Sdes bee lab manual

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CIRCUIT DIAGRAM:

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THEVININ’S THEOREM

AIM: Verification of Thevinin’s Theorem.

APPARATUS:

S. no Apparatus Type Range Quantity
1 Thevinin’s Theorem 1
Trainer it
2 Regulated power supply (0-40)V 1
3 Ammeter Digital (0-100)mA 1
4 Voltmeter Digital (0-15)V 1
5 Digital Multi meter 1
6 Different load resisters 75Ω,100 Ω,150 Ω
7 Connecting wires

STATEMENT:

Thevenins theorem states that any circuit having no, of sources, resistances and AB open
O/P terminals can be replaced by a simple equivalent circuit consisting of single voltage source
in series with a resistance where the values of voltage source is equal to the open circuit voltage
across the output terminals and series resistance is equal to the resistance seen in to the network
from output terminals with all sources are replaced by their internal resistance.

PROCEDURE:
1 . Connect the circuit as shown in the circuit diagram.
2. Measure the current through the load resistance and note down IL .
3. Remove the load resistance and measure the voltage across A,B which givn the Thevivins
voltage(VTH)
4. Measure the resistance between AB by short circuiting the voltage source which gives
Thevenins resistance (RTH).
5. Connect the circuit as shown in Fig.2 with VTH, RTH and the load resistance.
6. Measure the load current and compare with the current flowing through the R L in original
circuit.
7. Thus Thevinin,s theorem is verified.

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TABULAR COLUMN:
S.no. Vs(V) IL (mA) ILl (mA) Rth(Ω) Vth(V)

CALCULATIONS:

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RESULT:

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CIRCUIT DIAGRAM:

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MAXIMUM POWER TRANSFER THEOREM
AIM:
Verification of Maximum Power Transfer Theorem and find out the value of load
resistance when max power transferred to it.

APPARATUS:

S. no Apparatus Type Range Quantity
1 Max. Power Transfer 1
Theorem Trainer it
3 Ammeter Digital (0-200)mA 1
6 Connecting wires

STATEMENT:
Max. Power Transfer theorem states that in a DC Network Max. Power
Transferred from the source to the load when load resistance is equal to the load resistance.

PROCEDURE:
1. Connect the circuit as shown in figure.
2. Measure the current passing through the load resistance RL and voltage across it for a supply
voltage of V Volts
3. Now vary the load resistance RL and measure the value of IL and VL_
4. Tabulate all the values and find the power absorbed by the load. Resistance in each case.
5. Observe the load resistance for which Max. Power is transferred and compare with the source
resistance.
6. Hence Max. Power Transfer Theorem is verified.

.

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TABULAR COLUMN:

S.NO. Load Current Load Voltage Power (PL) Resistance(RL)
(mA) (mA) (mW) (mΩ)
1
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.

EXPECTED GRAPH:

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GRAPH:
A Graph is drawn by taking different values of load resistance on X-axis and the
respective powers on Y-axis

CALCULATIONS:

RESULT:

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CIRCUIT DIAGRAM:

m

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SUPER POSITION THEOREM

AIM: Verification of Super position Theorem.

APPARATUS:

S.no Apparatus Type Range Quantity
1 Superposition 1
Theorem Trainer kit
2 Regulated power supply (0-40V) 1
3 Ammeter Digital (0-200) mA 1
6 Connecting wires

STATEMENT:
Super position theorem states that in any linear bilateral network consisting of two or
more sources, the response in any element is equal to the algebraic sum of the responses caused
by individual source acting alone, while the other sources are non operative that is voltage source
are replaced by a short circuit and current sources replaced by a open circuit

PROCEDURE:
1. Connect the circuit as per the circuit diagram
2. Set V1=15V and V2=0Volts.
3. Measure the current flowing through the ammeter I1,I2,I3.
4. Now short circuit the voltage source V2 and Measure the current flowing through the
resistance I1'.
5. Short circuit the voltage source V1, reconnect the voltage source V2, Measure the current I1’’ .
6. It is found that I1= I1’+ I1’’
7. Hence super position theorem is verified.

