Hi,
It is the "Z" value you need to use. "R" is the resistive as you say, "X" is the reactive, and "Z" is the combined result, the impedance of the cable.
So, from the table it is 0.81
Therefore it would be current [100] x run length [25M] x .81 = 2025
Divide this by 1000 to get it back to volts from millivolts = 2.025V
That is the "between the lines" volt drop mind you, so this comes off the 400V you started with to give 397.97V
This is a drop of 0.5%, so, as you are allowed 5% for power or 3% for lighting you are very well within limits.
As i say, this is the "three phase" volt drop if you like. If you are going to split it into three separate phases at the end of your cable, you must divide the volt drop above by 1.732 to get the "single phase" volt drop.
This gives 1.169V as a "single phase" drop if you like.
If you divide this drop [1.169] by 230V [the "single phase" voltage] and multiply it by 100 you will come up with the drop in % again, and of course we are back to 0.5%
You could use smaller cable than 50mm with 100A, why you using such a big one?
john.
---------- Post Auto-Merged at 16:51 ---------- Previous post was made at 16:41 ----------
Hi again, should have said, the volt drop in the tables is based on the cable being fully loaded and at its max working temperature [70c or 90c depending on what cable you use]
As your cable will not be at "full capacity" and be somewhat cooler than this, you will have even less volt drop, as the resistance increases with temperature, so you will be even better off again!!
Make sure the chart on page 281 you have been looking at is the right one for your cable.
Ah, yes, i see why you have chosen this size cable and not the next size down, it is in a duct!
Teach me to read things properly the first time!!
john..