In a solid three-phase fault, how do the sequence currents relate to the fault current?

In a solid three-phase fault, how do the sequence currents relate to the fault current?

• A. I1 = I2 = I0 and If = 3*I1.
• B. I1, I2, I0 are all zero.
• C. I1 = I2 but not equal to I0.
• D. I1, I2, I0 are independent.

Answer: (A)

In a solid three-phase fault, the three phases are tied together in a perfectly symmetric short. That symmetry forces the positive-sequence, negative-sequence, and zero-sequence currents to be equal in magnitude and angle at the fault, so I1 = I2 = I0. The fault current is the sum of these sequence currents at the fault point, If = I1 + I2 + I0. Since all three are the same, If = 3*I1. This means the fault current is triple the positive-sequence current. The other possibilities describe scenarios with no fault, or an imbalance, which don’t match a balanced solid three-phase fault.

In a solid three-phase fault, the three phases are tied together in a perfectly symmetric short. That symmetry forces the positive-sequence, negative-sequence, and zero-sequence currents to be equal in magnitude and angle at the fault, so I1 = I2 = I0. The fault current is the sum of these sequence currents at the fault point, If = I1 + I2 + I0. Since all three are the same, If = 3*I1. This means the fault current is triple the positive-sequence current. The other possibilities describe scenarios with no fault, or an imbalance, which don’t match a balanced solid three-phase fault.