I understand the concept of voltage drop but I am worried about the sqrt 3 factor. To get voltage drop you have to know the current. In a 3 phase system that would be I = P/(sqrt(3)*V). In order to determine the voltage drop it's Vdrop = sqrt(3)*I*Z*L.
My question is, when figuring out the current, do you calculate it with the sqrt (3) prior to entering the I value into the Vdrop equation? If that's the case, the voltage drop equation would then become
Vdrop = 3*I*Z*L. The 3 coming from sqrt(3)*sqrt(3).
It depends on if you think of the phase-to-phase, or phase-to-neutral voltage in this calculation. The ratio between the two, in a standard WYE system, is sqrt(3).
When you use the phase-to-neutral voltage only (call it lowercase v), you don't even need to think about the square root of 3. In this example, I = P/(3*v), and Vdropratio = I*R/v, where R is the one-way resistance and v is the phase-to-neutral voltage. The power is carried by three circuits at voltage V, and the current travels out on the line and back on the neutral. Because the currents on the neutral add up to zero, the only current that generates a drop in voltage is the outgoing current on the line.
Note that combining these two equations to eliminate I, we get:
I = P/(3*v)
Vdropratio = I*R/v
Vdropratio = P*R/(3*v)
Now when we translate the formulas for capital V = phase-to-phase voltage instead, we introduce:
V = v*sqrt(3)
And substitute for v:
v = V/sqrt(3)
And then:
I = P/(3*(V/sqrt(3)))
Vdropratio = I*R/(V/sqrt(3))
Simplify:
I = P/(V*sqrt(3))
Vdropratio = I*R*sqrt(3)/V
If you eliminate current from the formula, you get:
Vdropratio = (P/(V*sqrt(3)))*R*sqrt(3)/V
Simplify:
Vdropratio = P*R/V^2
There isn't an "extra square root of 3" factor, because it ends up cancelling out of the equation.