E00627IR3IR3 = IH1 + IH2 + IH3 - IH3IR3 = IH1 + IH2 IH3 IH1 + IH2IH1 + IH2 + IH3jXc3Ia3Ib3IH2jXc2IR2Ia2Ib2IH1jXc1IR1Ia1Ib1Figure 48: Current distribution in an insulated system with C phase faultThe figure above shows the relays on the healthy motor feeders see the unbalance in the charging currents fortheir own feeder.The relay on the faulted feeder, however, sees the charging current from the rest of the system (IH1 and IH2 in thiscase), with its own feeders charging current (IH3) becoming cancelled out. This is further shown by the phasordiagrams in the next figure.P24xM Chapter 6 - Current Protection FunctionsP24xM-TM-EN-2.1 121