Home › Previous year questions › GATE 2015 EE (EE2) › Question 57GATE 2015 EE (EE2) – Question 57 Electrical Machines · Single phase transformer: equivalent circuit, phasor diagram, open circuit and short circuit tests, regulation and efficiency · 2 marks · Multiple choice
A three-winding transformer has N1=100,N2=N3=50 and negligible magnetizing current. With the shown polarities, $\bar I_2=2\angle30^\circ$ A and $\bar I_3=2\angle150^\circ$ A. Primary I1 is
$1\angle90^\circ$ A $1\angle270^\circ$ A $4\angle90^\circ$ A $4\angle270^\circ$ A Practise this question in The GATE Grind →
Show answer and explanation Correct answer: (A) $1\angle90^\circ$ A
Explanation The primary current enters its dotted terminal, while both load currents leave their dotted terminals. Ampere-turn balance gives N1I1=N2I2+N3I3.
Therefore I1=(I2+I3)/2. The two load phasors add to 2j A because their real components cancel and their imaginary components each equal 1 A.
Thus I1=j A=**1∠90° A, A**.
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