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Diib I now that lol just been working out thermal constraints on my design protect and noticing the lower the zs the higher the fault current was wondering if anyone could explain why other than ohms law
 
Diib I now that lol just been working out thermal constraints on my design protect and noticing the lower the zs the higher the fault current was wondering if anyone could explain why other than ohms law

Hey, think basic maths.....230V/23ohms = 10amps...... 230V/230ohms = 1 amp, notice that the larger the figure you divide by, the lower the answer..its really that simple .....ohms law;)
 
as above. current is inversely proportional to resistance, voltage being constant.
 
So if are zs are to good ie a very low impedance surely are high fault current could possibly be dangerous

You're looking at this from completely the wrong direction, if there's a L-E fault you want a low earth impedance for two reasons, firstly to prevent the chassis of fittings and appliances developing high touch voltages which could cause someone to get a fatal shock, secondly a low earth impedance means a very large fault current flows which in turn means the OCPD disconnects the faulty circuit very quickly.
 
But with very low external impedance, comes rather large fault currents which may cause a few problems with your earthing conductor and the breaking capacity of your ocpd's
 
Really, I think you'll find the general B type breakers are 6kA, although with back up protection of your main fuse you'll get away with up to 16kA
 
I bet they're type C, the usual "domestic" type B breaker is rated at 6kA....Icn & Ics
 
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Well all the MG ones we use are rated for 10kA anyways :p
Granted there are 6Ka out there to :p

As a general rule of thumb nearly all breakers designed for 3 phase DBs are 10kA, and nearly all designed for domestic CU ranges are 6kA. You will get interchangeability with most manufacturers though.
 

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