- HubPages
*»* - Technology
*»* - Computers & Software

# Introduction to transformers

## Introduction to transformers

Session 9

Contents

9.1 Principle operation of the transformer:

Aim:

To introduce the transformer and explain the working principles so that you will be able to understand the applications of it in general practice.

Specific Objectives:

At the end of the session you should be able-

- To identify the primary and secondary sides of the transformer
- To obtain the relationship between the induced e.m.f.s. and the number of turns.

Introduction:

You would have seen a transformer or at least heard the name. Could you say what a transformer is? If you say it is an electrical device, you are correct, but it is more than just that. You will learn more as you go along. How do we get most of the electrical energy in Sri Lanka? You might say that it is by using hydro or thermal power and it is true.

In relation to Sri Lanka, the power system may be denoted by Fig.9.1. The potential energy of the water is first converted into kinetic energy, and then into Electrical Energy through mechanical energy.

I expect you to know that you are getting electrical energy at a potential of about 230 V for your house.

Question:

Is 230 V the only voltage at which all the consumers are getting the supply in Sri Lanka?

Answer:

No. There are some other voltages also. Some industries are getting the supply at 3.3 kV and some at 11 kV.

Different consumers consume electrical energy at different potential levels. How does the Electricity Board give the supply at different potential levels as consumers’ request?

You might say that the Ceylon Electricity Board is generating electrical energy at different potential levels like 230 V, 3.3 kV, 11kV etc, but this is not correct. CEB generates electrical energy at about 11kV in most of the generating stations. You might know that the different potential levels used in Sri Lanka are about 220 V, 400 V, 3.3 kV, 6.6 kV, 11 kV, 33 kV, 132 kV & 220 kV. How do we get all these different potential levels, while we generate power only at about 11 kV? To answer this question one should know what a transformer is, and what its functions are. You have studied in your first lesson about the necessity of having a transmission system to link the generating stations and the load centres. The transmission system transfers the generated power to the load. While it transmits the power, it losses a little of it in the process. If the loss is high the efficiency of transmission is low. Therefore it is necessary to keep the loss at a minimum.

Question:

Can you remember from your schoolwork and say-what will be the loss when d.c. transmission system having a resistance R ohm carries a current of I amperes.

Answer:

From Ohm’s law you can find the voltage drop in the transmission system and it is (I x R)

V = I x R Volts.

The power loss can be obtained by multiplying V and I

Power loss =
I.R.I = I^{2}R Watts.

It is thus seen that the loss in a transmission system is proportional to the square of the current. The power can be related to the volt-ampere product. This product is known as apparent power. In an electrical power system, the generated volt-ampere product has to be transferred to the load centres.

Question:

Find the loss in a transmission system having a resistance of 10 ohm, when it transfers a volt-ampere product of 660 kVA at a potential of (a) 11kV (b) 132kV.

Answer:

(a) The volt- ampere product is transferred at 11 kV. The current in the system can be obtained by dividing the volt-ampere product by voltage and it is given by

I = 60 x 10^{3} / 11 x 10^{3} = 60A

As you will remember the loss in the system will be I^{2}.R

(i.e.)

The loss = (60)^{2}. 10 = 36,000 Watts

(b) The current in the system is 660/132 = 5A

Similar to part (a).

The loss = (5)^{2}. 10 = 250 Watts

From this example it is clear that the losses in a transmission system can be reduced much by transferring the Volt ampere product at high voltages.

Since the transmission losses are reduced at high voltages the efficiency of transmission increases. Therefore it is preferable to transfer the electrical energy from one place to another at high voltage.

Since we are generating at about 11 kV, it is necessary for a
device, which can increase the potential to a higher level. This device is
named as a __Transformer__. In this particular case you might have noticed
that the transformer is used to increase the potential and therefore it is
known as a __step-up transformer.__

At the end of transmission system we have the distribution system and the transferred energy has to be distributed among the consumers. The transmission voltage, generally 132 kV or 220 kV in Sri Lanka, is too high for distribution and therefore it is now necessary to reduce the voltage from the transmission voltage. This also can be done by transformers and these transformers are named as Step-down transformers.

