Need Help, About a CEPE test question.

[Carl_Jimmys]

Hi, there. I'm preparing the electrician red seal exam. I'm stocked with this question which says:" What is the minimum ampacity for a neutral conductor feeding a 700A load with 50A of fluorescent lighting?" And the answers you can choose are:" A.700A B.565A C.550A D.490A" ,the correct answer is B.565A. A reference is directed to Rule 4-018, and what it is as bellow:

4-018 Size of neutral conductor (see Appendix B)
1) The neutral conductor shall have sufficient ampacity to carry the unbalanced load.
2) The maximum unbalanced load shall be the maximum connected load between the neutral and
any one ungrounded conductor as determined by Section 8 but subject to the following:
a) there shall be no reduction in the size of the neutral for that portion of the load that consists
of
i) electric-discharge lighting; or
ii) non-linear loads supplied from a three-phase, 4-wire system; and
b) except as required otherwise by Item a), a demand factor of 70% shall be permitted to be
applied to that portion of the unbalanced load in excess of 200 A.
3) The size of a service neutral shall be not smaller than the size of a neutral selected in accordance
with Subrule 1) and shall
a) be not smaller than No. 10 AWG copper or No. 8 AWG aluminum; and
b) be sized not smaller than a grounded conductor as required by Rule 10-210 b), except in
service entrance cable or where the service conductors are No. 10 AWG copper or No. 8 AWG
aluminum.
4) In determining the ampacity of an uninsulated neutral conductor run in a raceway, it shall be
considered to be insulated with insulation having a temperature rating not higher than that of the
adjacent circuit conductors.

I think it could not make sense. All I can think the right answer may be C.550A. Because if the system is a 3-phase system. So 700A load of 50A lighting that is 14 lamps in that system, so let's assume that U-phase has ONE lamp, V-phase has ONE lamp, and W-phase has the remained 12 lamps. Now the system is at extremely unbalanced situation. At this time the current of lamps of U V W -phase has offset, so the neutral conductor's current is the W-phase remained 11 lamps' current. so 50A * 11 =550A.
I have been think about this question for about several days. So I need your help, thanks very much.

 

[synchro]

The problem is that they worded the question ambiguously so that it can be interpreted in more than one way.

In their question I put the words "a" and "with" in bold font to highlight what they probably intended for the question :
"What is the minimum ampacity for a neutral conductor feeding a 700A load with 50A of fluorescent lighting?"

When they say "a 700A load" that implies that there is one load (and on one phase), but it is a load "with 50A of fluorescent lighting." So that would mean that the 50A is part of the 700A load, with 650A remaining.
Therefore this 50A portion of the 700A load can not be reduced because of 4-018 2) a) i).
But 2) b) allows us to reduce the remaining 650A to:
200A + (650A - 200A) x 0.70 = 515A.
...↑ excluded ...................↑ reduction factor
Then we have to add back the 50A portion that is not subject to a reduction:
512A + 50A = 565A, which is the answer they wanted.

 

[Carl_Jimmys]

Really appreciate pal