Who told you this ? Do they understand electricity and can they point to any credible  reference? Films on Youtube do not count !

 The fault current is highest and most damaging to the breaker when the fault is as near as it credibly can be to the breaker.The breaker needs to match the supply PSSC, or perhaps slighty less if suitably de-rated by the energy limiting action of any up-stream fuse.

The idea of time synchronized  faults somewhere off down branches is a red herring, even if you could lay the two cables one over the other and fire a nail through both together, the breaker will already have begun to unlatch due to the first fault, before the nail shorts the second and fault current comes on .  And if they are not co-located how exactly are the two faults to be synchronized to well within the breaking time of the breaker - i.e. within perhaps a fraction of millisecond of each other....

15 B does not imply that at all. It is just an advisory drawing of one sort of circuit, Does this person also have an issue with ring spurs direct from the CU as well because they are not shown ?

It's the same.

EDIT

Rather like the cable installation methods in the annex, these are not the only wiring methods you can use of course - just the only ones that have been measured or calculated for which ratings are published. For other situations/ layouts one just has to interpret which standard case it is nearest to or do the design calcs yourself.

Mike

PS that nail gun thing...

*Assume the nail gun nail does as good as the best recorded here and shoots at 134 feet/second  that's 40 proper speed units (in technical works speed should be in m/s for any Americans reading)  this is  40mm per millisecond so perhaps it takes 1/4 millisecond to pierce the first T and E, - assuming no loss of speed in doing so. In reality it will be slower.

Compare this to the effective pseudo-speed of a 32A B type breaker  at 6kA fault ( assume an I2t of 29000 A2seconds from this Hagar datasheet)

time is 29000 / (6000.6000) = 0.008 seconds to break . By the time the nail hits the second cable the breaker has already started opening and is about 1/4 of the way through and once contacts begin to open the current is limited by the arc impedance.

This is not a real time, as really breaking completely takes a lot longer, but for most of the time the fault is not the current limiting factor.

Who told you this ? Do they understand electricity and can they point to any credible  reference? Films on Youtube do not count !

 The fault current is highest and most damaging to the breaker when the fault is as near as it credibly can be to the breaker.The breaker needs to match the supply PSSC, or perhaps slighty less if suitably de-rated by the energy limiting action of any up-stream fuse.

The idea of time synchronized  faults somewhere off down branches is a red herring, even if you could lay the two cables one over the other and fire a nail through both together, the breaker will already have begun to unlatch due to the first fault, before the nail shorts the second and fault current comes on .  And if they are not co-located how exactly are the two faults to be synchronized to well within the breaking time of the breaker - i.e. within perhaps a fraction of millisecond of each other....

15 B does not imply that at all. It is just an advisory drawing of one sort of circuit, Does this person also have an issue with ring spurs direct from the CU as well because they are not shown ?

It's the same.

EDIT

Rather like the cable installation methods in the annex, these are not the only wiring methods you can use of course - just the only ones that have been measured or calculated for which ratings are published. For other situations/ layouts one just has to interpret which standard case it is nearest to or do the design calcs yourself.

Mike

PS that nail gun thing...

*Assume the nail gun nail does as good as the best recorded here and shoots at 134 feet/second  that's 40 proper speed units (in technical works speed should be in m/s for any Americans reading)  this is  40mm per millisecond so perhaps it takes 1/4 millisecond to pierce the first T and E, - assuming no loss of speed in doing so. In reality it will be slower.

Compare this to the effective pseudo-speed of a 32A B type breaker  at 6kA fault ( assume an I2t of 29000 A2seconds from this Hagar datasheet)

time is 29000 / (6000.6000) = 0.008 seconds to break . By the time the nail hits the second cable the breaker has already started opening and is about 1/4 of the way through and once contacts begin to open the current is limited by the arc impedance.

This is not a real time, as really breaking completely takes a lot longer, but for most of the time the fault is not the current limiting factor.