INTRODUCTION TO TRANSFORMERS Introduction A transformer is a device which uses the phenomenon of mutual induction to change the values of alternating voltages and currents. In fact, one of the main advantages of a.c. transmission and distribution is the ease with which an alternating voltage can be increased or decreased by transformers. Losses in transformers are generally low and thus efficiency is high. Being static, they have a long life and are very stable. 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 1(a) as consisting of two electrical 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 1(b) S1 I1 + + E1 - E2 - N1 a t = N N S2 I2 N2 1 (b)20.2 Transformer principle Figure 1: A Basic single-phase two-winding transformer and its schematic equivalent circuit 2 Transformer principle of operation When the secondary is an open-circuit and an alternating voltage V1 is applied to the primary winding, a small current called the no-load current I0 – flows, which sets up a magnetic flux in the core. This alternating flux links with both primary and secondary coils and induces in them e.m.f.s of E1 and E2, respectively, by mutual induction. The induced e.m.f. E in a coil of N turns is given by where dΦ/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 E1/N1=E2/N2, i.e. the induced e.m.f. per turn is constant. Assuming no losses, E1=V1 and E2=V2. Hence V1/V2 is called the voltage ratio and N1/N2 the turns ratio, or the ‘transformation ratio’ of the transformer. If N2 is less than N1 then V2 is less than V1 and the device is termed a step-down transformer. If N2 is greater, then N1 then V2 is greater than V1 and the device is termed a step-up transformer. When a load is connected across the secondary winding, a current I2 flows. In an ideal transformer losses are neglected and a transformer is considered to be 100% efficient. Hence input power = output power, or V1 I1 = V2 I2, i.e. in an ideal transformer, the primary and secondary volt-amperes are equal. ππ πΌπΌ Thus, 1 = 2 (2) ππ2 πΌπΌ1 Combining equations (1) and (2) gives: ππ1 ππ1 πΌπΌ2 = = 3 ππ2 ππ2 πΌπΌ1 The rating of a transformer is stated in terms of the volt-amperes that it can transform without overheating. With reference to 1(a), the transformer rating is either V1I1 or V2I2, where I2 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. Solution For an ideal transformer, voltage ratio = turns ratio, i.e ππ1 ππ1 240 500 = , βππππππππ = ππ2 3000 ππ2 ππ2 Thus secondary voltage ππ2 = 3000 240 500 = 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. Solution 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, ππ1 2 = 7 ππ2 ππ ππ 2 240 For an ideal transformer, 1 = 1 ; βππππππππ = ππ2 ππ2 Thus the secondary voltage ππ2 = 7 240 2 7 = 840 V ππ2 Problem 3. An ideal transformer has a turns ratio of 8:1 and the primary current is 3A when it is supplied at 240 V. Calculate the secondary voltage and current. Solution ππ 8 A turns ratio of 8:1 means, 1 = , i.e. a step-down transformer. ππ2 Also, ππ1 ππ2 = 1 ππ1 ππ1 ππ2 = ππππ π π π π π π π π π π π π π π π π π π π£π£π£π£π£π£π£π£π£π£π£π£π£π£ ππ2 = ππ1 ππ2 ππ2 ππ1 πΌπΌ2 ; βππππππππ πΌπΌ1 π π π π π π π π π π π π π π π π π π ππππππππππππππ πΌπΌ2 = πΌπΌ1 ππ1 ππ2 = 3 8 1 1 = 240 8 = ππππ π¨π¨ = 30 ππππππππππ 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. Solution ππ 150 = 12.5 π΄π΄ ππ1 = 240 ππ, ππ2 = 12 ππ, πΌπΌ2 = = ππ2 12 ππ1 ππ1 240 = ππππ ππππππππππ ππππππππππ = = = 12 ππ2 ππ2 ππ1 πΌπΌ2 ππ2 12 = , ππππππππ π€π€π€π€π€π€π€π€, πΌπΌ1 = πΌπΌ2 = 12.5 240 ππ2 πΌπΌ1 ππ1 Hence, current taken from the supply, πΌπΌ1 = 12.5 20 = ππ. ππππππ π¨π¨ 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 and (c) the primary current at full load kVA Solution ππ1 10 (a) = ππππππ ππ1 = 2.5 ππππ = 2500 ππ ππ2 1 ππ1 ππ1 ππ2 1 = 250 ππ = , π π π π π π π π π π π π π π π π π π π£π£π£π£π£π£π£π£π£π£π£π£π£π£ ππ2 = ππ1 = 2500 10 ππ2 ππ2 ππ1 The transformer rating in volt-amperes = V2 I2 (at full