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IT Grounding System     161


                                     Equations (9.8) through (9.10) show that in the absence of ground
                                  faults (i.e., R G =∞), there is an identity between the system voltages
                                  at the supply and at the load.
                                     If R G equals zero (i.e., bolted fault to ground), we obtain

                                                       V 1G  = 0                       (9.11)
                                                       V   = V   − V                   (9.12)
                                                         2G    2N   1N
                                                       V   = V   − V                   (9.13)
                                                         3G    3N   1N
                                  In the above case, the voltage between each healthy phase conductor
                                  and the ground is the result of a vector difference, which assumes a
                                  magnitude as large as the line-to-line potential (e.g., 400 V vs. 230 V).
                                  The basic insulation of single-phase loads, eventually present when
                                  the neutral conductor is distributed, may be overstressed and punc-
                                  tured, if loads are not rated to withstand this overvoltage.
                                     With reference to Fig. 9.2, where we neglect the system leak-
                                  age resistance, we can now calculate the representative phasor I G
                                  of the ground current, together with its magnitude |I G |, by using
                                  Eq. (9.8):

                                            V 1G    j3 C 0 V  1N
                                        I =     =                                      (9.14)
                                        G
                                             R G   1 + j3 C 0 R G

                                             V        3 C 0 V

                                               1G            1N
                                        I G    =  =
                                              R G    1 + (3 C 0 R G ) 2

                                                     V                     V
                                                        1N                   1N
                                          =                       =                    (9.15)
                                            (1/3 C 0 ) 1 + (3 C 0 R G ) 2    1/(3 C 0 ) 2    + R G 2
                                  where |V 1G | is the magnitude of the phase-to-ground voltage of the
                                  system as per Eq. (9.8) in correspondence with a generic value of R G .


                             9.3 Resonant Faults in IT Systems
                                  A further technical drawback of IT systems, besides overvoltages, is
                                  the possible occurrence of resonant faults-to-ground. Such faults can
                                  cause very high earth potentials, thereby destroying the insulation of
                                  equipment, as well as originating fires.
                                     Resonant ground faults may be set off by contacts of the line con-
                                  ductor with ground by means of an inductance (e.g., the winding of a
                                  transformer). In this case, an inductive reactance X L =  L, where L is
                                  the self-inductance, is in parallel to the line capacitance of the faulty
                                  phase (Fig. 9.6).
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