2002
DOI: 10.1109/mper.2002.4312534
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An Analysis of the Five-Wire Distribution System

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“…To exemplify, let us apply the calculation of the Joule losses to Case A. The currents are (16) By denoting with the resistance of each phase conductor, the resistance of the neutral conductor and the ground resistance, the Joule losses in each physical conductor are given by (17) The total Joule losses are 76.8041 kW. For performing loss partitioning, the losses related to the neutral conductor and the ground have to be redistributed among the three phase currents.…”
Section: Relationships Among the Loss Partitioning Results And Thementioning
confidence: 99%
“…To exemplify, let us apply the calculation of the Joule losses to Case A. The currents are (16) By denoting with the resistance of each phase conductor, the resistance of the neutral conductor and the ground resistance, the Joule losses in each physical conductor are given by (17) The total Joule losses are 76.8041 kW. For performing loss partitioning, the losses related to the neutral conductor and the ground have to be redistributed among the three phase currents.…”
Section: Relationships Among the Loss Partitioning Results And Thementioning
confidence: 99%