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| Subject - Nuetral current on Wye system | |
| Ryan_J |
Determine the nuetral load on the following circuits: 1) line 1: 10A, line 2: 10A, line 3: 0A 2) Line 1: 18A, line 2: 43A, line 3: 29A |
| Scott Vickrey | This one has me doubting myself. 1) line 1: 10A, line 2: 10A, line 3: 0A Neutral current = 10A 2) Line 1: 18A, line 2: 43A, line 3: 29A Neutral current = 25A |
| iwire | I will take a shot I have not done this a while. 1) 12.6A 2)13.4A |
| David Hyatt | 1) 10 amps 2) 21.7 amps |
| frenchelectrician | i will try this one i haven't done this one for while.. #1 phase A] 10 amps , phase B] 10 amps, phase C] 0 amps,, i am leaning to 8 amps here . #2 phase A] 18 amps , phase B] 43 amps , phase C] 29 amps i am leaning more to 30 amps on netural line. bear in mind the wye system the line to netural do shift at 120 degrees apart so it will read the amp different than single phase will do. and i am sure this is not involed with 3Rd hamaromic at all. merci, marc |
| David Hyatt | Ryan, whats the answer?
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| Ryan_J | quote: You nailed it!!! The formula is the square root of: (line 1 sqaured+line 2 squared+line 3 sqaured]-[(line 1*line2)+(line 2*line 3)+(line 3*line 1)=nuetral current |
| David Hyatt | Tom Henry's books is the answer. |
| frenchelectrician | quote: ryan do the * mean dived or muliply ?? can you corret me ? i am try to follow the formula you have here but i got it part right but try figure out the * stand for so i don't want to repeat my mistake again. merci , marc |
| Ryan_J | Hi Marc. The * means multiply (times). Merci. |
| frenchelectrician | ryan : many thanks for corrceting me  
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| Scott Vickrey | Good question Ryan. Something else for my note book. I really don't think anyone has ever shown me this before. |
| Ryan_J | Thanks Scott. Glad I could help :) |