Bird J. Electrical Circuit Theory and Technology

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316 Electrical Circuit Theory and Technology

Transformers range in size from the miniature units used in electronic

applications to the large power transformers used in power stations. The

principle of operation is the same for each.

A transformer is represented in Figure 20.1(a) as consisting of two elec-

trical circuits linked by a common ferromagnetic core. One coil is termed

the primary winding which is connected to the supply of electricity, and

the other the secondary winding, which may be connected to a load. A

circuit diagram symbol for a transformer is shown in Figure 20.1(b).

Figure 20.1

20.2 Transformer

principle of operation

When the secondary is an open-circuit and an alternating voltage V

is applied to the primary winding, a small current—called the no-load

current I

—ﬂows, which sets up a magnetic ﬂux in the core. This alter-

nating ﬂux links with both primary and secondary coils and induces in

them e.m.f.’s of E

and E

respectively by mutual induction.

The induced e.m.f. E in a coil of N turns is given by

E DN

d

volts,

where d/dt is the rate of change of ﬂux. In an ideal transformer, the

rate of change of ﬂux is the same for both primary and secondary and

thus E

D E

,i.e.the induced e.m.f. per turn is constant.

Assuming no losses, E

D V

and E

D V

Hence

20.1

is called the voltage ratio and N

the turns ratio, or the ‘trans-

formation ratio’ of the transformer. If N

is less than N

then V

is less

than V

and the device is termed a step-down transformer.IfN

greater then N

then V

is greater than V

and the device is termed a

step-up transformer.

When a load is connected across the secondary winding, a current I

ﬂows. In an ideal transformer losses are neglected and a transformer is

considered to be 100% efﬁcient.

Transformers 317

Hence input power D output power, or V

D V

, i.e., in an ideal

transformer, the primary and secondary volt-amperes are equal.

Thus

20.2

Combining equations (20.1) and (20.2) gives:

20.3

The rating of a transformer is stated in terms of the volt-amperes that

it can transform without overheating. With reference to Figure 20.1(a),

the transformer rating is either V

or V

, where I

is the full-load

secondary current.

Problem 1. A transformer has 500 primary turns and 3000

secondary turns. If the primary voltage is 240 V, determine the

secondary voltage, assuming an ideal transformer.

For an ideal transformer, voltage ratio D turns ratio i.e.,

, hence

240

500

3000

Thus secondary voltage V

3000240

500

D 1440 V or 1.44 kV

Problem 2. An ideal transformer with a turns ratio of 2:7 is fed

from a 240 V supply. Determine its output voltage.

A turns ratio of 2:7 means that the transformer has 2 turns on the primary

for every 7 turns on the secondary (i.e. a step-up transformer). Thus,

For an ideal transformer,

; hence

240

Thus the secondary voltage V

2407

2

D 840 V

Problem 3. An ideal transformer has a turns ratio of 8:1 and the

primary current i

s 3 A when it is supplied at 240 V. Calculate the

secondary voltage and current.

318 Electrical Circuit Theory and Technology

A turns ratio of 8:1 means

, i.e. a step-down transformer.

, or secondary voltage V

D V





D 240





D 30 volts

Also,

I hence secondary current I

D I





D 3





D 24 A

Problem 4. An ideal transformer, connected to a 240 V mains,

supplies a 12 V, 150 W lamp. Calculate the transformer turns ratio

and the current taken from the supply.

D 240 V, V

D 12 V, I

150

D 12.5A

Turns ratio D

240

D 20

, from which, I

D I





D 12.5



240



Hence current taken from the supply, I

12.5

D 0.625 A

Problem 5. A 5 kVA single-phase transformer has a turns ratio of

10:1 and is fed from a 2.5 kV supply. Neglecting losses, determine

(a) the full-load secondary current, (b) the minimum load resistance

which can be connected across the secondary winding to give full

load kVA, (c) the primary current at full load kVA.

