Bird J. Electrical Circuit Theory and Technology
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336 Electrical Circuit Theory and Technology
Figure 20.12
The circuit diagram is shown in Figure 20.12.
(a) Turns ratio
N
1
N
2
D
V
1
V
2
D
220
1760
D
1
8
Equivalent input resistance of the transformer,
R
1
D
N
1
N
2
2
R
L
D
1
8
2
1.28 ð 10
3
D 20
Total input resistance, R
IN
D R C R
1
D 2 C 20 D 22
Primary current, I
1
D
V
1
R
IN
D
220
22
D 10 A
(b) For an ideal transformer
V
1
V
2
D
I
2
I
1
, from which I
2
D I
1
V
1
V
2
D 10
220
1760
D 1.25 A
Power dissipated in load resistor R
L
,P D I
2
2
R
L
D 1.25
2
1.28 ð 10
3
D 2000 watts or 2 kW
Problem 24. An a.c. source of 24 V and internal resistance 15 k
is matched to a load 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 Figure 20.13.
Figure 20.13
(a) For maximum power transfer R
1
needstobeequalto15k
R
1
D
N
1
N
2
2
R
L
, from which load resistance,
R
L
D R
1
N
2
N
1
2
D 15000
1
25
2
D 24 Z
Transformers 337
(b) The total input resistance when the source is connected to the
matching transformer is R
IN
C R
1
, i.e., 15 k +15k =30k
Primary current, I
1
D
V
30000
D
24
30000
D 0.8mA
N
1
N
2
D
I
2
I
1
, from which I
2
D I
1
N
1
N
2
D 0.8 ð 10
3
25
1
D 20 ð 10
3
A
Power dissipated in the load R
L
,P D I
2
2
R
L
D 20 ð 10
3
2
24
D 9600 ð 10
6
W D 9.6mW
Further problems on resistance matching may be found in Section 20.16,
problems 27 to 31, page 347.
20.11 Auto transformers
An auto transformer is a transformer which has part of its winding
common to the primary and secondary circuits. Figure 20.14(a) shows
the circuit for a double-wound transformer and Figure 20.14(b) that for
an auto transformer. The latter shows that the secondary is actually
part of the primary, the current in the secondary being (I
2
I
1
).
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 20.15 shows the circuit diagram symbol for an auto transformer.
Problem 25. 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.
Figure 20.14
Rating D 20 kVA D V
1
I
1
D V
2
I
2
Hence primary current, I
1
D
20 ð 10
3
V
1
D
20 ð 10
3
320
D 62.5A
and secondary current, I
2
D
20 ð 10
3
V
2
D
20 ð 10
3
250
D 80 A
Hence current in common part of the winding D 80 62.5 D 17.5A
The current flowing in each section of the transformer is shown in
Figure 20.16.
Figure 20.15
338 Electrical Circuit Theory and Technology
Figure 20.16
Saving of copper in an auto transformer
For the same output and voltage ratio, the auto transformer requires less
copper than an ordinary double-wound transformer. This is explained
below.
The volume, and hence weight, of copper required in a winding is
proportional to the number of turns and to the cross-sectional area of the
wire. In turn this is proportional to the current to be carried, i.e., volume
of copper is proportional to NI.
Volume of copper in an auto transformer / N
1
N
2
I
1
C N
2
I
2
I
1
see Figure 20.14(b)
/ N
1
I
1
N
2
I
1
C N
2
I
2
N
2
I
1
/ N
1
I
1
C N
2
I
2
2N
2
I
1
/ 2N
1
I
1
2N
2
I
1
since N
2
I
2
D N
1
I
1
Volume of copper in a double-wound transformer / N
1
I
1
C N
2
I
2
/ 2N
1
I
1
(again, since N
2
I
2
D N
1
I
1
)
Hence
volume of copper in an auto transformer
volume of copper in a double-wound transformer
D
2N
1
I
1
2N
2
I
1
2N
1
I
1
D
2N
1
I
1
2N
1
I
1
2N
2
I
1
2N
1
I
1
D 1
N
2
N
1
If
N
2
N
1
D x then
(volume of copper in an auto transformer)
= .1 − x/ (volume of copper in a double-wound transformer) (20.12)
If, say, x D
4
5
then
(volume of copper in auto transformer)
D
1
4
5
volume of copper in a double-wound transformer
D
1
5
volume in double-wound transformer
Transformers 339
i.e., a saving of 80%
Similarly, if x D
1
4
, the saving is 25%, and so on.
The closer N
2
is to N
1
, the greater the saving in copper.
Problem 26. Determine the saving in the volume of copper used
in an auto transformer compared with a double-wound transformer
for (a) a 200 V:150 V transformer, and (b) a 500 V:100 V trans-
former.
(a) For a 200 V:150 V transformer, x D
V
2
V
1
D
150
200
D 0.75
Hence from equation (20.12), (volume of copper in auto transformer)
D 1 0.75 (volume of copper in double-wound transformer)
D 0.25 (volume of copper in double-wound transformer)
D 25% of copper in a double-wound transformer
Hence the saving is 75%
(b) For a 500 V:100 V transformer, x D
V
2
V
1
D
100
500
D 0.2
Hence (volume of copper in auto transformer)
D 1 0.2 (volume of copper in double-wound transformer)
D 0.8 (volume in double-wound transformer)
D 80% of copper in a double-wound transformer
Hence the saving is 20%
Further problems on the auto-transformer may be found in Section 20.16,
problems 32 and 33, page 347.
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.
340 Electrical Circuit Theory and Technology
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 induc-
tion motors (see Chapter 22) and for interconnecting systems that are
operating at approximately the same voltage.
