3.4 analysing transformer

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3.4 Analysing Transformers Transformer:- Equipment used to raise or lower the potential difference of an alternating current supply Electrical appliances need different voltages (TV tubes -15 KV, radio – 12 V) Important in the transmission and distribution of electrical energy

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Principle, Function of Transformer

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Page 1: 3.4 analysing transformer

3.4 Analysing Transformers

Transformer:- Equipment used to raise or lower the potential difference of an alternating current supplyElectrical appliances need different voltages(TV tubes -15 KV, radio – 12 V)Important in the transmission and distribution of electrical energy

Page 2: 3.4 analysing transformer

Operating Principle of A transformer

It works on the principle of electromagnetic induction

It is made up of two coils wound on a soft iron core

The alternating current produces magnetic field lines in the primary coil and this magnetic field lines will cut the secondary coil.

A changing magnetic flux causes an induced e.m.f. across the secondary coil

Page 3: 3.4 analysing transformer
Page 4: 3.4 analysing transformer

When the current in the primary circuit increases and decreases this will cause a change of magnetic flux across the secondary coil and induces an e.m.f.

This induced e.m.f. also varies in magnitude and direction, as a result it is also a sinusoidal alternating voltage with the same frequency as the input source.

Page 5: 3.4 analysing transformer

Why does the transformer not work with a d.c. power supply?

On what condition current can be induced in the secondary coil using d.c.?

Answer:

A d.c. power supply will cause no changing magnetic flux in the secondary coil, because it gives a constant current in the primary circuit. Therefore electromagnetic induction does not take place

The primary current changes only at the instants when it is switched on and off. A current will be induced in the secondary coil if the switch in the primary coil is switched on and off continually.

Page 6: 3.4 analysing transformer

Types of TransformerStep-up Transformer

Step-down Transformer

Vs Vp

Ns Np

Vs Vp

Ns Np

Page 7: 3.4 analysing transformer

Experiment to understand step-up and step-down transformer

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Type of circuit Primary circuit Secondary circuit

Number of turns 120 400

Brightness of bulb Dim Bright

Voltmeter reading 2.0V 6.5 V

Type of circuit Primary circuit Secondary circuit

Number of turns 400 120

Brightness of bulb Bright Dim

Voltmeter reading 2.0 V 0.6 V

Step-up transformer

Step-down transformer

Page 9: 3.4 analysing transformer

To Investigate the relationship between Vp , Vs , Np , Ns

Np Ns Vp Vs Ns/Np Vs/Vp

300 600 2.0 3.8 2.00 1.90

300 900 2.0 5.9 3.00 2.95

900 600 10.0 6.6 0.67 0.66

900 300 10.0 3.2 0.33 0.32

Page 10: 3.4 analysing transformer

From the result of the experiment it was found that

The ratios Ns/ Np and Vs/ Vp for each pair of primary and secondary coils approximately equal

Taking into account experimental errors and the loss of power in the transformer, it can be inferred that

Ns / Np = Vs / Vp

The relationship between the number of turns in the primary coil, the number of turns in the secondary coil, the primary voltage and the secondary voltage is given by

Vs / Vp = Ns / Np

GET IT RIGHT: The step-down transformer steps down the input voltage only. The output current in the secondary circuit is greater than the primary current

Page 11: 3.4 analysing transformer

Example:

The figure below shows an ideal transformer

What is the voltage across the bulb?

Answer:Np =120 Ns =30 Vp =240 VNs / Np = Vs / Vp120/30/240/Vs Vs = 60 V

Page 12: 3.4 analysing transformer

Efficiency of a Transformer

The efficiency is defined as

E= [Output power/Input power ] x 100%

E= [ VsIs / Vp Ip ] x 100%.

