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When a line to ground fault occurs, the current in faulted phase is 100 A. Zero sequence current in this case will be
Zero
33.3 A
66.6 A
100 A
33.3 A
Given: Fault current in faulted phase (Ia) = 100 A
Fault current in faulted phase (Ia) = 100 A
I0=31Ia
Identify Fault Condition
In a Line-to-Ground (L-G) fault occurring on phase 'a', the fault current Ia is related to its symmetrical components (Ia0, Ia1, and Ia2).
Ia=Ia0+Ia1+Ia2
Apply Symmetrical Component Theory
For a single line-to-ground fault, the positive, negative, and zero sequence currents are equal: Ia0=Ia1=Ia2.
Ia=3Ia0
Calculate Zero Sequence Current
Given the total fault current Ia=100 A, solve for the zero sequence current I0.
I0=3100≈33.33 A
B is correct because in a single line-to-ground fault, the fault current is three times the zero sequence current, resulting in 33.3 A.
Remember that for L-G faults, the sequence networks are connected in series, whereas for L-L faults, they are connected in parallel, and for L-L-G faults, they are in parallel with specific impedance considerations.