In the circuit shown, the phase currents are IA = 572.812 + j50.115 A, IB = −254.525−j459.175 A, IC = −207.083+j444.091 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 Auxil iary Transformer (Yn) is 2:1 on every phase, the value of I_AR, rounded off to three decimal places, is:

The correct answer is 0.653 17.556° A.

Analysis of the Circuit Diagram:
Based on the provided circuit diagram, we can identify the following components and connections:
1. The three-phase lines carry primary phase currents $$_A, $$_B, and $$_C flowing downwards.
2. The main current transformers are labeled Main CT 300:5 with their secondary windings star-connected and the neutral point grounded.
3. The Auxiliary Transformer has a turns ratio of 2:1 on every phase. Its primary windings are star-connected (Yn) with the neutral point grounded, and its secondary windings are delta-connected (Δ).
4. The three load resistors are labeled R, and the current flowing through the phase A relay/resistor is designated as I_AR.

Step-by-Step Derivation and Calculation:
First, we write down the primary phase currents in rectangular form:
$I_A=572.812+j50.115\text{A}$
$I_B=-254.525-j459.175\text{A}$
$I_C=-207.083+j444.091\text{A}$

Next, we calculate the secondary currents of the Main CTs ($$_as, $$_bs, $$_cs) by scaling with the Main CT turns ratio of 300:5 (which is 60:1):
$I_{\text{as}}=\frac{I_A}{60}=9.547+j0.835\text{A}$
$I_{\text{bs}}=\frac{I_B}{60}=-4.242-j7.653\text{A}$
$I_{\text{cs}}=\frac{I_C}{60}=-3.451+j7.402\text{A}$

The auxiliary transformer is connected in a wye-delta (Yn-d) configuration. A closed-delta secondary winding offers a path for zero-sequence currents to circulate while acting as an open circuit (high impedance) to positive and negative sequence currents.
Thus, the zero-sequence secondary current of the main CT, $$_as0, flows entirely into the auxiliary transformer primary:
$I_{\text{as0}}=\frac{I_{\text{as}}+I_{\text{bs}}+I_{\text{cs}}}{3}$
Substituting the values:
$I_{\text{as0}}=\frac{(9.547-4.242-3.451)+j(0.835-7.653+7.402)}{3}$
$I_{\text{as0}}=\frac{1.854+j0.584}{3}=0.618+j0.195\text{A}$

Converting this zero-sequence phasor to polar form to find the magnitude and phase angle:
$\left|I_{\text{as0}}\right|=\sqrt{{0.618}^2+{0.195}^2}\approx 0.653\text{A}$
$\theta ={\tan }^{-1}\left(\frac{0.195}{0.618}\right)\approx 17.556°$
Thus, the value of the relay current $$_AR corresponding to the zero-sequence component is:
$I_{\text{AR}}=0.653\angle 17.556°\text{A}$