(a)

and V

D 2.5kVD 2500 V

Since

, secondary voltage V

D V





D 2500





D 250 V

The transformer rating in volt-amperes D V

(at full load),

i.e., 5000 D 250I

Transformers 319

Hence full load secondary current I

5000

250

D 20 A

(b) Minimum value of load resistance, R

250

D 12.5 Z

(c)

, from which primary current I

D I





D 20





D 2A

Further problems on the transformer principle of operation may be found

in Section 20.16, problems 1 to 9, page 344.

20.3 Transformer

no-load phasor diagram

(i) The core ﬂux is common to both primary and secondary windings in

a transformer and is thus taken as the reference phasor in a phasor

diagram. On no-load the primary winding takes a small no-load

current I

and since, with losses neglected, the primary winding is

a pure inductor, this current lags the applied voltage V

by 90

.In

the phasor diagram assuming no losses, shown in Figure 20.2(a),

current I

produces the ﬂux and is drawn in phase with the ﬂux.

The primary induced e.m.f. E

is in phase opposition to V

(by

Lenz’s law) and is shown 180

out of phase with V

and equal in

magnitude. The secondary induced e.m.f. is shown for a 2:1 turns

ratio transformer.

(ii) A no-load phasor diagram for a practical transformer is shown in

Figure 20.2(b). If current ﬂows then losses will occur. When losses

are considered then the no-load current I

is the phasor sum of

two components—(i) I

, the magnetizing component, in phase

with the ﬂux, and (ii) I

, the core loss component (supplying the

hysteresis and eddy current losses). From Figure 20.2(b):

No-load current, I



Y I

/, where I

= I

sinf

and

= I

cosf

Power factor on no-load D cos

The total core losses (i.e. iron losses) D V

cos

Problem 6. A 2400 V/400 V single-phase transformer takes a no-

load current of 0.5 A and the core loss is 400 W. Determine the

values of the magnetizing and core loss components of the no-

load current. Draw to scale the no-load phasor diagram for the

transformer.

Figure 20.2

320 Electrical Circuit Theory and Technology

Figure 20.3

D 2400 V, V

D 400 V, I

D 0.5A

Core loss (i.e. iron loss) D 400 D V

cos

i.e. 400 D 24000.5 cos

Hence cos

400

24000.5

D 0.3333



D arccos0.3333 D 70.53

The no-load phasor diagram is shown in Figure 20.3.

Magnetizing component, I

D I

sin

D 0.5sin70.53

D 0.471 A

Core loss component, I

D I

cos

D 0.5cos70.53

D 0.167 A

Problem 7. A transformer takes a current of 0.8 A when its

primary is connected to a 240 volt, 50 Hz supply, the secondary

being on open circuit. If the power absorbed is 72 watts, determine

(a) the iron loss current, (b) the power factor on no-load, and (c) the

magnetizing current.

D 0.8A,V D 240 V

(a) Power absorbed D total core loss D 72 D V

cos

Hence 72 D 240 I

cos

and iron loss current, I

D I

cos

240

D 0.30 A

(b) Power factor at no load, cos 

0.30

0.80

D 0.375

Pythagoras’ theorem, I

D I

C I

from which, magnetizing current, I



I

 I





0.80

 0.30



D 0.74 A

Further problems on the no-load phasor diagram may be found in

Section 20.16, problems 10 to 12, page 344.

20.4 E.m.f. equation of a

transformer

The magnetic ﬂux  set up in the core of a transformer when an alter-

nating voltage is applied to its primary winding is also alternating and is

sinusoidal.

Transformers 321

Let 

be the maximum value of the ﬂux and f be the frequency of

the supply. The time for 1 cycle of the alternating ﬂux is the periodic

time T, where T D 1/f seconds

The ﬂux rises sinusoidally from zero to its maximum value in

cycle,

and the time for

cycle is 1/4f seconds.

Hence the average rate of change of ﬂux D



1/4f

D 4f

Wb/s,

and since 1 Wb/s D 1 volt, the average e.m.f. induced in each turn

D 4f

volts.

