GATE EE 2026 Set 1 — Question 39

The main CTs have a turns ratio of $300:5$, i.e. a current ratio of $60$, so each CT secondary carries its line current divided by 60.Step 1 - zero-sequence component of the line currents$$3I_0 = I_A + I_B + I_C = (572.812 - 254.525 - 207.083) + j(50.115 - 459.175 + 444.091)$$$$3I_0 = 111.204 + j35.031\ \text{A} \implies I_0 = 37.068 + j11.677\ \text{A}$$Step 2 - what the auxiliary transformer doesThe three relay burdens $R$ are star-connected with an isolated star point, so the three relay currents must sum to zero: no zero-sequence current can flow through the relays. The auxiliary transformer is connected from the three CT secondary leads to earth, its primary star point being solidly earthed and its secondary forming a closed delta. For zero-sequence currents this is a complete path (equal currents in the three star windings return through the earthed neutral and are balanced by a current circulating in the closed delta), so the zero-sequence part of the CT secondary currents is diverted into the auxiliary transformer. For positive- and negative-sequence currents the closed delta offers no balancing path, so (with an ideal, unmagnetised core) the auxiliary transformer draws no such current and they pass on unchanged to the relays. The relay in phase A therefore carries the phase-A CT secondary current with its zero-sequence part removed:$$I_{AR} = \frac{I_A - I_0}{60}$$(The $2:1$ ratio only sets how large the circulating delta current is; it does not change $I_{AR}$.)Step 3 - evaluate$$I_A - I_0 = (572.812 - 37.068) + j(50.115 - 11.677) = 535.744 + j38.438\ \text{A}$$$$|I_A - I_0| = \sqrt{535.744^2 + 38.438^2} \approx 537.12\ \text{A}, \qquad \angle (I_A - I_0) = \tan^{-1}\frac{38.438}{535.744} \approx 4.10^\circ$$$$I_{AR} = \frac{537.12 \angle 4.10^\circ}{60} \approx 8.95 \angle 4.10^\circ\ \text{A}$$Carried to three decimals, the given data produce $8.952\angle 4.104^\circ$ A, which differs from option (D), $8.954 \angle 4.105^\circ$ A, only in the last digit, owing to rounding in the listed currents. Option (D) is the only choice with this magnitude and angle.The other options: (C), $537.240\angle 4.105^\circ$ A, is (approximately) this current before the CT ratio is applied; (B), $0.653\angle 17.556^\circ$ A, is close to the zero-sequence part $I_0/60 \approx 0.648\angle 17.49^\circ$ A, which goes into the auxiliary transformer and not the relay; and (A), $0$ A, would require the relay current to vanish, which happens only if $I_A$ were purely zero-sequence.Answer: (D) $8.954 \angle 4.105^\circ$ A