2 the IEE Regulations gives the value the
prospective fault current required operate the device within the various disconnection
times given.3).3 when the
protective device 30A semi-enclosed fuse. From Fig. The
small insert Table the top right Table 3.4s
from Regulation 411.
S
I t
k
S
2
2
2
2
200 4
115
1 10
(mm )
( s
mm
.
a
ation 543.3)
( s
mm
S
90 4
115
0 49
2
2
.
t Maximum operating time the protective device for circuit not exceeding 32A 0.
A 1.3.5mm2
PVC insulated copper cable
and incorporates 1.
So, this case the table states that 90A will trip 15A semi-enclosed fuse 0. Establish calculation that the CPC adequate size meet the
requirements Regulation 543.
.1.5 mm2
CPC acceptable since this the nearest standard-size conductor above the
minimum cross-sectional area 1.
. Verify that the 1.
.0 mm2
where mechanical protection not provided order comply with Regulation
544.
For final circuits less than 32A the maximum operating time the protective device 0.2.4s.3.
I
V
Z
Maximum fault current (A)
A
s
∴
230
1 15
200
.
.3. The characteristics the protective device are given IEE
Regulation Table 3.Overcurrent, short circuit and earth fault protection
251
as 1.4s.5mm2
where mechanical protection provided
or 4.1. The circuit correctly protected with 15A semi-enclosed
fuse 3036.5(b) you can see that the time taken clear fault of
200A about 0.5 mm2
CPC.
The CPC the cable greater than 0.4s.4s. 12.1.15Ω.2.10 mm2
found calculation.2A.
∴ I
t
k
S
I t
k
90
0 4
115 3
2
2
A
s
(from Table )
(mm (from Regul
.
Example 2
A supply feeds domestic immersion heater wired 2. 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.
k 115 (from Table 54.
From Table 3.49mm2
and therefore suitable.2.5mm CPC meets the requirements Regulation 543.2A can seen that current about 90A will trip the 15A fuse 0
