1
00:00:14,719 --> 00:00:40,030
A
single phase transformer like this is supplied
2
00:00:40,030 --> 00:01:03,659
from a constant voltage constant frequency
source from the primary side and connected
3
00:01:03,659 --> 00:01:17,210
to the load on the secondary side. In most
cases the load would like to see a constant
4
00:01:17,210 --> 00:01:28,670
voltage appearing across it independent of
the load current I 2; however, in a practical
5
00:01:28,670 --> 00:01:37,970
transformer it is usually not the case and
by it small there will be some variation in
6
00:01:37,970 --> 00:01:45,659
the load voltage depending on the load current.
This can be predicted from the approximate
7
00:01:45,659 --> 00:01:56,620
equivalent circuit which we draw referred
to the load side now; that is the result R
8
00:01:56,620 --> 00:02:31,110
e q X e q, the magnetizing branch, applied
voltage referred to the load side V 1 dash.
9
00:02:31,110 --> 00:02:50,959
This is the load impedance Z L, the load voltage
V 2, this is the load current I 2.
10
00:02:50,959 --> 00:03:08,660
Obviously, V 2 dashed is not same as V 1 dash
for all values of load current. This is due
11
00:03:08,660 --> 00:03:18,030
to the drop in the series impedances; it will
also depend not only on the magnitude of I
12
00:03:18,030 --> 00:03:28,280
2, it will also depend on the phase angle
of I 2 with respect to V 2. Now one important
13
00:03:28,280 --> 00:03:36,770
performance parameter of a single phase transformer
is its voltage regulation which is a measure
14
00:03:36,770 --> 00:03:44,460
of the variation in the load voltage with
variation of the load; that is percentage
15
00:03:44,460 --> 00:03:59,569
regulation, full load percentage regulation
of a single phase transformer is defined as
16
00:03:59,569 --> 00:04:24,830
V 2 at full load minus V 2 at no load divided
by V 2 at full load, or that is V 2 at no
17
00:04:24,830 --> 00:04:29,849
load minus V 2 at full load divided by V 2
at full load.
18
00:04:29,849 --> 00:04:39,031
That is how much the output voltage varies
as a percentage of its rated voltage from
19
00:04:39,031 --> 00:04:45,770
no load to full load. This is called the percentage
regulation of a single phase transformer,
20
00:04:45,770 --> 00:04:51,590
and it is a very important performance parameter
for a single phase transformer, because this
21
00:04:51,590 --> 00:05:00,110
indicates how constant the voltage will be
as the load varies. Now from the approximate
22
00:05:00,110 --> 00:05:09,530
equivalent circuit it is simple to calculate
this percentage regulation; since, we see
23
00:05:09,530 --> 00:05:18,310
that the voltage drop is entirely due to the
series impedance in order to calculate percentage
24
00:05:18,310 --> 00:05:21,919
variation we can neglect the shunt branch.
25
00:05:21,919 --> 00:05:38,690
Therefore, the equivalent circuit for calculating
percentage impedance will be R eq and j X
26
00:05:38,690 --> 00:06:01,930
eq; the applied voltage is V 1 dash, the load
voltage is V 2, the current is I 2. Let us
27
00:06:01,930 --> 00:06:18,650
draw the phasor diagram, say, this is the
load voltage V 2, load current I 2 at a power
28
00:06:18,650 --> 00:06:33,820
factor angle of phi. We can draw the phasor
of V 1 by noting that V 1 equal to V 2 plus
29
00:06:33,820 --> 00:07:04,610
I 2 R eq plus j X eq. This is I 2 R eq, this
is j I 2 X eq, this is V 1 dash; therefore,
30
00:07:04,610 --> 00:07:39,360
one can write V 1 dash cos delta is equal
to V 2 plus I 2 R eq cos phi plus I 2 X eq
31
00:07:39,360 --> 00:08:20,410
sine phi and V 1 dash sine delta equal to.
From the previous examples we have seen that
32
00:08:20,410 --> 00:08:28,009
the values of R eq and X eq are very small;
in the percentage or per unit term they are
33
00:08:28,009 --> 00:08:36,810
around 7 percent, 2 percent. So, the quantity
V 1 dashd sine delta, the angle delta is usually
34
00:08:36,810 --> 00:08:47,970
very close to 0 for a normal transformer even
at full load. Therefore, V 1 dashed cos phi
35
00:08:47,970 --> 00:09:01,170
cos delta is almost cos delta is approximately
equal to 1; hence, V 1 dashed is approximately
36
00:09:01,170 --> 00:09:29,019
equal to V 2 plus I 2 R eq cos phi plus I
2 X eq sine phi.
37
00:09:29,019 --> 00:09:44,399
However, so at a load current I 2 the load
voltage V 2 is given by V 2 is approximately
38
00:09:44,399 --> 00:10:04,250
equal to V 1 dash minus I 2 R eq cos phi minus
I 2 X eq sine phi, obviously, at no load.
