Ee Transformers

21Transformers

At the end of this chapter you should be able to:

understand the principle of operation of a transformer understand the term rating of a transformer use V1/V2 D N1/N2 D I2/I1 in calculations on transformers construct a transformer no-load phasor diagram and calculate magnetising and core

loss components of the no-load current state the e.m.f. equation for a transformer E D 4.44f8mN and use it in calculations construct a transformer on-load phasor diagram for an inductive circuit assuming

the volt drop in the windings is negligible describe transformer construction derive the equivalent resistance, reactance and impedance referred to the primary of

a transformer

understand voltage regulation describe losses in transformers and calculate efciency appreciate the concept of resistance matching and how it may be achieved perform calculations using R1 D N1/N22RL describe an auto transformer, its advantages/disadvantages and uses describe an isolating transformer, stating uses describe a three-phase transformer describe current and voltage transformers

21.1 Introduction

A transformer is a device which uses the phe-nomenon of mutual induction (see Chapter 9) tochange the values of alternating voltages and cur-rents. In fact, one of the main advantages of a.c.transmission and distribution is the ease with whichan alternating voltage can be increased or decreasedby transformers.

Losses in transformers are generally low and thusefciency is high. Being static they have a long lifeand are very stable.

Transformers range in size from the miniatureunits used in electronic applications to the largepower transformers used in power stations; the prin-ciple of operation is the same for each.

A transformer is represented in Fig. 21.1(a) asconsisting of two electrical circuits linked by acommon ferromagnetic core. One coil is termed the

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304 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

Figure 21.1

primary winding which is connected to the supplyof electricity, and the other the secondary winding,which may be connected to a load. A circuit diagramsymbol for a transformer is shown in Fig. 21.1(b)

21.2 Transformer principle ofoperation

When the secondary is an open-circuit and an alter-nating voltage V1 is applied to the primary wind-ing, a small current called the no-load currentI0 ows, which sets up a magnetic ux in thecore. This alternating ux links with both primaryand secondary coils and induces in them e.m.f.s ofE1 and E2 respectively by mutual induction.

The induced e.m.f. E in a coil of N turns is givenby E D Nd8/dt volts, where d8dt is the rateof change of ux. In an ideal transformer, the rateof change of ux is the same for both primaryand secondary and thus E1/N1 D E2/N2 i.e. theinduced e.m.f. per turn is constant.

Assuming no losses, E1 D V1 and E2 D V2

HenceV1N1

D V2N2

orV1V2

D N1N2

1

V1/V2 is called the voltage ratio and N1/N2the turns ratio, or the transformation ratio ofthe transformer. If N2 is less than N1 then V2 isless than V1 and the device is termed a step-downtransformer. If N2 is greater then N1 then V2 isgreater than V1 and the device is termed a step-uptransformer.

When a load is connected across the secondarywinding, a current I2 ows. In an ideal transformerlosses are neglected and a transformer is consideredto be 100 per cent efcient. Hence input power Doutput power, or V1I1 D V2I2 i.e. in an ideal

transformer, the primary and secondary ampere-turns are equal

ThusV1V2

D I2I1

2

Combining equations (1) and (2) gives:

V1V2

=N1N2

=I2I1

3

The rating of a transformer is stated in terms of thevolt-amperes that it can transform without overheat-ing. With reference to Fig. 21.1(a), the transformerrating is either V1I1 or V2I2, where I2 is the full-loadsecondary current.

Problem 1. A transformer has 500 primaryturns and 3000 secondary turns. If theprimary voltage is 240V, determine thesecondary voltage, assuming an idealtransformer.

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

V1V2

D N1N2

hence240V2

D 5003000

Thus secondary voltage

V2 D 2403000500 D 1440V or 1.44 kV

Problem 2. An ideal transformer with aturns ratio of 2:7 is fed from a 240V supply.Determine its output voltage.

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TRANSFORMERS 305

A turns ratio of 2:7 means that the transformerhas 2 turns on the primary for every 7 turns onthe secondary (i.e. a step-up transformer); thusN1/N2 D 2/7.

For an ideal transformer, N1/N2 D V1/V2hence 2/7 D 240/V2 Thus the secondary voltage

V2 D 24072 D 840V

Problem 3. An ideal transformer has aturns ratio of 8:1 and the primary current is3A when it is supplied at 240V. Calculatethe secondary voltage and current.

A turns ratio of 8:1 means N1/N2 D 1/8 i.e. astep-down transformer.(

N1N2

)D(

V1V2

)or secondary voltage

V2 D V1(

N1N2

)D 240

(18

)D 30 volts

Also,(

N1N2

)D(

I2I1

)hence secondary current

I2 D I1(

N1N2

)D 3

(81

)D 24A

Problem 4. An ideal transformer, connectedto a 240V mains, supplies a 12V, 150Wlamp. Calculate the transformer turns ratioand the current taken from the supply.

V1 D 240V, V2 D 12V, I2 D P/V2 D150/12 D 12.5A.

Turns ratio D N1N2

D V1V2

D 24012

D 20(

V1V2

)D(

I2I1

), from which,

I1 D I2(

V2V1

)D 12.5

(12240

)

Hence current taken from the supply,

I1 D 12.520 D 0.625A

Problem 5. A 12 resistor is connectedacross the secondary winding of an idealtransformer whose secondary voltage is120V. Determine the primary voltage if thesupply current is 4A.

Secondary current I2 D V2/R2 D 120/12 D10A.

V1/V2 D I2/I1, from which the primaryvoltage

V1 D V2(

I2I1

)D 120

(104

)D 300 volts

Problem 6. A 5 kVA single-phasetransformer has a turns ratio of 10 : 1 and isfed from a 2.5 kV supply. Neglecting losses,determine (a) the full-load secondary current,(b) the minimum load resistance which canbe connected across the secondary windingto give full load kVA, (c) the primary currentat full load kVA.

(a) N1/N2 D 10/1 and V1 D 2.5 kV D 2500V.Since

(N1N2

)D(

V1V2

), secondary voltage

V2 D V1(

N2N1

)D 2500

(110

)D 250V

The transformer rating in volt-amperes D V2I2(at full load) i.e. 5000 D 250I2Hence full load secondary current I2 D5000/250 D 20A.

(b) Minimum value of load resistance,RL D

(V2V1

)D(25020

)D 12.5Z.

