Experiments manual

Page No.1
Electrical Maintenance Lab Manual
Karnataka German Technical Training Institute
Gulbarga

Page No.2
Objective :
1.Study of Importance of Earthing Equipments Needed
1. Patch cords
Circuit diagram :
Figure 1.1
Procedure :
1) Make connections as shown in the above Figure 1.1. Connect the neutral point of the power
socket to the neutral point of the iron plug.
2) After that connect the phase point of the power socket to the phase point of the iron plug as
shown below Figure 1.2.
Figure 1.2
3) Here you will see that the device is “on”. Note that the earth connection is not required to make
a device on and keep modulatory that the earthing is just for our safety.

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4) Now connect a patch cord between body of iron and the hand of a human body. If there is any
current in the body of iron then it will be flowing through the human body, because the path for
electric current is getting completed through the human body.
Figure 1.3
5) We can prevent the flow of this current by putting some insulation in between.
6) Connect a patch cord between the Shoes and the feet of man. As shown in the following Figure
1.4.
7) Here you will see that the flow of current through the body will be stopped. So remember,
always wear shoes when working with any electric or electronic device.
8) Above condition can be occurred when earth connection is not there in our home.
Figure 1.4
9) Now we will see the effect of earth connection with the same circumstances.

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10) Remove the patch cord between shoes and the human feet. You will see that the flow of
current through the body is started again.
11) Now connect a patch cord between the earth point of the power socket and the earth point of
the iron plug, as shown below Figure 1.5.
12) Now you will see that the current on the body of iron will be flowing toward the earth through
the earth pin and the human body is safe now.
Figure 1.5
Fuse :
In electronics and electrical engineering a fuse (short for fusible link), is a type of
overcurrent protection device. Its essential component is a metal wire or strip that melts when too much
current flows, which breaks the circuit in which it is connected, thus protecting the circuit's other
components from damage due to excessive current.
A practical fuse was one of the essential features of Thomas Edison's electrical power
distribution system. An early fuse was said to have successfully protected an Edison installation from
tampering by a rival gas-lighting concern.
Fuses (and other over current devices) are an essential part of a power distribution system
to prevent fire or damage. When too much current flows through a wire, it may overheat and be damaged,
or even start a fire. Wiring regulations give the maximum rating of a fuse for protection of a particular
circuit. Local authorities will incorporate national wiring regulations as part of law. Fuses are selected to
allow passage of normal currents, but to quickly interrupt a short circuit or overload condition.
Characteristic Parameters : Rated current IN :
This is the maximum current that the fuse can continuously pass without interruption to
the circuit, or harmful effects on its surroundings.
The I 2 t value :
This is a measure of the energy required to blow the fuse element and is an important

Page No.5
characteristic of the fuse. It is an indication of the "let-through" energy passed by the fuse which
downstream circuit elements must withstand before the fuse opens the circuit.
Voltage drop :
The values of the voltage drop across a fuse are usually given by the manufacturer. A fuse
may become hot due to the energy dissipation in the fuse element at rated current conditions. The voltage
drop should be taken into account particularly when using a fuse in low-voltage applications.
Breaking capacity :
The breaking capacity is the maximum current that can safely be interrupted by the fuse.
Some fuses are designated High Rupture Capacity (HRC) and are usually filled with sand or a similar
material.
Voltage rating :
The voltage rating of a fuse should always be greater than or equal to the circuit voltage.
Low-voltage fuses can generally be used at any voltage up to their rating. Some medium-voltage and high-
voltage fuses used in electric power distribution will not function properly at lower voltages.
The speed at which a fuse operates depends on how much current flow through it and the material of
which the fuse is made. In addition, temperature influences the resistance of the fuse. Manufacturers of
fuses plot a time-current characteristic curve, which shows the time required melting the fuse and the
time required to clear the circuit for any given level of overload current.
Where several fuses are connected in series at the various levels of a power distribution
system, it is very desirable to clear only the fuse (or other over current devices) electrically closest to the
fault. This process is called "coordination" and may require the time-current characteristics of two fuses to
be plotted on a common current basis. Fuses are then selected so that the minor, branch, fuse clears its
circuit well before the supplying, major, fuse starts to melt. In this way only the faulty circuits are
interrupted and minimal disturbance occurs to other circuits fed by the supplying fuse.
Classification of fuse :
Fuses are often characterized as "fast-blow", "slow-blow" or "time-delay", according to
the time they take to respond to an over current condition. The selection of the characteristic depends on
what equipment is being protected. Semiconductor devices may need a fast or ultrafast fuse for protection
since semiconductors may have little capacity to withstand even a momentary overload. Fuses applied on
motor circuits may have a time-delay characteristic, since the surge of current required at motor start soon
decreases and is harmless to wiring and the motor.
According to the formula
P = VI
Where,
P = Power in Watt
V= Operating voltage in Volt
I = Current flowing through the load in Ampere
A 20A breaker should only carry 2400W continuously (if V is 110V).
However most breakers do not trip instantaneously when they reach 20A. Most are
thermal magnetic. This means that for a low value fault such at 30 or 40 amps the thermal portion of the
breaker comes into effect and eventually trips the breaker due to heat. If a hard fault of a few hundred
amps occurs the magnetic portion of the breaker will trip instantaneously.
Breakers are set up this way to allow for momentary overloads and to allow for motor

