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

A potential transformer is a device that transfers electrical power through inductive coupling. This means it uses electromagnetic fields to transfer energy from one circuit to another without direct electrical connection.

In contrast, an auto transformer can transfer power both conductively and inductively, as it has common windings between the input and output.

 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.

Option A is correct.

Assume k = V2/V1 where V1 is the higher terminal voltage and V2 the lower (tap) terminal voltage. Let the load apparent power be S = V2·I2.

The magnetically transformed (electromagnetic) VA in the autotransformer (the portion actually linked by flux) is S_m = (V1 - V2)·I2 = V1(1 - k)·I2.

Eliminating I2 between S and S_m gives S = (k/(1 - k))·S_m. This shows that for the same magnetically transformed VA rating (S_m) the autotransformer can deliver a larger terminal VA. Rearranging, the delivered kVA of the autotransformer is larger than the two-winding value by the factor 1/(1 - k); hence statement I is correct.

Copper loss in the portion that carries the transformed flux is proportional to the square of current in that portion. Because the autotransformer reduces the magnetically transformed VA (so that for the same delivered S the transformed portion carries smaller share of the VA), the copper loss of the autotransformer (for the same delivered output) is reduced roughly in proportion to the conducting/transforming portion. The standard comparison gives copper-loss (and active portion of winding loss that is avoided) reduced by the factor (1 - k) compared with the equivalent two-winding transformer; therefore statement II is correct.

The equivalent series impedance referred to the same terminals is reduced (not increased) for an autotransformer compared with the two-winding transformer because a portion of the voltage is supplied conductively. Thus the impedance drop seen at the terminals is smaller by approximately the factor (1 - k), so statement III, which gives 1/(1 - k), is incorrect.

Therefore statements I and II are correct and statement III is incorrect. Final answer: I and II (Option A).

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

Voltage regulation is independent of the size of the transformer.

Hysteresis losses occur maximum in the ferromagnetic material.

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

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

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

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

Voltage Regulation of Transformer at Lagging Power Factor

The phasor diagram for this case is 

Voltage Regulation of Transformer at Leading Power Factor

The phasor diagram for this case is