Unit Transformers

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1 Unit Transformers Prepared in Dec 1998 Second editing in march 2000

2 Learning objectives At the end of this unit you should be able to : 1. describe the structure and principle of operation of a basic iron-cored transformer as used for voltage transformations. 2. use the equations (V p / V s ) = (N p / N s ) V p I p = V s I s (for 100% efficiency)

3 Transformers Transformer is a device that changes a high alternating voltage at low current to a low alternating voltage at high current or vice-versa.

4 Principle of a Transformer Transfer electrical energy supplied from the primary coil to the secondary coil by electromagnetic induction between the two coils.

5 Structure of a Transformer

6 Structure of a Transformer Primary coil. Secondary coil. Laminated soft iron core. The primary and secondary coils are wound round the laminated soft iron core.

7 Function of Soft Iron Core is to link the magnetic fields produced by the primary coil to the secondary coil.

8 Needs of Alternating Current As an alternative continually opening and closing the circuit, we can use a.c. supply to the primary coil. The a.c. will create a continually changing magnetic field which will induce a current in the secondary coil.

9 Function of Laminated Iron Core is to reduce the eddy current induced in the iron core.

10 Operation of a Transformer

11 Operation of a Transformer At the primary coil applied alternating voltage sets up a changing magnetic field. The changing magnetic field passes through the soft core to the secondary coil. At the secondary coil an induced e.m.f. is produced.

12 Operation of a Transformer

13 Transformer Equation V V s = p N N s p V s is secondary output voltage (induced e.m.f.) V p is primary input voltage N s is the number of turns in the secondary coil N p is the number of turns in the primary coil N N s p is known as the turns ratio

14 Type of Transforms

15 Step-up Transformer secondary voltage V s is bigger than the primary voltage V p. V s > V p Number of turns in the secondary coil N s is bigger than the number of turns in the primary coil i.e N s > N p

16 Step-down Transformer secondary voltage V s is smaller than the primary voltage V p. V s < V p Number of turns in the secondary coil N s is smaller than the number of turns in the primary coil N s < N p

17 Efficiency

18 Ideal Transformer Efficiency = 100 %

19 Ideal Transformer For ideal transformer( efficiency = 100%) power in the primary coil = power in the secondary coil V p I p = V s I s where I p is the current in the primary coil I s is the current in the secondary coil V s is secondary output voltage (induced e.m.f.) V p is primary input voltage

20 Energy Loss in a Transformer Coils of wire have resistance energy is lost in the form of heat produced by current flowing in the coil Eddy currents are induced in the iron core because of the iron core is the region of changing magnetic field reduced by using a laminated iron core. Not all the magnetic field lines may be cut by the secondary coil

21 GCE O-LevelO Past Examination Paper Science (Physics) All rights go to University of Cambridge Examinations Syndicate and other sources

22 GCE O Nov What is the main function of a basic iron-cored transformer? A to change a.c. to d.c. B to change to a higher or lower a.c. voltage C to provide a constant voltage source D to store electrical energy B

23 GCE O Nov Which statement is true for an electrical transformer? A It will only work if a.c. is used. B It will work with either a.c. and d.c. C It is always used to increase voltage. D It should give out more energy than it receives. A

24 Nov A battery charger which works from the 240V main supply contains a transformer which provides an output of 15V. (a) There are 6400 turns on the primary coil of the transformer. Calculate the number of turns on the secondary coil. [2] Since N s / N p = V s / V p therefore N s = (15 x 6400) / 240 = 400 turns (continue on next slide)

25 Nov A battery charger which works from the 240V main supply contains a transformer which provides an output of 15V. (a) There are 6400 turns on the primary coil of the transformer. Calculate the number of turns on the secondary coil. [2] Since N s / N p = V s / V p therefore N s = (15 x 6400) / 240 = 400 turns (continue on next slide)

26 (Cont. ) Q 8 (b) Assuming that the transformer is 100% efficient, calculate the current flowing in the primary if the output current of the transformer is 2.0A. [2] Nov 1998 As efficiency = 100 % therefore Power input = Power output V p I p = V s I s Hint: P=VI 240 x I p = 15 x 2.0 I p = (15 x 2.0) / 240 = A

27 GCE O Nov (a) Describe the structure and principle of operation of a transformer which reduces voltage ( a step-down transformer ). Include a labelled diagram in your answer. [6] when a.c. flows through the primary coil, a magnetic field is produced in primary coil. The magnetic field is lined to secondary coil through the laminated soft iron core. (continue on next slide)

28 (Cont. ) Q. 13 (a) as the magnetic field cuts the secondary coil, hence current induced in the secondary coil. since a.c. is used, the current in the primary coil change continuously which produces a changing magnetic field and induced continuous current in secondary coil. as number of coils in primary is more than number of coils in secondary, therefore this is an step-up transformer. (continue on next slide)

29 (Cont. ) Q. 13 Nov (b) A certain transformer has 4000 turns on the primary and 200 turns on the secondary. The transformer was used to operate a device which drew a current of 0.5A. If the supply voltage to the transformer was 240V, calculate (i) the output voltage, N p / N s = V p / V s V s = (240 x 200) / 4000 = 12 V (ii) the current flowing in the primary. State the assumptions made in your calculation. Assume efficiency = 100 % V p I p = V s I s I p = (12 x 0.5) / 240 = A

30 GCE O Nov (a) With the aid of a labelled diagram, describe a step down transformer and explain how it works. [8] A step down transformer consists primary coil, secondary coil and laminated soft iron core. The primary coil has more number of turns than the secondary coil. The a.c. is supplied from the mains to the primary coil to induce continuous magnetic field in soft iron core. (continue in next slide)

31 (Cont. ) Q. 4 Nov (a) (Cont. ) The magnetic field will then induce current (or e.m.f.) as output of the secondary coil. The soft iron core are laminated with chemical on both sides of of iron plates for reducing the effect of Eddy s current. The following relation can be used as simple calculation of transformer. N p / V p = N s / V s (continue in next slide)

32 (Cont. ) Q. 4 Nov (b) Explain why the alternating voltage generated at a power station is transformed to a very high voltage before connection to the overhead transmission lines. [4] Usually power station transformed electricity at a very high voltage and very low current. The very low current supply through cables will then minimize the lost of energy during transmission.

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