Fig.9.2 indicates the position of a power transformer in a typical power distribution system.

Question:

Could you define the function of a transformer?

Answer:

From the information that you got from the session so far, you can define it as follows.

The function of a transformer is to absorb electrical energy at variable voltages and to deliver it at some other voltages more convenient for transmission to another place, or to distribute among the consumers.

Exercise:

In the following three examples, identify the correct statement from the given four statements.

1. Transformers are used to

(a) Convert mechanical energy into electrical energy

(b) Convert hydro power into electrical energy

(c) Change the energy level.

(d) Change the potential level.

Answer:

Transformers are used to change the potential level.

Exercise:

2. Step-up transformers are used to increase

(a) the electrical energy

(b) the electrical losses

(c) the potential level

(d) the mechanical energy

Answer:

(C) Step-up transformers are used to increase the potential level.

## 9.1 Principle operation of the transformer:

The operation of the transformer depends on the mutual induction between two coils.

Question:

What is mutual induction?

Answer:

You could have studied about mutual induction. If you can recall this, you may give the answer correctly, as indicated below.

When the coils A and C are placed close to each other along the same axis as indicated in Fig.9.3 and the switch S is closed, some of the flux produced by the current in A links with C. The e.m.f. thus induced in C circulates a momentary current through the galvanometer, G. Similarly when S is opened the collapse of the flux induces an e.m.f. in the reveres direction in C.

This shows a change of current in one coils is accompanied by a change of flux linked with the other coil and therefore an e.m.f. is induced in the latter. This is known as mutual induction.

Fig.9.4 shows an arrangement in which two coils P and S wound on an iron core.

Look at Fig. 9.4 and say what will happen to (a) coil P (b) iron core and (c) coil s when switch K is closed.

When the switch is closed an alternating source is coming in series with the coil P. Therefore there will be an alternating current passing through coil P.

This current will set up an alternating flux in the core. The coil S will be linked by this alternating flux and an e.m.f. of mutual induction will thereby be induced in it.

The current in the coil P will set up an alternating flux in the iron core. The mean path of this flux is indicted by the dotted line, in Fig. 9.5. This flux is linked by the coil P and therefore an e.m.f. will be induced in the coil P. If the whole of the flux produced by P passes through S, then the e.m.f induced per turn will be the same for P and S.

Question:

Looks at figure 9.5 and find the current in coil S.

Answer:

The coil S is not closed and therefore there will be no current in that coil.

If the coil P
has N_{1}, turns and the coil S has N_{2} turns, then the
induced e.m.f. per coil may be obtained as follows.

Total e.m.f.
induced in coil P = N_{1} x (e.m.f. per turn)

Total e.m.f. induced
in coil S = N_{2} x (e.m.f. per turn)

If the magnetic flux leakage is neglected the induced e.m.f. per turn for both coils P and S will be equal.

Total e.m.f.
induced in S/Total e.m.f. induced in P = N_{2} x e.m.f. per turn/N_{1}
x e.m.f. per turn

E_{2}/E_{1
}= N_{2}/N_{1} ------ (9.1)

Since the coil S is open-circuited its terminal voltage will be equal to the induced e.m.f. in coil S.

Question:

Look at Fig. 9.5 and find the relation between the applied voltage and the induced e.m.f. in coil P.

Answer:

You might remember your basic electricity and the Lenz’s law. Lenz’s law says that the direction of the magnetic flux on induced e.m.f is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that e.m.f.

Therefore it is very clear that the induced e.m.f. will oppose the applied voltage.

You could remember from your school Physics, Kirchoff’s second law. This says that in a closed circuit the algebraic sum of the products of the current and the resistance of each part of the circuit is equal to the resultant e.m.f. in the circuit. Apply this to the coil P. If the resistance of the coil is negligibly small, then the sum of resistance of each part of the circuit is zero and hence the resultant e.m.f. is also zero.

(i.e) V_{1}-E_{1} = 0

Therefore V_{1} is practically equal and
opposite to the induced e.m.f. in coil p

(i.e) V_{1}» E_{1} (in magnitude)

From equation 9.1

E_{1 }/ E_{2}» N_{1} / N_{2}

But E_{2} = V_{2} and E_{1}» V_{1} and opposite in direction.

