The current flowing to a balanced delta connected load through line 'a' is 15 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 = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A
• B. Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A
• C. Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A
• D. Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A

Answer: (A)

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

Ia1 = 1/3*(Ia + a*Ib + a²*Ic)
= 1/3*(15 + 15∟-120°)
= 7.5 + j*4.33 A
Ia2 = 1/3*(Ia + a²*Ib + a*Ic)
= (1/3)*(15 + 15∟120°)
= 7.5 - j*4.33 A
Iao = (1/3)*(Ia + Ib + Ic)
= (Ia - Ia)
= 0