Question 2.13: Neutral-Line Current. A four-wire, wye-connected balanced lo......
Neutral-Line Current. A four-wire, wye-connected balanced load has phase currents described by the following harmonics:
\begin{matrix} \hline \text{Harmonics} & \text{rms Current (A)} \\ \hline 1 & 100 \ \ \\ 3 & 50 \ \\ 5 & 20 \ \\ 7 & 10 \ \\ 9 & 8 \\ 11 & 4 \\ 13 & 2 \\ \hline \end{matrix}Find the rms current flowing in the neutral wire and compare it to the rms phase current.
Only the harmonics divisible by three will contribute to neutral line current, so this means that all we need to consider are the third and ninth harmonics.
Assuming the fundamental is 60 Hz, the harmonics contribute
Third harmonic: 3 × 50 = 150 A at 180 Hz
Ninth harmonic: 3 × 8 = 24 A at 540 Hz
The rms current is the square root of the sum of the squares of the harmonic currents, so the neutral wire will carry
I_{n} \ = \ \sqrt{150^{2} \ + \ 24^{2}} \ = \ 152 \ AThe rms current in each phase will be
I_{n} \ = \ \sqrt{100^{2} \ + \ 50^{2} \ + \ 20^{2} \ + \ 10^{2} \ + \ 8^{2} \ + \ 4^{2} \ + \ 2^{2}} \ = \ 114 \ AThe neutral current, rather than being smaller than the phase currents, is actually 33% higher!