(i.e) E_{1}
/ E_{2} = N_{1}/N_{2}» V_{1} / V_{2}

In an ideal transformer, the ratio of primary to secondary voltage is the ratio of primary to the secondary turns.

Now consider the
case in which the coil S is closed through the impedance Z_{2}.

Question:

Under the above conditions will there be any current in coil S?

Answer:

Since the coil S
is closed, the induced e.m.f. will circulate a current. If the winding
resistance is negligibly small then the voltage across the load Z_{2}
will be nearly equal to the induced e.m.f. in coil S.

Therefore, I_{2}» V_{2}/Z_{2}

Question:

What is the expression for power in a simple a.c. circuit?

Answer:

If the voltage and the current vectors are displaced by the phase angle f, cos f is known as power factor. In an a.c. circuit, if you could remember your basic electricity, the real power is obtained by multiplying the power or the volt ampere product by the power factor.

In the ideal transformer if we say that the losses are zero the power factor is unchanged then the power input must be equal to the power output.

(i.e.)
V_{1}.I_{1} Cos f = V_{2}.I_{2} Cos f

Therefore, V_{1}/
I_{2}» V_{2} / I_{1}

(i.e.) For an ideal transformer,

V_{1} /
V_{2 }» E_{1} / E_{2} = N_{1}/N_{2}» I_{2} / I_{1}

Self assessment question

1. In Fig.9.7 coil P is having 50 turns and the voltage across the terminals of coil S is measured as 360 V. Assuming that the winding resistances are negligibly small and no flux leakage, calculate (a) the number of turns of the coil S. (b) The induced e.m.f. per turn of coil P and (c) the induced e.m.f. per turn of coil S.

2. In the first problem, if the applied voltage (source voltage) varies from 180 V to 230 V then find the voltage variation across the terminals of the coil S.

3. Prove that the power taken by a coil of resistance R ohm and reactance X ohm is V.I. Cosf where, V is the source voltage; I is the current through the coil and, f is the angle between the voltage and current phases.

Let the number
of turns in P be N_{1} and in S be N_{2}. The coil P is
connected to a source and therefore there will be a current in coil P. This
current will set up a flux f in the
core.

If we neglect the flux leakage, the same flux will link both the coils P and S.

Therefore the e.m.f. per turn for both coils is same and let it be ‘e’.

The induced e.m.f. in P = N_{1}e = E_{1}

The induced e.m.f. in S = N_{2}e = E_{2}

Since there is no current in coil S, (because it is not closed)

E_{2} = V_{2} the coils have
negligible resistance.

By Kirchoff’s law, V_{1} – E_{1} =0

(i.e.) V_{1}
= E_{1}

Therefore,

V_{1} / V_{2
=} E_{1} / E_{2} = N_{1}e/N_{2}e

(i.e.) V_{1} / V_{2 =} N_{1}
/ N_{2}

(a) V_{1} = 200 V; V_{2} = 360 V; N_{1}= 50

Therefore,
N_{2 }= N_{1}. (V_{2}/V_{1})

= 50 x 360 / 200

\N_{2} = 90

(i.e.) the coils S have 90 turns.

(b) The induced e.m.f. in coils P.

= E_{1} = V_{1}

(i.e.) E_{1} = 200 = N_{1}e

Therefore,
e = E_{1}/ N_{1} = 200/50 = 4

(i.e.) Induced e.m.f. per turn in coil P is 4V.

(c) Since the induce e.m.f. per turn is same for both coils, the induce e.m.f. in coil S is also 4V.

2. Refer to Fig. 9.8, in the first problem it has been shown that,

V_{1} / V_{2 =} N_{1} / N_{2} (i.e.) V2 = (N_{1} / N_{2})
x V_{1}

When V_{1} = 180 V

V_{2} = (N_{2} / N_{1}) x V_{1} =
(90/50) x 180 = 324V

When V_{1} = 230 V

V_{2} = (N_{2} / N_{1}) x V_{1} =
(90/50) x 230 = 414V

Therefore if the source voltage varies from 180 V to 230 V, the voltage across the terminals of the coil S will vary from 324 V to 414 V.