Page No.6
starting. When motors start they draw about 6 times as much current as they do when running. If you had
a motor that ran at 12A it would run fine on a 20A breaker. But if this was an instantaneous breaker as
soon as you tried to start it the motor inrush would be about 72A and the breaker would trip. A thermal
magnetic breaker would run just fine and let the motor start.
The job of an Electrical Engineer involves selection of the right over current device to
properly protect equipment. One of the things you must also do is properly coordinate the breaker
protecting your equipment to the breaker that’s protecting the whole panel and then coordinate it to the
breaker feeding the panel then coordinate it to the breaker feeding the substation and on and on back to
the generators.
To do this manufacturers provide breaker (and fuse) trip curves that show the time-
current characteristic of a breaker. After you obtain all these curves you must select the proper settings
that will protect the equipment and keep from tripping out things higher up the chain.
Maximum prospective short circuit current :
The maximum prospective short circuit current is the maximum electrical current which
can flow in a particular electrical system under short circuit conditions. It is determined by the voltage and
impedance of the supply system.
Interrupting rating :
A fuse also has a rated interrupting capacity, also called breaking capacity, which is the
maximum current the fuse can safely interrupt. Generally this should be higher than the maximum
prospective short circuit current. Miniature fuses may have an interrupting rating only 10 times their rated
current. Fuses for small low-voltage wiring systems are commonly rated to interrupt 10,000 amperes.
Fuses for larger power systems must have higher interrupting ratings, with some low-voltage current-
limiting "high rupturing capacity" (HRC) fuses rated for 300,000 amperes. Fuses for high-voltage
equipment, up to 115,000 volts, are rated by the total apparent power (megavolt-amperes, MVA) of the
fault level on the circuit.
Voltage rating :
As well as a current rating, fuses also carry a voltage rating indicating the maximum circuit
voltage in which the fuse can be used. For example, glass tube fuses rated 32 volts should never be used in
line-operated (mains-operated) equipment even if the fuse physically can fit the fuse holder. Fuses with
ceramic cases have higher voltage ratings. Fuses carrying a 250 V rating may be safely used in a 125 V
circuit, but the reverse is not true as the fuse may not be capable of safely interrupting the arc in a circuit
of a higher voltage. Medium- voltage fuses rated for a few thousand volts are never used on low voltage
circuits, due to their expense and because they cannot properly clear the circuit when operating at very
low voltages.
Packages :
Fuses come in a vast array of sizes and styles to cater for the immense number of
applications in which they are used. While many are manufactured in standardised package layouts to
make them easily interchangeable, a large number of new styles are released into the market place every
year. Fuse bodies may be made of ceramic, glass, plastic, fiberglass, molded mica laminates, or molded
compressed fibre depending on application and voltage class.
Cartridge (ferrule) fuses have a cylindrical body terminated with metal end caps. Some
cartridge fuses are manufactured with end caps of different sizes to prevent accidental insertion of the
wrong fuse rating in a holder. An example of such a fuse range is the 'bottle fuse', which in appearance
resembles the shape of a bottle.
Fuses designed for soldering to a printed circuit board have radial or axial wire leads.
Surface mount fuses have solder pads instead of leads.

Page No.7
Fuses used in circuits rated 200-600 volts and between about 10 and several thousand
amperes, as used for industrial applications such as protection of electric motors, commonly have metal
blades located on each end of the fuse. Fuses may be held by a spring loaded clip or the blades may be
held by screws. Blade type fuses often require the use of a special purpose extractor tool to remove them
from the fuse holder.
Some most common packages are shown below Figures :
Figure 1.6 Figure 1.7 Figure 1.8
Figure 1.9 Figure 1.10 Figure 1.11

Page No.8
Experiment 2
Objective :
Study of importance and mechanism of Fuse Equipments Needed :
1. Digital multimeter (DMM)
Circuit diagram :
FIGURE 2.1
Procedure :
1. Make connections as shown in the above Figure 2.1. Connect Digital Ammeter (or Digital Multimeter)
in the Fuse Demonstration section.
2. Now remove the cap of fuse holder and put a fuse from given set of fuses.
3. Put the first toggle switch at normal condition and another at “off” condition.
4. In this section the power supply, ammeter, bulb, toggle switches and fuse are connected in series.
5. Now switch on the power supply using toggle switch.
6. Observe that the bulb is glowing, see the reading of ammeter.
7. In case of any type of short circuit lot of current flows through the load and the load may get
damaged. But when a fuse is connected with the load in series then the same current will be flowing
through the fuse also and the fuse will get damaged before load.
8. Here you can see the live demonstration of above statement. Switch the second toggle switch at
short circuit condition and see the affect on fuse.
9. Change the fuse and you can use the load again.

Page No.9
Circuit breaker :
A circuit breaker is an automatically-operated electrical switch designed to protect an
electrical circuit from damage caused by overload or short circuit. Unlike a fuse, which operates once and
then has to be replaced, a circuit breaker can be reset (either manually or automatically) to resume normal
operation. Circuit breakers are made in varying sizes, from small devices that protect an individual
household appliance up to large switchgear designed to protect high voltage circuits feeding an entire city.
When closing DC circuits, the current reaches about 95% of its final steady state value
after time 3τ, where τ is the time constant. Similarly, when an alternating current is closed, the current
reaches a steady value after a transient process. The time depends upon the resistive, inductive and
capacitive elements of the circuit. The highest switching current is achieved if switching is effected at zero
voltage (very high peak currents can develop if the switch is closed on short circuit conditions).
Below a threshold voltage, any circuit can be opened without any arc formation. In
practice, however, the commonly used switches do produce an arc while interrupting the current. The arc
must be either kept limited or extinguished at the earliest in order not to damage the contacts.
Operation :
All circuit breakers have common features in their operation, although details vary
substantially depending on the voltage class, current rating and type of the circuit breaker. The circuit
breaker must detect a fault condition; in low-voltage circuit breakers this is usually done within the breaker
enclosure. Large high-voltage circuit breakers have separate devices to sense an over-current or other
faults. Once a fault is detected, contacts within the circuit breaker must open to interrupt the circuit; some
mechanically stored energy within the breaker is used to separate the contacts, although some of the
energy required may be obtained from the fault current itself. When a current is interrupted, an arc is
generated - this arc must be contained, cooled, and extinguished in a controlled way, so that the gap
between the contacts can again withstand the voltage in the circuit. Finally, once the fault condition has
been cleared, the contacts must again be closed to restore power to the interrupted circuit.
Type of circuit breakers :
 Magnetic circuit breaker
 Thermal circuit breaker
 Thermomagnetic circuit breaker
Magnetic circuit breaker :
Magnetic circuit breakers use a solenoid (electromagnet) whose pulling force increases
with the current. The circuit breaker contacts are held closed by a latch. As the current in the solenoid
increases beyond the rating of the circuit breaker, the solenoid's pull releases the latch which then allows
the contacts to open by spring action. Some types of magnetic breakers incorporate a hydraulic time delay
feature using a viscous fluid. The core is restrained by a spring until the current exceeds the breaker rating.
During an overload, the speed of the solenoid motion is restricted by the fluid. The delay permits brief
current surges beyond normal running current for motor starting, energizing equipment, etc. Short circuit
currents provide sufficient solenoid force to release the latch regardless of core position thus bypassing the
delay feature. Ambient temperature affects the time delay but does not affect the current rating of a
magnetic breaker.
Thermal circuit breaker :
Thermal breakers use a bimetallic strip, which heats and bends with increased current,
and is similarly arranged to release the latch. This type is commonly used with motor control circuits.
Thermal breakers often have a compensation element to reduce the effect of ambient temperature on the
device rating.

