Can you show us how you get that please?

19 lig point 3amps

6 10 a double skt outlt
16 10a singl skt outlt 15 amps
3 strip heater 1000 watt

1 15amps skt outlet 10 amps
1 20amps skt outlet 15amps
3 exhaust fans 2amps
1 12.4 kw range 27 amps
1 4.6kw instant water heater 6.66amps
4 500watt tennis light 6.50amps

total 85.16 amps

You gave 2 different amounts

old one was mistyped mate.

Ok have been working through the next few shets of MD. Will put up these sheet by sheet and lets see if I'm right or if you have the same answers. Ok.
Q2 Same category.

Determine the maximum demand of a single phase 240V domestic installation.

28 Lighting points 3A+2a = 5A

8m Lighting track 8 x 2A = 16A

13 10A single socket outlets
+
9 10A double socket outlets
+
2 50W exhaust fans
+
1 1200Watts strip heaters = 13+9+2+1=25points so 10A + 5A = 15A

1 14.5KW range 14500/230 = 63 x 05 = 31.5A

1 3.4kW storage H.W.S. 3400/230 = 14.8A

1 3.0kW clothes drier N/A



Total = 82.3A



Right or wrong?

exactly same answer,Thanx

Question 3

Determine the maximum demand of the heaviest loaded phase in a domesticv electrical instalation having the following load-

21 Lighting points 3 + 2 = 5A

28 10A single socket outlets
+
8 10A socket outlets = 10A + 5A = 15A

1 range(2 hotplates 6kW each and 6.8kW oven) 18800/230=81.73x0.5=40.86A

1 air conditioner- (single phase 4.8kW) 4500/230 = 19.56 x 0.75= 14.67A

1 clothes dryer - ( single phase 4.8kW) N/A

1 washing machine - ( single phase self heating - 3.2kW) N/A

1 3phase instantaneous water heater (5.3kW per phase) 5300/230 = 23 x 0.333= 7.67A per phase

1 3 phase motor rated at 7.5A per phase = 7.5A per phase



Total 90.7A


Right or wrong?

For question 2 Max Demand I have got, total 68.3.

28 Lighting points 3A+2a = 5A

8m Lighting track 8 x 2A = 16A

2 50W exhaust fans = 2

ie total points is 28+ 16 + 2 = 46points

3A+2A+2A = 7A
see egin reg C2.3.2.1 load group A(1)

7A+15A+31.52A+14.78A

Total 68.3Amps

Please let me know if this correct.

The 16 for the track is not points but amps.

Please read note d on pg 358 for lighting

Also note in the question the voltage 240 is used.

for lighting track read d pg 358,

2 points per metre of track
ie

8 x 2 = 16 points

final answer [B]66.41 Amps[/B].

please check, I already check with one of my mate same answer as mine.

Thanx for correcting us askar.yea i hve also chkd again u r right.

Determine the maximum demand of a single phase 240V domestic installation.

28 Lighting points
+
8m Lighting track + 2 50W exhaust fans = 28+16+2= 46 points. Which is 3+2+2=7A

13 10A single socket outlets
+
9 10A double socket outlets + 1 1200 watts strip heaters = 13+9+1= 23 points so that would be 10+5=15A

2 50W exhaust fans

1 1200Watts strip heaters

1 14.5KW range 14500/240 = 60.41 x 0.5 = 30.2A

1 3.4kW storage H.W.S. 3400/240 = 14.16A

1 3.0kW clothes drier N/A



Total = 66.36A

This is what i get after revising the above. Just wondering where you get the extra 0.5A from?

Yes your answer is correct.

What about this one?