Not quite. First of all, we are talking about the wire impedance, not just resistance.
And a low PF makes the voltage drop a little less for small wire where the wires resistance is fairly large compared to the wires reactance.
But for larger wires, where reactance is larger, the voltage drop is less for higher power factors, with 2/0 wire actually having about the same drop for 95% PF and 80% PF.
Mike uses the formula 2xKxIxD/Vd to find the circular mils of the wire.
nec recommends no more the 3% so..
vd=6.9 voltage drop (3%of 230)
D=100' distance
I=30a current
K=12.9 for copper wire.
cmils= 2x12.9x30x100/6.9 = 11,217cmils
Chapter 9 table 8 back of book
10awg = 10380 =too small to limit vd to 3%
8awg = 16510= correct
VD= 2KIL/CM
(2 X 12.9 X 30 X 100) / 16510 = 4.68 Volts
230 X .03 = 6.9 volts allowed
B is correct
#10 would allow 7.45 volts
For balanced loads another way to look at is is Vd = IR in a single conductor one way regardless of service type (single or three phase) and then multiply by the line to phase voltage ratio (2 or sqrt3).
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