# Grounded conductor load calculation help



## Dennis Alwon (May 9, 2009)

I don't believe 70% is correct. You need to calculate the unbalanced load on the neutral and it may be as small as the egc if it is a feeder or as small as the gec if it is a service. I don't think your example fits 220.61


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## wildleg (Apr 12, 2009)

continuous load is 125%, so lighting is 1.25*30kw/(208*1.73) = 104amps
the other loads, non continuous are 15kw so thats 42 amps (oops, thats 72 amps) So I come up with 176 amps for the service, which is a 200 amp breaker ? (sorry, corrected my mistake)

neutral load is max unbalanced load (220.61 ?), which is if one leg has 15kw , so that is 72 amps. like Dennis said, no demand factor, but neutral is allowed to be size at 72 amps I think ( been a while)


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## Electro-fireman (Dec 3, 2012)

I failed the exam last year and did a review and all i retained was that question was make sure I add 125% to the neutral


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## Electro-fireman (Dec 3, 2012)

wildleg said:


> continuous load is 125%, so lighting is 1.25*30kw/(208*1.73) = 104amps
> the other loads, non continuous are 15kw so thats 42 amps. So I come up with 146 amps for the service, which is a 150 amp breaker ?


I thought the same as you but according to the state the total connected load is 208


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## Electro-fireman (Dec 3, 2012)

I know the 208 amp is right and that's the total connected load. I don't understand why they ve added 125% to the total calculation and not just the 30kw continuous load. It states the other 3 loads are non continuous. This question has me all confused.


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## wildleg (Apr 12, 2009)

dennis is right on the neutral, it's 220.61 B 2, only over 200a


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## Electro-fireman (Dec 3, 2012)

I know the total connected load is 208. I got that part right. I just don't understand .


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## Fibes (Feb 18, 2010)

Might they be wanting the answer as 

"SQRT (I²A + I²B + I²C) - (IA x IB) - (IB x IC) - (IC x IA) in lieu of a code answer?


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