39
00:10:04,250 --> 00:10:21,149
So, this is V 2 at load, and V 2 no load is
obtained by putting I 2 equal to 0 here. So,
40
00:10:21,149 --> 00:10:39,449
this is V 1 dash. Hence percentage regulation
is defined as V 2 no load
41
00:10:39,449 --> 00:10:56,970
minus V 2 at load divided by V 2 at load but
V 2 no load equal to V 1 dash; therefore,
42
00:10:56,970 --> 00:11:16,589
percentage regulation equal to V 1 dash minus
V 1 dash minus I 2 dashed R eq cos phi minus
43
00:11:16,589 --> 00:11:40,290
I 2 dashed X eq sin phi divided by V 2 at
load or percentage regulation into 100 because
44
00:11:40,290 --> 00:11:58,160
of the percentage.
45
00:11:58,160 --> 00:12:38,179
So, the per unit regulation per unit expressed
as per unit of rated voltage is equal to I
46
00:12:38,179 --> 00:13:11,560
2 R eq by V 2 at full load into cos phi plus
I 2 X eq by V 2 at full load into sin phi.
47
00:13:11,560 --> 00:13:25,920
This can be written as this is equal to I
2 divided by I 2 at full load into I 2 full
48
00:13:25,920 --> 00:13:52,249
load R eq by V 2 full load cos phi plus I
2 by I 2 full load into I 2 full load X eq
49
00:13:52,249 --> 00:14:17,829
divided by V 2 full load sin phi; however,
this quantity has earlier been defined as
50
00:14:17,829 --> 00:14:33,519
the per unit load x. This quantity has been
defined as R eq per unit there, and this quantity
51
00:14:33,519 --> 00:14:55,940
has been defined as X eq per unit.
Hence, the per unit regulation can be written
52
00:14:55,940 --> 00:15:22,970
as at any fractional loading x is equal to
x into R eq per unit into cos phi plus X eq
53
00:15:22,970 --> 00:15:39,529
per unit sin phi where the angle phase angle
phi is assumed to be positive if the load
54
00:15:39,529 --> 00:15:56,879
is lagging, and it is assumed to be negative
if it is leading from this expression. So,
55
00:15:56,879 --> 00:16:06,169
this is the general expression of per unit
regulation; for lagging load phi is assumed
56
00:16:06,169 --> 00:16:16,429
to be positive; for leading load phi is assumed
to be negative. So, it is operative that percentage
57
00:16:16,429 --> 00:16:25,980
regulation can be 0 at leading load. So, at
what value of cos phi the percentage regulation
58
00:16:25,980 --> 00:16:26,980
is 0?
59
00:16:26,980 --> 00:16:49,649
So, for zero percentage regulation zero regulation
we should have phi equal to X eq sin
60
00:16:49,649 --> 00:17:25,720
phi or tan phi equal to or the phase angle
at which percentage regulation will be 0 equal
61
00:17:25,720 --> 00:17:44,880
to tan phi tan inverse R eq per unit by X
eq per unit. This is leading; this is the
62
00:17:44,880 --> 00:17:54,240
leading angle. So, at this value of leading
phase angle the percentage regulation will
63
00:17:54,240 --> 00:18:06,789
be 0; that is at this leading phase angle
the percentage regulation at any load, percentage
64
00:18:06,789 --> 00:18:12,960
regulation at any load will be 0. So, the
terminal voltage under load condition will
65
00:18:12,960 --> 00:18:19,730
be same as the no load voltage, and if the
phase angle leading phase angle is larger
66
00:18:19,730 --> 00:18:30,610
than this then the percentage regulation will
be negative, because this term will be larger
67
00:18:30,610 --> 00:18:36,830
than this which means physically this will
mean under load condition the terminal voltage
68
00:18:36,830 --> 00:18:43,680
will be larger than the no load voltage.
