Secondary turns, = 96 turns 2
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  CHAPTER 21 TRANSFORMERS
Exercise 118, Page 342 1. A transformer has 600 primary turns connected to a 1.5 kV supply. Determine the number of secondary turns for a 240 V output voltage, assuming no losses. For a transformer, 
from which, secondary turns, 2. An ideal transformer with a turns ratio 2:9 is fed from a 220 V supply. Determine its output voltage.
from which, output voltage,  = 990 V
3. A transformer has 800 primary turns and 2000 secondary turns. If the primary voltage is 160 V, determine the secondary voltage assuming an ideal transformer.  and 
from which, output voltage,  = 400 V
4. An ideal transformer with a turns ratio 3:8 has an output voltage of 640 V. Determine its input voltage.
from which, input voltage,  = 240 V
5. An ideal transformer has a turns ratio of 12:1 and is supplied at 192 V. Calculate the secondary voltage.  and 
from which, output voltage,  = 16 V
6. A transformer primary winding connected across a 415 V supply has 750 turns. Determine how many turns must be wound on the secondary side if an output of 1.66 kV is required. from which, secondary turns,  = 3000 turns
7. An ideal transformer has a turns ratio of 15:1 and is supplied at 180 V when the primary current is 4 A. Calculate the secondary voltage and current. ,  and 
from which, output voltage,  = 12 V
 from which, secondary current,  = 60 A
8. A step-down transformer having a turns ratio of 20:1 has a primary voltage of 4 kV and a load of 10 kW. Neglecting losses, calculate the value of the secondary current.  and 
from which, output voltage,  = 200 V
Secondary power = from which, secondary current, necessary to supply a 240 V load. If the load current is 3 A determine the primary current. Neglect any losses.
supply. Neglecting losses, determine (a) the full load secondary current, (b) the minimum value of load resistance which can be connected across the secondary winding without the kVA rating being exceeded, and (c) the primary current. 10000 =  , and 
(a) 10000 VA = (b) Load resistance, (c) transformer whose secondary voltage is 150 V. Calculate the primary voltage and the turns ratio if the supply current is 5 A, neglecting losses.
losses, determine (a) the full load secondary current, and (b) the rating of the transformer. (a)  from which, full load secondary current,  = 20 A
(b) Transformer rating =  = 2000 VA = 2 kVA
or transformer rating =  = 2000 VA = 2 kVA
2. A 3300 V/440 V, single-phase transformer takes a no-load current of 0.8 A and the iron loss is 500 W. Draw the no-load phasor diagram and determine the values of the magnetizing and core loss components of the no-load current.
Core or iron loss = 500 = from which, The no-load phasor diagram is shown below. 
Magnetizing component,  = 0.786 A
Core loss component,  = 0.152 A
3. A transformer takes a current of 1 A when its primary is connected to a 300 V, 50 Hz supply, the secondary being on open-circuit. If the power absorbed is 120 watts, calculate (a) the iron loss current, (b) the power factor on no-load, and (c) the magnetizing current.
(a) Power absorbed = total core loss = 120 = i.e. 120 = (300) and iron loss current, (b) Power factor on no-load, (c) By Pythagoras, magnetizing current,  = 0.917 A
Exercise 120, Page 346 1. A 60 kVA, 1600 V/100 V, 50 Hz, single-phase transformer has 50 secondary windings. Calculate (a) the primary and secondary current, (b) the number of primary turns and (c) the maximum value of the flux.
(a) Transformer rating = hence, primary current, and secondary current, (b) (c) maximum flux,  = 9.0 mWb
2. A single-phase, 50 Hz transformer has 40 primary turns and 520 secondary turns. The cross- sectional area of the core is 270 cm2. When the primary winding is connected to a 300 volt supply, determine (a) the maximum value of flux density in the core, and (b) the voltage induced in the secondary winding (a) From equation (4), e.m.f. E1 = 4.44 f m N1 volts i.e. 300 = 4.44 (50) m (40) from which, maximum flux density, m = Hence, Bm 270 10-4 = 0.033784 from which, maximum flux density, Bm = (b) i.e. voltage induced in the secondary winding, V2 = (300) and an effective cross-sectional area of 60 secondary windings.
