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Test: Auto-Transformer - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: Auto-Transformer

Test: Auto-Transformer for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: Auto-Transformer questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Auto-Transformer MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Auto-Transformer below.
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Test: Auto-Transformer - Question 1

While comparing potential transformer to an auto transformer, a potential transformer transfers power ________

Detailed Solution for Test: Auto-Transformer - Question 1

Potential divider is resistance division and it does not take part in induction processes.

Test: Auto-Transformer - Question 2

The statements which support the points that auto transformers are disadvantageous as compared to 2-winding transformer
I. Weight of conductor reduces
II. Direct electrical contacts
III. Leakage reactance reduces
IV. Lower short-circuit current

Detailed Solution for Test: Auto-Transformer - Question 2

 Direct electrical contacts is a disadvantage to the auto transformer.
Short circuit current of the auto transformer is higher than the corresponding 2-winding transformer.

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Test: Auto-Transformer - Question 3

 I. KVA rating : 1/(1-k)
II. Losses : (1-k)
III. Impedance drop = 1/(1-k)
Which of the above are correct for an auto transformer when compared to the identical rating two winding transformer?

Detailed Solution for Test: Auto-Transformer - Question 3

 KVA(auto)/KVA(2-W) = V2*I2/V2*(I2-I1)
= 1/(1-I1/I2)
= 1/(1-k)
Losses(auto)= (1-k) *Losses(2-W)
Impedance drop(auto)= (1-k)*Impedance drop(2-W).

Test: Auto-Transformer - Question 4

The voltage regulation of a transformer at full-load 0.8 p.f leading is -2%. Its voltage regulation at full load 0.8 p.f lagging

Detailed Solution for Test: Auto-Transformer - Question 4

The leading p.f. has negative v.r. and lagging p.f. has major portion of positive voltage regulation.

Test: Auto-Transformer - Question 5

The voltage regulation of a transformer is not dependent on its

Detailed Solution for Test: Auto-Transformer - Question 5

Voltage regulation is independent of the size of the transformer.

Test: Auto-Transformer - Question 6

Three transformers having identical dimensions but with core of iron, aluminium and wood are wound with same number of turns and have same supply.Then choose the order for hysteresis losses.

Detailed Solution for Test: Auto-Transformer - Question 6

Hysteresis losses occur maximum in the ferromagnetic material.

Test: Auto-Transformer - Question 7

Maximum efficiency of a transformer for a constant load current , occurs at

Detailed Solution for Test: Auto-Transformer - Question 7

Efficiency = KVA*p.f/(KVA*p.f + Losses); So the efficiency is maximum at unity power factor.

Test: Auto-Transformer - Question 8

A 1-phase tranformer has a leakage impedance of 1+ j4 Ω for primary and 3+ j11 Ω for secondary windings. This transformer has

Detailed Solution for Test: Auto-Transformer - Question 8

The side which has lower impedance will have lower number of turns and so the low voltage side.

Test: Auto-Transformer - Question 9

If a transformer is at no load , then it will act like

Detailed Solution for Test: Auto-Transformer - Question 9

Transformer is nothing but the arranged windings which are magnetically coupled. The windings will inductive predominantly with very low resistance.

Test: Auto-Transformer - Question 10

Which of the following power factor gives positive voltage regulation in transformer?

Detailed Solution for Test: Auto-Transformer - Question 10

Concept:

The percentage change in output voltage from no-load to full-load is termed the voltage regulation of the transformer.

Voltage Regulation of Transformer at Unity Power Factor

The phasor diagram of this case is given

  • It shows a phasor diagram for the case of a resistive load (unity power factor) on the transformer 
  • Since the current I2 is in phase with the secondary voltage Vo (FL), the voltage drop across Re2 is also in phase with Vo (FL).
  • The drop across the leakage reactance Xe2 leads the secondary voltage V(FL) by 90o
  • The referred value of primary voltage Vo (NL) is beyond the arc, so it is bigger than the secondary voltage Vo (FL), which means the voltage regulation calculated is positive.

Voltage Regulation of Transformer at Lagging Power Factor

The phasor diagram for this case is 

  • It shows a phasor diagram for the case of an inductive load (lagging power factor) on the transformer (i.e., the load current lags the secondary voltage by 90o). The referred value of primary voltage Vo (NL) is beyond the arc,
  • So it is bigger than the secondary voltage Vo (FL), which means the voltage regulation calculated is positive for an inductive case.
  • For any current that lags the secondary voltage by 0 to 90o, the voltage regulation will be positive.

Voltage Regulation of Transformer at Leading Power Factor

The phasor diagram for this case is 

  • It shows a phasor diagram for the case of a capacitive load (leading power factor) on the transformer i.e., the load current leads the secondary voltage by 90o.
  • As a result, the referred value of primary voltage Vo (NL) is actually smaller than the secondary voltage Vo (FL), which means the voltage regulation calculated is negative for a capacitive case.
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