load), i.e. 5000 = 250 I2 500 Hence full load secondary current πΌπΌ2 = = ππππππ ππππππππππ (b) Minimum value of load resistance, (c) ππ1 ππ2 = πΌπΌ2 , from πΌπΌ1 250 ππ RL = 2 πΌπΌ2 = which primary current, πΌπΌ1 = 250 = ππππ. ππ ππ 20 ππ 1 πΌπΌ2 2 = 20 ππ1 10 = ππ π¨π¨ Transformer no-load phasor diagram 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 I0 and since, with losses neglected, the primary winding is a pure inductor, this current lags the applied voltage V1 by 90β¦. In the phasor diagram, assuming no losses, shown in Figure 2(a), current I0 produces the flux and is drawn in phase with the flux. The primary induced e.m.f. E1 is in phase opposition to V1 (by Lenz’s law) and is shown 180β¦ out of phase with V1 and equal in magnitude. The secondary induced e.m.f. is shown for a 2:1 turns ratio transformer. Figure 2 (ii) A no-load phasor diagram for a practical transformer is shown in Figure 2(b). If current flows then losses will occur. When losses are considered then the no-load current I0 is the phasor sum of two components – (a) IM, the magnetizing component, in phase with the flux, and (b) IC, the core loss component (supplying the hysteresis and eddy current losses). From Figure 2(b): No-load current,πΌπΌ0 = 2 πΌπΌππ + πΌπΌπΆπΆ2 , where π°π°π΄π΄ = π°π°ππ ππππππ ππππ ππππππ π°π°πͺπͺ = π°π°ππ ππππππ ππππ Power factor on no-load = ππππππ ππππ = π°π°πͺπͺ π°π°ππ The total core losses (i.e. iron losses) = ππ1 π°π°ππ ππππππ ππππ Problem 6. A 2400V/400V single-phase transformer takes a no-load current of 0.5A and the core loss is 400W. Determine the values of the magnetizing and core loss components of the no-load current. Draw to scale the noload phasor diagram for the transformer. Solution V1 = 2400 V, V2 = 400 V, I0 = 0.5A Core loss (i.e. iron loss) = 400 = V1 I0 ππππππ ππππ = (2400)(0.5) ππππππ ππππ 400 Hence πΆπΆπΆπΆπΆπΆ∅0 = = 0.3333 2400 0.5 ∅0 = Cos −1 0.3333 = 70.530 The no-load phasor diagram is shown in Figure 3. Magnetizing component, IM = I0 Sinφ0 = 0.5Sin70.53β¦ = 0.471A Core loss component, IC = I0 Cosφ0 = 0.5Cos70.53β¦ = 0.167A Figure 3: Phasor diagram Problem 7. A transformer takes a current of 0.8A 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 noload and (c) the magnetizing current. Solution I0 = 0.8 A, V1 = 240 V (a) Power absorbed = total core loss = 72 = V1 I0 cosφ0 Hence 72 = 240 I0 Cosφ0 and iron loss current, IC = I0 cosφ0 = 72/240 = 0.30A (b) Power factor at no load, cosφ0= IC/I0 = 0.30/0.80 = 0.375 (c) From the right-angled triangle in Figure 2(b) and using Pythagoras’ theorem, from which magnetizing current πΌπΌ0 = 2 πΌπΌππ + πΌπΌπΆπΆ2 , πΌπΌππ = πΌπΌ02 − πΌπΌπΆπΆ2 = 0.802 + 0.302 = 0.74A E.m.f. equation of a transformer The magnetic flux ππ set up in the core of a transformer when an alternating voltage is applied to its primary winding is also alternating and is sinusoidal. Let ππm 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 =1/f seconds. The flux rises sinusoidally from zero to its maximum value in 1/4 cycle, and the time for 1/4 cycle is 1/4 f seconds. ππ Hence the average rate of change of flux = ππ = 4ππππππ , and since 1Wb/s = 1 volt, the average e.m.f. 1/4ππ induced in each turn = 4ππππππ 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 = r.m.s. value average value = 1.11 Hence r.m.s. value = form factor × average value = 1.11×average value Thus r.m.s. e.m.f. induced in each turn = 1.11×4 f _m volts = 4.44 f ππππ volts Therefore, r.m.s. value of e.m.f. induced in primary, E1 = 4.44 fππππ N1 volts (4) and r.m.s. value of e.m.f. induced in secondary, E2 = 4.44 fππππ N2 volts (5) Dividing equation (4) by equation (5) gives: πΈπΈ1 ππ1 = , ππππ ππππππππππππππππππππ ππππππππππππππππ πΈπΈ2 ππ2 Problem 8. A 100 kVA, 4000 V/200 V, 50 Hz single-phase transformer 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. Solution V1 = 4000 V, V2 = 200 V, f = 50 Hz, N2 = 100 turns (a) Transformer rating= V1 I1 = V2 I2 = 100 000 VA Hence primary current, I1 = 100000/V1 = 100000/4000 = 25A and Secondary current, I2 = 100000/V2 = 100000/200 = 500 A (b) From ππ1 ππ2 = ππ1 , ππ2 primary turns ππ1 = ππ2 ππ1 ππ2 (c) From equation (5), E2 = 4.44 f ππππ N2, πΈπΈ 200 maximum flux ππππ = 4.44 2ππππ = ππππ ππππππ 1 = 100 ππ.