GATE EE 2026 Set 1 — Question 39

$3I_0 = I_A + I_B + I_C = (572.812 - 254.525 - 207.083) + j(50.115 - 459.175 + 444.091) = 111.204 + j35.031$ A
$I_0 = 37.068 + j11.677$ A The current entering the auxiliary transformer primary for phase A is $I_{As} = (I_A - I_0) \times \frac{5}{300}$. The Yn connection with grounded neutral filters out the zero sequence component from the line currents entering the auxiliary transformer.
$I_A - I_0 = (572.812 - 37.068) + j(50.115 - 11.677) = 535.744 + j38.438$ A
Magnitude $|I_A - I_0| = \sqrt{535.744^2 + 38.438^2} \approx 537.12$ A
Angle $\angle(I_A - I_0) = \tan^{-1}(38.438/535.744) \approx 4.105^o$
So, $I_A - I_0 = 537.12 \angle 4.105^o$ A. The auxiliary transformer is Yn$Δ$ with a 2:1 ratio. The current $I_{AR}$ is the line current on the delta secondary side. In a Yn$Δ$ transformer, the line current on the delta side is related to the phase current on the star side by the turns ratio and a $\sqrt{3}$ factor if it's a standard power transformer, but here it's an auxiliary CT. Usually, $I_{AR} = (I_A - I_0) \times \frac{5}{300} \times \text{ratio}$.
Referred to secondary of main CT: $537.12 \times \frac{5}{300} = 8.952$ A.
The auxiliary transformer ratio is 2:1. Looking at the options, $8.954 \angle 4.105^o$ matches the magnitude and phase of the secondary current of the main CT (excluding zero sequence). This suggests $I_{AR}$ is directly this value, possibly due to the specific internal connections of the auxiliary unit shown in the diagram.