Discuss Help with three phase power in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net
surely for design current
In = P/(400x1.732) = ?A
note: no power factor given
pure inductive load , and what about these super conductors in sern
Multiply 400 by 1.732 and you get 693.
Despite that the answer is right? Somehow?
:bucktooth:
my questioning comes from a lack of understanding on my part not questioning yours but as i was taught
Ib = P/(400x1.732xpf) example 10kW so 10,000/(400x1.732x0.8)= 18.04
selection of protective device
In= 20A
then It = In/rating factors ca cg etc
then on to VD
mVxIxLoR/1000
Oh come on. something that you freely admit you don't understand but you're still prepared to work on it and use that to further your knowledge?i dont understand 3 phase at all.... il have to start off like for like swaps and build on it...
i dont understand 3 phase at all....
Reply to Help with three phase power in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net
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