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TABULAR COLUMN:

I1 (ma)
V (VOLTS) THERITICAL PRACTICAL
V1=
V2=

V1=
V2=

V1=
V2=

CALCULATIONS:

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RESULT:

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CIRCUIT DIAGRAM:

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RECIPROCITY THEOREM

AIM: Verification of Reciprocity Theorem.

APPARATUS:

S no Apparatus Type Range Quantity
1 Reciprocity Theorem 1
Trainer it
3 Ammeter Digital (0-200)ma 1
6 Connecting wires

STATEMENT:
Reciprocity Theorem States that in any linear Bilateral network if a single voltage source
Va in Branch a produce a current Ib in Branch b, then the removal of voltage source from
branch a and its insertion in branch b will produce a current Ib in Branch a.

PROCEDURE:
1. Connect the circuit as per the diagram.
2. Set the voltage of power supply of 10V.and connect across the terminals A&B
3. A milli Ammeter connected to a terminal A&B and note down the current I.
4. Now interchange the position of ammeter and voltage source and note down the current value
let it be I’.
5. it is found that both the currents are equal that is I=I’.
6. Calculate the ratio of Voltage to current in both the cases.
7. It is found that both are Equal.
8. Hence Reciprocity theorem is verified.

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TABULAR COLUMN:

Voltage Current (mA) Voltage/Current
V1=
V2=
V1=
V2=

CALCULATIONS:

RESULT:

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CYCLE-II

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CIRCUIT DIAGRAM:

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Magnetization Characteristics of a D.C. Shunt Generator
Aim:
To draw the magnetization characteristics of a DC shunt generator to determine the
critical resistance (Rc) and critical speed (Nc).

Name Plate Details:
Motor Generator

Power = KW Power = KW
Armature voltage = volts Speed = rpm
Field voltage = volts Armature voltage = volts
Field current = amps Armature current = amps
Speed = rpm Field voltage = volts
Armature current = amps Field current = amps
Wound = Shunt Wound = Shunt

Apparatus Required:

S no Apparatus Type Range Quantity
1 Rheostat Wire wound 2A/200Ω 2
2 Ammeter Moving coil 0-2A 1
3 Volt meter Moving coil 0-300V 2
4 Tachometer Digital 0-10000 rpm 1

Theory:
Magnetization Curve:
The graph between the field current and corresponding flux per pole is
called the magnetization characteristic of the machine. This is same as B-H curve of the material
used for the pole construction.

In a d.c. generator, for any given speed, the induced e.m.f in the armature is directly
proportional to the flux per pole.

ZN P
Eg = X
60 A
Where, Φ is the flux per pole in Weber’s, Z is the no. of conductors in the armature, N is
the speed of the shaft in rpm, P is the no. of poles and A is the no .of parallel paths.

A = 2 (wave)
A = P (lap)

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Observation Table:
Vs=220V
S.No. If (Field) Eg Eg Eg
amps (increasing) (decreasing) (Average)
volts volts volts
1

2

3

4

5

6

7

8
Critical Resistance Calculations

Critical speed calculations

Sl.No. Speed (rpm) Induced
emf(volts)

1

2

3

4

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Open – Circuit Characteristics:

The armature is driven at a constant speed and the field current is increased gradually
from zero to its rated value. The terminal voltage (VL) at no-load condition is measured at
different If values. The graph, VL vs. If is called open-circuit characteristic. VL differs from Eg
due to (a) Armature reaction (b) voltage drop in the armature circuit. Ia is very small at no-load
condition, these effects are negligible. Hence, VL = Eg at no-load condition. Thus, the open
circuit characteristic is same as magnetization curve.
As shown in the figure

Critical Field Resistance (RC):

Critical Field Resistance is defined as the maximum field circuit resistance at which
the shunt generator would just excite at any given speed. At this value the generator will just
excites. If the field circuit resistance is increased beyond this value, the generator will fail to
excite.
Rc is given by initial slope value of the O.C.C. curve in the linear region (AB) passing
through the origin for the speed at which data is obtained.
If the field circuit resistance (Rf) is increased to RC, the machine fail to excite and no
e.m.f. is induced in the generator. For exiting the generator, Rf < RC.

Critical Speed:

For any given field circuit resistance, the speed above which the generator builds up
an appreciable voltage is called critical speed.

