According to IEEE 141/1993 cap. 3.11 Calculation of voltage drops
V = IRcosf + IX sinf where:
V is the voltage drop in circuit, line to neutral
I is the current flowing in conductor
R is the line resistance for one conductor, in ohms
X is the line reactance for one conductor, in ohms
f is the angle whose cosine is the load power factor
cosf is the load power factor, in decimals
sinf is the load reactive factor, in decimals
For three-phase you have to multiply by 1.73
Let's take an example.
From NEC Table 9 500 kcmil copper R=0.027 ohm/kft
dcond=0.789
dins.core=1”
overall dia=2.13”
However, the reactance is lower for three core cable than three single core cables in cradle configuration as per NEC. Then we have to calculate it.
X=2*pi*f*(0.1404*log10(DMD/DMR)/10^3 ohm/kft
For 3 core 3 phase cable the reactance is
X=2*pi*f*(0.1404*log10(2*S/dia)+0.0153)/10^3 =0.0271 ohm/kft
where S=ins.core dia and dia=dcond.
For 3 core cable used as single phase
the distance=overall dia [when touching] but usually the distance is 1 ft=12”.
DMD=(2^(1/3)*distance[geometric mean of distances in flat configuration]=2.68 or 15.11 at 12”.
DMR=(ins.core dia^2*0.7788*cond.dia/2)^(1/3)= (1^2*0.7788*0.789/2)^(1/3)= 0.675
X=0.0317 when touching and X=0.07147 at 12” apart.
If there are three three phase cables parallel then the resistance and reactance will be 1/3
If there are three one phase three core cables the resistance will be 1/3 but the reactance does not.
So eventually in first case R=0.027/3 and X=0.0271/3
and in the second case
R=0.027/3 and X=0.0317 [or 0.07147]
If the total current will be 1000 A cosf=0.85 and sinf=0.5267 then in first case
V=1.73*1000*(0.85*0.009+0.5267*0.00903)=21.46 V
In second case
V=1.73*1000*(0.85*0.009+0.5267*0.0317)=42.12 V[or 78.4 V]