Basic Electrical Installation Work

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1.1. 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 2.2. Example 2 A supply feeds domestic immersion heater wired 2. From Table 3.4s.5mm2 PVC insulated copper cable and incorporates 1. Establish calculation that the CPC adequate size meet the requirements Regulation 543.3 when the protective device 30A semi-enclosed fuse. I V Z Maximum fault current (A) A s ∴ 230 1 15 200 . t Maximum operating time the protective device for circuit not exceeding 32A 0.2.4s from Regulation 411.5 mm2 CPC acceptable since this the nearest standard-size conductor above the minimum cross-sectional area 1. So, this case the table states that 90A will trip 15A semi-enclosed fuse 0.2A can seen that current about 90A will trip the 15A fuse 0.3). The characteristics the protective device are given IEE Regulation Table 3. A 1.Overcurrent, short circuit and earth fault protection 251 as 1. ∴ I t k S I t k 90 0 4 115 3 2 2 A s (from Table ) (mm (from Regul . k 115 (from Table 54. For final circuits less than 32A the maximum operating time the protective device 0. From Fig. 12. .3.5mm CPC meets the requirements Regulation 543. Verify that the 1.4s.1.5(b) you can see that the time taken clear fault of 200A about 0.3) ( s mm S 90 4 115 0 49 2 2 . a ation 543. .3.4s. The small insert Table the top right Table 3.5mm2 where mechanical protection provided or 4.0 mm2 where mechanical protection not provided order comply with Regulation 544. The CPC the cable greater than 0.3. The circuit correctly protected with 15A semi-enclosed fuse 3036.4s.49mm2 and therefore suitable.15Ω. .5 mm2 CPC.2 the IEE Regulations gives the value the prospective fault current required operate the device within the various disconnection times given.2. . S I t k S 2 2 2 2 200 4 115 1 10 (mm ) ( s mm .2A.10 mm2 found calculation