As E α N, the value of critical speed, Nc can be given as Nc = (B/A)*N

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EXPECTED GRAPH:

Rsh
critical field resistance
X C
Eg volts

Y
B
A

Z
O

If amps

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Procedure:

 Note down the ratings of the d.c .shunt motor and the d.c .shunt generator.
 Connect the circuit as shown in the diagram.
 Keep the generator field rheostat at maximum resistance position and motor field
rheostat in minimum resistance position.
 Now start the motor using a 3-point starter
 Adjust the motor field rheostat to bring the motor speed to rated value.
 Now decrease the field rheostat of generator and note down If and Eg up to the rated-
voltage of the generator.
 The experiment is repeated for decreasing order of If
 Maintain the speed of the motor (Prime Mover) at a constant value during the
experiment.
 Plot the magnetization curve

Graphs:
Draw the graph for (1) Eg Vs If & (2) Eg Vs N

PRECAUTION:

 The motor initially should be started without any load.
 The rotor resistance starter should be in the maximum resistance position while starting.

Result:

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CIRCUIT DIAGRAM:

,

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Swinburne’s Test

Aim:
To pre-determine the efficiency of a D.C. shunt machine considering it as a generator
or as a motor by performing Swinburne’s test on it.

Name plate details:
Motor

Power = hp Speed = rpm
Armature voltage = volts Field voltage = volts

Apparatus Required:

SL NO Apparatus Type Range Quantity
1 Voltmeter Moving coil 0-300V 1
4 Rheostat Wire wound 1.5/ 300 Ω 1
5 Tachometer Digital (0-10000 )rpm 1

Theory:

Testing of D.C .machines can be divided into three methods: (i) direct, (ii) Regenerative
and (iii) indirect.

Swinburne’s Test is an indirect method of testing a dc machine. In this method, the
constant losses of the D.C. machine are calculated at no-load. Hence, its efficiency either as
motor or as a generator can be pre-determined. In this method, the power requirement is very
small. Hence, this method can be used to pre-determine the efficiency of higher capacity dc
machines as a motor and as a generator.

Disadvantages:
(i) Efficiency at actual load is not accurately known (ii) Temperature rise on
load is not known and (iii) Sparking at commutator on load is not known.
Power input at No-load = Constant losses + Armature copper losses (which is negligible)
Power input at No-load = Constant losses
Power input = Va Ia + Vf If

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Observation table:

S.No VL(V) IL(A) If(A) Stray losses Fixed losses

As a motor:
Sl. IL Power Copper Total Power Efficiency
No. Input Loss Loss Output
1.
2.
3.
4.
5.

As a Generator:
Sl IL Power Copper Total Power Efficiency
No. Input Loss Loss Output
1.
2.
3.
4.
5.

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Losses in a DC machine:
The losses in a D.C. machine can be divided as 1) Constant losses 2) Variable losses,
which changes with the load.

Constant losses:
Mechanical Losses:
Friction and Wind age losses are called mechanical losses. They depend upon the speed.
A dc shunt machine is basically a constant speed machine both as a generator and as a motor.
Thus, the mechanical losses are constant.

Iron Losses:
For a dc shunt machine, the field current hence the flux per pole is constant (Neglecting
the armature reaction which reduces the net flux in the air gap). Hence, hysteresis and eddy
current losses (which are also called as iron losses) remains constant.

Field Copper Losses:
Under normal operating conditions of a D.C. shunt machine, the field current remains
constant. Thus, power received by the field circuit (which is consumed as field copper losses) is
constant.
Constant losses in a dc shunt machine=Mechanical + losses Iron losses+ Field cu. Losses.

Variable Losses:
The power lost in the armature circuit of a dc machine increases with the increase in load.
Thus, the armature copper loss is called as variable losses.

Procedure:

 Note down the ratings of the dc shunt motor
 Connect the circuit as shown in the diagram.
 Keep motor field rheostat in minimum resistance position.
 Now start the motor using a 3-point starter
 Adjust the motor field rheostat to bring the motor speed to rated value.
 Run the machine as a motor at no-load.
 Note down the voltage and current readings of the motor and generator at no-load.
 Calculate the efficiency of the machine working as motor and generator after taking
the values of field and armature circuit resistances.

