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GATE 2026 EE – Question 39

Power Systems · Principles of over-current, differential, directional and distance protection; Circuit breakers · 2 marks · Multiple choice

In the circuit shown, the phase currents are

$I_A=572.812+j50.115$ A
$I_B=-254.525-j459.175$ A
$I_C=-207.083+j444.091$ 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 Auxiliary Transformer (YnΔ) is 2:1 on every phase, the value of $I_{AR}$, rounded off to three decimal places, is:

Diagram for GATE 2026 EE question 39
  1. 0 A
  2. $0.653\angle17.556^\circ$ A
  3. $537.240\angle4.105^\circ$ A
  4. $8.954\angle4.105^\circ$ A

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Correct answer: (D) $8.954\angle4.105^\circ$ A

Explanation

The main CTs (ratio 60:1) give secondary currents $I_A/60,\ I_B/60,\ I_C/60$. The Yn–Δ auxiliary transformer traps the zero-sequence current in its delta, so only the non-zero-sequence part reaches the relay branch: $I_{AR}=\dfrac{I_A-I_0}{60}$ with $I_0=\dfrac{I_A+I_B+I_C}{3}$. $I_A+I_B+I_C=111.204+j35.031$, so $I_0=37.068+j11.677$. $I_A-I_0=535.744+j38.438$ A. Dividing by 60: $8.929+j0.641=8.954\angle4.105^\circ$ A.