what is max current for this cable please

[flameport]

Another problem with a 1.5mm cable of that length will be the total impedance, which is likely to be too high for use with a 20A circuit breaker.

If the impedance is too high, the circuit breaker (or fuse etc) would not disconnect quickly enough in the event of a fault. Or in some circumstances, the circuit would never be disconnected, resulting in the cable overheating / fires / people being electrocuted / other bad things.

 

[ColJack]

ColJack's way of looking at it is correct, if starting from scratch, i.e. working out what size cable to use in a given situation, but if you're trying to determine the current carrying capacity of an existing cable then the type of protective device does affect that.

no it doesn't..
for arguments sake lets say a cable is rated at 10A..
then it can safely take 10A before it starts to get too hot and eventually melts

this choice of OCPD has no affect on this at all, it can still take 10A..
the fact that some 10A OCPD's let more than 10A flow for a long duration before they operate makes the cable unsuitable for use with the OCPD or the OCPD unsuitable for use with the cable, but the cable is STILL rated at 10A..

if the cable is existing and you're chosing an OCPD for it then you have to derate the OCPD if needs be to suit the MAX rating of the cable..

 

[ban-all-sheds]

Ib ≤ In ≤ Iz

You start with your It and apply factors for installation method, grouping, ambient temperature etc, all of which chip away at your Iz.

If the OPD is a 3036 you have to multiply It x 0.725 as well.

If, after all of the installation-type factors are applied your cable ends up at 10A then if it's on a 3036 the figure you plug into Ib ≤ In ≤ Iz is 7.25A, not 10A, because In is what it is - the 5A/15A/30A etc ratings of fusewire have not had the 0.725 factor applied to them.

I know what you mean when you say that the cable can carry 10A whatever the OPD type is, but when people ask what the capacity of a cable is, and you start asking how it's installed, is it bunched with others and so on what you do is apply Ca, Ci, Cg etc to It as multipliers, because you want to establish what the Iz is so that you can advise on the maxima for Ib & In. Hence you have to include, if relevant, a 0.725 multiplier to It along with Ca, Ci, Cg etc.

Using the factor the way that you describe is the way you find out what It you need by dividing 0.725, Ca, Ci, Cg etc into In.