side would be splitting between 2.7k and the load resistance back to N..

Its one of those things that if you draw it out with the load and fault as simple resistors on a basic series/parallel circuit, and blocks indicating the supply/RCD/MCB etc..

It can make more sense than your brains logic saying Oh I have a low resistance on my IR tests this should trip the RCD!!!

Guinness

 
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