= 464 turns  from which, secondary turns, 
= 58 turns 4. A 3.3 kV/110 V, 50 Hz, single-phase transformer is to have an approximate e.m.f. per turn of 22 V and operate with a maximum flux of 1.25 T. Calculate (a) the number of primary and secondary turns, and (b) the cross-sectional area of the core
Hence primary turns, N1 = and secondary turns, N2 = (b) E.m.f. E1 = 4.44 f m N1 from which, m =  =  = 0.0991 Wb
Now flux, m = Bm A, where A is the cross-sectional area of the core, hence area, A =  =  = 0.07928  or 792.8 
Exercise 121, Page 347 1. A single-phase transformer has 2400 turns on the primary and 600 turns on the secondary. Its no-load current is 4 A at a power factor of 0.25 lagging. Assuming the volt drop in the windings is negligible, calculate the primary current and power factor when the secondary current is 80 A at a power factor of 0.8 lagging. Let  be the component of the primary current which provides the restoring m.m.f.
Then i.e. If the power factor of the secondary is 0.8, then from which, If the power factor at no load is 0.25, then from which, In the phasor diagram shown below, shown anti-phase to 
The no load current, Current Total horizontal component, = (4)(0.25) + (20)(0.8) = 1 + 16 = 17 A Total vertical component, = (4)(sin 75.52) + (20)(sin 36.87) = 15.87 A Hence, magnitude of and Hence, power factor = cos = cos 43.03 = 0.73 Exercise 122, Page 350 1. A transformer has 1200 primary turns and 200 secondary turns. The primary and secondary resistances are 0.2 and 0.02 respectively and the corresponding leakage reactances are 1.2 and 0.05 respectively. Calculate (a) the equivalent resistance, reactance and impedance referred to the primary winding, and (b) the phase angle of the impedance. (a) Equivalent resistance,  = 0.92
Equivalent reactance, = 3.0 Equivalent impedance,  = 3.138 or 3.14
(b) Exercise 123, Page 350 1. A 6 kVA, 100 V/500 V, single-phase transformer has a secondary terminal voltage of 487.5 V when loaded. Determine the regulation of the transformer. Regulation = 
= 2. A transformer has an open circuit voltage of 110 volts. A tap-changing device operates when the regulation falls below 3%. Calculate the load voltage at which the tap-changer operates. Regulation =
Hence, 3 = from which, and Exercise 124, Page 352 1. A single-phase transformer has a voltage ratio of 6:1 and the h.v. winding is supplied at 540 V. The secondary winding provides a full load current of 30 A at a power factor of 0.8 lagging. Neglecting losses, find (a) the rating of the transformer, (b) the power supplied to the load, (c) the primary current.  and  hence,  = 90 V and = 30 A
(a) Rating of transformer = (b) Power supplied to load = V I cos = (2700)(0.8) since power factor = cos = 0.8 = 2.16 kW (c) 800 W and iron losses of 500 W. Determine the transformer efficiency at full load and 0.8 power factor.