20.12 Isolating
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 transformer is a
1:1 ratio transformer with several important applications, including bath-
room shaver-sockets, portable electric tools, model railways, and so on.
20.13 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 20.17, which gives a considerable saving in the amount of iron
used. The primary and secondary windings in Figure 20.17 are wound on
top of each other in the form of concentric cylinders, similar to that shown
in Figure 20.6(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.
A delta-connection is shown in Figure 20.18(a) and a star-connection
in Figure 20.18(b).
Problem 27. 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.
(a) For a star-connection, V
L
D
p
3V
p
(see Chapter 19)
Primary phase voltage, V
p1
D
V
L1
p
3
D
2400
p
3
D 1385.64 volts
For a delta-connection, V
L
D V
p
N
1
N
2
D
V
1
V
2
, from which,
Transformers 341
Figure 20.17
Figure 20.18
342 Electrical Circuit Theory and Technology
secondary phase voltage, V
p2
D V
p1
N
2
N
1
D 1385.64
50
500
D 138.6 volts
(b) For a delta-connection, V
L
D V
p
,
hence primary phase voltage V
p1
D 2.4kVD 2400 volts
Secondary phase voltage, V
p2
D V
p1
N
2
N
1
D 2400
50
500
D 240 volts
For a star-connection, V
L
D
p
3V
p
,
hence the secondary line voltage D
p
3240 D 416 volts
A further problem on the three-phase transformer may be found in
Section 20.16, problem 34, page 347.
20.14 Current
transformers
For measuring currents in excess of about 100 A a current transformer
is normally used. With a d.c. moving-coil ammeter the current required
to give full scale deflection is very small— typically a few milliamperes.
When larger currents are to be measured a shunt resistor is added to the
circuit (see Chapter 10). 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 a double-wound transformer:
I
1
I
2
D
N
2
N
1
from which, secondary current I
2
= I
1
N
1
N
2
In current transformers the primary usually consists of one or two turns
whilst the secondary can have several hundred turns. A typical arrange-
ment is shown in Figure 20.19.
If, for example, the primary has 2 turns and the secondary 200 turns,
then if the primary current is 500 A,
secondary current, I
2
D I
1
N
1
N
2
D 500
2
200
D 5A
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,2Aor5A.
Figure 20.19
Transformers 343
Figure 20.20
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 short-circuit 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 20.20.
Problem 28. 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 . The resistance of the secondary winding is 0.25 . 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.
(a) Reading on the ammeter, I
2
D I
1
N
1
N
2
D 300
1
60
D 5A
(b) P.d. across the ammeter D I
2
R
A
, where R
A
is the ammeter
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 secondary D 50.40 D 2.0V
Total load on secondary D 2.05 D 10 VA
A further problem on the current transformer may be found in
Section 20.16, problem 35, page 348.
20.15 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 20.21.
Since
V
1
V
2
D
N
1
N
2
the secondary voltage, V
2
= V
1
N
2
N
1
Thus if the arrangement in Figure 20.21 has 4000 primary turns and 20
secondary turns then for a voltage of 22 kV on the primary, the voltage
Figure 20.21
344 Electrical Circuit Theory and Technology
on the secondary,
V
2
D V
1
N
2
N
1
D 22000
20
4000
D 110 volts
20.16 Further problems
on transformers
Principle of operation
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. [96]
2 An ideal transformer with a turns ratio of 2:9 is fed from a 220 V
supply. Determine its 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. [400 V]
4 An ideal transformer has a turns ratio of 12:1 and is supplied at
192 V. Calculate the secondary voltage. [16 V]
5 An ideal transformer has a turns ratio of 12:1 and is supplied at
180 V when the primary current is 4 A. Calculate the secondary
voltage and current. [15 V, 48 A]
6 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. [50 A]
7 A transformer has a primary to secondary turns ratio of 1:15. Calcu-
late the primary voltage necessary to supply a 240 V load. If the load
current i
s 3 A determine the primary current. Neglect any losses.
[16V,45A]
8 A 10 kVA, single-phase transformer has a turns ratio of 12:1 and is
supplied from a 2.4 kV 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.
[(a) 50 A (b) 4 (c) 4.17 A]
9A20 resistance is connected across the secondary winding of a
single-phase power transformer whose secondary voltage is 150 V.
Calculate the primary voltage and the turns ratio if the supply current
is 5 A, neglecting losses. [225 V, 3:2]
No-load phasor diagram
10 (a) Draw the phasor diagram for an ideal transformer on no-load.
Transformers 345
(b) A 500 V/100 V, single-phase transformer takes a full load
primary current of 4 A. Neglecting losses, determine (a) the full
load secondary current, and (b) the rating of the transformer.
[(a) 20 A (b) 2 kVA]
11 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. [0.786 A, 0.152 A]
12 A transformer takes a current o
f 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) 0.4 A (b) 0.4 (c) 0.92 A]
E.m.f equation
13 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) 37.5 A, 600 A (b) 800 (c) 9.0 mWb]
14 A single-phase, 50 Hz transformer has 40 primary turns and 520
secondary turns. The cross-sectional area of the core is 270 cm
2
.
When the primary winding is connected to a 300 volt supply, deter-
mine (a) the maximum value of flux density in the core, and (b) the
voltage induced in the secondary winding.
[(a) 1.25 T (b) 3.90 kV]
15 A single-phase 800 V/100 V, 50 Hz transformer has a maximum
core flux density of 1.294 T and an effective cross-sectional area of
60 cm
2
. Calculate the number of turns on the primary and secondary
windings. [464, 8]
16 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.
[(a) 150, 5 (b) 792.8 cm
2
]
Transformer on-load
17 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.
[23.26 A, 0.73]