As the ﬂux  varies sinusoidally, then a sinusoidal e.m.f. will be

induced in each turn of both primary and secondary windings.

For a sine wave, form factor D

rms value

average value

D 1.11 (see Chapter 14)

Hence rms value D form factor ð average value

D 1.11 ð average value

Thus rms e.m.f. induced in each turn D 1.11 ð4f

volts

D 4.44f

volts

Therefore, rms value of e.m.f. induced in primary,

= 4.44f 8

volts 20.4

and rms value of e.m.f. induced in secondary,

= 4.44f 8

volts 20.5

Dividing equation (20.4) by equation (20.5) gives:

, as previously obtained in Section 20.2.

Problem 8. A 100 kVA, 4000 V/200 V, 50 Hz single-phase trans-

former has 100 secondary turns. Determine (a) the primary and

secondary current, (b) the number of primary turns, and (c) the

maximum value of the ﬂux.

D 4000 V, V

D 200 V, f D 50 Hz, N

D 100 turns

(a) Transformer rating D V

D V

D 100000 VA

Hence primary current, I

100000

4000

D 25 A

and secondary current, I

100000

200

D 500 A

322 Electrical Circuit Theory and Technology

(b) From equation (20.3),

from which, primary turns, N





N

 D



4000

200



100

i.e., N

= 2000 turns

D 4.44f

from which, maximum ﬂux 

4.44fN

200

4.4450100

(assuming E

D V

)

D 9.01

× 10

−3

Wb or 9.01 mWb

[Alternatively, equation (20.4) could have been used,

where E

D 4.44f

from which, 

4.44fN

4000

4.44502000

(assuming E

D V



D 9.01 mWb, as above]

Problem 9. A single-phase, 50 Hz transformer has 25 primary

turns and 300 secondary turns. The cross-sectional area of the core

is 300 cm

. When the primary winding is connected to a 250 V

supply, determine (a) the maximum value of the ﬂux density in the

core, and (b) the voltage induced in the secondary winding.

(a) From equation (20.4), e.m.f. E

D 4.44f

volts i.e.,

250 D 4.4450

25

from which, maximum ﬂux density, 

250

4.445025

D 0.04505 Wb

However, 

D B

ð A, where B

D maximum ﬂux density in the

core and A D cross-sectional area of the core (see Chapter 7)

Hence B

ð 300 ð10

4

D 0.04505

from which, maximum ﬂux density, B

0.04505

300 ð 10

4

D 1.50 T

(b)

, from which, V

D V





i.e., voltage induced in the secondary winding,

Transformers 323

D 250



300



D 3000 V or 3 kV

Problem 10. A single-phase 500 V/100 V, 50 Hz transformer has

a maximum core ﬂux density of 1.5 T and an effective core cross-

sectional area of 50 cm

. Determine the number of primary and

secondary turns.

The e.m.f. equation for a transformer is E D 4.44f

and maximum ﬂux, 

D B ð A D 1.550 ð 10

4

 D 75 ð 10

4

Since E

D 4.44f

then primary turns, N

4.44f

500

4.445075 ð 10

4



D 300 turns

Since E

D 4.4f

then secondary turns, N

4.4f

100

4.445075 ð 10

4



D 60 turns

Problem 11. A 4500 V/225 V, 50 Hz single-phase transformer is

to have an approximate e.m.f. per turn of 15 V and operate with a

maximum ﬂux of 1.4 T. Calculate (a) the number of primary and

secondary turns and (b) the cross-sectional area of the core.

(a) E.m.f. per turn D

D 15

Hence primary turns, N

4500

D 300

and secondary turns, N

225

D 15

(b) E.m.f. E

D 4.44f

from which, 

4.44fN

4500

4.4450300

D 0.0676 Wb

Now ﬂux 

D B

ð A, where A is the cross-sectional area of the

core, hence

area A D



0.0676

1.4

D 0.0483 m

or 483 cm

324 Electrical Circuit Theory and Technology

Further problems on the e.m.f. equation may be found in Section 20.16,

problems 13 to 16, page 345.