In the process of transfer electrical energy from one circuit to another a faction of it is lost as heat energy

Efficiency of transformer is normally less than 100%

Ideal Transformer

No energy loss, all the energy supplied to the primary coil will be transferred to the secondary coil

Has efficiency of 100%

Output power = Input power

VsIs = Vp Ip or Vs / Vp = Ip / Is

Page 13: 3.4 analysing transformer

Example:

Figure below shows a transformer with a bulb at its output terminals. The bulb lights up with normal brightness.

(a) Determine the value of Np (b) Calculate the efficiency of the transformer.Answer:(a) Np/ 150 = 240/12 Np =3000(b) Efficiency = Output power/Input power x 100% = 24/0.15 x 240 x 100% = 66.7 %

Page 14: 3.4 analysing transformer

Example:

Figure above shows 2 bulbs of the same rating connected to the output terminals of a transformer which has an efficiency of 80%. When an a.c. power supply of 240V is connected to its input terminals, both bulbs light up with normal brightness. Determine (a) The input power of the transformer(b) The turns ratio of the transformer. (Ns /Np )(c) The current in the primary coil

Page 15: 3.4 analysing transformer

Answer:(a) Efficiency= [Output power / Input power ] x 100% 80% = [24/ Input power ]x 100% Input power = 100/80 x 24= 30 W

( Total output power = V I =24 (I1+I2) = 24(12/24+12/24) =24x1=24WOR total output power is= 12W+12W=24 W

(b) Vs /Vp =Ns /Np Ns/Np =24/240 =1/10

(c)P = VI I = P/V =30/240 = 1/8 A

Page 16: 3.4 analysing transformer

Factors that affect the efficiency of a transformer and ways to improve the efficiency of a transformer

1.Resistance of the coil

A very long wire is required to make coils, hence it has electrical resistance. Heat energy is lost in the coil because of the heating effect of current flow in a conductor. The energy loss is equal to I2 R.

Use thicker wire to reduce resistance of the coil

2. Magnetisation and demagnetisation of the core (Hysteresis loss)

The core is continually magnetised and demagnetised by the changing magnetic field. The energy required for this process is converted into heat energy in the core.

Using soft iron to reduce the energy loss. (It loses almost all its magnetism when it is demagnetised and it requires little energy to be magnetised)

Page 17: 3.4 analysing transformer

3. Eddy currents in the iron core

The core is itself a conductor, so the changing magnetic flux in the iron core induces current in it. These circulating eddy currents generate heat in the core and cause further loss of power.

Using laminated core. It is made from thin, insulated sheet of iron (insulated with enamel paint) to increase the resistance of flow of eddy currents. With less eddy currents, energy loss is reduced.

Page 18: 3.4 analysing transformer

4. Leakage of magnetic flux

The magnetic flux produced by the primary current is not completely linked to the secondary coil. The leakage means that some of the input energy is not transferred to the output of the transformer.

The loss can be reduced by winding the secondary coil on top of the primary coil

Page 19: 3.4 analysing transformer

Example:Refer to the diagrams belowa). Name the type of transformer.b). What is the reading on the voltmeter when the primary coil

terminals are connected to of 120 V a .c power supply?c). Two identical bulbs with the same rating of 12V, 6W are connected

to the output terminals as shown in the right diagram.(i) Will each bulb light up with normal brightness? Explain.(ii) Find the current flowing through each bulb. (ii) How much is the current flowing through the primary coil, if the

efficiency of the transformer is 90%?

Page 20: 3.4 analysing transformer

Answer:a). Step-down transformer, because Ns Np

b). Vs /Vp =Ns /Np Vs =Ns/Np x Vp =120/600 x120 =24 V

c). i) Yes, each bulb will light up with normal brightness, because the potential difference across each bulb is 12 V.

ii). From P = VI , hence 6 = 12 x Is Is = 0.5Aiii). Efficiency= [ VsIs / Vp Ip ] x 100% 90% = Vs Is/VpIp x 100% Ip = [24 x0.5/120 ]x 100/90 =0.111 A