EXPERIMENT NO.-1
AIM: - To verify Thevenin’s theorem and study of Norton’s theorem.
APPRATUS REQUIRED:1. Resistors
2. Multimeter
3. DC Supply
4. Breadboard
5. Connecting wires
THEORY:Sometimes we want to determine the response (current,
voltage and power delivered) in a single load resistance in network, then
Thevenin’s and Norton’s theorem enables us to replace the reminder of the
network by a simple equivalent circuit determining the response in the load
resistance becomes much easier. The use of these theorem is very helpful
and time saving when we have to find the response in any branch of given
network for different values of load resistances.
According to the Thevenin’s theorem “ In any linear network
containing energy sources (generator) and resistances (impedances) can be
replaced by an equivalent circuit consisting of a voltage source V th in
series with Rth. The value of Vth is the open circuit voltage between the load
terminals of the network and Rth is the impedance measured between the
terminals with all energy sources replaced by their internal impedances”.
Current across the load resistances RL is given by the equation
IL=Vth/ (Rth + RL)
Vth =open circuit voltage across the terminals
Rth= equivalent across the terminals
RL =load resistances
Original circuit diagram
Thevenin’s Equivalent circuit
According to Norton’s theorem “ in any linear network containing
energy sources (generator ) and resistances (impedances) can be replaced by
an equivalent circuit consisting of a current source I n in parallel with RN
across the load terminals . The value of In is the short circuit current through
the load terminals after short circuiting it and RN is the impedance measured
between the load terminals (by removing the load impedance)with all energy
sources replaced by their internal impedances”.
Load current IL= In x Rn/ (Rn + RL)
Norton equivalent circuit
OBSERVATION TABLE:Thevenin’s equivalent circuit
Original circuit
S. No.
Supply
voltage
Load
resistance
Load
current
Thevenin’s Thevenin’s
Equivalent Equivalent
source
resistor
Load
current
RESULT:PRECAUTIONS:1. Circuit implementation should be accurate in breadboard.
2. Make sure that terminals of the connecting wires and resistors should
be properly attached with the holes of bred board.
3. Carefully vary supply variable DC voltage.
EXPERIMENT NO.-2
AIM: - Measurement of power in three phase circuit by two wattmeter
method.
APPRATUS REQUIRED:1. Dynamometer type Wattmeter- 5/10A,125/250/500V
2. MI Voltmeter 0-500v
3. MI Ammeter 0-10A
4. Three phase resistive load
THEORY:A wattmeter has two coils a current coil & a Potential coil. A single phase
wattmeter is connected in the circuit to measure the power
P = V I cos 
Where in the load current passes through the current coil & the voltage is
applied across the voltage coil.
Instead of using three wattmeter to measure 3 phase power, it is possible and
economical to use two wattmeter to measure 3- phase power.
The phasor diagram of this circuit , assuming balanced lagging load has been
shown in fig. as such , rms values of currents, I R ,IY ,IB are take equal in
magnitude and lagging by an angle  with respect to its own phase voltage ,
similarly, rms values of phase voltage are also equal in magnitude but
displaced by 1200 . The phase sequence has been assumed as RYB. Based of
the phasor diagram, power consumed and the power facto r of load can be
calculated from the reading of two watt meters w1and w2 as explained
below:-
Power consumed by the load:Current through the current coil of wattmeter, W1 = IR
Voltage across the pressure coils of the wattmeter, W1 = VRB
Phase difference between IR and VRB (ref. phase diagram) = 30-
Hence reading of wattmeter, W1=IR VRB cos (30- ) ------------ (i)
Similarly, reading of wattmeter, W2= IY VYB cos (30+) --------- (ii)
Moreover, IR = IY = IB = IL (line current) -------------------- (iii)
Also, VRY = VRB = VBR =VL (line voltage) ----------------- (iv)
Substituting equations (iii) and (iv) in to eqns. (i) and (ii) and then adding these,
W1 + W2 = 3 VL IL cos (power draw by 3 phase load) ------------ (v)
Hence, the sum of two wattmeter reading is equal to the total power by a 3
phase balanced load.
Power factor of the load:Subtracting eqns. (ii) from eqn. (i)
W1 – W2 = VI IL cos (30- ) – VL IL cos (cos +)
= VL IL sin 
------------------ (vi)
Dividing eqn. (vi) be eqn. (v)
Tan  =
3
W1 W 2
W1 W 2
1
Thus power factor of the load, =
1  3(W 1  W 2 /W 1  W 2) 2
Hence, the power factor of the load can also be calculated from the observed
reading of the two wattmeter.