So, this is a condition against which we must
69
00:18:43,680 --> 00:18:51,940
guard, because the equipment connected to
the transformer will in that case see a voltage
70
00:18:51,940 --> 00:18:59,230
which is even larger than the no load voltage,
and it may damage the equipment. Similarly
71
00:18:59,230 --> 00:19:05,750
it will be interesting to find out at what
load power factor angle the percentage regulation
72
00:19:05,750 --> 00:19:13,120
is maximum; this is obtained by differentiating
the expression. So, for maximum percentage
73
00:19:13,120 --> 00:19:40,669
regulation we will have d d phi of cos phi
plus X eq per unit d d phi of sin phi should
74
00:19:40,669 --> 00:20:06,210
be equal to 0, or X eq cos phi should be equal
to R eq sin phi, or tan phi again equal to
75
00:20:06,210 --> 00:20:30,809
X eq by R eq or phi equal to tan inverse X
eq by R eq, but this is lagging.
76
00:20:30,809 --> 00:20:51,110
So, if you draw the percentage regulation
versus phi this is the lagging direction;
77
00:20:51,110 --> 00:21:13,799
this is the leading direction. Please note
that at unity power factor percentage regulation
78
00:21:13,799 --> 00:21:21,140
is not 0; it is some positive. It is 0 at
some leading power factor; let us say at this
79
00:21:21,140 --> 00:21:31,279
power factor. This value of phi is equal to
we have found this value of phi to be tan
80
00:21:31,279 --> 00:21:40,019
inverse R eq by X eq. Similarly, there is
a power factor phi the leading zone where
81
00:21:40,019 --> 00:21:56,250
the regulation is maximum is this value. So,
the percentage regulation if you draw appears
82
00:21:56,250 --> 00:22:07,679
somewhat like this. At this point it is maximum
at this point it is 0.
83
00:22:07,679 --> 00:22:23,230
This value of phi equal to tan inverse X eq
by R eq, and this value of phi equal to minus
84
00:22:23,230 --> 00:22:37,920
tan inverse R eq by X eq. This is how the
percentage regulation of a transformer which
85
00:22:37,920 --> 00:22:48,610
is a measure of how much the output voltage
varies depending on the load current can be
86
00:22:48,610 --> 00:22:55,009
plotted against load power factor angle. And
of course, it is proportional at any given
87
00:22:55,009 --> 00:23:03,980
power factor angle it is proportional to the
percentage loading. Now let us look at a problem
88
00:23:03,980 --> 00:23:13,919
of how to find out the percentage regulation
and the output voltage of a transformer.
89
00:23:13,919 --> 00:23:33,139
Let us consider an 11 kV by 440 volt single
phase transformer 50 hertz 11 kV by 440 volt
90
00:23:33,139 --> 00:24:26,120
single phase transformer. This transformer
from the short circuit test data and 75 kV.
91
00:24:26,120 --> 00:24:39,200
A short circuit test data gave
the applied voltage V sc equal to 310 volts,
92
00:24:39,200 --> 00:25:07,559
I sc equal to 11 ampere, and W sc equal to
1.6 kilowatt. In order to find out percentage
93
00:25:07,559 --> 00:25:18,669
regulation we need the series equivalence
resistance and the reactance; these are obtained
94
00:25:18,669 --> 00:25:42,309
as first we find out Z sc. Z sc equal to V
sc by this.
95
00:25:42,309 --> 00:26:03,549
This is rated current that is not 11 ampere;
this is 11 kV 75 KVA by 11 kV V sc by I sc
96
00:26:03,549 --> 00:26:35,700
is equal to 310 volts divided by I sc which
is 45.47 ohm. I
97
00:26:35,700 --> 00:26:50,690
R eq, therefore, equal to W sc divided by
I sc square which is 1.6 into 10 to the power
98
00:26:50,690 --> 00:27:20,950
3 into 11 square divided by 75 square. This
comes to 34.42 ohm. X eq equal to Z sc square
99
00:27:20,950 --> 00:27:41,980
minus R eq square and comes to 29.71 ohm.
Now we have to convert them to per unit values.
100
00:27:41,980 --> 00:27:55,380
So, referred to the short circuit test is
done on the HV side. So, V base equal to 11
101
00:27:55,380 --> 00:28:23,990
kV and KVA base of course, is equal to 75
KVA.
