Answer / arvind

the conductor cross sectional area will be decrease by n^2
times

Answer / snseng

Depending the Conductor material (copper, AL, Etc..),length
Vdn=0.04 * nVs (V)
VLn= nVs - Vd (V)
In = P / VL (A)=1/n(I)
Rn= Vd / I (Ω)=n(R)
An= 2*φ * L / R (mm²)= 2/n(A)

Answer / xerxes

To supply same power
P=V'I'
I'=I/n
nV/R'=V/nR
n²=R'/R
n²=A/A'
A'=A/n²

Answer / brijesh

Not Important If youi are Increas the current so You have
to Be Increas The area

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