The line currents in amperes in phases a, b and c respectively are 500 +j150, 100 – j600 and – 300 + j600 referred to the same reference vector with phase sequence of abc. Find the symmetrical component of currents.

• A. Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A
• B. Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A
• C. Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A
• D. Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A

Answer: (A)

Calculation:

Ia1 = 1/3*(Ia + a*Ib + a²*Ic)
= 500 + j150 + (- 0.5 + j*0.866)*(100 – j6000) + (-0.5 - j*0.866)*(– 300 + j600)
= (546.41 + j*156.47) A

Ia2 = 1/3*(Ia + a²*Ib + a*Ic)
= 500 + j150 + (- 0.5 - j*0.866)*(100 – j6000) + (-0.5 + j*0.866)*(– 300 + j600)
= (-146.41 - j*56.47) A

Iao = 1/3*(Ia + Ib + Ic)
= 1/3*(500 + j150 + 100 – j6000 – 300 + j600)
= (100 + j*50) A