Answer / sree

presuming, the unit of load is in HP(It is out put power)
output in KW= 735.5xHP(A).
Input in KW= A/Effy.(B).
Current= B/((1.732 * V * 0.7) (for 3 Phase star connected
load)(C)
Starting current= (1.5 to 2 times of full load current
(approx.)(D)

Based on cable catalague the size of cable will be chosen
as per above current rating(D).

or

Presuming the load is in KW(It is input power)
current=KW/((1.732 * V * 0.7) (for 3 Phase star connected
load)(A).
Starting current= (1.5 to 2 times of full load current
(approx.)(B)

Based on cable catalague, the size of cable will be chosen
as per above current rating(B).

Answer / vinay

Depending on amps rating.

A=P/V (for 1phase. P=Power in W, V=Voltage)

A=P/(1.732 * V * 0.7) (for 3 Phase)

Answer / jagjeet

Dear Mannu/ Vinay
Efficiency has to be considered for the system also.

what is the exact direction of current? as we know current travels from higher potential to lower potential,so if we take battery +ve to -ve implies current direction is from positive terminal to negative terminal.,whereas electrons flow from negative to positive as they are attracted by positive ions.current will be the plow of charge but not flow of positive ions.please clarify my doubt friends.

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