(c)(

N1N2

)D(

I2I1

)from which primary current

I1 D I2(

N1N2

)D 20

(110

)D 2A

Now try the following exercise

Exercise 114 Further problems on thetransformer principle of operation1 A transformer has 600 primary turns

connected to a 1.5 kV supply. Determine thenumber of secondary turns for a 240V outputvoltage, assuming no losses. [96]

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306 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

2 An ideal transformer with a turns ratio of 2:9is fed from a 220V supply. Determine itsoutput voltage. [990V]

3 A transformer has 800 primary turns and2000 secondary turns. If the primary voltageis 160V, determine the secondary voltageassuming an ideal transformer. [400V]

4 An ideal transformer with a turns ratio of 3:8is fed from a 240V supply. Determine itsoutput voltage. [640V]

5 An ideal transformer has a turns ratio of12:1 and is supplied at 192V. Calculate thesecondary voltage. [16V]

6 A transformer primary winding connectedacross a 415V supply has 750 turns.Determine how many turns must be woundon the secondary side if an output of 1.66 kVis required. [3000 turns]

7 An ideal transformer has a turns ratio of 12:1and is supplied at 180V when the primarycurrent is 4A. Calculate the secondaryvoltage and current. [15V, 48A]

8 A step-down transformer having a turns ratioof 20:1 has a primary voltage of 4 kV anda load of 10 kW. Neglecting losses, calculatethe value of the secondary current. [50A]

9 A transformer has a primary to secondaryturns ratio of 1:15. Calculate the primaryvoltage necessary to supply a 240V load. Ifthe load current is 3A determine the primarycurrent. Neglect any losses. [16V, 45A]

10 A 10 kVA, single-phase transformer has aturns ratio of 12:1 and is supplied from a2.4 kV supply. Neglecting losses, determine(a) the full load secondary current, (b) theminimum value of load resistance which canbe connected across the secondary windingwithout the kVA rating being exceeded, and(c) the primary current.

[(a) 50A (b) 4 (c) 4.17A]11 A 20 resistance is connected across the

secondary winding of a single-phase powertransformer whose secondary voltage is150V. Calculate the primary voltage andthe turns ratio if the supply current is 5A,neglecting losses. [225V, 3:2]

21.3 Transformer no-load phasordiagram

The core ux is common to both primary andsecondary windings in a transformer and is thustaken as the reference phasor in a phasor diagram.On no-load the primary winding takes a small no-load current I0 and since, with losses neglected, theprimary winding is a pure inductor, this current lagsthe applied voltage V1 by 90. In the phasor diagramassuming no losses, shown in Fig. 21.2(a), currentI0 produces the ux and is drawn in phase withthe ux. The primary induced e.m.f. E1 is in phaseopposition to V1 (by Lenzs law) and is shown 180out of phase with V1 and equal in magnitude. Thesecondary induced e.m.f. is shown for a 2:1 turnsratio transformer.

A no-load phasor diagram for a practical trans-former is shown in Fig. 21.2(b). If current owsthen losses will occur. When losses are consideredthen the no-load current I0 is the phasor sum oftwo components (i) IM, the magnetising compo-nent, in phase with the ux, and (ii) IC, the coreloss component (supplying the hysteresis and eddycurrent losses). From Fig. 21.2(b):

No-load current, I0 D

I 2M C I 2C whereIM D I0 sinf0 and IC D I0cosf0.

Power factor on no-load D cos0 D IC/I0.The total core losses (i.e. iron losses)

D V1I0 cos0

Problem 7. A 2400V/400V single-phasetransformer takes a no-load current of 0.5Aand the core loss is 400W. Determine thevalues of the magnetising and core losscomponents of the no-load current. Draw toscale the no-load phasor diagram for thetransformer.

V1 D 2400V,V2 D 400V and I0 D 0.5A Core loss(i.e. iron loss) D 400 D V1I0 cos0.i.e. 400 D 24000.5 cos0Hence cos0 D 400

24000.5D 0.3333

0 D cos1 0.3333 D 70.53The no-load phasor diagram is shown in Fig. 21.3

Magnetising component,IM D I0 sin0 D 0.5 sin 70.53 D 0.471A.Core loss component, IC DI0 cos0 D0.5 cos 70.53D 0.167A

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TRANSFORMERS 307

Figure 21.2

Figure 21.3

Problem 8. A transformer takes a current of0.8A when its primary is connected to a 240volt, 50Hz supply, the secondary being onopen circuit. If the power absorbed is72 watts, determine (a) the iron loss current,(b) the power factor on no-load, and (c) themagnetising current.

I0 D 0.8A and V D 240V(a) Power absorbed D total core loss D 72 D

V1I0 cos0. Hence 72 D 240I0 cos0 and ironloss current, Ic D I0 cos0 D 72/240 D 0.30A

(b) Power factor at no load,

cos0 D ICI0

D 0.30.8

D 0.375

(c) From the right-angled triangle in Fig. 21.2(b)and using Pythagoras theorem, I20 D I2C C I2Mfrom which, magnetising current,

IM D

I20 I2C Dp

0.82 0.32 D 0.74A

Now try the following exercise

Exercise 115 Further problems on theno-load phasor diagram1 A 500V/100V, single-phase transformer takes

a full load primary current of 4A. Neglectinglosses, determine (a) the full load secondarycurrent, and (b) the rating of the transformer.

[(a) 20A (b) 2 kVA]2 A 3300V/440V, single-phase transformer

takes a no-load current of 0.8A and theiron loss is 500W. Draw the no-load phasordiagram and determine the values of themagnetising and core loss components of theno-load current.

[0.786A, 0.152A]3 A transformer takes a current of 1A when

its primary is connected to a 300V, 50Hzsupply, the secondary being on open-circuit.

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308 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

If the power absorbed is 120 watts, calculate(a) the iron loss current, (b) the power factoron no-load, and (c) the magnetising current.

[(a) 0.4A (b) 0.4 (c) 0.92A]

21.4 E.m.f. equation of a transformer

The magnetic ux 8 set up in the core of a trans-former when an alternating voltage is applied to itsprimary winding is also alternating and is sinusoidal.

Let 8m be the maximum value of the ux and fbe the frequency of the supply. The time for 1 cycleof the alternating ux is the periodic time T, whereT D 1/fseconds

The ux rises sinusoidally from zero to its max-imum value in (1/4) cycle, and the time for (1/4)cycle is 1/4f seconds. Hence the average rate ofchange of ux D 8m/1/4f D 4f8m Wb/s, andsince 1Wb/s D 1 volt, the average e.m.f. induced ineach turn D 4f8m volts. As the ux 8 varies sinu-soidally, then a sinusoidal e.m.f. will be induced ineach turn of both primary and secondary windings.

For a sine wave,

form factor D r.m.s. valueaverage value

D 1.11 (see Chapter 14)Hence r.m.s. value D form factoraverage value D1.11 average value Thus r.m.s. e.m.f. induced ineach turn

D 1.11 4f8m voltsD 4.44f8m volts

Therefore, r.m.s. value of e.m.f. induced in primary,

E1 = 4.44 f8mN1 volts 4

and r.m.s. value of e.m.f. induced in secondary,

E2 = 4.44 f8mN2 volts 5

Dividing equation (4) by equation (5) gives:(

E1E2

)D(

N1N2

),

as previously obtained in Section 21.2

Problem 9. A 100 kVA, 4000V/200V,50Hz single-phase transformer has 100secondary turns. Determine (a) the primaryand secondary current, (b) the number ofprimary turns, and (c) the maximum value ofthe ux.