ππππ 4000 200 = ππππππππ ππππππππππ ππππππππππππππππ πΈπΈ1 = πΈπΈ2 = 9.01×10−3 Wb or 9.01 mWb 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 cm2. 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. Solution (a) From equation (4), e.m.f. E1 = 4.44 f ππππ N1 volts. This implies 250 = 4.44(50) ππππ (25) Hence, maximum flux density, ππππ = 250 (4.44)(50)(25) ππππ = 0.04505 ππππ However, ππππ = Bm × A, where Bm = maximum flux density in the core and A = cross-sectional area of the core. Hence Bm × 300 × 10−4 = 0.04505 from which, maximum flux density, π΅π΅ππ = 0.04505 300 ∗10−4 = 1.50ππ (b) ππ1 ππ2 = ππ1 , ππ2 from which, voltage induced in the secondary winding ππ2 = ππ1 ππ2 ππ1 300 ππ2 = 250 = ππππππππ π½π½ ππππ ππππππ 25 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 cm2. Determine the number of primary and secondary turns Solution The e.m.f. equation for a transformer is E =4.44 f ππππ N and maximum flux, ππππ = B × A = (1.5)(50 ×10−4) = 75 × 10−4 Wb Since E1 = 4.44 f ππππ N1 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 density of 1.4 T. Calculate (a) the number of primary and secondary turns and (b) the crosssectional area of the core. Solution Transformer on-load phasor diagram If the voltage drop in the windings of a transformer are assumed negligible, then the terminal voltage V2 is the same as the induced e.m.f. E2 in the secondary. Similarly, V1 = E1. Assuming an equal number of turns on primary and secondary windings, then E1 E2, and let the load have a lagging phase angle φ2. In the phasor diagram of Figure 4, current I2 lags V2 by angle φ2. When a load is connected across the secondary winding a current I2 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 hap- pen since reduction of the core flux reduces E1, hence a reflected increase in primary current I1÷ occurs which provides a restoring mmf. Hence at all loads, primary and secondary mmfs are equal, but in opposition, and the core flux remains constant. I1÷ is sometimes called the ‘balancing’ current and is equal, but in the opposite direction, to current I2, as shown in Figure 4. I0, shown at a phase angle φ0 to V1, is the no-load current of the transformer. The phasor sum of I1÷ and I0 gives the supply current I1, and the phase angle between V1 and I1 is shown as φ1 Figure 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. Solution Let πΌπΌ1′ be the component of the primary current which provides the restoring mmf. Then In the phasor diagram shown in Figure 5, I2=100 A is shown at an angle of φ2=31.8β¦ to V2 and πΌπΌ1′ = 40A is shown in anti-phase to I2 Figure 5 Transformer construction There are broadly two types of single-phase double-wound transformer constructions – the core type and the shell type, as shown in Figure 6. The low- and high-voltage windings are wound as shown to reduce leakage flux Figure 6 (i) For power transformers, rated possibly at several MVA and operating at a frequency of 50 Hz, 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. (ii) 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. (vi) Radio frequency (r.f.) transformers, operating in the MHz frequency region, have either an air core, a ferrite core or a dust core. Ferrite is a ceramic material having magnetic properties similar to silicon steel with a high resistivity. Dust cores consist of fine particles of carbonyl iron or permalloy (i.e. nickel and iron), each particle of which is insulated from its neighbour. Radio frequency transformers are used in radio and television receivers. (vi) Transformer windings are usually of enamel-insulated copper or aluminium. (vii) Cooling is achieved by air in small transformers and oil in