One can make allowance in the volt drop calculation for load power factor to permit the use of 10mm2 

mv|A|m (max) = 6*10³/47.6*0.85*36* = 4.12

Reference to Table 4D4B column 4 indicates that mv|A|m value of 10mm² is 3.4mv so 10mm² can be used to satisfy the 6v specification.

(Table 4D4B is used rather than 4E2B as the cable is not being run at 90C, however, applying the load power factor to the voltage drop calculation would still permit the use of 10mm2 if the candidate was referring to the latter table.)

Actual voltage drop = 3.4*0.85*47.6*36/1000 = 4.95v.

Actual voltage drop = 4*0.85*47.6*36/1000 = 5.8v using Table 4E2B.

Likewise 10mm2 can be used to satisfy current carrying capacity PROVIDING one selects an appropriate value of Ca (which was the crux of my original question).

Totally agree that there will be ambiguity.

The Chief Examiner constantly alludes to the need for candidates to set out their assumptions "at this level".

Interesting approach.

Why not put the pf in the volt drop calc.

Doesn't the cables being thermosetting or thermoplastic make a difference in there performance?

I don't have a cable calc software package, I wonder what Amtech would come up with?

Interesting approach.

Why not put the pf in the volt drop calc.

Doesn't the cables being thermosetting or thermoplastic make a difference in there performance?

I don't have a cable calc software package, I wonder what Amtech would come up with?