V1 D 4000V, V2 D 200V, f D 50Hz, N2 D 100turns

(a) Transformer ratingDV1I1 DV2I2 D1 00 000VAHence primary current,

I1 D 1 00 000V1

D 1 00 0004 000

D 25A

and secondary current,

I2 D 1 00 000V2

D 1 00 000200

D 500A

(b) From equation (3), V1V2

D N1N2

from which, pri-mary turns,

N1 D(

V1V2

)N2D

(4000200

)100D2000 turns

(c) From equation (5), E2 D 4.44f8mN2 fromwhich, maximum ux,

8m D E4.44fN2D 200

4.4450100(assuming E2 D V2

D 9.01 103 Wb or 9.01mWb

[Alternatively, equation (4) could have been used,where

E1 D 4.44f8mN1from which,

8m D 40004.44502000

(assuming E1 D V1

D 9.01mWb as above]

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TRANSFORMERS 309

Problem 10. A single-phase, 50Hztransformer has 25 primary turns and 300secondary turns. The cross-sectional area ofthe core is 300 cm2. When the primarywinding is connected to a 250V supply,determine (a) the maximum value of the uxdensity in the core, and (b) the voltageinduced in the secondary winding.

(a) From equation (4),e.m.f. E1 D 4.44f8mN1 voltsi.e. 250 D 4.44508m(25) from which, maxi-mum ux density,

8m D 2504.445025

Wb D 0.04505Wb

However, 8m D BmA,where Bm D maximumux density in the core and A D cross-sectionalarea of the core (see Chapter 7). HenceBm 300 104 D 0.04505 from which,

maximum ux density, Bm D 0.04505300 104D 1.50T

(b) V1V2

D N1N2

from which, V2 D V1(

N2N1

)i.e.

voltage induced in the secondary winding,

V2 D 250(300

25

)D 3000V or 3 kV

Problem 11. A single-phase 500V/100V,50Hz transformer has a maximum core uxdensity of 1.5 T and an effective corecross-sectional area of 50 cm2. Determine thenumber of primary and secondary turns.

The e.m.f. equation for a transformer is E D4.44f8mN and maximum ux, 8m D B A D1.550 104 D 75 104 Wb

Since E1 D 4.44f8mN1 then primary turns,N1 D E14.44f8m D

5004.445075 104

D 300 turnsSince E2 D 4.4f8mN2 then secondary turns,

N2 D E24.44f8m D100

4.445075 104D 60 turns

Problem 12. A 4500V/225V, 50Hzsingle-phase transformer is to have anapproximate e.m.f. per turn of 15V andoperate with a maximum ux of 1.4 T.Calculate (a) the number of primary andsecondary turns and (b) the cross-sectionalarea of the core.

(a) E.m.f. per turn D E1N1

D E2N2

D 15

Hence primary turns, N1 D E115 D450015

D 300

and secondary turns, N2 D E215 D25515

D 15

(b) E.m.f. E1 D 4.44f8mN1 from which,

8mE1

4.44fN1D 4500

4.4450300D 0.0676Wb

Now ux, 8m D Bm A, where A is the cross-sectional area of the core,

hence area, A D(

8mBm

)D(0.0676

1.4

)

D 0.0483m2 or 483 cm2

Now try the following exercise

Exercise 116 Further problems on thetransformer e.m.f. equation1 A 60 kVA, 1600V/100V, 50Hz, single-phase

transformer has 50 secondary windings. Cal-culate (a) the primary and secondary current,(b) the number of primary turns, and (c) themaximum value of the ux

[(a) 37.5A, 600A (b) 800 (c) 9.0mWb]2 A single-phase, 50Hz transformer has 40 pri-

mary turns and 520 secondary turns. Thecross-sectional area of the core is 270 cm2.When the primary winding is connected to a300 volt supply, determine (a) the maximumvalue of ux density in the core, and (b) thevoltage induced in the secondary winding

[(a) 1.25 T (b) 3.90 kV]

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310 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

3 A single-phase 800V/100V, 50Hz trans-former has a maximum core ux density of1.294 T and an effective cross-sectional areaof 60 cm2. Calculate the number of turns onthe primary and secondary windings.

[464, 58]4 A 3.3 kV/110V, 50Hz, single-phase trans-

former is to have an approximate e.m.f. perturn of 22V and operate with a maximumux of 1.25 T. Calculate (a) the number ofprimary and secondary turns, and (b) the cross-sectional area of the core

[(a) 150, 5 (b) 792.8 cm2]

21.5 Transformer on-load phasordiagram

If the voltage drop in the windings of a transformerare assumed negligible, then the terminal voltage V2is the same as the induced e.m.f. E2 in the secondary.Similarly, V1 D E1. Assuming an equal numberof turns on primary and secondary windings, thenE1 D E2, and let the load have a lagging phaseangle 2

Figure 21.4

In the phasor diagram of Fig. 21.4, current I2lags V2 by angle 2. When a load is connectedacross the secondary winding a current I2 owsin the secondary winding. The resulting secondarye.m.f. acts so as to tend to reduce the core ux.

However this does not happen since reduction of thecore ux reduces E1, hence a reected increase inprimary current I01 occurs which provides a restoringm.m.f. Hence at all loads, primary and secondarym.m.f.s are equal, but in opposition, and the coreux remains constant. I01 is sometimes called thebalancing current and is equal, but in the oppositedirection, to current I2 as shown in Fig. 21.4. I0,shown at a phase angle 0 to V1, is the no-loadcurrent of the transformer (see Section 21.3)

The phasor sum of I01 and I0 gives the supplycurrent I1 and the phase angle between V1 and I1 isshown as 1

Problem 13. A single-phase transformer has2000 turns on the primary and 800 turns onthe secondary. Its no-load current is 5A at apower factor of 0.20 lagging. Assuming thevolt drop in the windings is negligible,determine the primary current and powerfactor when the secondary current is 100A ata power factor of 0.85 lagging.

Let I01 be the component of the primary currentwhich provides the restoring m.m.f. Then

I01N1 D I2N2i.e. I012000 D 100800

from which, I01 D100800

2000D 40A

If the power factor of the secondary is 0.85, thencos2 D 0.85, from which, 2 D cos1 0.85 D 31.8If the power factor on no-load is 0.20, thencos0 D 0.2 and 0 D cos1 0.2 D 78.5

In the phasor diagram shown in Fig. 21.5, I2 D100A is shown at an angle of D 31.8 to V2 andI01 D 40A is shown in anti-phase to I2

The no-load current I0 D 5A is shown at an angleof 0 D 78.5 to V1. Current I1 is the phasor sumof I01 and I0, and by drawing to scale, I1 D 44Aand angle 1 D 37.

By calculation,

I1 cos1 D 0a C 0bD I0 cos0 C I01 cos2D 50.2 C 400.85D 35.0A

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TRANSFORMERS 311

Figure 21.5

and I1 sin1 D 0c C 0dD I0 sin0 C I01 sin2D 5 sin 78.5 C 40 sin 31.8D 25.98A

Hence the magnitude of I1 Dp

35.02 C 25.982 D43.59A and tan 1 D 25.98/35.0 from which,f1 D tan1 25.98/35.0 D 36.59 Hence thepower factor of the primary D cos1 D cos 36.59 D0.80

Now try the following exercise

Exercise 117 A further problem on thetransformer on-load1 A single-phase transformer has 2400 turns on

the primary and 600 turns on the secondary.Its no-load current is 4A at a power factor of0.25 lagging. Assuming the volt drop in thewindings is negligible, calculate the primarycurrent and power factor when the secondarycurrent is 80A at a power factor of 0.8 lagging.

