a
ation 543.3)
( s
mm
S
90 4
115
0 49
2
2
.2 the IEE Regulations gives the value the
prospective fault current required operate the device within the various disconnection
times given.
From Table 3.3.
∴ I
t
k
S
I t
k
90
0 4
115 3
2
2
A
s
(from Table )
(mm (from Regul
.10 mm2
found calculation.
S
I t
k
S
2
2
2
2
200 4
115
1 10
(mm )
( s
mm
.5(b) you can see that the time taken clear fault of
200A about 0.2A.15Ω.
So, this case the table states that 90A will trip 15A semi-enclosed fuse 0.
The CPC the cable greater than 0.
I
V
Z
Maximum fault current (A)
A
s
∴
230
1 15
200
.
k 115 (from Table 54.4s.
A 1. The characteristics the protective device are given IEE
Regulation Table 3.
Example 2
A supply feeds domestic immersion heater wired 2.49mm2
and therefore suitable.4s.3). Verify that the 1. Establish calculation that the CPC adequate size meet the
requirements Regulation 543. The
small insert Table the top right Table 3.
.
.
.5 mm2
CPC acceptable since this the nearest standard-size conductor above the
minimum cross-sectional area 1.1. 12. From Fig.2.
.0 mm2
where mechanical protection not provided order comply with Regulation
544.3 when the
protective device 30A semi-enclosed fuse.4s
from Regulation 411.1.
t Maximum operating time the protective device for circuit not exceeding 32A 0.4s.3.4s. the protective
conductor separate conductor, that is, does not form part cable this example
and not enclosed wiring system Example the cross-section the protective
conductor must not less than 2.5mm CPC meets the requirements Regulation 543.5mm2
PVC insulated copper cable
and incorporates 1.2.5 mm2
CPC. The circuit correctly protected with 15A semi-enclosed
fuse 3036.1.Overcurrent, short circuit and earth fault protection
251
as 1.3.2A can seen that current about 90A will trip the 15A fuse 0.
For final circuits less than 32A the maximum operating time the protective device 0.2.5mm2
where mechanical protection provided
or 4
