# Electrical Machines Questions and Answers – Autotransformers

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This set of Electrical Machines Multiple Choice Questions & Answers (MCQs) focuses on “Autotransformers”.

1. While comparing potential transformer to an auto transformer, a potential transformer transfers power ________
a) conductively
b) inductively
c) both conductively as well as inductively
d) electromagnetic induction

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

2. The statements which support the points that auto transformers are advantageous?

```I. Weight of conductor reduces
II. Ohmic losses reduces
III. Leakage reactance reduces
IV. Lower short-circuit current
```

a) I, II, III
b) II, III, IV
c) I, II, III, IV
d) I, IV

Explanation: Short circuit current of the auto transformer is higher than the corresponding 2-winding transformer.
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3. The statements which support the points that auto transformers are advantageous?

```I. Weight of conductor reduces
II. Direct electrical contacts
III. Leakage reactance reduces
IV. Lower short-circuit current
```

a) I, III
b) II, III
c) I, II, III, IV
d) I, IV

Explanation: 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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4. 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
```

a) I, III
b) II, III
c) II, IV
d) I, II, IV

Explanation: 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.

5. Which of the above are correct for an auto transformer when compared to the identical rating two winding transformer?

```I. KVA rating : 1/(1-k)
II. Losses : (1-k)
III. Impedance drop = 1/(1-k)
```

a) I, II
b) II, III
c) 1, III
d) I, II, III, IV

Explanation: 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).

6. 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 _____________
a) will be positive
b) will be negative
c) may be positive
d) may be negative

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

7. The voltage regulation of a transformer depends on its __________

```(A) Equivalent reactance
(B) Equivalent resistance
(D) Transformer size
```

a) A, B, C, E
b) A, B, C, D, E
c) A, B, D, E
d) A, B, C, D

Explanation: Voltage regulation is independent of the size of the transformer.

8. 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.
a) wood > aluminium > iron
b) aluminium > iron > wood
c) iron > wood > aluminium
d) iron > aluminium > wood

Explanation: Hysteresis losses occur maximum in the ferromagnetic material.

9. 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 eddy current losses.
a) wood > aluminium > iron
b) aluminium > iron > wood
c) iron > wood > aluminium
d) iron > aluminium > wood

Explanation: The eddy current losses are dependent on resistance offered to the currents. Wood is an insulator so it will get heated up most.

10. Maximum efficiency of a transformer for a constant load current, occurs at __________
a) at any p.f
c) zero p.f lagging
d) unity p.f

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

11. A 1-phase tranformer has a leakage impedance of 1+ j4 Ω for primary and 3+ j11 Ω for secondary windings. This transformer has __________
a) H.V primary
b) Medium voltage primary
c) L.V primary
d) L.V secondary

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

12. If a transformer is at no load, then it will act like __________
a) a resistor at p.f = 0
b) an inductive reactor at 0.2 lagging
c) a capacitive reactor at0.2 leading
d) an inductive reactor at 0.8 lagging

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

13. The tappings on the transformers is provided on __________
a) h.v side at one end of the winding
b) l.v. side at one end of the winding
c) h.v side at middle
d) l.v side at middle

Explanation: High voltage side winding has more voltage so the variation of the voltage will be more and better applicable. And it is better to place at the middle because it gives uniform voltage distribution by eliminating axial forces.

14. In a transformer, we place the tapping windings at the middle because __________
a) of radial forces on the windings
b) axial forces on the windings
c) insulation level of the winding
d) to provide a mechanical balance to the windings

Explanation: Due to two different voltages, different currents will flow in the windings and it will cause the axial forces due to currents interaction appear causing damage.

15. When a transformer winding suffers a short circuit, then inter turns of the same winding experience ______ forces.
a) an attractive
b) a repulsive
c) no force
d) may be attractive or repulsive type

Explanation: As the current will flow in one direction, attractive force will appear across the inter turns as per Biot-savart’s law.

16. Two tranformer having same voltage but different KVA are operating in parallel. For a good load sharing __________
a) impedance must be equal
b) pu impedances must be equal
c) pu impedance and X/R ratios must be equal
d) impedance and X/R ratios must be equal

Explanation: Both the conditions are to be met in order to have an efficient operation while in parallel condition.

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