[23.26A, 0.73]

21.6 Transformer construction(i) There are broadly two types of single-phase

double-wound transformer constructions thecore type and the shell type, as shown in

Fig. 21.6. The low and high voltage windingsare wound as shown to reduce leakage ux.

Figure 21.6

(ii) For power transformers, rated possibly atseveral MVA and operating at a frequency of50Hz in Great Britain, the core material usedis usually laminated silicon steel or stalloy,the laminations reducing eddy currents andthe silicon steel keeping hysteresis loss to aminimum.

Large power transformers are used in themain distribution system and in industrialsupply circuits. Small power transformers havemany applications, examples including weldingand rectier supplies, domestic bell circuits,imported washing machines, and so on.

(iii) For audio frequency (a.f.) transformers, ratedfrom a few mVA to no more than 20VA, andoperating at frequencies up to about 15 kHz, thesmall core is also made of laminated siliconsteel. A typical application of a.f. transformersis in an audio amplier system.

(iv) Radio frequency (r.f.) transformers, operat-ing in the MHz frequency region have eitheran air core, a ferrite core or a dust core. Ferriteis a ceramic material having magnetic proper-ties similar to silicon steel, but having a highresistivity. Dust cores consist of ne particlesof carbonyl iron or permalloy (i.e. nickel andiron), each particle of which is insulated fromits neighbour. Applications of r.f. transformersare found in radio and television receivers.

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312 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

(v) Transformer windings are usually of enamel-insulated copper or aluminium.

(vi) Cooling is achieved by air in small transform-ers and oil in large transformers.

21.7 Equivalent circuit of atransformer

Figure 21.7 shows an equivalent circuit of a trans-former. R1 and R2 represent the resistances of theprimary and secondary windings and X1 and X2 rep-resent the reactances of the primary and secondarywindings, due to leakage ux.

The core losses due to hysteresis and eddy cur-rents are allowed for by resistance R which takes acurrent IC, the core loss component of the primarycurrent. Reactance X takes the magnetising compo-nent Im. In a simplied equivalent circuit shown inFig. 21.8, R and X are omitted since the no-loadcurrent I0 is normally only about 35 per cent ofthe full load primary current.

It is often convenient to assume that all of theresistance and reactance as being on one side of

the transformer. Resistance R2 in Fig. 21.8 can bereplaced by inserting an additional resistance R02 inthe primary circuit such that the power absorbed inR02 when carrying the primary current is equal to thatin R2 due to the secondary current, i.e.

I21R02 D I22R2

from which, R02 D R2(

I2I1

)2D R2

(V1V2

)2

Then the total equivalent resistance in the primarycircuit Re is equal to the primary and secondaryresistances of the actual transformer.

Hence Re D R1 C R02

i.e. Re = R1 + R2(V1

V2

)26

By similar reasoning, the equivalent reactance in theprimary circuit is given by Xe D X1 C X02

i.e. Xe = X1 + X2(V1

V2

)27

Figure 21.7

Figure 21.8

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TRANSFORMERS 313

The equivalent impedance Ze of the primary andsecondary windings referred to the primary isgiven by

Ze =

R2e + X 2e 8

If e is the phase angle between I1 and the volt dropI1Ze then

cosfe =ReZe

9

The simplied equivalent circuit of a transformer isshown in Fig. 21.9

Problem 14. A transformer has 600 primaryturns and 150 secondary turns. The primaryand secondary resistances are 0.25 and0.01 respectively and the correspondingleakage reactances are 1.0 and 0.04respectively. Determine (a) the equivalentresistance referred to the primary winding,(b) the equivalent reactance referred to theprimary winding, (c) the equivalentimpedance referred to the primary winding,and (d) the phase angle of the impedance.

(a) From equation (6), equivalent resistance

Re D R1 C R2(

V1V2

)2

i.e. Re D 0.25 C 0.01(600150

)2

D 0.41Z since N1N2

D V1V2

(b) From equation (7), equivalent reactance,

Xe D X1 C X2(

V1V2

)2

i.e. Xe D 1.0 C 0.04(600150

)2D 1.64Z

(c) From equation (8), equivalent impedance,Ze D

R2e C X2e D

p0.412 C 1.642 D 1.69Z

(d) From equation (9),cose D Re

ZeD 0.41

1.69

Hence fe D cos1 0.411.69 D 75.96

Now try the following exercise

Exercise 118 A further problem on theequivalent circuit of a transformer1 A transformer has 1200 primary turns and 200

secondary turns. The primary and secondaryresistances are 0.2 and 0.02 respectivelyand the corresponding leakage reactancesare 1.2 and 0.05 respectively. Calculate(a) the equivalent resistance, reactance andimpedance referred to the primary winding,and (b) the phase angle of the impedance.

[(a) 0.92, 3.0, 3.14 (b) 72.95]

21.8 Regulation of a transformerWhen the secondary of a transformer is loaded,the secondary terminal voltage, V2, falls. As the

Figure 21.9

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314 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

power factor decreases, this voltage drop increases.This is called the regulation of the transformerand it is usually expressed as a percentage ofthe secondary no-load voltage, E2. For full-loadconditions:

Regulation =(

E2 V2E2

) 100% 10

The fall in voltage, (E2 V2), is caused by theresistance and reactance of the windings. Typicalvalues of voltage regulation are about 3% in smalltransformers and about 1% in large transformers.

Problem 15. A 5 kVA, 200V/400V,single-phase transformer has a secondaryterminal voltage of 387.6 volts when loaded.Determine the regulation of the transformer.

From equation (10):

regulationD

No load secondary voltage terminal voltage on loadno load secondary voltage

100%

D(400 387.6

400

) 100%

D(12.4400

) 100%

D 3.1%

Problem 16. The open circuit voltage of atransformer is 240V. A tap changing deviceis set to operate when the percentageregulation drops below 2.5%. Determine theload voltage at which the mechanismoperates.

RegulationD

No load secondary voltageterminal voltage on loadno load secondary voltage

100%

Hence 2.5 D(240 V2

240

) 100%

2.5240

100D 240 V2

i.e. 6 D 240 V2from which, load voltage, V2 D 2406 D 234 volts

Now try the following exercise

Exercise 119 Further problems onregulation1 A 6 kVA, 100V/500V, single-phase trans-

former has a secondary terminal voltage of487.5 volts when loaded. Determine the reg-ulation of the transformer. [2.5%]

2 A transformer has an open circuit voltageof 110 volts. A tap-changing device operateswhen the regulation falls below 3%. Calculatethe load voltage at which the tap-changer oper-ates.

[106.7 volts]

21.9 Transformer losses and efciencyThere are broadly two sources of losses in trans-formers on load, these being copper losses and ironlosses.