OBSERVATION TABLE:-
S. No.
V
I
W1
W2
W1+W2 W1-W2
RESULT: - Power factor for resistive load is almost unity.
PRECAUTIONS:1) Read the manual before work on panel.
2) All connections should be tight.
3) Take reading carefully.
4)
Never touch live wire.
COS
EXPERIMENT NO.-3
AIM: - Measurement of efficiency of a single phase transformer by load
test.
APPRATUS REQUIRED:1. Single Phase Transformer (1KVA)
2. Control Panel
(a) MI Volt Meter 96 x 96 mm sq 0-300 V
(b) MI Ammeter 96 x 96 mm sq 0-10A
(c) Watt Meter (single phase dynamo type)
(d) Single phase Variac (Auto Transformer) 0-270V
(e) Suitable Electric load
THEORY:A transformer is an electromagnetic device which transforms electrical energy
from one circuit to another circuit without change in frequency. In the simplest
version it consists of two windings wrapped around a magnetic core; windings
are not electrically connected, but they are coupled by the magnetic field, as it
shown. When one winding is connected to the AC electric power, current is
generated. This winding is called the primary winding. The primary current
produces the magnetic field and the magnetic flux links to another winding,
called the secondary winding. The AC flux through the secondary winding
produces an AC voltage, so that if some impedance is connected to the
terminals, an AC electric current is supplied.
EFFICIECY OF A TRANSFORMER:The efficiency of a transformer at a particular load and power factor is defined
as the ratio of output to input.
output
Efficiency () =
input
Efficiency ()
Efficiency ()
Efficiency ()
output
output  cu..loss  iron..loss
input  losses
=
input
losses
= 1- input
=
LOAD TEST:In load test, we put a resistive load on secondary side of the transformer.
Performance of the transformer can be determined as follows from the
observations of the load test.
Power input to the transformer = W1 (reading of wattmeter)
Power output of transformer = V2I2 watts (cos  being unity for lamp load)
V 2I 2
Thus, efficiency at a particular load,  =
x 100%
W1
OBSERVATION TABLE:S No.
W1
V2
I2
RESULT: - The efficiency of the transformer is
V2 I2
%
PRECAUTIONS:1.
2.
3.
4.
All connections should be tight.
The circuit should be according to the circuit diagram.
Take all readings carefully.
Read the manual before work on the panel.
EXPERIMENT NO.-4
AIM:- Determination the parameters and losses in a single phase transformer
by open circuit and short circuit test.
APPRATUS REQUIRED:1. Single Phase Transformer (1KVA)
2. Control Panel
(a) MI Volt Meter 96 x 96 mm sq 0-300 V
(b) MI Ammeter 96 x 96 mm sq 0-10A
(c) Watt Meter (single phase dynamo type) 2.5/5 Amp
75,150,300V portable UPF type
(d) Watt Meter (single phase dynamo type) 1/2 Amp
75,150,300V portable LPF type
(e) Single phase Variac (Auto Transformer) 0-270V
THEORY:(A) Open circuit test :
In this test low voltage winding is connected to the supply as per rating of
transformer and the high voltage winding is left open as shown in fig. The
primary winding draw very low current hardly 2% to 5% of full load
current(may be up to 10% for very small rating transformer used for laboratory
purpose) under this condition. As such copper losses in the primary winding
will be negligible. Thus mainly iron losses occur in the transformer under no
load or open circuit condition, which are indicated by the wattmeter connected
in the circuit.
Hence, total iron losses = W0 (reading of wattmeter)
From the observation of this test, the parameter R0 and Xm of the parallel
branch of the equivalent circuit can also be calculated, following the step give
below:
Power drawn , W= V0 I0 cos  0
Thus , no load power factor ,cos 0 =
W0
V 0I 0
Core loss component of no load current , Iw = I0 cos 0
Magnetizing component of no load current, Im = I0 sin 0
V0
Equivalent resistance representing the core loss, R0 = Iw
Magnetizing reactance representing the magnetizing current X0 =
V0
Im
(B) Short circuit test :
In this test , low voltage winding is short circuited and a low voltage hardly
5% to 8% of the rated voltage of the high voltage winding is applied to this
winding . this test is performed at rated current flowing in both the winding .
Iron losses occurring in the transformer under this condition is negligible,
because of very low applied voltage. Hence the total losses occurring under
short circuit are mainly copper losses.
