.Basic Electrical Installation Work
248
TN-S (cable sheath earth) supplies. 12.5mm2
PVC cable incorporating 1.51 103
Ω/m. take account this, must multiply the value cable resistance the factor
Supply
company’s
installation
Consumer’s
installation
Line
conductor
R1
R2
Load
Consumer’s
isolation and
protection
equipment
Supply company’s
metering and
protection equipment
on consumer’s premises
Protective
conductor
Neutral
conductor
Fault
Secondary
winding
of supply
transformer
ZE
FIGURE 12. Also the above formula, the
resistance the line conductor and the resistance the earth
conductor.
We have:
Z R
Z
S E
E
( )
(value given the question)
1 2
0 5
Ω
. The complete earth fault loop path shown Fig.
Values have been calculated for copper and aluminium conduc-
tors and are given Table the Site Guide shown Table 12.
Example
A 20A radial socket outlet circuit wired 2.5Ω.5mm2
CPC.1 a
2. The earth fault loop impedance the
supply 0.
For cable m
R R
1 2
3
19 585
.5mm phase conductor with 1. Ω
From the value given Table the Site Guide and reproduced Table 12.7
Earth fault loop path for TN-S system. Calculate the total earth fault loop impedance ZS, and establish that the value
is less than the maximum value permissible for this type circuit.5mm protective conductor has (R1 R2) value of
19.
Ω Ω
However, under fault conditions, the temperature and therefore the cable resistance will
increase.1
of this book.7.
.
The cable length 30m installed ambient temperature 20°C and the consumer’s
protection 20A MCB Type 60898
