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4.1.4 Step 2

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4.1.4 Step 2

  1. 1. Demand Description Current allowance (Amps) Load Group of load per unit Calculation Circuit No (column 1) (given list) (column 2) L1 L2 L3 3 A (1st 20) 1 A (i) 13 Lighting points na 3.0 2 A (next 20) 2 A (i) 14 Lighting points 3 A (1st 20) na 3.0 3 A (i) 15 Lighting points 3 A (1st 20) na 3.0 8 + (4 x 2) = 16 8 single + socket points + 4 B (i) 10 A (1st 20) 10.0 1 (pool filter 4 double outlets pump) = 17 7 single + 7 + (5 x 2) = 5 B (i) 10 A (1st 20) 10.0 17 socket points 5 double outlets One 15 A 10 A B (ii) single phase (See note on group na 10.0 6 outlet in Table C1, B (ii)) 50% 7 C Cook top 4200/240 x 0.5 8.8 (or x 0.5) 50% 8 C Oven rating 5200/240 x 0.5 10.8 (or x 0.5) Pump rating = 4.3 A so socket outlet to be used for pump supply. 9 B (i) Pool filter pump na Socket will be part of circuit L1 so Included with circuit 4 calculations Air conditioner 75% 10 D 23.6 x 0.75 17.7 17.7 17.7 (23.6 A) (or x 0.75) 11a OPHWS F Full load current 15.0 bottom element 3600/240 11b OPHWS Included F Full load current 0.0 top element above Total demand current per phase 40.7 39.5 46.5

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