(a) Copper losses are variable and result in a heat-ing of the conductors, due to the fact that theypossess resistance. If R1 and R2 are the primaryand secondary winding resistances then the totalcopper loss is I21R1 C I22R2

(b) Iron losses are constant for a given value offrequency and ux density and are of twotypes hysteresis loss and eddy current loss.(i) Hysteresis loss is the heating of the core as

a result of the internal molecular structurereversals which occur as the magnetic uxalternates. The loss is proportional to thearea of the hysteresis loop and thus low lossnickel iron alloys are used for the core sincetheir hysteresis loops have small areas.(SeeChapters 7)

(ii) Eddy current loss is the heating of thecore due to e.m.f.s being induced not onlyin the transformer windings but also in thecore. These induced e.m.f.s set up circulat-ing currents, called eddy currents. Owing tothe low resistance of the core, eddy currentscan be quite considerable and can cause a

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TRANSFORMERS 315

large power loss and excessive heating of thecore. Eddy current losses can be reduced byincreasing the resistivity of the core mate-rial or, more usually, by laminating the core(i.e. splitting it into layers or leaves) whenvery thin layers of insulating material canbe inserted between each pair of laminations.This increases the resistance of the eddy cur-rent path, and reduces the value of the eddycurrent.

Transformer efciency,

D output powerinput power

D input power - lossesinput power

i.e. h = 1 lossesinput power

11

and is usually expressed as a percentage. It is notuncommon for power transformers to have efcien-cies of between 95% and 98%

Output power D V2I2 cos2.Total losses D copper loss C iron losses,and input power D output power C losses

Problem 17. A 200 kVA rated transformerhas a full-load copper loss of 1.5 kW and aniron loss of 1 kW. Determine the transformerefciency at full load and 0.85 power factor.

Efciency, D output powerinput power

D input power lossesinput power

D 1 lossesinput power

Full-load output power D VI cos D 200 0.85D 170 kW.

Total losses D 1.5 C 1.0 D 2.5 kWInput power D output power C lossesD 170 C 2.5 D 172.5 kW.

Hence efciency D(1 2.5

172.5

)D 1 0.01449

D 0.9855 or 98.55%

Problem 18. Determine the efciency ofthe transformer in Problem 17 at halffull-load and 0.85 power factor.

Half full-load power output D 1/22000.85D 85 kW.

Copper loss (or I2R loss) is proportional to cur-rent squared. Hence the copper loss at half full-loadis:( 12)2

1500 D 375WIron loss D 1000W (constant)Total losses D 375C1000D1375W or 1.375 kW.Input power at half full-loadD output power at half full-load C lossesD 85 C 1.375 D 86.375 kW. Hence

efciency D 1 lossesinput power

D(1 1.375

86.375

)

D 1 0.01592D 0.9841 or 98.41%

Problem 19. A 400 kVA transformer hasa primary winding resistance of 0.5 anda secondary winding resistance of 0.001.The iron loss is 2.5 kW and the primary andsecondary voltages are 5 kV and 320V respec-tively. If the power factor of the load is 0.85,determine the efciency of the transformer(a) on full load, and (b) on half load.

(a) Rating D 400 kVA D V1I1 D V2I2. Henceprimary current,

I1 D 400 103

V1D 400 10

3

5000D 80A

and secondary current,

I2 D 400 103

V2D 400 10

3

320D 1250A

Total copper loss D I21R1 C I22R2, (whereR1 D 0.5 and R2 D 0.001

D 8020.5 C 125020.001D 3200 C 1562.5 D 4762.5 wattsOn full load, total loss D copper lossC iron loss

D 4762.5 C 2500 D 7262.5W D 7.2625 kWTotal output power on full load

D V2I2 cos2 D 400 1030.85 D 340 kW

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316 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

Input power D output power C lossesD 340 kW C 7.2625 kW D 347.2625 kW

Efciency, D(1 losses

input power

) 100%

D(1 7.2625

347.2625

) 100%

D 97.91%(b) Since the copper loss varies as the square of the

current, then total copper loss on half loadD 4762.5 ( 12

)2 D 1190.625W. Hence totalloss on half load D 1190.625 C 2500 D3690.625W or 3.691 kW.Output power on half full load D ( 12

)340

D 170 kW.Input power on half full loadD output power C losses

D 170 kW C 3.691 kWD 173.691 kWHence efciency at half full load,

D(1 losses

input power

) 100%

D(1 3.691

173.691

) 100% D 97.87%

Maximum efciencyIt may be shown that the efciency of a transformeris a maximum when the variable copper loss (i.e.I21R1 C I22R2) is equal to the constant iron losses.

Problem 20. A 500 kVA transformer has afull load copper loss of 4 kW and an ironloss of 2.5 kW. Determine (a) the output kVAat which the efciency of the transformer is amaximum, and (b) the maximum efciency,assuming the power factor of the load is 0.75

(a) Let x be the fraction of full load kVA at whichthe efciency is a maximum. The correspond-ing total copper loss D 4 kWx2. At maxi-mum efciency, copper loss D iron loss. Hence4x2 D 2.5 from which x2 D 2.5/4 andx D p2.5/4 D 0.791.Hence the output kVA at maximumefciency D 0.791 500 D 395.5 kVA.

(b) Total loss at maximum efciencyD2 2.5D5 kWOutput power D 395.5 kVA p.f.D 395.5 0.75 D 296.625 kWInput power D output power C lossesD 296.625 C 5 D 301.625 kWMaximum efciency,

D(1 losses

input power

) 100%

D(1 5

301.625

) 100% D 98.34%

Now try the following exercise

Exercise 120 Further problems on lossesand efciency1 A single-phase transformer has a voltage ratio

of 6:1 and the h.v. winding is supplied at540V. The secondary winding provides a fullload current of 30A at a power factor of 0.8lagging. Neglecting losses, nd (a) the ratingof the transformer, (b) the power supplied tothe load, (c) the primary current

[(a) 2.7 kVA (b) 2.16 kW (c) 5A]2 A single-phase transformer is rated at 40 kVA.

The transformer has full-load copper losses of800W and iron losses of 500W. Determinethe transformer efciency at full load and 0.8power factor [96.10%]

3 Determine the efciency of the transformerin problem 2 at half full-load and 0.8 powerfactor [95.81%]

4 A 100 kVA, 2000V/400V, 50Hz, single-phasetransformer has an iron loss of 600W and afull-load copper loss of 1600W. Calculate itsefciency for a load of 60 kW at 0.8 powerfactor. [97.56%]

5 Determine the efciency of a 15 kVA trans-former for the following conditions:(i) full-load, unity power factor(ii) 0.8 full-load, unity power factor(iii) half full-load, 0.8 power factorAssume that iron losses are 200W and the full-load copper loss is 300W

[(a) 96.77% (ii) 96.84% (iii) 95.62%]

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TRANSFORMERS 317

6 A 300 kVA transformer has a primary wind-ing resistance of 0.4 and a secondarywinding resistance of 0.0015. The ironloss is 2 kW and the primary and secondaryvoltages are 4 kV and 200V respectively. Ifthe power factor of the load is 0.78, determinethe efciency of the transformer (a) on fullload, and (b) on half load.

[(a) 96.84% (b) 97.17%]7 A 250 kVA transformer has a full load copper

loss of 3 kW and an iron loss of 2 kW. Calcu-late (a) the output kVA at which the efciencyof the transformer is a maximum, and (b) themaximum efciency, assuming the power fac-tor of the load is 0.80

[(a) 204.1 kVA (b) 97.61%]

21.10 Resistance matchingVarying a load resistance to be equal, or almostequal, to the source internal resistance is calledmatching. Examples where resistance matching isimportant include coupling an aerial to a transmitteror receiver, or in coupling a loudspeaker to anamplier, where coupling transformers may be usedto give maximum power transfer.

