Double- Line -to Ground Fault
The faulted segment for a 2LG fault is shown in Fig. 8.7 where it is assumed that the fault has occurred at node k of the network. In this the phases b and c got shorted through the impedance Zf to the ground. Since the system is unloaded before the occurrence of the fault we have the same condition as (8.8) for the phase-a current. Therefore
 |
(8.17) |
Fig. 8.7 Representation of 2LG fault.
Also voltages of phases b and c are given by
 |
(8.18) |
Therefore
 |
(8.19) |
We thus get the following two equations from (8.19)
 |
(8.20) |
 |
(8.21) |
Substituting (8.18) and (8.20) in (8.21) and rearranging we get
 |
(8.22) |
Also since I fa = 0 we have
 |
(8.23) |
The Thevenin equivalent circuit for 2LG fault is shown in Fig. 8.8. From this figure we get
 |
(8.24) |
The zero and negative sequence currents can be obtained using the current divider principle as
|
(8.25) |
 |
(8.26) |
Fig. 8.8 Thevenin equivalent of a 2LG fault.
Example 8.3 |