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Conclusion:

 The power required to conduct the test is very less as compared to the direct loading test.
 Constant losses are calculated from this method are used to compute the efficiency of a
dc machine as a generator and as a motor without actually loading it.
Hence, this is an economic method

RESULT:

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CIRCUIT DIAGRAM:

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Brake test on a DC Shunt Motor
Aim:
To obtain the performance characteristics of a DC Shunt motor by a load test.

1) Armature current Vs Speed
2) Armature current Vs Torque
3) Armature current Vs Induced emf
4) Armature current Vs Flux per pole
5) Torque Vs Speed
6) Output Vs Efficiency

Name plate details:
Motor

Power = hp Speed = rpm
Armature voltage = volts Field voltage = volts

Apparatus require:

Si no Equipment Range Type Quantity
1 Volt meter 0-300V Moving coil 1
2 Ammeter 0-2A Moving coil 1
3 Ammeter 0-20A Moving coil 1
4 Rheostat Wire wound 1.5/300Ω 1
5 Tachometer Digital 10000 rpm 1

Theory:

This is a direct method of testing a dc machine. It is a simple method of measuring motor
output, speed and efficiency etc., at different load conditions A rope is would round the pulley
and its two ends are attached to two spring balances S1 andS2. The tensions provided by the
spring balances S1 and S2 are T1 and T2 the tensions of the rope can be adjusted with the help of
swivels.

The force acting tangentially on the pulley is equal to the difference between the readings
of the two spring balances in kg- force.

The induced voltage Eb =V-Ia Ra and Eb= KΦN, Thus, KΦ=Eb /N
V= applied voltage, Ia =armature current, Ra =Armature resistance.
Total power input to the motor Pin =Field circuit power + Armature power

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Observation table:
Armature voltage =
Field voltage =
Field current =
No load speed =

Sl. Ia N T1 T2 Input Shaft ω Shaft % E K
No. amps rpm kg kg (Pin) Torque (rad/sec) Output η (volts) Vs/r
watts (j/rad) (watts)
1.
2.

3.
4.
5.
6.

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= VfIf + Va Ia

If ‘r’ is the radius of the pulley , then torque at the pulley is given by

Tshaft = 9.81 (T1~T2 )r = 1.5 (T1~T2) N-m
2 N
 = is the angular velocity of the pulley, in rad/sec.
60
2 N
Motor output power Pout =Tshaft *  =1.5 (T1~T2)
60

Pout
% Efficiency = X 100
Pin
A dc shunt motors rotates due to the torque developed in the armature when the armature and
field terminals are connected to the dc supply. The direction of rotation can be explained with the
help of Fleming’s left hand principle.

A counter emf or back emf (Eb) is induced in the armature conductors while the armature
(rotor) rotating in the magnetic field. The direction of the induced emf can be explained with the
help of Fleming’s right hand principle and Lenz’s law. The is induced emf is also called as back
emf Eb.

ZN P
The equation of the motor is V= Eb + Ia Ra Where Eb = X
60 A
V  Eb
Ia =
Ra

The value of Eb is zero while starting the motor. Hence the voltage across the armature has to
be increase gradually.

2 N
The power developed in the rotor (armature) = EbIa = T ω Where ω =
60

In a dc motor T α Φ Ia where Φ= Flux produced by the shunt field per pole
Ia = Armature current

The torque developed in the motor is opposed by the torques due to (a) Friction and windage
(b) eddy currents and hysterisis and (c) mechanical load connected at the shaft. The motor runs at
a stable speed when the developed torque and resisting torques balance each other.

Let a small load be increased, then the resisting torque increases and motor speed falls. The
V  Eb
back emf reduces due to the fall in the speed. Hence, the armature current increases (Ia = )
Ra

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If Φ is assumed constant, (i.e. neglecting the armature reaction) the torque developed by the
mot or increases and a new stable speed is reached at which the developed torque equals the
resisting torque.

Armature Current ~ Speed characteristics:
The armature current Ia increases with increase in the load at the shaft. Hence Ia Ra drop
increases and counter emf (Eb) decreases.

Eb = V-IaRa where Ra is armature resistance and Eb α ΦN, if Φ is constant in the shunt motor
by neglecting the armature reaction; the speed falls as Eb falls.

In a dc motor Ra is very small, hence Ia Ra is a small value and fall in Eb with increase in load
is small. Thus, the speed falls slightly as Ia increases.