Total full load copper losses = Wsc (reading of wattmeter)
The equivalent resistance Req and reactance Xeq referred to a particular winding
can also be calculated from the observation of this test , following the steps
give below:
Equivalent resistance referred to H.V. winding ,
W sc
Req = I 2 sc
Also , equivalent impedance referred to H.V. winding , Zeq =
Thus the equivalent reactance referred to H.V. winding
Xeq =
V sc
I sc
Z 2 eq  R 2 eq
OBSERVATIN TABLE:
S.No.
Open circuit test
V0
I0
Short circuit test
W0
VSC
ISC
WSC
CALCULATION:
Cos 0
S. No.
Rc
Xm
Re
RESULT:-
PRECAUTIONS:1.
2.
3.
4.
All connections should be tight.
Take reading carefully.
Never touch live wire.
Read the manual before work on the panel.
Xe
Ze
EXPERIMENT NO:-5
AIM: - Study of characteristics of DC Motor.
(a) Armature current vs. Speed
(b) Armature current vs. Torque
(c) Torque vs. Speed
APPARATUS REQUIRED:1. MC Voltmeter 96  96 mm flush mounted
2. MC Ammeter 96  96 mm flush mounted
3. Indicating Light
4. DPIC Switch 16 amp, 240volt
5. DC Starter face plate type suitable for above motor.
Specification
:
Motor
Type
Insulation
Type of loading
:
:
:
SPDP (Screen Protected)
Class ‘B’
Mechanical
Specification
:
Mechanical loading
Type
:
Mechanical loading through belt pulley
:
CI type drum pulley suitable for water cooling
d= 185mm approx.
Round dial type
arrangement.
Diameter of
Drum pulley
Spring balance
:
THEORY:-
The load test on a dc motor is performed to obtain its various
performance Characteristics Including efficiency. The motor can be loaded by a belt
and pulley arrangement as shown in figure. If w1 and w2 be the tension in kg
indicated by the two spring balances provided on the two sides of the belt, then the
load torque o the motor is given by,
Load torque, T = (w1- w2)  r kg – m
The speed of the dc motor is given by the following expression,
Speed of rotation, N = k
V  IaRa

Where,
r – Radius of the pulley in meter.
N – Speed of the motor in RPM.
V – Voltage applied to the motor.
I – input current drawn by the motor.
Where the applied voltage V is constant, more over the flux  is nearly
constant for shunt motor increase, because of the increase in the armature voltage
drop, Ia Ra. The drop I speed from no load to full load operation is hardly 4 to 5
percent of the rated speed; as such DC shunt motor is regarded as constant speed
motor for all practical purpose.
Figure:
Load Test on DC motor.
OBSERVATION TABLE:S No.
V
W1(Kg.) W2(Kg.)
Ia
N
W1- W2
RESULT:-
PRECAUTIONS:1.
2.
3.
4.
All connections should be tight.
The circuit should be according to the circuit diagram.
Take all readings carefully.
Read the manual before work on the panel.
T
EXPERIMENT NO:-6
AIM: - Study of characteristics of AC motor.
APPARATUS REQUIRED:AC Motor (Synchronous Motor)
THEORY: The action of the synchronous motor can be explained
through the 3 phase, 2 pole synchronous machine. The rotor has two poles
and the stator has two poles per phase. In practice it is usual to have salient
poles on the rotor, but the armature winding is housed in slots in the
concave periphery of the stator. In the fig salient type stator poles have
been shown for convenience only.
As the rotor is excited from DC supply, so the poles of the rotor retain
the same polarity throughout but the polarity of the stator poles changes
as it connected to an AC 3- phase supply mains .
First of all let us consider the rotor as stationary and in the position shown
in fig -a, at this instant the rotor S-pole is attracted to the stator N-pole and
therefore tends to rotate in clockwise direction .after half of period (i.e.
after T/2 second whereas T=1/f)
The polarity of stator poles is reversed but the polarity of the rotor poles
remains the same (as in fig -b). at this instant ,the rotor S-pole is
repelled by the stator S-pole, being similar in nature and therefore rotor
tends to rotate in counter –clockwise direction .Thus the torque acting on
the rotor of a synchronous motor is not unidirectional but pulsating one
due to inertia of the rotor ,it will not move in any direction .so the
synchronous motor has got no self starting torque.
Now let us consider the rotor rotating in clockwise direction, which is
possible by external means (by using damper winding or any external
arrangement etc.) and in the position shown in fig–a. At this instant as
mentioned above, the rotor S-pole is attracted to stator N-pole so the
torque act on the rotor in clockwise direction .after a half period the stator
polarity is reversed i.e. stator S-pole becomes N-pole and vice versa but if
the rotor is rotated at such a speed by some external arrangement at the
starting moment that rotor S-pole advances by a pole pitch (180 degree)
so that it is again under the influence of stator N pole as shown in fig –
c so the torque acting in on the rotor will be again clockwise. Hence a
continuous unidirectional torque is produced. Now if rotor in running
condition if we remove external means is removed the rotor will
continue to rotate in clockwise direction under the influence of
clockwise continuous torque acting on the rotor .
RESULT: -
Study of synchronous motor successfully has done.
PRECAUTIONS: -
EXPERIMENT NO:-7
AIM: - Study of characteristic of DC generator.
APPARATUS REQUIRED:1.
2.
3.
4.
5.
Volt Meter 0-300 Volt
Ammeter 0-10 Amp
DC shunt generator 1.5-H.P/Kw,6 Amp1440 Rpm ,220volt
Tubular rheostat 1.2A, 260,
DPIC Switch 16A, 240V
THEORY:-
The external characteristic of DC generator represent the
graphical relationship between the terminal voltage and the load current, the
generator being operated at constant rated speed and with the same excitation
as under the no load conditions. The nature of the characteristic depends upon
the following factors.
1. Voltage drop in the armature winding, inter pole and
compensating winding.
2. Voltage drop at the brush contact.
3. Voltage drop due to armature reaction.
External characteristic of the generator indicates the fall in the
terminal voltage as the load on the generator increases. External characteristic
of a shunt generator is more drooping compared to that of separately excited
generator. The fall in the terminal voltage due to increases in load can be
compensated by an additional winding on the field system and connected in
series with the armature winding i.e. by proving series field winding, thus
making the generator a compound generator.
EXTERNAL CHARACTERISTIC OF A DC SHUNT GENERATOR
For a shunt generator the external characteristic is obtained
with constant resistance in the exciting circuit, the connections necessary for
determining the characteristic experimentally. For the normal test the resistance
in the field regulator is adjusted as that normal e m f is generated on open
circuit and the resistance is not varied during the test which again is carried out
at constant speed and with the machine at its usual working temperature.
With the shunt generator the fall that takes place in the terminal
voltage when load is put on, is more marked then with the separately excited
machine. The increased drop is the fact that a drop in the terminal voltage
results in a fall in the value of the exciting current with a consequent fall in the
value of the flux. When load is put on the shunt generator its voltage thus tends
to fall by a process of exactly a reverse nature to that of “building up”
External characteristics of a DC Shunt Generator
WORKING We have selected instruments for machine of following
rating:1. Prime Mover : D.C. shunt wound, 3/2 HP, 220,230 V , 1500 RPM
2. DC Generator : Self excited – shunt wound 1.5/1.0 Kw, 230 v, 1500
rpm.
Both the machines are flexibly coupled and mounted on common m.s. channel
base.
OBSERVATION TABLE:-
MOTOR
VS
GENERATOR
IS
VG
IG
RPM
RESULT:PRECAUTIONS:1. Never start the dc shunt motor against full load of generator.
2. When the dc machines are put is use after a gap, clean the commutator of
DC machines by means of soft sand paper.
3. The position of rocker arm of machines should only be shifted, when so
required, to run the machine spark free.
4. Check the proper tension of spring on the carbons holder. If the carbons
are too short replace it immediately with the new ones.
EXPERIMENT NO:-8
AIM: - Speed control of DC shunt motor.
APPARATUS REQUIRED:1.
2.
3.
4.
5.
6.
7.
MC Voltmeter 96 x 96 mm
MC Ammeter 96 x 96 mm
DC shunt motor( 2/3HP, 230V, 1500 RPM)
Tubular Rheostat 260 , 1.4 A
Tachometer
DPIC Switch
DC starter face plate type
THEORY: - The back emf for a DC motor is given by,
Back emf, Eb = P ф N Z / 60A
The number of poles P, the armature conductors Z, and the number of parallel
paths A are constant for a particular machine.
The speed of a DC motor is given by the relationship:
N=K (V- Ia Ra / ф )_
This shows that the speed of D.C. series motor depends on the supply voltage
V, the armature Circuit resistance Ra and field flux which is produced by field
current. Depending upon the above factors there are two methods of speed
control of D.C. series motor.