With d.c. generators or secondary cells, the inter-nal resistance is usually very small. In such cases,if an attempt is made to make the load resistance assmall as the source internal resistance, overloadingof the source results.

A method of achieving maximum power trans-fer between a source and a load (see section 13.9,page 179), is to adjust the value of the load resis-tance to match the source internal resistance. Atransformer may be used as a resistance matchingdevice by connecting it between the load and thesource.

The reason why a transformer can be used for thisis shown below. With reference to Fig. 21.10:

RL D V2I2

and R1 D V1I1

For an ideal transformer,

V1 D(

N1N2

)V2

and I1 D(

N2N1

)I2

Figure 21.10

Thus the equivalent input resistance R1 of thetransformer is given by:

R1 D V1I1

D

(N1N2

)V2

(N2N1

)I2

D(

N1N2

)2 (V2I2

)D(

N1N2

)2RL

i.e. R1 =(N1

N2

)2RL

Hence by varying the value of the turns ratio,the equivalent input resistance of a transformer canbe matched to the internal resistance of a load toachieve maximum power transfer.

Problem 21. A transformer having a turnsratio of 4:1 supplies a load of resistance100. Determine the equivalent inputresistance of the transformer.

From above, the equivalent input resistance,

R1 D(

N1N2

)2RL

D(41

)2100 D 1600Z

Problem 22. The output stage of anamplier has an output resistance of 112.Calculate the optimum turns ratio of atransformer which would match a loadresistance of 7 to the output resistance ofthe amplier.

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318 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

Figure 21.11

The circuit is shown in Fig. 21.11The equivalent input resistance, R1 of the trans-

former needs to be 112 for maximum powertransfer.

R1 D(

N1N2

)2RL

Hence(

N1N2

)2D R1

RLD 112

7D 16

i.e.N1N2

Dp

16 D 4

Hence the optimum turns ratio is 4:1

Problem 23. Determine the optimum valueof load resistance for maximum powertransfer if the load is connected to anamplier of output resistance 150 througha transformer with a turns ratio of 5:1

The equivalent input resistance R1 of the transformerneeds to be 150 for maximum power transfer.

R1 D(

N1N2

)2RL

from which, RL D R1(

N2N1

)2

D 150 ( 15)2 D 6Z

Problem 24. A single-phase, 220V/1760Videal transformer is supplied from a 220Vsource through a cable of resistance 2. Ifthe load across the secondary winding is1.28 k determine (a) the primary currentowing and (b) the power dissipated in theload resistor.

The circuit diagram is shown in Fig. 21.12

Figure 21.12

(a) Turns ratio(

N1N2

)D(

V1V2

)D( 2201760

)D(18

)

Equivalent input resistance of the transformer.

R1 D(

N1N2

)2RL D

(18

)21.28 103 D 20

Total input resistance,RIN D R C R1 D 2 C 20 D 22Primary current,

I1 D V1RIN

D 22022

D 10A

(b) For an ideal transformerV1V2

D I2I1

from which,

I2 D I1(

V1V2

)D 10

(2201760

)D 1.25A

Power dissipated in load resistor RL,

P D I22RL D 1.2521.28 103D 2000 watts or 2 kW

Problem 25. An a.c. source of 24V andinternal resistance 15 k is matched to aload by a 25:1 ideal transformer. Determine(a) the value of the load resistance and(b) the power dissipated in the load.

The circuit diagram is shown in Fig. 21.13

(a) For maximum power transfer R1 needs to beequal to 15 k.

R1 D(

N1N2

)2RL

from which, load resistance,

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TRANSFORMERS 319

Figure 21.13

RL D R1(

N2N1

)2D 15 000

(125

)2D 24Z

(b) The total input resistance when the source isconnected to the matching transformer is RIN CR1 i.e. 15 k C 15 k D 30 k.Primary current,

I1 D V30000 D24

30000D 0.8mA

N1/N2 D I2/I1 from which, I2 D I1N1/N2 D0.8 10325/1 D 20 103 A.Power dissipated in the load RL,

P D I22RL D 20 103224D 9600 106 W D 9.6mW

Now try the following exercise

Exercise 121 Further problems onresistance matching1 A transformer having a turns ratio of

8:1 supplies a load of resistance 50.Determine the equivalent input resistance ofthe transformer. [3.2 k]

2 What ratio of transformer is required to makea load of resistance 30 appear to have aresistance of 270? [3:1]

3 Determine the optimum value of loadresistance for maximum power transfer if theload is connected to an amplier of outputresistance 147 through a transformer witha turns ratio of 7:2 [12]

4 A single-phase, 240V/2880V ideal trans-former is supplied from a 240V sourcethrough a cable of resistance 3. If theload across the secondary winding is 720determine (a) the primary current owing and

(b) the power dissipated in the load resistance.[(a) 30A (b) 4.5 kW]

5 A load of resistance 768 is to be matchedto an amplier which has an effective outputresistance of 12. Determine the turns ratioof the coupling transformer. [1:8]

6 An a.c. source of 20V and internal resistance20 k is matched to a load by a 16:1 single-phase transformer. Determine (a) the value ofthe load resistance and (b) the power dissi-pated in the load.

[(a) 78.13 (b) 5mW]

21.11 Auto transformersAn auto transformer is a transformer which haspart of its winding common to the primary andsecondary circuits. Fig. 21.14(a) shows the circuitfor a double-wound transformer and Fig. 21.14(b)that for an auto transformer. The latter shows thatthe secondary is actually part of the primary, thecurrent in the secondary being (I2 I1). Sincethe current is less in this section, the cross-sectionalarea of the winding can be reduced, which reducesthe amount of material necessary.

Figure 21.14

Figure 21.15 shows the circuit diagram symbolfor an auto transformer.

Figure 21.15

Problem 26. A single-phase autotransformer has a voltage ratio 320V:250Vand supplies a load of 20 kVA at 250V.Assuming an ideal transformer, determinethe current in each section of the winding.

Rating D 20 kVA D V1I1 D V2I2.

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320 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

Hence primary current,

I1 D 20 103

V1D 20 10

3

320D 62.5A

and secondary current,

I2 D 20 103

V2D 20 10

3

250D 80A

Hence current in common part of the windingD 80 62.5 D 17.5A

The current owing in each section of thetransformer is shown in Fig. 21.16

Figure 21.16

Saving of copper in an auto transformerFor the same output and voltage ratio, the autotransformer requires less copper than an ordinarydouble-wound transformer. This is explained below.

The volume, and hence weight, of copper requiredin a winding is proportional to the number of turnsand to the cross-sectional area of the wire. In turnthis is proportional to the current to be carried, i.e.volume of copper is proportional to NI.