Armature current ~ Torque characteristics:
If Φ is constant, developed torque increases with increase in Ia
T= KΦ Ia
In actual condition, Φ slightly falls withy increase in Ia due to the effect of armature reaction.

Armature current ~ induced emf (back emf):

Induced emf (back emf Eb ) falls slightly with increase in Ia as per the equation Eb =V-Ia Ra

Armature current ~ Flux per pole:

The resultant Flux per pole decreases with the increase in Ia due to the effect of armature
react ion and amp-turns of the file d is constant at any load.

Torque ~ Speed:

With increase in load, Ia and Ta increases since the shunt field Φ is constant. The fall in speed
Eb
is very small as the Ia Ra drop is very small compared to V. In a dc shunt motor N α


Output ~ Efficiency

The graph between Output ~ Efficiency indicates that max torque occurs when armature
copper losses is equal to the constant losses. (Sum of field copper losses, mechanical losses and
Iron losses)

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Procedure:
1. Note down the name plate details.
2. Connect the circuit as shown in the diagram.
3. Keep the motor field rheostat in minimum resistance position.
4. Loosen the rope on the brake drum and put some water inside the rim of the brake drum
5. Now start the motor using a 3-point starter
6. Adjust the motor field rheostat to bring the motor speed to rated value.
7. Record the readings of the meters at no-load condition.
8. Gradually, increase the load on the brake drum and record the readings as per the given
table.
9. Do not exceed the armature current more than its rated value.
10. Gradually, reduce the load and switch off the supply.

PRECAUTION:


RESULT:

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OPEN CIRCUIT TEST:

SHORT CIRCUIT TEST:

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O.C. TEST AND S.C. TEST ON A SINGLE PHASE
TRANSFORMER
AIM:
1. To obtain the equivalent circuit of the transformer (ref LV & HV side).
2. To predetermine the efficiency and voltage regulation at various assumed
3. To verify the predetermined results by a direct load test.
NAME PLATE DETAILS:
Rating: ____________KVA
Primary Voltage: ____________ Volts
Secondary Voltage _____________Volts
APPARATUS:
S.N. Components Type Specifications Quantity
1 Ammeter MI 0 – 20 A& 2A 1+1 No
2 Voltmeter MI 0 – 300 V&75V 1+1NO
3 Wattmeter LPF (Dynamo Type) 3KW, 0 – 300V,2 A 1 No.
4 Wattmeter UPF(Dynamo Type) 3KW, 0 – 300 V, 10A 1 No.

THEORY:
A Transformer is a static device which transfers the electrical energy from one circuit
to a nother circuit with changes in voltages and current but without any change in the
frequency. The transformer works on the principle of electromagnetic induction between two
windings placed on a common magnetic circuit. The two windings are electrically insulated
from each other and also from the core. The approximate equivalent circuit of the transformer is
shown in figure –2
The losses in transformer are (i) magnetic losses and ohmic losses or copper losses. These
can be determined by performing (a) open circuit test and given transformer can be
predetermined at any given load (b) short circuit test. From the above tests, the efficiency and
voltage regulation of a given transformer can be predetermined at any given load. The power
consumed during these tests is very small compared to that in a load test. Another parting what
follows, LV side parameters are denoted by suffix 1 and HV side parameters by suffix 2
OPEN CIRCUIT TEST;
In the open circuit test, HV side is usually kept open and supply given to the LV side, as
Shown in the figure; when rated voltage is applied to the LV side, the ammeter reads the no-load
current I0 is 2 to 5% of full load current. Hence the copper losses at no-load are negligible. W0
represents the iron or core losses. Iron losses are the sum of hysteresis and eddy current losses.