large transformers. Equivalent circuit of a Transformer Figure 7 shows an equivalent circuit of a transformer. R1 and R2 represent the resistances of the primary and secondary windings and X1 and X2 represent the reactances of the primary and secondary windings, due to leakage flux. The core losses due to hysteresis and eddy currents are allowed for by resistance R which takes a current IC, the core loss component of the primary current. Reactance X takes the magnetizing component IM. In a simplified equivalent circuit shown in Figure 8, R and X are omitted since the no-load current I0 is normally only about 3–5% of the full load primary current. It is often convenient to assume that all of the resistance and reactance is on one side of the transformer. Resistance R2 in Figure 8 can be replaced by inserting an additional resistance π π 2′ in the primary circuit such that the power absorbed in π π 2′ when carrying the primary current is equal to that in R2 due to the secondary current, i.e. πΌπΌ12 π π 2′ = πΌπΌ22 π π 2′ . Thus, π π 2′ = πΌπΌ22 π π 2 πΌπΌ2 1 2 = ππ12 π π 2 ππ 2 2 2 Then the total equivalent resistance in the primary circuit Re is equal to the primary and secondary resistances of the actual transformer. Hence 2 2 ππ 1 π π ππ = π π 1 + π π 2′ = π π 1 + π π 2 2 ππ2 Figure 8 By similar reasoning, the equivalent reactance in the primary circuit is given by ππππ = ππ1 + 2 ′ ππ1 ππ2 ππ 2 2 2 (7) The equivalent impedance Ze of the primary and secondary windings referred to the primary is given by: π π ππ2 + ππππ2 (8) ππππ = If φe is the phase angle between I1 and the volt drop I1 Ze then π π πΆπΆπΆπΆπΆπΆπΆπΆππ = ππ ππππ (9) The simplified equivalent circuit of a transformer is shown in Figure 9. Figure 9 Problem 13. A transformer has 600 primary turns and 150 secondary turns. The primary and secondary resistances are 0.25 β¦ and 0.01β¦, respectively, and the corresponding leakage reactances are 1.0 β¦ and 0.04 β¦, respectively. Determine (a) the equivalent resistance referred to the primary winding, (b) the equivalent reactance referred to the primary winding, (c) the equivalent impedance referred to the primary winding and (d) the phase angle of the impedance. Solution Regulation of a transformer When the secondary of a transformer is loaded, the secondary terminal voltage, V2, falls. As the power factor decreases, this voltage drop increases. This is called the regulation of the transformer and it is usually expressed as a percentage of the secondary no-load voltage, E2. For full-load conditions. πΈπΈ2 − ππ2 π π π π π π π π π π π π π π π π π π π π = π₯π₯ 100% 10 πΈπΈ2 The fall in voltage, (E2−V2), is caused by the resistance and reactance of the windings. Typical values of voltage regulation are about 3% in small transformers and about 1% in large transformers. Problem 14. A 5kVA, 200V/400V, single-phase transformer has a secondary terminal voltage of 387.6 volts when loaded. Determine the regulation of the transformer. Solution Problem 15. The open-circuit voltage of a transformer is 240V. A tap-changing device is set to operate when the percentage regulation drops below 2.5%. Determine the load voltage at which the mechanism operates. Solution Transformer losses and efficiency There are broadly two sources of losses in transformers on load, viz: copper losses and iron losses. (A) Copper losses are variable and result in a heating of the conductors, due to the fact that they possess resistance. If R1 and R2 are the primary and secondary winding resistances then the total copper loss is I 2 R1 + I 2 R2 (B) Iron losses are constant for a given value of frequency and flux density and are of two types – hysteresis loss and eddy current loss. (i) Hysteresis loss is the heating of the core as a result of the internal molecular structure reversals which occur as the magnetic flux alternates. The loss is proportional to the area of the hysteresis loop and thus low loss nickel iron alloys