Chapter A General rules of electrical installation design

G - Sizing and protection of conductors
3 Determination of voltage drop
Example 2 (see Fig. G30)
A 3-phase 4-wire copper line of 70 mm 2 c.s.a. and a length of 50 m passes a current
of 150 A. The line supplies, among other loads, 3 single-phase lighting circuits, each
of 2.5 mm 2 c.s.a. copper 20 m long, and each passing 20 A.
It is assumed that the currents in the 70 mm 2 line are balanced and that the three
lighting circuits are all connected to it at the same point.
What is the voltage drop at the end of the lighting circuits?
Solution:
b Voltage drop in the 4-wire line:
∆
∆U%
100 U
=
Un
Figure G28 shows 0.55 V/A/km
ΔU line = 0.55 x 150 x 0.05 = 4.125 V phase-to-phase
4 . 125
which gives: = 2.38 V phase to neutral.
3
b Voltage drop in any one of the lighting single-phase circuits:
ΔU for a single-phase circuit = 18 x 20 x 0.02 = 7.2 V
The total voltage drop is therefore
7.2 + 2.38 = 9.6 V
9.6 V
x 100 = 4.2%
230 V
This value is satisfactory, being less than the maximum permitted voltage drop of 6%.
Fig. G30 : Example 2
Schneider Electric - Electrical installation guide 2008
50 m / 70 mm2 Cu
IB = 150 A
20 m / 2.5 mm2 Cu
IB = 20 A
G23
© Schneider Electric - all rights reserved