20.5 Transformer

on-load phasor diagram

If the voltage drop in the windings of a transformer are assumed negli-

gible, then the terminal voltage V

is the same as the induced e.m.f. E

the secondary. Similarly, V

D E

. Assuming an equal number of turns

on primary and secondary windings, then E

D E

, and let the load have

a lagging phase angle 

In the phasor diagram of Figure 20.4, current I

lags V

by angle 

When a load is connected across the secondary winding a current I

ﬂows

in the secondary winding. The resulting secondary e.m.f. acts so as to tend

to reduce the core ﬂux. However this does not happen since reduction of

the core ﬂux reduces E

, hence a reﬂected increase in primary current I

occurs which provides a restoring mmf. Hence at all loads, primary and

secondary mmf’s are equal, but in opposition, and the core ﬂux remains

constant. I

is sometimes called the ‘balancing’ current and is equal, but

in the opposite direction, to current I

as shown in Figure 20.4. I

, shown

at a phase angle 

to V

, is the no-load current of the transformer (see

Section 20.3).

The phasor sum of I

and I

gives the supply current I

and the phase

angle between V

and I

is shown as 

Figure 20.4

Problem 12. A single-phase transformer has 2000 turns on the

primary and 800 turns on the secondary. Its no-load current is 5 A

at a power factor of 0.20 lagging. Assuming the volt drop in the

windings is negligible, determine the primary current and power

factor when the secondary current is 100 A at a power factor of

0.85 lagging.

Let I

be the component of the primary current which provides the

restoring mmf. Then

D I

i.e., I

2000 D 100800

from which, I

100800

2000

D 40 A

If the power factor of the secondary is 0.85

then cos

D 0.85, from which, 

D arccos0.85 D 31.8

If the power factor on no-load is 0.20,

then cos

D 0.2and

D arccos0.2 D 78.5

In the phasor diagram shown in Figure 20.5, I

D 100 A is shown at an

angle of 

D 31.8

to V

and I

D 40 A is shown in anti-phase to I

Figure 20.5

Transformers 325

The no-load current I

D 5 A is shown at an angle of 

D 78.5

to V

Current I

is the phasor sum of I

and I

and by drawing to scale,

D 44 A and angle 

D 37

By calculation, I

cos

D oa C ob

D I

cos

C I

cos

D 50.2 C 400.85

D 35.0A

and I

sin

D oc C od

D I

sin

C I

sin

D 5 sin 78.5

C 40 sin31.8

D 25.98 A

Hence the magnitude of I

35.0

C 25.98

 D 43.59 A

and tan 



25.98

35.0



, from which, 

D arctan



25.98

35.0



D 36.59

Hence the power factor of the primary D cos

D cos36.59

D 0.80

A further problem on the transformer on-load may be found in

Section 20.16, problem 17, page 345.

20.6 Transformer

construction

(i) There are broadly two types of single-phase double-wound trans-

former constructions— the core type and the shell type, as shown

in Figure 20.6. The low and high voltage windings are wound as

shown to reduce leakage ﬂux.

(ii) For power transformers, rated possibly at several MVA and

operating at a frequency of 50 Hz in Great Britain, the core material

used is usually laminated silicon steel or stalloy, the laminations

reducing eddy currents and the silicon steel keeping hysteresis loss

to a minimum.

Large power transformers are used in the main distribution

system and in industrial supply circuits. Small power transformers

have many applications, examples including welding and rectiﬁer

supplies, domestic bell circuits, imported washing machines, and

so on.

(iii) For audio frequency (a.f.) transformers, rated from a few mVA

to no more than 20 VA, and operating at frequencies up to about

15 kHz, the small core is also made of laminated silicon steel.

A typical application of a.f. transformers is in an audio ampliﬁer

system.

Figure 20.6