Page No.10
Thermomagnetic circuit breaker :
Thermomagnetic circuit breakers, which are the type found in most distribution boards,
incorporate both techniques with the electromagnet responding instantaneously to large surges in current
(short circuits) and the bimetallic strip responding to less extreme but longer-term over-current conditions.
Low voltage circuit breakers :
Small circuit breakers are either installed directly in equipment, or are arranged in a
breaker panel.
Figure 2.2
The 10 ampere thermal-magnetic miniature circuit breaker is the most common style in
modern domestic consumer units and commercial electrical distribution boards. The design includes the
following components:
1. Actuator lever : Used to manually trip and reset the circuit breaker. Also indicates the status of the
circuit breaker (On or Off/tripped). Most breakers are designed so they can still trip even if the lever
is held or locked in the on position. This is sometimes referred to as "free trip" or "positive trip"
operation.
2. Actuator mechanism : Forces the contacts together or apart.
3. Contacts : Allow current to flow when touching and break the flow of current when moved apart .
4. Terminals
5. Bimetallic strip
6. Calibration screw : Allows the manufacturer to precisely adjust the trip current of the device after
assembly.
7. Solenoid
8. Arc divider/Extinguisher
Rated Current :
International Standards define the rated current In of a circuit breaker for household
applications as the current that the breaker is designed to carry continuously (at an ambient air
temperature of 30 °C). The commonly-available preferred values for the rated current are 1 A, 6 A, 10 A,
13 A, 16 A, 20 A, 25 A, 32 A, 40 A, 50 A, 63 A, 80 A and 100 A. The circuit breaker is labeled with the rated
current in ampere with letter "B", "C" or "D" that indicates the instantaneous tripping current, that is the
minimum value of current that causes the circuit-breaker to trip without intentional time delay (i.e., in
less than 100 ms), expressed in terms of In:

Page No.11
Type Instantaneous Tripping Current
B Above 3 In up to and including 5 In
C Above 5 In up to and including 10 In
D Above 10 In up to and including 20 In
K
Above 8 In up to and including 12 In
For the protection of loads that cause frequent short duration (approximately
400 ms to 2 s) current peaks in normal operation.
Above 2 In up to and including 3 In for periods in the order of tens of seconds.
Z For the protection of loads such as semiconductor devices or measuring
circuits using current transformers.

Page No.12
Experiment 3
Objective :
Study of importance and mechanism of MCB Equipments Needed :
1. A load (of more than 1000W)
Circuit diagram :
Figure 3.1
Procedure :
1. Connect the mains cord to the trainer; connect the load in the power socket given on the trainer.
2. Move the lever of MCB at “on” position.
3. Switch on the load; switch on the supply by rotating the potentiometer.
4. Here you will observe a deflection in the ammeter. Ammeter is showing the current flowing in the
circuit.
5. Now increase the current slowly by rotating the potentiometer in clockwise direction.
6. Mounted MCB is of C type with current rating 1A that means it will trip instantaneously at 5A.
7. When the current is lower then 5A it will take some time until an enough magnetic field is generated
in the coil (7), see above diagram. And when enough magnetic field is generated in the coil, the core
inside the coil (7) will be pushed by the magnetic field of coil, and MCB will be tripped.
8. If you switch on the MCB again it will not be getting on because of residual magnetic field inside the
coil. Wait for few seconds and do the experiment again.

Page No.13
Experiment 4
Objective:
Study of the resistances individually, as well as in series & in parallel connections
Theory:
Resistors are identified or values can be known by their colour codes. There is a set of
co-axial rings or bands printed on the resister. In order to understand the significance of these rings, we
have to go through the following table.
Figure 4.1
For example: If any unknown resistor has following colour bands.
1. Brown 2. Black 3. Black 4. Black 5. Brown
So the value can be determined as 100 x 100 ± 1% 100 ± 1% 0
The first three rings or strips from one end give the first three significant figures of
resistance in ohm. The fourth ring indicates the decimal multiplier. The last ring indicates the tolerance in
percent above the indicated values.
Resistance in Series:
Resistance are said to be connected in series between two points, if they provide only a
signal path between the two points. When resistors are connected in series some current flows through
each resistor when some potential difference is applied across the combination. It means that the
equivalent resistance of any number of resistors is equal to the sum of their individual resistances.
Figure 4.2
RS = R1 + R2 + R3

