Answer Posted / abokamal

cable design.
first we have to choose CB.
(I)load=P/(V*PF)......... assume pf=0.8
(I)=746/(230*0.8)=4.1 Ampere
(I)cb=(I)load *s.f (FASETY FACTOR)=4.1*1.25=5.1 Ampere
(I)cable=(I)cb/D.F (DERATING FACTOR)=5.1/0.8=6.33 Ampere
from catalouge you can select the cable according to the
current.
you have to see (volt/ampere/meter) to see the voltage drop
limit which according to +-5%

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