1. Armature Resistance control: In this method a variable series resistance
is connected in series with the armature circuit. In this case the
current and hence the flux are affected by the variation of armature circuit
resistance The voltage drop in the Rse reduces the voltage applied to
armature and so speed of motor is reduced.
2. Variation of field flux: Since the flux is produced by the field current so
the control of speed in this method is done by control of field current. A
variable resistance Rd is connected across the series field, which
is called a diverter. A part of main circuit is diverted through Rd. Thus the
current flow through the field winding is reduced. This reduces the field flux
and speed of motor is increased.
OBSERVATION TABLE:Va1 =
S.No.
Field Current
If (A)
Va2 =
Speed
N (rpm)
Field Current
If (A)
Va3 =
Speed
N (rpm)
RESULT:PRECAUTIONS:1.
2.
3.
4.
Make sure that all connections are tight.
Vary the value of rheostat slowly.
Take the readings carefully and accurately.
Don’t touch the live connection.
Field Current
If (A)
Speed
N (rpm)
EXPERIMENT NO:-9
AIM: - To perform running and reversing of three phase induction motor.
APPARATUS REQUIRED:1.
2.
3.
4.
5.
6.
7.
3 Phase induction Motor
Voltmeter
Ammeter
DOL starter
Tacho Motor
DP Reversing Switch
Phase Sequence Indicator
THEORY:
Three phase induction motor is the most widely used motor for industrial purpose as
well as for traction purpose. It is highly recommended for its excellent running
performance at a constant speed. Induction motor is based on the principle of
electromagnetic induction. The induction motor essentially consists of two parts:
1. Stator
2. Rotor
The supply is connected to the stator and the rotor received power by induction
caused by the stator rotating flux, hence the motor obtains its name –induction motor.
The stator consists of a cylindrical laminated & slotted core placed in a frame of
rolled or cast.
When three phase windings of stator are supplied by three phase ac supply, it
produces a rotating magnetic field. This rotating field cuts the rotor conductors and
induced a current in the conductors. Due to the interaction of rotor current and stator
magnetic field a torque is developed which tries to rotate the rotor in the direction of
rotating magnetic field. Rotor is free to move therefore it starts to rotate at a fixed
speed. The direction of rotor can be reversed by changing any two terminals of AC
input.
RESULT:PRECAUTIONS:1. Do not press the switch of PSI for more than 5 Sec. Immediately release the
push button after checking the direction.
2. All connection should be tight.
3. Before changing the connection, switch off the supply properly.
4. Always use the starter of proper ratings.
EXPERIMENT NO.-10
AIM: - Study of single phase energy meter.
APPRATUS REQUIRED:1. Wattmeter
2. Control Panel
a. MI Volt Meter 96 x 96 mm sq 0-300 V
b. MI Ammeter 96 x 96 mm sq 0-10A
c. Suitable Electric load
d. Energy Meter
THEORY:A single phase energy meter consists of two electromagnets, an aluminum
disc, a counter for recording the number of revolutions. The upper magnet
caries a coil known as the voltage coil. The center limb also carries a friction
compensator. The power factor compensator consists of a copper strip around
the center limb. The lower magnet is a ‘U’ shaped core and carries the current
coil. The braking magnets consists of two anisotropic magnet located above
and below the disc. For getting a fine adjustment a steel screw, shunting part of
the magnetic flux, is used. The rotor disc has a magnetic suspension.
Voltage and current coil magnets produce two alternating fluxes proportional
to the supply voltage and current. These fluxes which setup eddy currents. Set up
a torque on the rotor disc causing it to rotate .The speed is proportional to the
power being measured by an energy meter installed. To prevent the disc from
rotating (when load is removed) an auto creep device is provided. This is done
by cutting two radial slots in the periphery of the disc diametrically opposite to
each other.
Schematic diagram for single phase energy meter connection
Single phase energy meter
FOMULA USED:-
R / K  kWT
x 100
kWT
kW = Power in kilo watt as computed from wattmeter reading.
R = No. of revolution made by disc of meter (blinks LED) in t sec.
K = Revolution kWh from name plate of the meter.
Meter error in % =
R
K

Energy recorded by meter under test =

Energy computed by reading of wattmeter = kw  t
in kwh.
OBSERVATION TABLE:S.
No
Wattmete
r Reading
M
F
Revoluti K
Final
on
wattmeter
(R)
reading(kW
)
t
RESULT:PRECAUTIONS:1. All connections should be tight.
2. Take reading carefully.
3. Never touch live wire.
4. Read the manual before work on the panel.
R/K
kWt
%
error