102
00:28:23,990 --> 00:28:49,049
Therefore, R eq per unit equal to 34.42 into
KVA base divided by voltage base square. This
103
00:28:49,049 --> 00:29:12,330
gives 0.0213. X eq per unit equal to 29.71
into KVA base into 75 KVA divided by 11 kV
104
00:29:12,330 --> 00:29:31,850
square 0.0184, and cos phi let us say this
transformer is loaded at load with a power
105
00:29:31,850 --> 00:29:50,490
factor angle of 0.8 lagging, and at half load
that is x equal to 0.5. So, regulation per
106
00:29:50,490 --> 00:30:28,360
unit p. u should be equal to x into R eq cos
phi plus X eq per unit X eq sin phi equal
107
00:30:28,360 --> 00:30:57,269
to 0.5 into 0.0213 into 0.8 plus 0.0184 into
0.6. This then comes to 0.01404. So, percentage
108
00:30:57,269 --> 00:31:11,830
regulation equal to 1.404 percent.
109
00:31:11,830 --> 00:31:33,159
Which means the 1.40 percentage regulation
is by definition V 2 at no load minus V 2
110
00:31:33,159 --> 00:31:57,309
at load divided by V 2 at full load. So, V
2 at load will be V 2 at no load minus V 2
111
00:31:57,309 --> 00:32:21,789
at full load into percentage regulation. Now
V 2 at no load is given as 440 volts; V 2
112
00:32:21,789 --> 00:32:39,789
at full load at the same power factor can
be obtained from the same formula into 0.01404;
113
00:32:39,789 --> 00:32:52,860
however, it is not necessary, because this
is multiplied by a very small factor. So,
114
00:32:52,860 --> 00:33:14,460
changing V 2 at full load actually will be
given by V 2 at no load minus V 2 drop, but
115
00:33:14,460 --> 00:33:20,429
the magnitude of V 2 drop is much smaller
compared to V 2 at no load. And when it is
116
00:33:20,429 --> 00:33:25,779
multiplied by this regulation this becomes
even smaller; therefore, in all practical
117
00:33:25,779 --> 00:33:28,440
purpose these can be assumed to be equal to
this.
118
00:33:28,440 --> 00:33:50,769
So, V 2 at load equal to 440 into 1 minus
0.01404. So, this is how from the percentage
119
00:33:50,769 --> 00:33:58,369
regulation the given short circuit test data
the percentage regulation of a single phase
120
00:33:58,369 --> 00:34:08,230
transformer can be found, and with that we
can also find what will be the output voltage
121
00:34:08,230 --> 00:34:21,490
at any given load. So, far we have seen that
most of the parameters equivalent circuit
122
00:34:21,490 --> 00:34:27,750
parameters and performance parameters of a
single phase transformer can be found from
123
00:34:27,750 --> 00:34:36,790
short circuit and no load test data, and we
have also seen how these tests are to be done.
124
00:34:36,790 --> 00:34:44,869
It is important to choose the right equipment
for performing these tests. So, let us see
125
00:34:44,869 --> 00:34:58,380
an example how we will choose the meters and
other equipment for performing these tests.
126
00:34:58,380 --> 00:35:16,619
Let us say we have a 200 volt by 400 volt
single phase transformer and rated at 4 KVA.
127
00:35:16,619 --> 00:35:27,120
I want to perform no load and short circuit
test on this transformer. So, what should
128
00:35:27,120 --> 00:35:37,790
be the rating of different equipments that
I should choose? Now for this transformer
129
00:35:37,790 --> 00:36:10,940
the rated voltages are for the LV side 200
volts, for the HV side 400 volts; rated currents
130
00:36:10,940 --> 00:36:42,950
for the LV side it is 4 KVA by rated voltage
equal to 20 amperes. On the HV side the rated
131
00:36:42,950 --> 00:37:06,369
voltage is 4 KVA divided by rated voltage
equal to 10 amperes. So, for the LV side ratings
132
00:37:06,369 --> 00:37:25,300
are 200 volt 20 amperes; for the HV side rated
voltages are 400 volts 10 amperes.
133
00:37:25,300 --> 00:37:45,020
Let us just look at first the short circuit
test; this is done on the HV side. So, it
134
00:37:45,020 --> 00:38:09,440
will be done on the 400 volt side with the
200 volt side shorted. The measurements required
135
00:38:09,440 --> 00:38:31,700
are we have to measure the voltage; you will
have to measure the power, and you will have
136
00:38:31,700 --> 00:38:46,930
to measure the current. Normally when the
transformer is shorted it requires very little
137
00:38:46,930 --> 00:38:54,140
current to very little voltage to circulate
the full load current; therefore, this is
138
00:38:54,140 --> 00:39:15,100
supplied from a single phase variac. So, the
equipment required are, this is the rated
139
00:39:15,100 --> 00:39:38,760
220 volt or 230 volt single phase supply.