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Resistance in Parallel:
Resistance are said to connected in parallel if the potential difference across each of them
is the same and is equal to the applied potential difference.
1/RP = 1/R1 + 1/R2 + 1/R3
Figure 4.3
For any number of resistors connected in parallel the reciprocal of the equivalent
resistance is equal to the sum of the reciprocals of their individual resistances.
Procedure:
1. Take any three resistors of same value and enter the colour codes of resistors into the table.
2. Enter the value of resistors according to the colour code in the table.
3. Enter the tolerances in the table.
4. Now measure the resistance of all three resistors with the help of multimeter. Enter the measured
value in the table.
Resistor
Colours Coded
Tolerance
Measured
1st 2nd 3rd 4th Resistance Resistance
1
2
3
5. Now connect the three resistors in series on the trainer board. Measure the resistance of the
combinations as shown in the following diagram by connecting the leads of the multimeter between
the points at the ends of shown arrows. You can calculate the resistances by using formulas and can
compare with the multimeter readings.
Figure 4.4

Page No.15
Calculate:
R12 = ……
R23 = ……
R123 = …..
It is clear that R123 = R1 + R2 + R3
6. Now connect the resistors in parallel as shown in figure and measure the resistance of different
combinations.
Figure 4.5
Calculate :
R12 = ……
R23 = ……
R123 = …..
It is clear that 1/R123 = 1/R1 + 1/R2 + 1/R3
Compare the result of formulas with the result of multimeter.
7. Now you can take different types of resistor values and repeat the above steps.
Figure 4.6
8. Construct a combination of series and parallel connections of resistors.
Calculate :
R1 = …… R23 = …… R123 = …..

Page No.16
Experiment 5
Objective:
Study of the ohm’s law mathematical relationship between three variables voltage (V),
current (I) and resistance (R)
Theory:
We know that electric current is proportional to drift velocity which is turn in proportional
to electric field strength. The electric field strength is proportional to potential difference. So the electric
current is proportional to potential difference, which is ohm’s law. If V is the potential difference and I is
the current, then V = IR , where R = resistance
Since current I is proportional to the potential difference V therefore the graph between V
and I is a straight line.
V – I graph for an ohmic conductor
Figure 5.1
Procedure:
1. Take any one resistor from given component box. You can detect its value from last experiment and
record it in table.
2. Now connect the circuit as shown in figure 5.2.
Figure 5.2
a. Connect a resister’s one end to an ammeter and other end to –ve terminal of DC power supply.
b. Connect positive of DC power supply to other end of ammeter.
3. Switch ‘On’ the trainer board.

Page No.17
4. Set the multimeter to the appropriate range and measure the current flowing through the
resistance. Record this value of current in the table.
Note: when current is to be measure we have to connect ammeter (multimeter) in series.
5. Now disconnect the above setup & connect the new circuit as shown in figure 5.3.
Figure 5.3
a Connect a resister across the DC power supply.
b Connect a voltmeter across the resistance.
6. Now measure the voltage across the resistor (Note that voltage is to be measure in parallel).
7. Record the corresponding voltages/current in table for different resistors. You can start with the
lower values of resistors.
Resistance Current Voltage Voltage/
(R ) (A) (V) Resistance
8. Draw graph between current (vertical axis) and resistance (horizontal axis). It is clear that current is
inversely proportional to resistance.
9. Now you can compare the values of V/R with current. According to the ohm’s law current is given by
ratio of voltage to the resistance.

Page No.18
Experiment 6
Objective:
Study of the voltage and current flowing into the circuit
Theory :
In any series circuit the voltage is distributed according to the size of the resistor. It means
that for high value of resistance a high voltage drop will be there and for a low value resistance low voltage
drop will be there.
In any parallel circuit the voltage is the same across all the elements of parallel
combination. It means parallel resistances represent a single resistance and that’s why the same voltage
drop is there. In combination of series and parallel circuits the parallel resistors were actually one resistor,
which is then in series with the first then the rules are same as above.
In series circuit the current is same for all values of resistance. But in parallel circuit
current is not same in all the branches it will be different for different resistances. In any resistance circuit
series or parallel or both the voltage, current and resistance are related by ohm’s law. i.e. V = IR This can be
observed in results.
Procedure :
1. Connect the three resistors of same value in the series circuit as shown in figure 10, connect the DC
power supply across it.
Figure 6.1
2. Measure the voltage drop across the resistance and series combinations with the help of multimeter.
R1 = V1 =
R2 = V2 =
R3 = V3 =
R12 = V12 =
R23 = V23 =
R123 = V123 =
Repeat the steps with different values of resistors.
3. Connect the parallel circuit and measure the voltage across each of the resistor and combination.

Page No.19
Figure 6.2
R1 = V1 =
R2 = V2 =
R3 = V3 =
R123 = V123 =
4. Connect the combination of series and parallel circuit.
Figure 6.3
R1 = V1 =
R23 = V23 =
R123 = V123 =
5. Now repeat the above steps for different values of resistances.
6. Again connect the same values of resistors in series, connect the DC power supply across it.
7. As we know that for current measurement we have to connect the ammeter in series. So make
connections according to figure 6.4.
Figure 6.4
Now measure the current from multimeter it is the current between supply and 1st resistance say I1.

Page No.20
Experiment 7
Objective:
Study of the behavior of current when light bulbs are connected in series/parallel circuit
Procedure:
1. Connect the circuit as shown in following figure 7.1.
Figure 7.1
a. Connect two light bulbs in parallel.
b. Connect third light bulb in series with parallel combination of two bulbs.
c. Connect DC/AC power supply between one end of parallel bulbs and other end of series bulb.
2. Now as you switch ‘On’ the trainer board the current flows in the circuit. It is equally divided in the
two parallel branches, but flows in full amount through the series bulb.
3. Result of this is, parallel bulbs are glowing with low intensity and series is glowing with full light.

