In the circuit shown, the phase currents are
\[
I_A=572.812+j50.115\text{ A}
\]
\[
I_B=-254.525-j459.175\text{ A}
\]
\[
I_C=-207.083+j444.091\text{ A}
\]
Given that the CTs are ideal with no saturation, and the turns ratio of the Main CT is \(300:5\) and that of the Auxiliary Transformer (\(Yn\Delta\)) is \(2:1\) on every phase, the value of \(I_{AR}\), rounded off to three decimal places, is:
