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GATE 2025 EE – Question 43

Electrical Machines · Three-phase transformers: connections, vector groups, parallel operation · 2 marks · Multiple choice

The transformer connection given in the figure is part of a balanced 3-phase circuit where the phase sequence is “$abc$”. The primary to secondary turns ratio is 2:1. If ($I_a+I_b+I_c=0$), then the relationship between $I_A$ and $I_{ad}$ will be

Diagram for GATE 2025 EE question 43
  1. $\dfrac{|I_A|}{|I_{ad}|}=\dfrac1{2\sqrt3}$ and $I_{ad}$ lags $I_A$ by 30°.
  2. $\dfrac{|I_A|}{|I_{ad}|}=\dfrac1{2\sqrt3}$ and $I_{ad}$ leads $I_A$ by 30°.
  3. $\dfrac{|I_A|}{|I_{ad}|}=2\sqrt3$ and $I_{ad}$ lags $I_A$ by 30°.
  4. $\dfrac{|I_A|}{|I_{ad}|}=2\sqrt3$ and $I_{ad}$ leads $I_A$ by 30°.

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Correct answer: (A) $\dfrac{|I_A|}{|I_{ad}|}=\dfrac1{2\sqrt3}$ and $I_{ad}$ lags $I_A$ by 30°.

Explanation

The primary is star-connected and the secondary delta-connected. With a 2:1 turns ratio the secondary winding current is $|I_a|=2|I_A|$ (ampere-turns balance). In the delta, the line current is $\sqrt3$ times the winding current, so $|I_{ad}|=\sqrt3\cdot2|I_A|=2\sqrt3|I_A|$, giving $|I_A|/|I_{ad}|=\dfrac1{2\sqrt3}$. At terminal $a$ the line current is the difference of two winding currents, $I_{ad}=I_a-I_c$, and for the $abc$ sequence $I_a-I_c=\sqrt3\,I_a\angle-30^\circ$: the line current lags by 30°.