Page No.21
Experiment 8
Objective:
Study of the Kirchhoff’s Law for electrical circuits
Theory:
Kirchhoff’s laws are simply the expression of conservation of electric charge and of energy.
There are two famous rules developed by Gustav Robert Kirchhoff in the year 1842. After him these rules
are known as Kirchhoff’s rules.
Kirchhoff First Law or Kirchhoff’s current law or junction rule:
In any electrical network the algebraic sum of currents meeting at a junction is always zero.
I = 0
The currents directed towards the junction are taken as positive while those directed
towards away from the junction are taken as negative.
I1 + I2 – I3 – I4 – I5 = 0
I1 + I2 = I3 – I4 – I5
Figure 8.1
From above expression we can say that the sum of current flowing towards the junction
is equal to the sum of currents leaving the junction.
Kirchhoff’s Second Law or Kirchhoff’s voltage law or loop rule:
The algebraic sum of all the potential drops around a closed loop is equal to the sum of
the voltage sources of that loop. This voltage law gives the relationship between the ‘voltage drops’
around any closed loop in a circuit and the voltage sources in that loop. The total of these two quantities is
always equal. Equation can be given by E source = E1 + E2 + E3 = I1R1 + I2 R2 + I3 R3
E = IR
i.e. Kirchhoff’s voltage law can be applied only to closed loop. A closed loop must meet two conditions.
1. It must have one or more voltage sources.
2. It must have a complete path for current flow from any point, around the loop and back to that point.
Procedure:
1. Connect the following circuit.
Figure 8.2

Page No.22
a. Connect R1 (330 ) to +ve end of supply and its other end to one ends of R2 (220 ) and R5 (100 ).
b. Connect other end of R2 to one ends of R3 (200 ) and R6 (100 )
c. Now connect other end of R3 to one ends of R4 (100 ) and R7 (100 )
d. Connect other end of R4 to one end of R8 (100 )
e. Connect other end of R8 to one end of R9 (100 )
f. Connect other end of R9 to other ends of R7, R6, R5 & and to one end of R10 (100 )
g. Connect other end of R10 to –ve end of power supply.
2. Now for testing the KCL at node ‘B’ we have to measure the following.
3. First remove supply connection to R1, now measure incoming current Iin between +ve end of source
and first end of resistor R1 i.e., connect the multimeter as ammeter between these two points.
Note: whenever we have to measure the current in the branch, we have to connect the ammeter in
series and after measuring the current disconnect the ammeter and make the connection as previous.
4. Measure outgoing current I1 between second end of R5 (when this terminal is not connected to trainer)
and common end of R10 and R9.
5. Measure outgoing current I2 between second end of R2 (when this terminal is not connected to trainer)
and junction of R3 and R6.
6. Check whether the incoming current Iin is equal to the sum of outgoing current (I1 & I2).
7. Repeat the procedure for junction point C, D, G, H, I.
To test the KVL in the loop ABIJ measure the following:
1. Measure current Iin following through resistor of 330 with the help of ammeter as step (3).
2. Measure current I1 as step 4.
3. Measure current Iout between point J and one end of resistor R10.
4. Calculate the different IR drops in the ABIJ loop (sign of IR drops should be given after considering
direction of current.)
5. Measure the sum of IR drop with their sign.
6. Equate the sum of all IR drops (with their sign) and sum of the source voltage of that particular loop it
should be equal.
7. In case of no voltage source in loop take the sum of all voltage source equal to zero.
Repeat above procedure for loop BCHI, CDGH, DEFG.

Page No.23
Experiment 9
Objective:
Study of the characteristics of:-
Semiconductor diode Schottky Diode
Theory:
Semiconductor Diode: A diode is an electrical device allowing current to move through it in one direction
with far greater ease than in the other. The most common type of diode in modem circuit design is the
semiconductor diode, although other diode technologies exist. Semiconductor diodes are symbolized in
schematic diagrams as shown below
Figure 9.1
When placed in a simple battery-lamp circuit, the diode will either allow or prevent current
through the lamp, depending on the polarity of the applied voltage:
Figure 9.2
When the polarity of the battery is such that electrons are allowed to flow through the
diode, the diode is said to be forward-biased. Conversely, when the battery is "backward" and the diode
blocks current, the diode is said to be reverse biased. A diode may be thought of as a kind of switch:
"closed" when forward-biased and "open" when reverse-biased.
V-I Characteristic:
The static voltage-current characteristics for a P-N Junction diode are shown in figure
Forward Characteristic:
When the diode is in forward-biased and the applied voltage is increased from zero, hardly
any current flows through the device in the beginning. It is so because the external voltage is being
opposed by the internal barrier voltage VB whose value is 0.7 V for Si and 0.3 V for Ge. As soon as VB is
neutralized, current through the diode increases rapidly with increasing applied supply voltage. It is found
that as little a voltage as 1.0 V produces a forward current of about 50mA.
Reverse Characteristic:
When the diode is reverse-biased, majority carrier are blocked and only a small current
(due to minority carrier) flows through the diode. As the reverse voltage is increased from zero, the reverse
current very quickly reaches its maximum or saturation value Io which is also known as leakage current. It
is of the order of nanoamperes (nA) and microamperes ( A) for Ge.

Page No.24
Procedure:
1. Connect the circuit as shown in figure 9.3.
Figure 9.3
a. Connect +ve terminal of DC power supply to one end of potentiometer & Connect middle terminal
of potentiometer to one end of resistor 1K.
b. Connect other terminal of resister to +ve end (Black end) of diode IN4007 & Connect –ve terminal
of DC power supply to other end of potentiometer.
2. Now rotate the potentiometer fully clockwise position.
3. Connect one multimeter (at voltmeter range) point A & B it means across the diode.
4. Now connect the multimeter (as ammeter) between –ve terminal of diode & ground, it means we
have to connect it in series with diode.
Figure 9.4
5. Switch ‘On’ the power supply.
6. Vary the pot so as to increase the value of diode voltage VD in the steps of 100mV.
7. Now record the diode current ID (mA) with corresponding diode voltage VD in table.
S. No. Diode Voltage VD Diode Current ID (mA)
8. Plot a curve between diode voltage VD and current ID as shown in figure (1st quadrant) using suitable
scale with the help of observation table. This curve is required forward characteristics of Si diode.
Figure 9.5

Page No.25
For Reverse Characteristics:
For this experiment only the polarity of diode will be reversed.
1. Measure the diode voltage VD in steps of 1 Volt and corresponding diode current same as the
previous given procedure.
2. Plot a curve between diode voltage VD and diode current ID as shown in figure 9.5 (3rd quadrant). This
curve is required characteristics of Si diode.
Schottky Diode: is a diode with a low forward voltage drop and a very fast switching action. A Schottky
diode is a special type of diode with a very low forward-voltage drop. When current flows through a diode
there is a small voltage drop across the diode terminals. A normal silicon diode has a voltage drop between
0.6–1.7 volts while a Schottky diode voltage drop is between approximately 0.15–0.45 volts. This lower
voltage drop can provide higher switching speed and better system efficiency. The symbol of Schottky
diode is shown as under:
Figure 9.6
A contact between a metal and a semiconductor is typically a Schottky barrier contact.
However, if the semiconductor is very highly doped, the Schottky barrier depletion region becomes very
thin. At very high doping levels, a thin depletion layer becomes quite transparent for electron tunneling.
This suggests that a practical way to make a good ohmic contact is to make a very highly doped
semiconductor region between the contact metal and the semiconductor.
V-I Characteristic:
The static voltage-current characteristics for a Schottky diode are shown in figure
Forward Characteristic:
Under forward bias (metal connected to positive in an n-doped Schottky), there are many
electrons with enough thermal energy to cross the barrier potential into the metal. Once the applied bias
exceeds the built-in potential of the junction the forward current If will increase rapidly with increase in Vf.
Reverse Characteristic:
When the Schottky diode is reverse biased, the potential barrier for electrons becomes
large; hence there is a small probability that an electron will have sufficient thermal energy to cross the
junction. The reverse leakage current will be in the nano ampere range.
Procedure:
1. Connect the circuit as shown in figure 9.7.
Figure 9.7

Page No.26
a. Connect +ve terminal of DC power supply to one end of potentiometer.
b. Connect middle terminal of potentiometer to one end of resistor 1K.
c. Connect other terminal of resister to +ve end (Orange end) of diode BAT85.
d. Connect –ve terminal of DC power supply to other end of potentiometer.
2. Now rotate the potentiometer fully clockwise position.
3. Connect one multimeter (at voltmeter range) point A and B it means across the diode.
4. Now connect the multimeter (as ammeter) between –ve terminal of diode and ground, it means
we have to connect it in series with diode.
Figure 9.8
6. Vary the potentiometer so as to increase the value of diode voltage VD in the steps of 10mV.
7. Now record the diode current ID (mA) with corresponding diode voltage VD in table.
S. No. Diode Voltage VD Diode Current ID (mA)
8. Plot a curve between diode voltage VD and current ID as shown in figure (1st quadrant) using
suitable scale with the help of observation table. This curve is required forward characteristics of
Schottky diode.
Figure 9.9
For Reverse Characteristics:
For this experiment only the polarity of diode will be reversed.
1. Measure the diode voltage VD in steps of 1 Volt and corresponding diode current (A) same as the
previous given procedure.
2. Plot a curve between diode voltage VD and diode current ID as shown in figure 33 (3rd quadrant). This
curve is required characteristics of Schottky diode.

Page No.27
Experiment 10
Objective:
Study of the characteristics of a transistor
Theory:
Transistor characteristics are the curves, which represent relationship between different
dc currents and voltages of a transistor. These are helpful in studying the operation of a transistor when
connected in a circuit. The three important characteristics of a transistor are:
1. Input characteristic.
2. Output characteristic.
3. Constant current transfer characteristic.
Input Characteristic:
In common emitter configuration, it is the curve plotted between the input current (IB)
verses input voltage (VBE) for various constant values of output voltage (VCE).
The approximated plot for input characteristic is shown in figure 34. This characteristic reveal that for fixed
value of output voltage VCE, as the base to emitter voltage increases, the emitter current increases in a
manner that closely resembles the diode characteristics.
Figure 10.1
Output Characteristic:
This is the curve plotted between the output current IC verses output voltage VCE for
various constant values of input current IB.
The output characteristic has three basic region of interest as indicated in figure 35. The
active region, cutoff region and saturation region.
In active region the collector base junction is reverse biased while the base emitter
junction if forward biased. This region is normally employed for linear (undistorted) amplifier.
In cutoff region the collector base junction and base emitter junction of the transistor
both are reverse biased. In this region transistor acts as an ‘Off’ switch.

Page No.28
Figure 10.2
In saturation region the collector base junction and base emitter junction of the transistor
both are forward biased. In this region transistor acts as an ‘On’ switch.
Constant current transfer Characteristics:
This is the curve plotted between output collector current IC verses input base current IB
for constant value of output voltage VCE. The approximated plot for this characteristic is shown in figure
36.
Figure 10.3
Procedure:
Circuit used to plot different characteristics of transistor is as follows.
Common emitter (NPN) configuration
Figure 10.4

Page No.29
Input Characteristics:
1. For these experiments you must have an external power supply of +12V.
2. Connect +ve end of +5V supply to one end of 10K potentiometer.
3. Connect center end of pot P1 to a resistor 5K.
4. Connect BC547 transistor to the trainer board.
5. Connect other end of resistor to base of transistor.
6. Connect emitter of transistor to –ve end of +5V supply and to third end of potentiometer.
7. Rotate the potentiometer in fully counter clockwise position.
8. Now connect collector of transistor to a resistor 100 and other end of resistor to center end of other
10K pot. (which is supplied in component box)
9. Connect a external +12V supply to other ends of potentiometer P2 and switch ‘On’ the trainer board.
10. Connect a voltmeter across collector and emitter of transistor and set VCE at some constant value say
1V with the help of P2.
11. Now vary the potentiometer P1 so as to increase the value of I/P voltage VBE (across base and emitter
of transistor) from 0 to 0.8V in steps and measure the corresponding values of I/P current IB by
connecting a ammeter in series between resistor 5K and base of transistor.
12. Repeat the above procedure for different constant values of VCE.
13. Now plot a curve between I/P voltage VBE and I/P current IB as shown in theory
Observation Table:
Input voltage
Input current IB (uA) at constant
S. No. value of output voltage
VBE
VCE = 1V VCE = 3V VCE =5V
1. 0.0V
2. 0.1V
3. 0.2V
4. 0.3V
5. 0.4V
6. 0.5V
7. 0.6V
8. 0.7V
9. 0.8V
Output Characteristics:
1. Switch ‘Off’ the supply.
2. Rotate both the potentiometer P1 and P2 in fully counter clockwise position.
4. Vary the potentiometer and set a value of I/P current IB at some constant value say (0 A, 10 A, …..100
A) by connecting a meter between resistor 5K and base of transistor.
5. Vary the potentiometer P2 so as to increase the value of O/P voltage VCE from 0 to maximum value in
steps and measure the output current IC by connecting ammeter between collector of transistor and
100 resistor.

Page No.30
6. Repeat the procedure for different values of I/P current IB.
7. Plot a curve between O/P voltage VCE and O/p current IC as shown in theory.
Observation Table:
Output voltage
Output current IC (mA) at constant value of
S.no. input current
VCE
IB = 0uA IB =10uA IB =20uA IB =30uA IB =40uA
1. 0.0V
2. 0.5V
3. 1.0V
4. 2.0V
5. 3.0V
6. 4.0V
7. 5.0V
8. 6.0V
9. 7.0V
10. 8.0V

Page No.31
Experiment 11
Objective:
Understanding the Faraday’s Law of electromagnetic induction
Whenever there is a change in the magnetic flux linked with a circuit an emf and
consequently a current is induced in the circuit. However, it lasts only so long as the magnetic flux is
changing.
Procedure:
1. Take any coil from the given coil set and let the two ends of the coil be connected to the two terminal
of galvanometer.
Figure 11.1
2. Now take a bar magnet and keep its north pole stationary near one end of coil as shown in figure 11.1.
The galvanometer shall not show any deflection. When magnet is stationary.
3. When the magnet is moved toward the coil, the galvanometer shows deflection as shown in figure
11.2.
Figure 11.2
4. When the magnet is moved away from the coil, the galvanometer again shows deflection but in
opposite direction.
5. Similar results are obtained when the magnet is kept stationary and the coil is moved.
6. When the magnet is moved slowly the deflection in meter is small, but when the magnet is moved fast
the deflection is large.
7. It is clear from above experiment that when magnetic flux. Changes through a coil, a current are
induced in the coil.
8. Observe the results for different coils.

Page No.32
Experiment 12
Objective:
Study of the behavior of current when inductance is introduced in the circuit
Procedure:
1. Connect the following circuit.
Figure 12.1
a. Connect AC power supply with a one end of coil (800 turn) and other end to a light bulb.
b. Connect other end of light bulb to other end of AC power supply.
2. Now as you switch ‘On’ the trainer board, you can observe that light bulb is glowing with good
intensity.
3. Take I-core and insert in the coil, result will be the light of decreased intensity.
4. The glow of the bulb will decrease because, as the iron rod is inserted in the coil its inductance
increases so inductive reactance increase. This result in an increase in impedance of the circuit.
Consequently, the current in the circuit decreases and hence the glow of the bulb decreases.

Page No.33
Experiment 13
Objective:
Study of the Lenz’s Law and effect of eddy current
The induced current produced in the conductor always flows in such a direction that the
magnetic field it produces will oppose the change that is producing it.
Lenz’s law states that in a given circuit with an induced emf caused by a change in a
magnetic flux, the induced emf causes a current to flow in the direction that oppose the change in flux.
That is if a decreasing magnetic flux induces an emf, the resulting current will oppose a further decrease in
magnetic flux. Likewise for an emf induced by an increasing magnetic flux, the resulting current flows in a
direction that opposes a further increase in magnetic flux. when the north pole of the magnet is moved
towards the coil, the direction of the induced current in the coil will be such that the upper face of the coil
acquired north polarity. So the coil repels the magnet. In other words, the coil oppose the motion of the
magnet towards itself which is really the cause of the induced current in the coil. Similarly, if the south
pole of a magnet is moved towards the coil, the upper face of the coil will acquire south polarity there by
opposing the motion of the magnet.
Eddy Current:
Eddy current may be defined as current induced in a thick conductor when the conductor
is placed in a changing magnetic field. Consider a metal block placed in a continuously varying magnetic
field. The magnetic field can be changed either by having a permanent magnetic field and moving the block
in and out of it or by keeping the block fixed and changing the magnetic field with the help of an
alternating current. Due to the continuous change of magnetic flux linked with the metal block, induced
current will be setup in the body of the metal block itself.
These current assume a circular path and their direction is given by Lenz’s law. These
current look like eddies in a fluid and hence called eddy current. Since the resistance of this conductor is
quite low therefore the eddy current are generally quite large in magnitude and produces heating effect. In
following experiment, we will see its effect.
Procedure:
1. Take a 400 turn coil.
2. Fix a U-core into the bracket given on the trainer board.
3. Insert the coil in any end of U core.
4. Connect one end of coil to positive terminal of DC power supply and other to the one terminal of
switch.
Figure 13.1

Page No.34
5. Connect –ve terminal of power supply to other end of switch.
6. Let the toggle switch in ‘Off’ (upward direction if first two terminals are used) condition
7. Take a soft iron square piece from the accessories and put it on the upper base of U-core where coil
is connected.
8. Switch ‘On’ the trainer board. As you switch ‘On’ the toggle switch the metallic piece is thrown up.
Because when the current begins to grow through the coil, the magnetic flux through the core and
hence metallic piece begins to increase. This sets up eddy currents in the metallic piece. If the
upper face of the core acquires N polarity in it then the lower face of the metallic piece also
acquires N polarity according to the Lenz’s law. Due to force of repulsion between same poles, the
metallic piece is thrown up.

