Answer / habeebm

Yes we have formula.....

Power P =400KW, Length L =50 Mtr, Voltage V =415V

ANSWER:-------

First find current,

Current I= 695.65A

Apply the factors in the below mentioned formula:-

R = pL/a

Where R=Resistance, p=Resistivity, L=Length, a=Area

Since p=12 for Aluminiun and 14 for Copper.

Then, R= pL/a

We know that V=IR

Apply "R" in Ohms Law

V = IR

V = I pL/a

Therefore ,

Area "a"= (IpL/V) / 1.732 (since 3 phase)

a = (695.65 x 12 x 50 / 415)/1.732

a = 580.69 ( Approximately 600 Sq.mm )

a = 600 ( 2 x 300 Sq.mm)

Therefore 2R, 3.5C X 300 Sq.mm cable is required.

Answer / noob

wat does 2R mean?
how did u get V = 415v?
50 meters is the length of wat?
the wire? the load?

we all know that in capacitor there will be no ay physical connection between the two electrodes of the capacitor ,and the charge is stored between these electrodes during charging , so with out any physical connection how these charge is transferred from one plate to the another plate during the discharge?????

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