Volume of copper in an auto transformer/ N1 N2I1 C N2I2 I1

see Fig. 21.14(b)/ N1I1 N2I1 C N2I2 N2I1/ N1I1 C N2I2 2N2I1/ 2N1I1 2N2I1 (since N2I2 D N1I1

Volume of copper in a double-wound transformer/ N1I1 C N2I2 / 2N1I1

(again, since N2I2 D N1I1). Hencevolume of copper inan auto transformer

volume of copper in adouble-wound transformer

D 2N1I1 2N2I12N1I1

D 2N1I12N1I1

2N2I12N1I1

D 1 N2N1

If N2/N1 D x then

(volume of copper in an auto transformer)= .1 x/ (volume of copper in a double-

wound transformer) .12/

If, say, x D 4/5 then (volume of copper in autotransformer)

D (1 45) (volume of copper in a

double-wound transformer)D 15 (volume in double-wound transformer)

i.e. a saving of 80%.Similarly, if x D 1/4, the saving is 25 per cent,

and so on. The closer N2 is to N1, the greater thesaving in copper.

Problem 27. Determine the saving in thevolume of copper used in an auto transformercompared with a double-wound transformerfor (a) a 200V:150V transformer, and (b) a500V:100V transformer.

(a) For a 200V:150V transformer,

x D V2V1

D 150200

D 0.75Hence from equation (12), (volume of copper inauto transformer)

D 1 0.75 (volume of copper indouble-wound transformer)D 0.25 (volume of copper indouble-wound transformer)D 25% (of copper in adouble-wound transformer)Hence the saving is 75%

(b) For a 500V:100V transformer,

x D V2V1

D 100500

D 0.2

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TRANSFORMERS 321

Hence, (volume of copper in auto transformer)

D 1 0.2 (volume of copper indouble-wound transformer)D 0.8 (volume in double-wound transformer)D 80% of copper in a double-wound transformerHence the saving is 20%.

Now try the following exercise

Exercise 122 Further problems on theauto-transformer1 A single-phase auto transformer has a volt-

age ratio of 480V:300V and supplies a loadof 30 kVA at 300V. Assuming an ideal trans-former, calculate the current in each section ofthe winding.[I1 D 62.5A, I2 D 100A, (I2 I1 D 37.5A]

2 Calculate the saving in the volume ofcopper used in an auto transformer comparedwith a double-wound transformer for (a)a 300V:240V transformer, and (b) a400V:100V transformer [(a) 80% (b) 25%]

Advantages of auto transformersThe advantages of auto transformers over double-wound transformers include:

1 a saving in cost since less copper is needed (seeabove)

2 less volume, hence less weight3 a higher efciency, resulting from lower I2R

losses4 a continuously variable output voltage is achiev-

able if a sliding contact is used5 a smaller percentage voltage regulation.

Disadvantages of auto transformersThe primary and secondary windings are not electri-cally separate, hence if an open-circuit occurs in thesecondary winding the full primary voltage appearsacross the secondary.

Uses of auto transformersAuto transformers are used for reducing the voltagewhen starting induction motors (see Chapter 23) and

for interconnecting systems that are operating atapproximately the same voltage.

21.12 Isolating transformersTransformers not only enable current or voltageto be transformed to some different magnitudebut provide a means of isolating electrically onepart of a circuit from another when there isno electrical connection between primary andsecondary windings. An isolating transformer isa 1:1 ratio transformer with several importantapplications, including bathroom shaver-sockets,portable electric tools, model railways, and so on.

21.13 Three-phase transformersThree-phase double-wound transformers are mainlyused in power transmission and are usually of thecore type. They basically consist of three pairsof single-phase windings mounted on one core, asshown in Fig. 21.17, which gives a considerablesaving in the amount of iron used. The primary andsecondary windings in Fig. 21.17 are wound on topof each other in the form of concentric cylinders,similar to that shown in Fig. 21.6(a). The windingsmay be with the primary delta-connected and thesecondary star-connected, or star-delta, star-star ordelta-delta, depending on its use.

A delta-connection is shown in Fig. 21.18(a) anda star-connection in Fig. 21.18(b).

Problem 28. A three-phase transformer has500 primary turns and 50 secondary turns. Ifthe supply voltage is 2.4 kV nd thesecondary line voltage on no-load when thewindings are connected (a) star-delta, (b)delta-star.

(a) For a star-connection, VL Dp

3Vp (see Chap-ter 20). Primary phase voltage,

Vp D VL1p3 D2400p

3D 1385.64 volts.

For a delta-connection, VL D Vp. N1/N2 DV1/V2 from which, secondary phase voltage,

Vp2 D Vp1(

N2N1

)D 1385.64

( 50500

)

D 138.6 volts

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322 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

Figure 21.17

Figure 21.18

(b) For a delta-connection, VL D Vp hence, primaryphase voltage Vp1 D 2.4 kV D 2400 volts.Secondary phase voltage,

Vp2 D Vp1(

N2N1

)D 2400

( 50500

)D 240 volts

For a star-connection, VL Dp

3Vp hence, thesecondary line voltage, VL2 D

p3240

D 416 volts.

Now try the following exercise

Exercise 123 A further problem on thethree-phase transformer1 A three-phase transformer has 600 primary

turns and 150 secondary turns. If the supplyvoltage is 1.5 kV determine the secondary linevoltage on no-load when the windings areconnected (a) delta-star (b) star-delta

[(a) 649.5V (b) 216.5V]

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TRANSFORMERS 323

21.14 Current transformersFor measuring currents in excess of about 100Aa current transformer is normally used. With ad.c. moving-coil ammeter the current required togive full scale deection is very small typicallya few milliamperes. When larger currents are to bemeasured a shunt resistor is added to the circuit (seeChapter 10). However, even with shunt resistorsadded it is not possible to measure very largecurrents. When a.c. is being measured a shunt cannotbe used since the proportion of the current whichows in the meter will depend on its impedance,which varies with frequency.

In a double-wound transformer:I1I2

D N2N1

from which,

secondary current I2 = I1(N2

N1

)

In current transformers the primary usually consistsof one or two turns whilst the secondary can haveseveral hundred turns. A typical arrangement isshown in Fig. 21.19

Figure 21.19

If, for example, the primary has 2 turns and thesecondary 200 turns, then if the primary current is500A,

secondary current, I2 D I1(

N2N1

)D 500

(2

200

)

D 5ACurrent transformers isolate the ammeter from the

main circuit and allow the use of a standard range

of ammeters giving full-scale deections of 1A, 2Aor 5A.

For very large currents the transformer core canbe mounted around the conductor or bus-bar. Thusthe primary then has just one turn.

It is very important to short-circuit the secondarywinding before removing the ammeter. This isbecause if current is owing in the primary,dangerously high voltages could be induced in thesecondary should it be open-circuited.

Current transformer circuit diagram symbols areshown in Fig. 21.20

Figure 21.20

Problem 29. A current transformer has asingle turn on the primary winding and asecondary winding of 60 turns. Thesecondary winding is connected to anammeter with a resistance of 0.15. Theresistance of the secondary winding is0.25. If the current in the primary windingis 300A, determine (a) the reading on theammeter, (b) the potential difference acrossthe ammeter and (c) the total load (in VA) onthe secondary.

(a) Reading on the ammeter,

I2 D I1(

N1N2

)D 300

(160

)D 5A.