W0 = VLV I0 Cos 0
Cos0 = W0 / VLV I0, I = I0 Sin 0 , Iw = I0 Cos0
R01 and X01 is ref LV.
R0= VLV / Iw, X0 = VLV / I

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OBSERVATIONS:
1. O.C Test:

O.C. Voltage (V) O.C Current (A) No-Load Power (w)

2. S.C Test:

S.C. Voltage (V) S.C Current (A) Power (w)

EQUIVALENT CIRCUIT FOR 1Ø TRNSFORMER:

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This test is performed to determine t he equivalent resistance and leakage reactance of
the transformer and copper losses at full-load condition.
In this test, usually LV side is shorted and meters are connected on HV side. A variable low
voltage is applied to the HV winding with the help of an auto-transformer. This voltage is varied
till the rated current flows in the HV side and LV side. The voltage applied is 5 to 10 percentage
of rated voltage, while the rated current flows in the windings. The wattmeter indicated the full
load copper losses and core losses at Vsc. But the core losses at this low voltage are negligible as
compared to the iron losses at the rate voltage.
2 2
Hence, Wsc = Full load copper losses = I 2 R2eq = I 2R02
2 2
Z02 = Vsc / Isc and X02 = Z 02 – R 02
Req2 and Xeq2 are referred to HV side. The same parameters, referred to LV side, will be 1 /
aeq2 and 1/eq2

PROCEDURE:

OPEN CIRCUIT TEST:
1. Connect the circuit diagram as shown in the figure 1.1.
2. Gradually increase the voltage using the auto-transformer till the voltmeter reads 230V
3. Record the voltmeter, ammeter and L.P.F. wattmeter readings.
4. The ammeter indicates the no-load current and wattmeter indicates the iron losses
5. Switch off the supply and set the auto-transformer at zero position.

S.C. TEST:
1, Connect the circuit diagram as shown in the figure 1.2
2. Gradually increase the voltage using the auto-transformer till the ammeter reads 4.8 amps (the
rated current of the transformer on HV side)
3. Record the voltmeter, ammeter and U.P.F. wattmeter readings.
4. The ammeter indicates Isc, Voltmeter indicates Vsc and wattmeter indicates Wsc copper losses
of the transformer at full load condition.
5. Switch off the supply and set the auto-transformer at zero position.

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CIRCUIT DIAGRAM:

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BRAKE TEST ON 3-Ph SQUIRREL DAGE INDUCTION MOTOR

AIM: To conduct the load test on three phase squirrel cage induction motor and to draw the
performance characteristics curve.

NAME PLATE DETAILS:

3Ø INDUCTION MOTOR
3Ø AUTO TRANSFORMER

APPARATUS:

Sl No Name of Type Range Quantity
Apparatus
1 Ammeter MI (0-10A) 1
2 Volt meter MI (0-600A) 1
3 Watt Meter UPF(Dynamo (600V,10A) 2
Type)
4 Tachometer digital (0-10000)rpm 1

PROCEDURE:
 Connections are given as per the circuit diagram.
 The TPSTS is closed and the motor is started with the help of rotor resistance starter.
Where the rotor resistance starter is turned on from maximum resistance to minimum
resistance position to run the motor at its rated speed.
 At No load condition the speed, current, voltage and power are noted down.
 By applying the load with the help of spring balance and brake drum arrangement the
speed, current, voltage, power and spring balance readings or noted for various values of
load up to the rated currents.
 The load is later released gradually and the Rotor resistance starter is brought to the
original position before switching off the motor.
 The motor is switched off.

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TABUL AR FORM

S No Line Load Watt meter Readings Spring Balance Speed
Voltage Current Readings
(VL) (IL) (W1) (W2) (S1) (S2) (rpm)
1
2
3
4
5
6

CALCULATIONS TABLE;

S.N Current Input Torque Output Power p.f ἠ=O/P/I/P
(Amps) Power (S1-S2)(R=t/2) 2NT/60
(W1+W2) (9.81)
1
2
3
4
5
6
7

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FORMULAE:

Torque= (S1-S2) (R + t/2)*9.81 N-m

S1, S2 –spring balance readings in Kg

R- Radius of the brake drum in m.

T- Thickness of the belt in m.
Output power =2πNT/60 watts

N- Rotor speed in rpm.
T- Torque in N-m.
Input Power = (W1+W2) Watts.

W1, W2 – Wattmeter readings in Watts.
Percentage of Efficiency= (output power)/Input power*100
Percentage of slip = (Ns-N) Ns*100.
Ns-Synchronous speed in rpm
N-Speed of the motor in rpm
Power Factor = (W1+W2) √3VLIL

PRECAUTION:


RESULT:

fr

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GRAPH for Mechanical Characteristics.

GRAPH for Mechanical Characteristics.

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Sdes bee lab manual