Equipment required first is single phase variac;
140
00:39:38,760 --> 00:40:04,670
its input voltage is 230 volts. The output
it should be able to carry at least the rated
141
00:40:04,670 --> 00:40:21,780
full load current of the single phase transformer.
So, its output current should be at least
142
00:40:21,780 --> 00:40:28,520
10 ampere.
Then we need an ammeter which should be able
143
00:40:28,520 --> 00:40:40,040
to measure ac. So, this is moving iron type
ammeter. This should be able to measure 10
144
00:40:40,040 --> 00:40:55,589
amperes of current. Next I need a wattmeter
whose current coil should be able to carry
145
00:40:55,589 --> 00:41:12,039
at least 10 amperes; what about the voltage
coil potential coil? Now for a single phase
146
00:41:12,039 --> 00:41:22,960
transformer if we look at the equivalent circuit
single phase transformer. When we short the
147
00:41:22,960 --> 00:41:29,960
secondary side the required voltage to circulate
full load current is the series impedance
148
00:41:29,960 --> 00:41:37,680
multiplied by the full load current which
under usually is in the range of 5 to 10 percent
149
00:41:37,680 --> 00:41:46,970
of the rated voltage.
Therefore, for the high voltage side the rated
150
00:41:46,970 --> 00:41:57,299
voltage is 400 volts and at best 10 percent
of it which would be required. So, the applied
151
00:41:57,299 --> 00:42:05,380
required voltage in no case would be possibly
exceeds 50 ampere. So, let us choose a wattmeter
152
00:42:05,380 --> 00:42:15,099
whose potential coil can read at, say, 100
volts; it may be even lower; it can be even
153
00:42:15,099 --> 00:42:20,710
50 volts possibly, but it is difficult to
get wattmeter of that rating. So, it has to
154
00:42:20,710 --> 00:42:30,260
use 100 volt. A 75 volt is a normal rating
that is usually available. So, it may be even
155
00:42:30,260 --> 00:42:44,000
75 volts. Then we need a voltmeter moving
iron voltmeter. Since the potential coil of
156
00:42:44,000 --> 00:42:52,859
the wattmeter and the voltmeter are connected
to the same points the drop across the current
157
00:42:52,859 --> 00:43:03,470
coil being negligible; this can also be either
75 volt or 100 volt reading. So, these are
158
00:43:03,470 --> 00:43:10,839
the list of equipments required for the short
circuit test. Now let us look at the open
159
00:43:10,839 --> 00:43:27,890
circuit test.
160
00:43:27,890 --> 00:44:00,579
The open circuit test is done one the LV side
with the HV side open circuited. Here again
161
00:44:00,579 --> 00:44:29,970
we need to measure the applied voltage, power
162
00:44:29,970 --> 00:44:52,700
and the current. Now open circuit tests are
done at rated voltage. Since open circuit
163
00:44:52,700 --> 00:44:58,559
tests are done at rated voltage it could have
been directly connected to the source, but
164
00:44:58,559 --> 00:45:05,289
the voltage rating of the LV side may not
exactly match with the voltage available;
165
00:45:05,289 --> 00:45:13,770
therefore, here also normally a variac will
be used.
166
00:45:13,770 --> 00:45:27,330
For example, the available single phase supply
may be of 230 volt, whereas, our transformer
167
00:45:27,330 --> 00:45:48,220
rated voltage is just 200 volt. So, we need
a variac. So, now let us see what is needed.
168
00:45:48,220 --> 00:46:14,359
One is single phase variac; its input voltage
will be 230 volts means all variac will give
169
00:46:14,359 --> 00:46:20,750
you up a voltage at least up to input voltage.
So, this voltage rating is sufficient. The
170
00:46:20,750 --> 00:46:27,359
current drawn from this variac will be the
no load current of the transformer which is
171
00:46:27,359 --> 00:46:33,540
very small; on the LV side the rated current
itself is 20 ampere, the no load current of
172
00:46:33,540 --> 00:46:42,329
a transformer will be around 5 percent of
that. So, the no load current will be within
173
00:46:42,329 --> 00:46:51,609
1 ampere. So, the variac that can give 1 ampere
output current should be sufficient for this
174
00:46:51,609 --> 00:46:57,369
test.
However, if we do both the single phase the
175
00:46:57,369 --> 00:47:03,510
open circuit and the short circuit test I
can use the same variac that was used in case
176
00:47:03,510 --> 00:47:09,200
of short circuit test for doing this test.
Now there the output current was specified
177
00:47:09,200 --> 00:47:14,580
as 10 ampere which is sufficient for to be
used for the no load test. So, here also we
178
00:47:14,580 --> 00:47:23,560
use the same variac with an input of 230 volt
and output current of 10 ampere. It should
179
00:47:23,560 --> 00:47:33,049
be emphasized that this 10 ampere output current
is not required for this no load test. However,
180
00:47:33,049 --> 00:47:39,420
in order to reduce the duplication of the
equipment we are using the same variac that
181
00:47:39,420 --> 00:47:45,940
we used for the short circuit test. Hence
this is sufficient; however, if the short
182
00:47:45,940 --> 00:47:51,599
circuit test is not required, only open circuit
test is required then we can choose a variac
183
00:47:51,599 --> 00:48:00,910
of much lower current rating.
Second is the ammeter; again here also we
184
00:48:00,910 --> 00:48:21,400
will need an ammeter moving iron ammeter.
As we have already mentioned the current drawn
185
00:48:21,400 --> 00:48:27,310
will be simply the no load current of the
transformer; for this particular case on the
186
00:48:27,310 --> 00:48:40,450
LV side the rated current is 20 amperes, no
load current I 0 is usually within 5 percent
187
00:48:40,450 --> 00:48:55,059
of I fl. So, I 0 may be 1 ampere. So, again
whichever ammeter we have chosen for short
188
00:48:55,059 --> 00:49:04,750
circuit test could have been sufficient. However,
this current rating 10 ampere is too large
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for performing the open circuit test. The
problem is the higher the maximum range of
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the meter the lower its accuracy, becomes
particularly with moving iron equipment at
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the lower end the accuracy suffers. Therefore,
it will be necessary to choose a lower rated
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ammeter.
So, let us choose something like 3 ampere
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rated ammeter for this test. Since, the same
10 ampere ammeter will possibly give erroneous
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result. Sometimes meters with multiple ranges
are available. For example, we can get an
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ammeter of 2.5 amperes, 5 amperes and 10 amperes
in which case the 2.5 ampere range can be
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used, same ampere can be used for both of
the tests which during open circuit test we
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use the 2.5 ampere range; for the short circuit
test we use the 10 ampere range. Similarly,
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for here also we will need a wattmeter, there
also we will need a wattmeter. The wattmeter
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in this case we will have a current coil rating
which is same as the ammeter current coil
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rating. So, preferably 2.5 ampere, 3 ampere
current coil rating would have been good.
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Obviously, this same wattmeter current coil
will not be useful here, because here again
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we will have the same problem of accuracy;
we cannot use a wattmeter with 10 ampere rated
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current. We will have to use a much lower,
2.5 ampere would be ideal; however, it is
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difficult to get a wattmeter. So, let us choose
either a 2.5 ampere if available or a 5 ampere
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rating for the current coil. For the potential
coil though here the applied voltage is same
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as the rated voltage of the LV side which
is 200 volt in this case. So, we will have
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to choose a wattmeter whose potential coil
is rated at least at 200 volts it should be
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more. So, we choose a wattmeter with a potential
coil rating of let us, say, 300 hundred volts.
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Most importantly for getting accurate result
we will have to be careful about the type
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of the wattmeter; at no load the power factor
is very poor. It is close to 0, it is less
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than 0.2. So, this must be a low power factor
wattmeter. In the short circuit test also
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the power factor may be poor, but usually
it is not so poor as in the case of open circuit
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test. So, possibly a normal wattmeter will
be sufficient for short circuit test, but
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for no load test a special low power factor
wattmeter which measures accurately below
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a power factor of 0.2 must be used. Finally,
we will need an ac voltmeter which is a moving
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iron voltmeter, and since they are connected
the same points as the potential coil of the
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wattmeter this should also be rated at 300
volts.
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So, to summarize to do this test the following
equipments will be required. One is a single
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phase variac 230 volt 10 ampere MI A 0 - 2.5
- 5 - 10 ampere if it is available then 1;
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otherwise, you will have to get one 0 to 10
ampere 1 and 0 to 2.5 ampere 1. Voltmeter
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you can get again choose a 0 - 100 0 - 75
– 150 - 300 volts. This is a normally available
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voltmeter. You will need two types of wattmeter.
One is LPF current coil rating 0 to 2.5 amperes,
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potential coil rating 300 volts. Another normal
wattmeter the current coil rating 0 to 10
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amperes, potential coil rating 75 volts. These
are the list of equipments that would be necessary
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for performing the short circuit and the open
circuit test on this particular transformer.