1
GATE EE 2014 Set 2
MCQ (Single Correct Answer)
+2
-0.6
For a single phase, two winding transformer, the supply frequency and voltage are both increased by 10%. The percentage changes in the hysteresis loss and eddy current loss, respectively are
A
10 and 21
B
-10 and 21
C
21 and 10
D
-21 and 10
2
GATE EE 2013
MCQ (Single Correct Answer)
+2
-0.6
The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage VWX1 = 100V is applied across WX to get an open circuit voltage VYZ1 across YZ. Next, an ac voltage VYZ2 =100V is applied across YZ to get an open circuit voltage VWX2 across WX. Then, $$\frac{V_{YZ1}}{V_{WX1}}$$, $$\frac{V_{WX2}}{V_{YZ2}}$$ are respectively, GATE EE 2013 Electrical Machines - Transformers Question 58 English
A
125/100 and 80/100
B
100/100 and 80/100
C
100/100 and 100/100
D
80/100 and 80/100
3
GATE EE 2012
MCQ (Single Correct Answer)
+2
-0.6
A single phase 10 kVA, 50 Hz transformer with 1 kV primary winding draws 0.5 A and 55 W, at rated voltage and frequency, on no load. A second transformer has a core with all its linear dimensions $$\sqrt2$$ times the corresponding dimensions of the first transformer. The core material and lamination thickness are the same in both transformers. The primary windings of both the transformers have the same number of turns. If a rated voltage of 2 kV at 50 Hz is applied to the primary of the second transformer, then the no load current and power, respectively, are
A
0.7 A, 77.8 W
B
0.7 A, 155.6 W
C
1 A, 110 W
D
1 A, 220 W
4
GATE EE 2010
MCQ (Single Correct Answer)
+2
-0.6
A balanced star-connected and purely resistive load is connected at the secondary of a star-delta transformer as shown in the figure. The line-to-line voltage rating of the transformer is 110 V/220 V. Neglecting the non-idealities of the transformer, the impedance 'Z' of the equivalent star-connected load, referred to the primary side of the transformer, is GATE EE 2010 Electrical Machines - Transformers Question 56 English
A
(3 + j0) Ω
B
(0.866 − j0.5) Ω
C
(0.866 + j0.5) Ω
D
(1 + j0) Ω
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