 WAZIPOINT Engineering Science & Technology: Electrical Transformer Working Principle

## Thursday, February 2, 2023

### Electrical Transformer Working Principle

According to the rules of Electro Magnetic Induction of Transformers Faraday works on the principle of Mutual Induction. If Rakhi is supplied in either of the two coils nearby, EMF Induce from the magnetic flux to the other coil also. The transformer works on this basis.

## Mutual Induction of a Transformer

The mutual induction between the two coils is called the property of the coil. When two coils are together, the induction process is feasible. If one of the two coils placed together is connected to a voltage, magnetic flux will be generated in the surrounding area.

The magnetic flux that arises is also EMF Induce in the side coils from contact with the coils placed together. Which is called mutual induction.

Go to the main page of the Transformer.

### Basic MCQ Question about Transformer

1. A transformer transforms
(a) frequency
(b) voltage
(c) current
(d) voltage and current.

2. Which of the following is not a basic element of a transformer?
(a) core
(b) primary winding
(c) secondary winding
(d) mutual flux.

3. In an ideal transformer,
(a) windings have no resistance
(b) core has no losses
(c) core has infinite permeability
(d) all of the above.

4. The main purpose of using a core in a transformer is to
(a) decrease iron losses
(b) prevent eddy current loss
(c) eliminate magnetic hysteresis
(d) decrease reluctance of the common magnetic circuit.

5. Transformer cores are laminated in order to
(a) simplify its construction
(b) minimize eddy current loss
(c) reduce cost
(d) reduce hysteresis loss.

6. A transformer having 1000 primary turns is connected to a 250-V a.c. supply. For the secondary voltage of 400 V, the number of secondary turns should be
(a) 1600 (b) 250
(c) 400 (d) 1250

7. The primary and secondary induced e.m.fs. E1
and E2 in a two-winding transformer are always
(a) equal in magnitude
(b) antiphase with each other
(c) in phase with each other
(d) determined by the load on the transformer secondary.

8. A step-up transformer increases
(a) voltage (b) current
(c) power (d) frequency.
9. The primary and secondary windings of an
ordinary 2-winding transformers always have

(a) different number of turns
(b) same size of copper wire
(c) a common magnetic circuit
(d) separate magnetic circuits.

10. In a transformer, the leakage flux of each winding is proportional to the current in that winding because
(a) Ohm’s law applies to magnetic circuits
(b) leakage paths do not saturate
(c) the two windings are electrically isolated
(d) mutual flux is confined to the core.

11. In a two-winding transformer, the e.m.f. per turn in the secondary winding is always.......the induced e.m.f. power turn in the primary.
(a) equal to K times
(b) equal to 1/K times
(c) equal to
(d) greater than.

12. In relation to a transformer, the ratio 20 : 1 indi- cates that
(a) there are 20 turns on primary one turn on secondary
(b) secondary voltage is 1/20th of primary voltage
(c) the primary current is 20 times greater than the secondary current.
(d) for every 20 turns on primary, there is one turn on the secondary.

13. In performing the short circuit test of a transformer
(a) high voltage side is usually short-circuited
(b) low voltage side is usually short-circuited
(c) any side is short-circuited with preference
(d) none of the above.

14. The equivalent resistance of the primary of a transformer having K = 5 and R1 = 0.1 ohm when referred to as secondary becomes.......ohm.
(a) 0.5
(b) 0.02
(c) 0.004
(d) 2.5

15. A transformer has negative voltage regulation when its load power factor is
(a) zero
(b) unity
(d) lagging.

16. The primary reason why the open-circuit test is performed on the low-voltage winding of the transformer is that it
(a) draws sufficiently large on-load current for
(b) requires the least voltage to perform the test
(c) needs a minimum power input
(d) involves less core loss.

17. No-load test on a transformer is carried out to determine
(a) copper loss
(b) magnetizing current
(c) magnetizing current and no-load loss
(d) the efficiency of the transformer.

18. The main purpose of performing open-circuit test on a transformer is to measure its
(a) Cu loss
(b) core loss
(c) total loss
(d) insulation resistance.

19. During short-circuit test, the iron loss of a transformer is negligible because
(a) the entire input is just sufficient to meet Cu losses only
(b) flux produced is a small fraction of the normal flux
(c) the iron core becomes fully saturated
(d) supply frequency is held constant.

20. The iron loss of a transformer at 400 Hz is 10 W. Assuming that eddy current and hysteresis losses vary as the square of flux density, the iron loss of the transformer at rated voltage but at 50 Hz would be....... watt.
(a) 80 (b) 640
(c) 1.25 (d) 100

21. In operating a 400 Hz transformer at 50 Hz
(a) only voltage is reduced in the same proportion as the frequency
(b) only kVA rating is reduced in the same pro- portion as the frequency
(c) both voltage and kVA ratings are reduced in the same proportion as the frequency
(d) none of the above.

22. The voltage applied to the h.v. side of a transformer during short-circuit test is 2% of its rated voltage. The core loss will be.......percent of the rated core loss.
(a)4 (b) 0.4
(c) 0.25 (d) 0.04

23. Transformers are rated in kVA instead of kW because
(a) load power factor is often not known
(b) kVA is fixed whereas kW depends on load p.f.
(c) total transformer loss depends on volt-ampere
(d) it has become customary.

24. When a 400-Hz transformer is operated at 50
Hz its kVA rating is
(b) increased 8 times
(c) unaffected
(d) increased 64 times.

25. At relatively light loads, transformer efficiency is low because
(a) secondary output is low
(b) transformer losses are high
(c) fixed loss is high in proportion to the output
(d) Cu loss is small.

26. A 200 kVA transformer has an iron loss of 1 kW and a full-load Cu loss of 2kW. Its load kVA corresponding to maximum efficiency is ....... kVA.
(a) 100 (b) 141.4
(c) 50 (d) 200

27. If Cu loss of a transformer at 7/8th full load is 4900 W, then its full-load Cu loss would be.......watt.
(a) 5600 (b) 6400
(c) 375 (d) 429

28. The ordinary efficiency of a given transformer is maximum when
(a) it runs at half full-load
(c) its Cu loss equals iron loss

29. The output current corresponding to maximum efficiency for a transformer having core loss of 100 W and equivalent resistance referred to secondary of 0.25 Ω is ....... ampere.
(a) 20
(b) 25
(c) 5
(d) 400

30. The maximum efficiency of a 100-kVA transformer having iron loss of 900 kW and F.L. Cu loss of 1600 W occurs at ....... kVA.
(a) 56.3 (b) 133.3
(c) 75 (d) 177.7

1. d 2. d 3. d 4. d 5. b 6. a 7. c 8. a 9. c 10. b 11. c 12. d 13. b 14. d
15. c 16. a 17. c 18. b 19. b 20. b 21. b 22. d 23. c 24. a 25. c 26. b 27. b 28. c 29. a 30. c 31. d 32. d 33. c 34. a 35. b 36. a 37. d 38. c