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And now a little puzzle

Discuss And now a little puzzle in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

the pict

the pict

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If any one like me has to do the full 18th course on line version from the NICEIC or if not here is a sample of the Challenges
A load of 750 watt 2 correction factors 1.03 and .65 what is the answer in Amps and how did you arrive at it, I will give my reasons later
Pict
 
750 / 230 = 3.2608695652

1.03 x .65 = 0.6695




edit

Ib = 3.2608695652
In = 6A

6/ 0.6695 = 8.9619118745

Iz = 8.96A
 
Last edited:
the pict said:
A load of 750 watt 2 correction factors 1.03 and .65 what is the answer in Amps and how did you arrive at it, I will give my reasons later
Pict
Click to expand...

This question makes no sense at all, I'm not even sure that there is a question here?
What do you want calculated, what voltage is this at, what do the correction factors relate to?
 
I will put up a screenshot of the actual question but will need to see about copyright, but basically the example equation was presented in an obscure manner as was the wording, whether or not this is meant to be "trick question" or an error I am not sure
 
the pict said:
If any one like me has to do the full 18th course on line version from the NICEIC or if not here is a sample of the Challenges
A load of 750 watt 2 correction factors 1.03 and .65 what is the answer in Amps and how did you arrive at it, I will give my reasons later
Pict
Click to expand...

I would say, what is the apparent design current.
So, 750/230/1.03/0.65 = 4.87A
 
If using a 4A mcb then:


Ib = 3.2608695652
In = 4A

4/ 0.6695 = 5.9746079164


Iz = 5.97A
 

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