GATE 2015 EE (EE2) – Question 53

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

A positive-sequence balanced source feeds a star load through the shown star–delta transformer. Line ratings are 230 V on the star side and 115 V on the delta side. Neglect magnetizing current. If $\bar I_s=100\angle0^\circ$ A, primary current Ip is

Diagram for GATE 2015 EE (EE2) question 53

$50\angle30^\circ$ A
$50\angle-30^\circ$ A
$50\sqrt3\angle30^\circ$ A
$200\angle30^\circ$ A

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Show answer and explanation

Correct answer: (A) $50\angle30^\circ$ A

Explanation

The line-voltage ratio is 2:1, so conservation of three-phase apparent power gives primary line-current magnitude half the secondary: 50 A.

For the shown positive-sequence delta connection and dot polarities, secondary line current is the difference of two winding currents, giving the standard 30° shift. Tracing the dotted winding currents in the indicated directions makes primary current lead Is by 30°.

Therefore **Ip=50∠30° A, A**. The sign depends on the actual delta interconnection, which is why the source diagram is retained.

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