Full-load output power = V I cos = (40)(0.8) = 32 kW Total losses = 800 + 500 = 1.3 kW Input power = output power + losses = 32 + 1.3 = 33.3 kW Hence, efficiency, Copper loss (or Hence, copper loss at half full load = Iron loss == 500 W (constant) Total loss = 200 + 500 = 700 W or 0.7 kW Input power at half full load = output power at half full load + losses = 16 + 0.7 = 16.7 kW Hence, efficiency, load copper loss of 1600 W. Calculate its efficiency for a load of 60 kW at 0.8 power factor. Efficiency = 
Full-load output power = V I cos = (100)(0.8) = 80 kW Load power = 60 kW Hence the transformer is at Hence, copper loss at 3/4 load = Total losses = 900 + 600 = 1.5 kW Input power = output power + losses = 60 + 1.5 = 61.5 kW Hence, efficiency, 5. Determine the efficiency of a 15 kVA transformer 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 factor. Assume that iron losses are 200 W and the full-load copper loss is 300 W
Losses = 200 + 300 = 500 W or 0.5 kW Input power at full load = output power + losses = 15 + 0.5 = 15.5 kW Hence, efficiency, (ii) At 0.8 full load, unity power factor, output power = 0.8 15 = 12 kW Losses = Input power at 0.8 full load = output power at 0.8 full load + losses = 12 + 0.392 = 12.392 kW Hence, efficiency, (iii) At 0.5 full load and 0.8 power factor, output power = 0.5 15 0.8= 6 kW Losses = Input power at 0.5 full load = output power at 0.5 full load + losses = 6 + 0.275 = 6.275 kW Hence, efficiency, 6. A 300 kVA transformer has a primary winding resistance of 0.4 and a secondary winding resistance of 0.0015 . The iron loss is 2 kW and the primary and secondary voltages are 4 kV and 200 V respectively. If the power factor of the load is 0.78, determine the efficiency of the transformer (a) on full load, and (b) on half load. (a) Rating = 300 kVA = V1 I1 = V2 I2 Hence primary current, I1 =  =  = 75 A
and secondary current, I2 = Total copper loss = I12 R1 + I22 R2, (where R1 = 0.4 and R2 = 0.0015 )
= (b) Since the copper loss varies as the square of the current, then total
Hence total loss on half load = 1406.25 + 2000 = 3406.25 W or 3.40625 kW Output power on half full load =  (234) = 117 kW
Input power on half full load = output power + losses = 117 kW + 3.40625 kW = 120.40625 kW Hence efficiency at half full load, =  100%
= 7. A 250 kVA transformer has a full load copper loss of 3 kW and an iron loss of 2 kW. Calculate (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 0.80.
The corresponding total copper loss = At maximum efficiency, copper loss = iron loss Hence, Thus, the output kVA at maximum efficiency = 0.8165 250 = 204.1 kVA (b) Total loss at maximum efficiency = 2 2 = 4 kW Output power = 204.1 0.8 = 163.3 kW Input power = output power + losses = 163.3 + 4 = 167.3 kW Hence, maximum efficiency, equivalent input resistance of the transformer. The equivalent input resistance, R1 =  =  (50) = 3200 Ω = 3.2 k
2. What ratio of transformer turns is required to make a load of resistance 30 appear to have a resistance of 270  i.e. 270 = 
from which, i.e. the turns ratio required is 3:1
connected to an amplifier of output resistance 147 through a transformer with a turns ratio of 7:2
R1 = 4. A single-phase, 240 V/2880 V ideal transformer is supplied from a 240 V source through a cable of resistance 3 . If the load across the secondary winding is 720 determine (a) the primary current flowing and (b) the power dissipated in the load resistance.
(a) Equivalent input resistance, Total input resistance, Hence, primary current, (b) Power dissipated in load, P =  = 4500 W or 4.5 kW
5. A load of resistance 768 is to be matched to an amplifier which has an effective output resistance of 12 . Determine the turns ratio of the coupling transformer. hence 12 =  and 
and 
Hence, the turns ratio of the coupling transformer is 1:8 6. An a.c. source of 20 V and internal resistance 20 k is matched to a load by a 16:1 single-phase transformer. Determine (a) the value of the load resistance and (b) the power dissipated in the load. The circuit is shown below. 
(a) For maximum power transfer, from which, load resistance,  = 78.13
(b) Total input resistance when source is connected to the matching transformer is i.e. 20 k + 20 k = 40 k Primary current, from which,  = 8 mA
Power dissipated in load, P =  = 5 mW
Exercise 126, Page 357 1. A single-phase auto transformer has a voltage ratio of 480 V:300V and supplies a load of 30 kVA at 300 V. Assuming an ideal transformer, calculate the current in each section of the winding.