are used for the core since their hysteresis loops have small areas. (ii) Eddy current loss is the heating of the core due to e.m.f.s being induced not only in the transformer windings but also in the core. These induced e.m.f.s set up circulating currents, called eddy currents. Owing to the low resistance of the core, eddy currents can be quite considerable and can cause a large power loss and excessive heating of the core. Eddy current losses can be reduced by increasing the resistivity of the core material or, more usually, by laminating the core (i.e. splitting it into layers or leaves) when very thin layers of insulating material can be inserted between each pair of laminations. This increases the resistance of the eddy cur- rent path, and reduces the value of the eddy current.m ππππππππππππ ππππππππππ ππππππππππ ππππππππππ − ππππππππππππ = ππππππππππππππππππππππ πΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈπΈ, ππ % = ππππππππππ ππππππππππ ππππππππππ ππππππππππ ππππππππππππ 11 = 1 − ππππππππππ ππππππππππ It is not uncommon for power transformers to have efficiencies of between 95% and 98%. Output power= V2 I2 cosφ2 total losses = copper loss + iron losses and input power = output power + losses Problem 16. A 200 kVA rated transformer has a full-load copper loss of 1.5 kW and an iron loss of 1kW. Determine the transformer efficiency at full load and 0.85 power factor. Solution Problem 17. Determine the efficiency of the transformer in worked Problem 16 at half full load and 0.85 power factor. Solution Problem 18. A 400 kVA transformer has a primary winding resistance of 0.5 K and a secondary winding resistance of 0.001 K. The iron loss is 2.5 kW and the primary and secondary voltages are 5 kV and 320 V, respectively. If the power factor of the load is 0.85, determine the efficiency of the transformer (a) on full load, and (b) on half load. Solution Maximum efficiency It may be shown that the efficiency of a transformer is a maximum when the variable copper loss (i.e. I 2 R1 + I 2 R2) is equal to the constant iron losses. Problem 19. A 500 kVA transformer has a full load copper loss of 4 kW and an iron loss of 2.5 kW. Determine (a) the output kVA at which the efficiency of the transformer is a maximum, and (b) the maximum efficiency, assuming the power factor of the load is 0.75 Solution Autotransformers An auto transformer is a transformer which has part of its winding common to the primary and secondary circuits. Figure 14(a) shows the circuit for a double- wound transformer and Figure 14(b) that for an auto transformer. The latter shows that the secondary is actually part of the primary, the current in the secondary being (I2 – I1). Since the current is less in this section, the cross-sectional area of the winding can be reduced, which reduces the amount of material necessary. Figure 14 Figure 15 shows the circuit diagram symbol for an auto transformer. Figure 15: Circuit diagram symbol for an auto transformer Advantages of auto transformers The advantages of auto transformers over double-wound transformers include: 1. A saving in cost since less copper is needed (see above) 2. Less volume, hence less weight 3. A higher efficiency, resulting from lower I 2 R losses 4. A continuously variable output voltage is achievable if a sliding contact is used 5. A smaller percentage voltage regulation Disadvantages of auto transformers The primary and secondary windings are not electrically separate, hence if an open-circuit occurs in the secondary winding the full primary voltage appears across the secondary. Uses of auto transformers Auto transformers are used for reducing the voltage when starting induction motors and for interconnecting systems that are operating at approximately the same voltage Problem 20. A single-phase auto transformer has a voltage ratio 320 V: 250 V and supplies a load of 20 kVA at 250 V. Assuming an ideal transformer, determine the current in each section