Page No.35
Experiment 14
Objective:
Study of the relay and construction of a switching circuit by using relay
A relay is an electrical switch that opens and closes under control of another electrical
circuit. In the original form the switch is operated by an electromagnet to open or close one or many sets
of contacts. Generally relay is having following terminals and contacts.
 Input Coil: Operating voltage for relay is feeded to it.
 Normally closed (NC) Contact: It disconnect the circuit when the relay is activated.
 Normally Open Contact (NO): It connects the circuit when the relay is activated.
 Pole: It is the common terminal between NC and NO.
When a current flows through the coil, the resulting magnetic field attracts an armature
that is mechanically linked to a moving contact. The movement either makes or breaks a connection with a
fixed contact. When the current to the coil is switched off, the armature is returned by a force that is half
as strong as the magnetic force to its relaxed position.
Figure 14.1
Procedure:
Connect the circuit as shown in following figure 14.2.
Figure 14.2
1. Connect positive terminal of DC power supply to one end of coil and negative terminal to other end
of coil through a toggle switch.
2. Connect pole to any one terminal of AC power supply.
3. Connect NO terminal to one end of Buzzer and NC terminal of relay to one end of a light bulb.
4. Connect other end of AC power supply with other ends of Buzzer and light bulb.
5. Keep the toggle switch in off condition.
6. Now switch ‘On’ the power supply. In this condition relay coil is not getting supply voltage so the
pole and NC terminals are shorted with each other and since we have connected a light bulb with
this terminals in series with a AC power supply so it gets lightened.
7. Now as you turn ‘On’ the toggle switch, coil of relay will get supply voltage =5V and hence the NC
point of relay is separated from pole and NO point of relay attracted towards pole and make
contact with pole. Since we have connected a buzzer with these terminals in series with an AC
power supply so buzzer gives the sound of frequency 50 Hz.

Page No.36
Experiment 15
Objective:
Construction and study of the step down transformer with the help of given coils and
cores
Theory:
Transformer is working on the principle of electromagnetic induction i.e. when current in
one circuit changes, an induced current is set up in the neighboring circuit. It consists of two coils primary
and secondary. These coils are insulated from each other and wound over the same core. In order to avoid
eddy currents the core is laminated. The alternating electrical energy is supplied to the primary coil. The
output electrical energy is drawn from the 2nd coil. In the step down transformer the primary coil consists
of a large number of turns of fine insulated copper wire. The secondary coil consists of few turns of thick
insulated copper wire. In step up transformers the primary coil consists of few turns of thick insulated
copper wire and the secondary coil consists of large number of turns of fine insulated copper wire. In these
types of transformers the core is largely surrounded by coils.
The transformer makes use of faraday’s law and Ferro magnetic properties, of an iron
core to efficiently raise or lower AC voltages. It of course cannot increase the power so that if the voltage is
raised the current is proportionally lowered and vice-versa.
Figure 15.1
For an Ideal transformer the voltage ratio is equal to the turns ratio and power in equals
to the power out. It means
1. Vs = Ns
Vp Np
Where,
Vs = Secondary voltage
Vp = Primary voltage
Ns = No. of turns in secondary coil
Np = No. of turns in primary coil
2. Pp = VpIp = VsIs = Ps.
Procedure:
1. Take two 400 turn coils from the coil set.
2. Take U shaped core and fit it on the trainer board with the help of screws.
3. Now insert one 400 turn coil in U core as a primary coil and other as a secondary coil. Also put I
shaped core on the U core to complete the flux linkages.
4. Connect primary coil to the DC power supply.
5. Measure the secondary voltage with the help of multimeter.
6. Observe the result there will be no output voltage.
7. Now change the power supply from DC to AC.

Page No.37
8. Now measure the secondary voltage in multimeter there will be some reading.
Note: These are not ideal transformers. Since coils are not coaxially wound on the same core the losses are
more or the voltage transformation ratio is proportionately below the ideal values based on number
of turns per coil. But effective quantitative investigation can be done with the help of this set upto
20% loss may be obtained from desire voltage.
9. Note the reading (ideally it should be same as input voltage because number of turns are same). Also
note the percentage of losses. It will be helpful for further pairs of coils and for their calculations.
10. Now connect 200 turn coil in secondary and measure the voltage. You can observe the voltage is
lowered.
11. In the same way you can connect any large number of coil as a primary and can record the results for
different lower turns secondary coils in the table.
12. Now you can change the type of core and observe the results.
No. of turns
Type of AC
O/P voltage
Core I/P Voltage
Primary Coil Secondary Coil
400 400 6V
400 200 6V
800 400 6V
800 200 6V
1600 800 6V
1600 400 6V
1600 200 6V
3200 1600 6V
3200 800 6V
3200 400 6V
3200 200 6V

Page No.38
Experiment 16
Objective:
Construction and study of the step up transformer
Procedure:
1. Repeat step 1 to 9 from the previous experiment of step down transformer.
2. Now connect 800 turn coil in the circuit as secondary coil. