National Institute of Technology Calicut

Department of Electrical Engineering

EE3091D
Electrical Machines Lab – 1
S5 EEE
 List of Experiments

1) Open circuit characteristics of a self excited dc shunt generator

2) Load test on dc shunt generator
3) Break test on dc shunt and series motors
4) Load Test on a single-phase transformer
5) Polarity test on single –phase transformers and three-phase connection of
single phase transformers
6) O.C and S.C tests on a single phase transformer
7) Hopkinson test
8) Retardation Test
9) Separation of losses in a single phase transformer
10) Scott Connection
11) Sumbners test

Note:

• Detailed objectives of experiments are along with the questions of each

experiment.
• Students shall answer all the question in the record book.
• Inferences based on the results obtained, shall be given at the end of each
experiment.
1)OPEN CIRCUIT CHARACTERISTIC OF
A DC SHUNT GENERATOR

DC SHUNT MOTOR DC SHUNT

(PRIME MOVER) GENERATOR

NAME PLATE DETAILS:

DC generator: 1500 rpm, 220 V, 15.9 A, 3.5 kW
DC motor: 1500 rpm, 220 V, 15.9 A, 3.5 kW

PROCEDURE
1. Take down the machine details and select the meters / rheostat properly.
2. Keep the position of starting resistance ‘Rs’ and field-controlled resistances ‘Rfm’ and
‘Rfg’ properly
3. Start the prime mover (DC motor). Slowly cut down the starting resistance Rs.
Adjust Rfm for the rated speed of the generator.
4. Note down the generated voltage due to residue magnetism. Close the switch ‘S’ and
very the field current of the generator in steps and note down the open circuit voltage.
Increase the field current up to a value such that the complete range of OC is obtained
including saturation. Ensure the constancy of speed throughout out
SAMPLE CALCULATION:
E1/E2 = N1/N2

For 1200 rpm; E2 = N2*E1/N1

Critical resistance=

OBSERVATION TABLE:

S. No If (A) No Load Voltage(V) No Load Voltage(V) No Load Voltage(V)

at N rpm (1500) at N1 rpm (1200) at N2 rpm (1800)
1. 0 10
2. 0.1 60
3. 0.2 100
4. 0.3 130
5. 0.4 154
6. 0.5 172
7. 0.6 187
8. 0.7 200
9. 0.8 210
10. 0.9 218
11. 1.3 240
12. 1.48 246
QUESTIONS
1. Plot the OCC at the rated speed and determine the critical resistance. Find the critical
resistance at 80% ratted speed form the above.
2. State the condition of building up of voltage in a self excited generator.
3. A separately excited generator can operator can operate over the entire range of
magnetization curve, where as self excited can operate only a small range. Explain.
4. What is meant by “Critical speed”?
5. Why the initial portion OCC is straight?
6. Why “Series” generators are not practicable?
 2)LOAD TEST ON DC SHUNT
GENERATOR
CIRCUIT DIAGRAMS:
Fig (a) Load Test

Fig (b) Measurement of Armature Resistance:

NAME PLATE DETAILS:

DC generator: 1500 rpm, 230 V, 15.2 A

Induction generator: 950 rpm, 5 hp, 7.5 A

PROCEDURE:
1. Take down the name plate details of the machine and select proper rheostats, meters and
suitable loads. (RL1- Loading rheostats, RL2- 15 Ohm/10A resistor).
2. Set up the circuit as shown in the figure (a). Keep he switch S3 and S4 open. Adjust the
positions of R, RL1 and RL2 properly. Close switch S1 and start the prime over using
the Star delta starter.
3. Adjust the field current of the generator for the rated output voltage. Close switch S3 and
apply load when the voltage drops to a sufficiently low value close switch S4 and apply
load using RL2.
4. Set up the circuit as shown in figure (b) keeping switch S5 open. Adjust R1 properly and
close S5. Measure the reading and calculate the armature resistance.
5. Plot the required characteristics.

SAMPLE CALCULATIONS:
Id = IL + If =

Ia =

Eg = V + Ia*Ra =

ra= Voltage/ Current

OBSERVATION TABLE:

S. No Voltage (V) Current (A) ra (ohm)

1. 1 0.5
2. 2 1
3. 3 1.5
S. No If (A) Voltage (V) IL (A) Ia (A) Eg (V)

1. 0.46 230 0 0.46

2. 0.45 224 0.4 0.85
3. 0.42 215 1.2 1.62
4. 0.4 203 2 2.4
5. 0.38 192 2.7 3.08
6. 0.36 184 3.2 3.56
7. 0.34 168 3.7 4.04
8. 0.28 144 4.3 4.58
9. 0.18 90 4.5 4.68
10. 0.13 66 3.9 4.03
11. 0.1 48 3.7 3.8
12. 0.08 30 3.1 3.18
13. 0.05 2 2.7 2.75

QUESTIONS
1. Plot the external characteristics. Deduce the internal characteristics and armature reaction
curve.
2. Explain why the voltage drops on loading? Will the voltage drop same for separately
excited machines?
3. What are interpoles? Whether it is providing in all machines?
4. What is the value of voltage regulation for an ideal DC generator?
5. During armature resistance measurement, if armature rotates whether the reading will be
correct?
6. What is meant by computing? Is it possible to get constant voltage characteristics by
computing?
 3)BRAKE TEST ON SHUNT AND
SERIES MOTORS

OBJECTIVE: Conduct load test on DC shunt series motors and obtain performance
characteristics.
SHUNT MOTOR:

SERIES MOTOR:
 NAME PLATE DETAILS:

Shunt Motor: 3 hp, 220 V, 1500 rpm, 12 A

Series Motor: 3.8 kW, 220 V, 970-2900 rpm

THEORY: This load test or brake test is a direct method of testing motors. A belt
around the water cooled pulley has its ends attached to the spring balances W1 and W2.
The spring balances are calibrated in Kg. Then motor output is given by 2∏N/60(W1-
W2)*9.81 watts, where W1 and W2 are the readings of spring balance and r, the effective
radius of brake drum in meters and N, the speed in rpm.
Efficiency is calculated as the ratio of output to input. Input is obtained by V1 watts.

PROCEDURE:

1. Take the name plate and details of motor and select meters and rheostat properly.
Connections are made as shown in the diagram.
2. Shunt motor should be started no load so loose the belt completely. Keep Re field control
rheostat in minimum resistance position and starter in initial start position. Start the motor
by moving slowly starter handle to the fully ON position. Adjust Rf to get the rated load.
Now the machine running under No-load. Take the readings. Apply load in steps up to
the full load by tightening the belt and corresponding reading shall be noted. During
loading, cool down the brake drum by water. Remove the load by loosening the belt after
the test. Then switch off the motor.
3. Series motor shall be started with some lead only. Adjust the belt for sometimes.
Basically, the motor under test is a compound motor with weak shunt field. Keep Re field
control rheostat in minimum resistance position and starter in the initial state position.
Start the motor by moving slowly the starter handle to the fully ON position. Adjust Rf to
the rated speed (specified). Apply load in steps up to full load by tightening the belt and
corresponding reading shall be noted. During, cool the drum by water, switch off the
motor after reducing the load.
 SAMPLE CALCULATION:

(For shunt machine)

Pin = V*I =
Po = (2*pie*N*T)/60 =
T = (W1-W2) * 9.81 * R Efficiency = (Po/Pin) *100

(For series Motor)

Pin = V*I =
Po = (2*pie*N*T)/60 =
T = (W1-W2) * 9.81 * R = Efficiency = (Po/Pin) *100=

OBSERVATION TABLES:

Shunt Machine:
S. Voltage (V) Current Pin (W) W1 W2 Nrpm Po (W) Torque Efficie
No (A) (Nm) ncy
1. 212 0.8 169.5 0 0 1500 0
2. 210 2 420 2 0.4 1460 183.47
3. 208 4 832 6.2 0.8 1428 605.63
4. 206 6 1236 10 1.3 1390 950.5
5. 204 8 1632 12.5 1 1365 1233.6
6. 202 10 2020 15.5 1.7 1360 1474
Series Machine:
S. Voltage Current Pin W1 W2 Nrpm Po Torque Efficiency
No (V) (A) (W) (W) (Nm) (%)
1. 206 6.5 1339 5.5 0.5 2000 1093
2. 206 7 1442 6.5 0.5 1995 1309.6
3. 206 10 2060 9 1 1978 1731.2
4. 206 12.5 2575 12.5 2 1940 2228.2
5. 206 15 3090 14.5 2.5 1930 2533.8
6. 206 17.5 3605 17.5 2.5 1924 3157.5
7. 206 20 4120 19 2.5 1910 3447.9

QUESTIONS
1. Plot the following performance characteristics.
(1) Efficiency Vs output (2) Vs output (3) speed Vs torque (4) Armature current Vs
output (5) Torque Vs Line current
2. Why the speed falls on loading?
3. State the merit and demerit s of brake test.
4. What are the problems with spring dynamometer measurement?
5. What will happen if the field circuit gets opened for a dc motor under load?
6. What will happen if the field terminals are reversed under rated load running at a rated
speed?
7. How do you change the direction of rotation of a dc motor?
8. Are DC motors still in use? State the merits of Dc motor.
9. DC series motor shall always be started with some load. Why?
10. Compare the starting torque of dc shunt and series motor of same rating
4)LOAD TEST ON 1 PHASE TRANSFORMERS
CIRCUIT DIAGRAMS

NAME PLATE DETAILS

Transformer: 1.25 kVA, 220/115 V, 1 phase

Auto Transformer: (0-230) V, 50 Hz

PROCEDURE
1. Take down the transformer details and select the meters properly.
2. Mark the polarities of transformers.
3. For load test- With the switch S in open position in the load circuit. No load secondary
voltage (HV side) can be measured. Adjust the loading rheostat and note down the
readings up to the rated current.
For three phase connections- Connect the three single phase transformers in YY0 position
with proper polarity. Measure the line and phase voltages with and without neutral. The n
connects in YD1 position, keep the switch S2 open (without neutral) and measure the
voltage across S3. This gives three times harmonic voltage per phase. Close the switch
 S2, and now measure the voltage across S3. If this voltage is zero close S3. Check the
secondary line and phase voltages.

SAMPLE CALCULATION:
i/p power(Pin) =

o/p power(Po) = V2 * I2 =

Efficiency = (Po/ Pin)*100

V.R = (VNL – VL)/VNL =

OBSERVATION TABLE (for load test)

S. 10 10 Pin 20 20 %
No Current Voltage Power Current Voltage Po (W) %VR Efficiency
(A) (V) (W) (A) (V)
1 1 115 0 0 192
2 2 112 120 0.1 186
3 3.4 102 300 1 184
4 4.8 100 440 1.6 178
5 6 98 452.14 2.5 172
6 7.4 96 720 4 170
7 8.5 95 840 4.5 166
8 9.7 94 960 5.2 164

QUESTIONS
1. Determine and plot (1) efficiency-output (2) Regulation- output characteristics by load
test.
2. Define regulation. How does it vary load current.
3. What are the losses in a transformer? What is the normal range of efficiency of a
transformer?
4. Generally a transformer with high efficiency and low regulation is preferred why?
5. From the efficiency characteristics, find the load at which maximum efficiency occurs.
Determine the copper losses at that load.
6. Draw the phasor diagram showing all the details for transformer under UPF and 0.8 lag
ang lead condition of load.
 5) POLARITY TEST ON SINGLE –PHASE
TRANSFORMERS AND THREE-PHASE
CONNECTION OF SINGLE PHASE
TRANSFORMERS

CIRCUIT DIAGRAMS
Fig (a) Polarity Test Transformer.

Fig ( b ) Y-Y-0 Connection Fig ( c ) Y-Y-6 Connection

Fig (d) Y-D-1 Connection

NAME PLATE DETAILS

Transformer: 1.25 kVA, 220/115 V, 1 phase

Auto Transformer: (0-230) V, 50 Hz

PROCEDURE
1. Take down the transformer details and select the meters properly.
2. Mark the polarities of transformers.
3. Connect single phase transformer in cumulative and differential mode and verify using
and verify using wattmeter

OBSERVATIONS

Connection V1(V) V2(V) VT(V)

Cumulative 230 124
Differential 230 124

SAMPLE CALCULATION

Cumulative
VT = V1+ V2

Differential

VT = V1 - V2

QUESTIONS
1. Why do we conduct polarity test?
2. Measure the line and phase voltages in three phase connection. Find the third harmonic
voltage in YD1 connections.
3. What are the different methods of connection of single phase transformers for 3-phase
applications?
4. Draw the phasor diagram for YY0 and YD1 connections.
5. What is third harmonic voltage? How it is taken in delta connections. What are the
problems induced harmonics in supply systems?
6. State the merits and demerits of 3- phase connection of single phase transformer.
 6)OPEN CIRCUIT AND SHORT CIRCUIT TESTS
ON SINGLE PHASE TRANSFORMER
CIRCUIT DIAGRAMS:
Fig (a) O.C. Test

Fig (b) S.C. Test

NAME PLATE DETAILS

1-Phase Transformer: 230 V/115 V, 1.25 kVA

PROCEDURE
 1. Take down the name plate details of the machine and select properly the meters and other
components.
2. Set up the circuit as in fig (a) and keeping the auto transformer in proper position close
switch SI. Adjust for the rated voltage and take the readings of the meters.
3. Set up the circuit as in fig (b) and keeping the auto transformers in proper position close
switch S2. Adjust for the rated current and take readings of the meters.
4. Predetermine the parameter of the equivalent circuit and plot the required characteristics.
5. Using the equivalent circuit determine primary current, regulation and efficiency when an
impedance of 9+j12 ohm is connected across the secondary winding (115V side).
6. Predetermine the upf load at which the efficiency is maximum.

PROCEDURE
1. Take down the name plate details of the transformer and select the meters perfectly for
the Test.
2. Hint (1) for O.C. test - no load current will be about 5% of the rated current.
Being magnetizing correct, the power factor will be low. So Wattmeter shall be suitable
for LPF-compensation.
Hint (II) For SC test -5 to 10% of the rated voltage is sufficient to circulate rated current
under the test.
3. S.C. Test- Adjust the auto transformer very carefully till the rated current of high voltage
side is shown in the ammeter. Note down the readings.

SAMPLE CALCULATION:

Wo =

Vo =

Io =
𝐶𝑂𝑆∅𝑜 = Wo /(Vo*Io) =, Ro= V0/Iw =

Iw = Io* 𝐶𝑂𝑆∅𝑜 = X0 = V0/Iu =

Iu = IO * 𝑆𝑖𝑛∅ =

(For SC Test)
Zsc = Vsc/Isc=

Ro2 = Wsc/Isc2 = X01 =

Ro1 = X02 =

For pf = 0.8 and 100% load (=> x=1)

P0 = x*S* 𝐶𝑂𝑆∅𝑜 =

Efficiency = P0/( P0+(x2*copperloss)+ironloss)

𝑥 𝐼1
Voltage Regulation(V. R) = ∗ (𝑅01 𝐶𝑂𝑆∅𝑜 ± 𝑋01 𝑆𝑖𝑛∅𝑜 )
𝑉1

OBSERVATION TABLES:

S. No Test Voltage (V) Current (A) Wattmeter (W)

1. O.C 115 1 20.5

2. S.C 10.8 5.68 50

p.f = 0.5 p.f = 0.7 p.f = 0.8 p.f = upf

% Load
Po (W) Effic. Po (W) Effic. Po (W) Effic. Po (W) Effic.
(%) (%) (%) (%)
10
20
40
60
80
100

% p.f = 0.5 p.f = 0.7 p.f = 0.8 0 p. f UPF

Load lag lead lag lead Lag lead lag lead
10
20
40
60
80
100
Tabular Column for VR

QUESTIONS:
1. Determine and draw the equivalent circuits of the transformer referred to high voltage
side. How it differs when referred to low voltage side.
2. Using the equivalent circuit parameters, draw (a) efficiency – output characteristic for
UPF and 0.8 lag PF conditions (b) regulation – PF characteristics for full load.
3. Find out the current at the input side, regulation and efficiency when a load having
impedance 9+J12 ohm is connected across the LV side.
4. Predetermine the UPF load at which the efficiency is maximum?
5. Why do we conduct OC test on the low voltage side and SC test on high voltage side?
6. Why the SC Test is called impedance test?
7. Have you measured the frequency during the test? Comment on the importance of
frequency for the test.
8. Explain how the nature of load affects the voltage regulation?
9. Why transformers are specified in KVA rating?
 7)HOPKINSON TEST

CIRCUIT DIAGRAM

NAME PLATE DETAILS:

Generator: 230V, If = 0.68A, Ia =14.84A, 3.7kW, N=1500 rpm

Motor: 230V, If = 0.4A, Ia = 19.2A, 3.7kW, N=1500 rpm

PROCEDURE

1. Take down the name plate details of the machines and select proper
rheostats, meters and other components.
2. Set up the circuit as in figure. Keep the position of Rl,R2 and R3 properly. Keeping
the switch S2 open, close switch S1 and run the machine M as motor a t rated speed
by adjusting Rl.
 3. Adjust R3 such that the voltage across S2 becomes zero. Close S2 parallel t o the
machine G.
4. Adjusting R3 properly, obtain different load currents. For each case takedown the
5. readings at all meters. The rated speed is to be maintained through the experiment.

6. Measure the resistance of armature windings by proper method

SAMPLE CALCULATION:

(For motor)

Wc = [V*I1 – (I1 + I2)2*rm – I22*rg] / 2

i/p = V*(I1 + I2 + I3) =

o/p = i/p – Wc – V*I3 – (I1+I2)2*rm =

Efficiency = (o/p / i/p) *100 =

(For generator)

o/p = V*(I1 + I2 + I3) =

i/p = o/p + Wc +V*I3 + (I1+I2)2*rm

Efficiency = (o/p / i/p) *100=

OBSERVATION TABLE:
I1 I2 I3 I4 V Wc Motor Generator

(A) (A) (A) (A) (V) (W)

i/p o/p Eff. o/p i/p Eff.

(W) (W) (%) (W) (W) (%)

1.5 2 0.42 0.3 210 151

1.6 4 0.42 0.3 210 149

1.8 6 0.42 0.32 210 150

2.2 8 0.42 0.35 210 163.8

2.5 10 0.42 0.35 210 160

3 12 0.42 0.37 210 167.4

3.6 14 0.42 0.4 210 175.7

Machine V (V) I (A) Rd (ohm) Mean

0.8 1 0.8
Motor 1.2 1.5 0.8
1.6 2 0.8
0.8 1 0.8
Generator 1.2 1.5 0.8
1.6 2 0.8

Questions
1. calculate various losses and pre-determine efficiency of the machine at 25% 50% 75%
90% and 100% rated load. plot efficiency characteristics of both machines.
2. Why is the Hopkinson’s test called regenerative test?
3. What are the merits of Hopkinson’s test?
4. What are the components of stray losses?
5. In the Hopkinson’s test after the two machines are parallel
a) Which of the machine is motoring? Why?
b) Is it possible to run the motoring machine as a generator? How?
c) Is it possible to run both machines as motors? How?
d) Can both machines be made to run as generates?
6 Can the Hopkinson’s test be conducted on two identical mechanically coupled DC
compound machines?
7 Is the Hopkinson’s test more accurate than the Swinburne's test? If so, why?
8 why the voltage across the switch S2 main zero before closing
 8)RETARDATION TEST

CIRCUIT DIAGRAM

NAME PLATE DETAILS:

Motor: 230V, 11kW, 1.15A, 1500 rpm

PROCEDURE
1. Take down the name plate details of the machines and select proper rheostats, meters
and other components.
2. Set up the circuit as in figure. The switch S2 is to be thrown to position (1).adjusting R1
properly, choose switch S1. Run machine at rated speed at rated voltage
3. Adjust R1 properly to increase the speed slightly (say 1600 rpm). Open the switch S1 and
throw the switch S2 from the position (1) to (0); Note down the time for the speed to fall
from 1550 to 1450 rpm.
4. Repeat step 2. Adjust for a speed of 1600 rpm. Throw the switch S2 from the position (1) to
position (0). Note down the time for dropping speed from 1550 to 1450 rpm.
5. Repeat step 2. Adjust for a speed of 1600 rpm. keeping S1 closed through switch S2 from
6. position (1) to position (2). note down the time taken for the reduction in speed from 1500
rpm 1450rpm. also note down the wattmeter reading corresponding to the speed 1550 rpm
and 1450 rpm

SAMPLE CALCULATION:
Wc = (W1 + W2 )/2 =

2𝜋 𝑑𝑁
Wm + Wi = ( ) 2 𝐽 𝑁 ( ) (1)
60 𝑑𝑡 2

2𝜋 𝑑𝑁
Wm + Wi + Wc = ( 60 ) 2 𝐽 𝑁 ( 𝑑𝑡 ) (2)
3

(2) – (1) = Wc =

=> J =

Wm =

Wi =

Wc =

OBSERVATION TABLE:

S1 Position S2 Position Time dN/dt Wattmeter Reading

initial final Initial final Taken (s) (rpm/s) initial final

Close Open Close Open 5 40 - -

Close Close Close Open 4.72 42.3 - -
Close Close Close Close 2.25 88.8 320 250

Questions
1. Determine and separate out the losses in the DC machine in to iron losses and mechanical
losses
2. Find the moment of inertia of the Rotor
3. Explain Windage losses stray load losses
4. Compare different methods of finding separating losses in DC machine
5. Is the wattmeter reading constant during retardation? Explain
6. Status the assumptions/ limitations of the test
7. Explain the variations of friction-windage-hysteresis-eddy current-armature copper losses
with
a) speed
b) load
c) voltage
d) frequency
 9)SEPARATION OF LOSSES IN A SINGLE-PHASE
TRANSFORMER

CIRCUIT DIAGRAM

NAME PLATE DETAILS:

DC Motor: 230V, 15.2A, 3.5kW, 1500 rpm

Alternator: 3-phase upf, 230/400V,10.8/6.3A, 4kW, 1500 rpm, 50Hz

Rotor: 300V, 7.8 A

PROCEDURE
1. Take down the name plate details of the machines and select proper meters and other
components.
 2. Determine the value of V/F corresponding to rated frequency and voltage and also
determine the voltages to be maintained at different frequencies, say 40,45,50 and 55 Hz
to keep V/F ratio constant.
3. Set the circuit as in figure. Keep the switch S2 open. Adjust the position of R1 and auto
transformer properly. Close switch S1 and run the alternator at rated speed (for 50 Hz
frequency)
4. Excite the alternator till the alternator terminal voltage reaches the rated value (say 230
V). Close the switch S2 and note down the wattmeter reading (W).
5. Run the alternator at speeds corresponding to different frequencies 40,45,50,55 Hz and
adjust the corresponding terminal voltage (as calculated in step 2) by varying the
excitation to the alternator. Note down the wattmeter reading (W).

SAMPLE CALCULATIONS:

V/f = = constant

 Voltage at new f = V/f *fnew =

 N = (120 * fnew)/P =
 P=4
W/f = = A + B*fnew

A=

B=

Wh = A * fnew =

We = B * fnew2 =

 Wi = Wh+ We
OBSERVATION TABLE:
Speed Frequency Voltage Wattmeter V/f W/f Wh We Wi
(rpm) f (V) (W) (W) (W) (W)
(Hz)
1500 50(f) 220 34
1470 49 215.6 32
1440 48 211.2 30
1410 47 206.8 30

Questions
1. Determine the core loss at various frequencies and hence to separate its components at
a) rated voltage and rated frequency?
b) rated voltage and different frequencies in the range 45Hz to 55Hz?
2. What steps are taken in the design of a transformer to minimize its iron losses?
3. How does the iron loss vary with frequency? Explain?
4. Why the core is laminated in plane parallel to the flux path?
5. Can a 50 Hz transformer be used for 25Hz? Explain?
6. While separating core loss components by variable frequency method, why do you keep
the ratio (V/f) constant?
7. What is the order of magnitude of the exciting current in terms of rated current in power
transformers?
8. Is the area of hysteresis loop dependent upon the frequency of alteration of flux?
9. If a transformer is operated at rated frequency but at a voltage higher than
10. the rated value, how do you expect the following quantities to change
a) No load current
b) Hysteresis loss
c) Eddy current loss
d) Copper loss
 10)SUMPNER'S TEST

CIRCUIT DIAGRAM

NAME PLATE DETAILS:

Auto-Transformer: 250V, 15A, 1-phase

Transformer: 1.25 kVA, 200/115 V

PROCEDURE
1. Take down the name plate details of the machines and select proper meters and other
2. components.
3. Set up the circuit as shown in fig.
 4. Keeping the switch S3 open, close switch S1. Note down the reading of the voltmeter
connected across S3. If it is zero proceed to step 4 or get the connections corrected.
5. Close the switch S3 and switch on S2 keeping the position of the autotransformer
property.
6. Adjust the autotransformer so that rated current flows through the secondaries of the
Transformer.
7. Note down the readings of all meters.

SAMPLE CALCULATION:
WO = WC = Wi/2 = Wsc = W2/2 =
Vo = , Io = I1/2 = Vsc = V2/2 =
𝑐𝑜𝑠∅ = Wo / (Vo*Io) = Isc =
Iw = Io *𝑐𝑜𝑠∅ = Z02 = Vsc/Isc =
Iu = Io *𝑠𝑖𝑛∅ = R02 = Wsc/Isc2 =
 Xo = Vo/Iu = Xo2 = (Z022 - R022)1/2 =
 ro = Vo/Iw = R01 = R02/k2 =
 Xo’ = Xo * k2 = X01 = X02/k2 =
 ro’ = ro * k2 =
(Voltage Regulation)
𝑥 𝐼𝑟
Voltage Regulation(V. R) = ∗ (𝑅01 𝐶𝑂𝑆∅𝑜 ± 𝑋01 𝑆𝑖𝑛∅𝑜 )
𝑉𝑟

Lag =>+
Lead => -
(Efficiency)
P0 = x*S* 𝐶𝑂𝑆∅𝑜 =

Efficiency = P0/( P0+(x2*copperloss)+ironloss)

OBSERVATION TABLE:

I1 (A) I2 (A) V1 (V) V2 (V) W1 (W) W2 (W)

1.5 5.7 115 3.6 44 160

Voltage Regulation
o/p rated 0.5 pf 0.8 pf 0 pf Unity pf
I (A)
lag lead lag lead lag lead
(*10-3) (*10-3) (*10-3) (*10-3) (*10-3) (*10-3)
0.1

0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1

Efficiency
Rated 0.5 pf 0.8 pf u.p.f
Current
(A) o/p i/p Eff. o/p i/p Eff. o/p i/p Eff.
(W) (W) (%) (W) (W) (%) (W) (W) (%)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1

Questions

1. Pre-determine the efficiency and regulation at 25%, 50%, 75% and 100% rated load
under 0.5 lag & lead,0.8 lag & lead and upf conditions
2. Draw
a. efficiency-output characteristic for UPF and 0.8 lag PF conditions
b. regulation-PF characteristics for full load.
3. Determine the equivalent circuit of the transformer?
4. Compare the above results with that obtained under OC&SC tests, If the experiments are
done on the same transformer.
5. What are the relative merits of Sumpner's test?
6. In the back-to-back test, will the wattmeter reading (W1) and ammeter reading (I1)
change, when current is injected into the secondary circuit?
Explain.
 11)SCOTT CONNECTION

CIRCUIT DIAGRAM

NAME PLATE DETAILS:

Transformer: 400/230 V, 2 kVA

PROCEDURE
1. Take down the nameplate details of the machines and select proper meters and other
components.
2. Keep switch S2 and S3 open. Close switch S1 and measure the output voltages.
3. Close switch S3. Vary the load RL.2 and take down the readings for different positions.
4. Open switch S3 and close switch S2. Vary the load RL1 and take down the readings for
different load positions.
5. Close switches S2 and S3 and vary the loads RL1 and RL2 for balanced and unbalanced
conditions.

SAMPLE CALCULATION:
Im = IT = Vm = VT =
Pi = WT + Wm =
Po = Vm * Im + VT * IT =
Efficiency = (Po / Pin) *100 =

OBSERVATION TABLE:

Switch IR IY IB IT Im VT Vm WT Wm Po P1 Eff.
position (A) (A) (A) (A) (A) (V) (V) (W) (W) (W) (W) (%)
S2, S3 0 0 0 0 0 240 240 0 0 0 0 0
open
3.65 2.1 1.4 5 0 240 240 310 0
S2 closed 4.6 2.8 1.8 6.5 0 238 240 395 0
S3 open 3.4 2.05 1.3 4.9 0 234 240 295 0
4.1 2.35 1.6 5.9 0 232 240 350 0
0.5 1.2 1.4 0 1.4 240 236 0 140
S2 open 0.5 1.8 2 0 2.9 240 236 0 200
S3 closed 0.5 2.3 .6 0 4 240 232 0 260
0.5 3.4 3.5 0 5.9 240 232 0 375
S2, S3 1.5 2 2 2 2.9 238 236 115 200
Closed 2.15 1.8 1.6 3 2 238 236 180 160
Unbalanced 3.4 2.65 2.4 4.9 2.9 234 236 295 200
load 4.05 3.25 3.1 5.8 4 232 232 245 270
S2, S3 1.5 1.5 1.5 2 2 238 236 120 140
Closed 2.15 2.15 2.15 3 3 238 236 180 200
Balanced 2.8 2.8 2.8 4 4 235 236 240 265
load 4.15 4.15 4.15 5.9 5.9 232 232 350 380

Questions

1. Determine the efficiency of conversion from 3Ø to 2Ø at various Upf loads

With
a) both main & teaser transformers loaded-- balanced loading
b) both main & teaser transformers loaded - unbalanced loading
c) main transformer alone is loaded
d) teaser transformer alone is loaded
Plot the efficiency characteristics for all the cases
2. What are the practical applications of Scott connection?
3. Can Scott connection provide 3Ø to 3Ø transformation?
4. Draw a phasor diagram showing the natural position on the three-phase side for (a) when
the two-phase side is balanced.
5. In Scott connected transformers are the iron losses for the main and teaser transformers
equal? If so, why?
6. Prove that the secondary voltage of the teaser and main transformers are in quadrature.
Also verify this through measurements
7. Prove that the main transformer rating is greater than that of teaser Transformer. Find
how much % greater.
8. Why the number of turns of the teaser primary is chosen equal to 86.6 percent of the
number of turns of main primary?