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The sequence components of the fault current are as follows: Ipositive = j1.5 pu, I negative = – j0.5 pu, Izero = – j1 pu. The type of fault in the system is
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- LG
- LL
- LLG
- LLLG
Correct Option: C
Given, Ia1 = j 1.5 p.u.
Ia2 = – j 0.5 p.u. and, | ![]() | Ia | ![]() | = | ![]() | 1 | 1 | 1 | ![]() | ![]() | Ia0 | ![]() | ||
Ib | 1 | α² | α | Ia1 | ||||||||||
Ic | 1 | α | α² | Ia2 |
Ia0 = – j 1.0 p.u.
then, Ia = Ia1 + Ia2 + Ia0 = 0.
In case of LLG faults at terminal b, c, the conditions at the faults are characterized by Vb = 0, Vc = 0, Ia = 0, as shown in figure also
