Harmonics in neutral conductor

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akyip

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The neutral conductor will only carry the third-harmonic currents. This is because:

Positive-sequence (and negative-sequence) balanced currents have the same magnitude but are displaced by 120 degrees from each other. Due to the 120-degree displacement, they add up to 0 at the neutral connection point.

1<0 + 1<-120 + 1<+120 = 0

48<0 + 48<-120 + 48<+120 = 0

Zero-sequence currents have both the same magnitude and same phase angles, so they are additive at the neutral connection point. So in this problem:

10 A + 10 A + 10 A = 30 A
 
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