(b) P.d. across the ammeter D I2RA, (where RA is theammeter resistance D 50.15 D 0.75 volts.(c) Total resistance of secondary circuit D

0.15 C 0.25 D 0.40.Induced e.m.f. in secondaryD 50.40D2.0V.Total load on secondary D 2.05 D 10VA.

Now try the following exercise

Exercise 124 A further problem on thecurrent transformer1 A current transformer has two turns on the

primary winding and a secondary winding of

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324 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

260 turns. The secondary winding is connectedto an ammeter with a resistance of 0.2.The resistance of the secondary winding is0.3. If the current in the primary windingis 650A, determine (a) the reading on theammeter, (b) the potential difference acrossthe ammeter, and (c) the total load in VA onthe secondary [(a) 5A (b) 1V (c) 7.5VA]

21.15 Voltage transformers

For measuring voltages in excess of about 500V itis often safer to use a voltage transformer. Theseare normal double-wound transformers with a largenumber of turns on the primary, which is connectedto a high voltage supply, and a small number of turnson the secondary. A typical arrangement is shownin Fig. 21.21

Figure 21.21

SinceV1V2

D N1N2

the secondary voltage,

V2 D V1N2V1

Thus if the arrangement in Fig. 21.21 has 4000primary turns and 20 secondary turns then for avoltage of 22 kV on the primary, the voltage on thesecondary,

V2 D V1(

N2N1

)D 22 000

( 204000

)D 110 volts

Now try the following exercises

Exercise 125 Short answer questions ontransformers1 What is a transformer?2 Explain briey how a voltage is induced in

the secondary winding of a transformer3 Draw the circuit diagram symbol for a

transformer4 State the relationship between turns and volt-

age ratios for a transformer5 How is a transformer rated?6 Briey describe the principle of operation of

a transformer7 Draw a phasor diagram for an ideal trans-

former on no-load8 State the e.m.f. equation for a transformer9 Draw an on-load phasor diagram for an ideal

transformer with an inductive load10 Name two types of transformer construction11 What core material is normally used for

power transformers12 Name three core materials used in r.f. trans-

formers13 State a typical application for (a) a.f. trans-

formers (b) r.f. transformers14 How is cooling achieved in transformers?15 State the expressions for equivalent resis-

tance and reactance of a transformer, referredto the primary

16 Dene regulation of a transformer17 Name two sources of loss in a transformer18 What is hysteresis loss? How is it minimised

in a transformer?19 What are eddy currents? How may they be

reduced in transformers?20 How is efciency of a transformer calcu-

lated?21 What is the condition for maximum ef-

ciency of a transformer?22 What does resistance matching mean?

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TRANSFORMERS 325

23 State a practical application where matchingwould be used

24 Derive a formula for the equivalent resistanceof a transformer having a turns ratio ofN1 : N2 and load resistance RL

25 What is an auto transformer?

26 State three advantages and one disadvantageof an auto transformer compared with adouble-wound transformer

27 In what applications are auto transformersused?

28 What is an isolating transformer? Give twoapplications

29 Describe briey the construction of a three-phase transformer

30 For what reason are current transformersused?

31 Describe how a current transformer operates

32 For what reason are voltage transformersused?

33 Describe how a voltage transformer operates

Exercise 126 Multi-choice questions ontransformers (Answers on page 376)1 The e.m.f. equation of a transformer of

secondary turns N2, magnetic ux densityBm, magnetic area of core a, and operatingat frequency f is given by:(a) E2 D 4.44N2Bmaf volts

(b) E2 D 4.44N2Bmfa

volts

(c) E2 D N2Bmfa

volts

(d) E2 D 1.11N2Bm a f volts2 In the auto-transformer shown in Fig. 21.22,

the current in section PQ is:(a) 3.3A (b) 1.7A (c) 5A (d) 1.6A

3 A step-up transformer has a turns ratio of10. If the output current is 5A, the inputcurrent is:(a) 50A (b) 5A (c) 2.5A (d) 0.5A

V1

V2

1.7A

3.3A

Q

p

Figure 21.22

4 A 440V/110V transformer has 1000 turns onthe primary winding. The number of turns onthe secondary is:(a) 550 (b) 250 (c) 4000 (d) 25

5 An advantage of an auto-transformer is that:(a) it gives a high step-up ratio(b) iron losses are reduced(c) copper loss is reduced(d) it reduces capacitance between turns6 A 1 kV/250V transformer has 500 turns on

the secondary winding. The number of turnson the primary is:(a) 2000 (b) 125 (c) 1000 (d) 250

7 The core of a transformer is laminated to:(a) limit hysteresis loss(b) reduce the inductance of the windings(c) reduce the effects of eddy current loss(d) prevent eddy currents from occurring

8 The power input to a mains transformer is200W. If the primary current is 2.5A, thesecondary voltage is 2V and assuming nolosses in the transformer, the turns ratio is:(a) 40:1 step down (b) 40:1 step up(c) 80:1 step down (d) 80:1 step up

9 A transformer has 800 primary turns and 100secondary turns. To obtain 40 V from thesecondary winding the voltage applied to theprimary winding must be:(a) 5V (b) 320V(c) 2.5V (d) 20VA 100 kVA, 250V/10 kV, single-phase trans-former has a full-load copper loss of 800Wand an iron loss of 500W. The primary wind-ing contains 120 turns. For the statements in

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326 ELECTRICAL AND ELECTRONIC PRINCIPLES AND TECHNOLOGY

questions 10 to 16, select the correct answerfrom the following list:(a) 81.3 kW (b) 800W (c) 97.32%(d) 80 kW (e) 3 (f) 4800(g) 1.3 kW (h) 98.40% (i) 100 kW(j) 98.28% (k) 200W (l) 101.3 kW(m) 96.38% (n) 400W

10 The total full-load losses11 The full-load output power at 0.8 power factor12 The full-load input power at 0.8 power factor13 The full-load efciency at 0.8 power factor14 The half full-load copper loss15 The transformer efciency at half full-load,

0.8 power factor16 The number of secondary winding turns17 Which of the following statements is false?

(a) In an ideal transformer, the volts per turnare constant for a given value of primaryvoltage

(b) In a single-phase transformer, the hystere-sis loss is proportional to frequency

(c) A transformer whose secondary current isgreater than the primary current is a step-up transformer

(d) In transformers, eddy current loss isreduced by laminating the core

18 An ideal transformer has a turns ratio of 1:5and is supplied at 200V when the primarycurrent is 3 A. Which of the following state-ments is false?

(a) The turns ratio indicates a step-up trans-former

(b) The secondary voltage is 40V(c) The secondary current is 15A(d) The transformer rating is 0.6 kVA(e) The secondary voltage is 1 kV(f) The secondary current is 0.6A

19 Iron losses in a transformer are due to:(a) eddy currents only(b) ux leakage(c) both eddy current and hysteresis losses(d) the resistance of the primary and secondary

windings

20 A load is to be matched to an amplierhaving an effective internal resistance of 10via a coupling transformer having a turnsratio of 1:10. The value of the load resistancefor maximum power transfer is:(a) 100 (b) 1 k(c) 100m (d) 1m

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