The current flowing in to a balanced delta connected load through line 'a' is 10 A when the conductor of line ’b’ is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of ’abc’.

• A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A.
• B. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A.
• C. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A.
• D. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A.

Answer: (B)

For the given question,
Ib = 0
Ia = -Ic
Calculation:

Ia1 = 1/3*(Ia + a*Ib + a²*Ic)
= 1/3*(10 + 10∟-120°)
= 5 + j*2.88 A
Ia2 = 1/3*(Ia + a²*Ib + a*Ic)
= (1/3)*(10 + 10∟120°)
= 5 - j*2.88 A
Iao = (1/3)*(Ia + Ib + Ic)
= (Ia - Ia)
= 0