of the winding. Solution The current flowing in each section of the transformer is shown in Figure 16. Figure 16 Three-phase transformers Transformers not only enable current or voltage to be transformed to some different magnitude, but provide a means of isolating electrically one part of a circuit from another when there is no electrical connection between primary and secondary windings. An isolating trans- former is a 1:1 ratio transformer with several important applications such as bathroom shaver-sockets, portable electric tools and model railways. Three-phase transformers Three-phase, double-wound transformers are mainly used in power transmission and are usually of the core type. They basically consist of three pairs of single- phase windings mounted on one core, as shown in Figure 17, which gives a considerable saving in the amount of iron used. The primary and secondary wind- ings in Figure 17 are wound on top of each other in the form of concentric cylinders, similar to that shown in Figure 6(a). The windings may be with the primary delta-connected and the secondary star-connected, or star–delta, star–star or delta–delta, depending on its use. Figure 20.17 A delta connection is shown in Figure 18(a) and a star connection in Figure 18(b) Problem 21. A three-phase transformer has 500 primary turns and 50 secondary turns. If the supply voltage is 2.4 kV, find the secondary line voltage on no-load when the windings are connected (a) star–delta, (b) delta–star. Solution (a) For a star connection, VL = 3Vp Current transformers For measuring currents in excess of about 100 A, a current transformer is normally used. With a d.c. movingcoil ammeter the current required to give full-scale deflection is very small – typically a few milli-amperes. When larger currents are to be measured a shunt resistor is added to the circuit. However, even with shunt resistors added it is not possible to measure very large currents. When a.c. is being measured a shunt cannot be used since the proportion of the current which flows in the meter will depend on its impedance, which varies with frequency. In current transformers, the primary usually consists of one or two turns whilst the secondary can have several hundred turns. A typical arrangement is shown in Figure 19. Figure 19 For example, if the primary has 2 turns and the secondary 200 turns, then if the primary current is 500A, Current transformers isolate the ammeter from the main circuit and allow the use of a standard range of ammeters giving full-scale deflections of 1A, 2A or 5A. For very large currents the transformer core can be mounted around the conductor or bus-bar. Thus the primary then has just one turn. It is very important to shortcircuit the secondary winding before removing the ammeter. This is because if current is flowing in the primary, dangerously high voltages could be induced in the secondary should it be open-circuited. Current transformer circuit diagram symbols are shown in Figure 20. Figure 20 Problem 22. A current transformer has a single turn on the primary winding and a secondary winding of 60 turns. The secondary winding is connected to an ammeter with a resistance of 0.15 K. The resistance of the secondary winding is 0.25 K. If the current in the primary winding is 300 A, determine (a) the reading on the ammeter, (b) the potential difference across the ammeter and (c) the total load (in VA) on the secondary. Solution Voltage transformers For measuring voltages in excess of about 500 V it is often safer to use a voltage transformer. These are normal double-wound transformers with a large number of turns on the primary, which is connected to a high voltage supply, and a small number of turns on the secondary. A typical arrangement is shown in Figure 21. Figure 21 Thus, if the arrangement in Figure 21 has 4000 primary turns and 20 secondary turns then for a voltage of 22 kV on the primary, the voltage on the secondary.