Hence, primary current, and secondary current, Hence, current in common part of winding = 2. Calculate the saving in the volume of copper used in an auto transformer compared with a double-wound transformer for (a) a 300 V:240 V transformer, and (b) a 400 V:100 V transformer.
From equation (20.12), volume of copper in auto transformer = (1 – 0.80)(volume of copper in a double-wound transformer) = (0.20)(volume of copper in a double-wound transformer) Hence, saving is 80% (b) For a 400 V:1000 V transformer, x = From equation (20.12), volume of copper in auto transformer = (1 – 0.25)(volume of copper in a double-wound transformer) = (0.75)(volume of copper in a double-wound transformer) Hence, saving is 25% Exercise 127, Page 358 1. A three-phase transformer has 600 primary turns and 150 secondary turns. If the supply voltage is 1.5 kV determine the secondary line voltage on no-load when the windings are connected (a) delta-star, (b) star-delta.
hence, primary phase voltage, Secondary phase voltage, For a star connection, hence, secondary line voltage = (b) For a star connection, Primary phase voltage, For a delta connection, from which, secondary phase voltage, 
= 216.5 V = secondary line voltage Exercise 128, Page 353 1. A current transformer has two turns on the primary winding and a secondary winding of 260 turns. The secondary winding is connected to an ammeter with a resistance of 0.2 , the resistance of the secondary winding is 0.3 . If the current in the primary winding is 650 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.
(b) P.d. across ammeter = (c) Total resistance of secondary circuit = 0.2 + 0.3 = 0.5 Induce e.m.f. in secondary = (5)(0.5) = 2.5 V Total load on secondary = (2.5)(5) = 7.5 VA © John Bird Published by Taylor and Francis Directory: 9780415662857 Download 90.8 Kb. Share with your friends:
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= 96 turns
and 
and 
= 10000 i.e. 200
= 10000
= 50 A
i.e. 
then primary voltage, 
= 16 V
then primary current, 
= 45 A
= 200 V
= 4 
= 4.17 A
= 7.5 A , 
and 
= 150 V
= 225 V
= 1.5 or 
or 3:2
, 
and 
i.e. 500 = 
= 0.1894 and 
and 
= 
= 0.40 A 
= 0.40
from which, 
, 
, f = 50 Hz, 
= 37.5 A
= 600 A
from which, primary turns, 
= 800 turns
Wb = 0.033784 Wb
= 1.25 T
= 
from which, V2 = V1
= 3900 V or 3.90 kV
. Calculate the number of turns on the primary and 
then 
= 7.764 mWb
from which, primary turns, 
= 
= 22
= 
= 150
= 
= 5
from which, 
= 20 A
to 
and is 
= 4 A is shown at an angle 
to 
is the phasor sum of 
= 
= 23.26 A
and 
and phase angle of impedance, 
= 72.95
= 2.5%
= 106.7 V = voltage at which the tap-changer operates.
= 2700 VA or 2.7 kVA
= 5 A
= 0.961 or 96.10%
= 16 kW
loss) is proportional to current squared
= 200 W
= 0.9581 or 95.81%
full load
= 900 W
= 0.9756 or 97.56%
= 0.9677 or 96.77%
= 392 W or 0.392 kW
= 0.9684 or 96.84%
= 275 W or 0.275 kW
= 0.9562 or 95.62%
= 
= 1500 A
100% = 96.84%
= 1406.25 W
100% = 97.17%
from which, 
and x = 
= 0.8165
= 0.9761 or 97.61%
and 
= 147
= 12 
= 5
= 3 + 5 = 8 
= 30 A
= 2.5 A
needs to be 20 k
,
= 0.5 mA
= 62.5 A
= 100 A
= 100 – 62.5 = 37.5 A
= 0.80
= 0.25
= 1.5 kV = 1500 V
= 
= 375 V
= 649.5 V
= 866.0 V
= 5 A
= 1 V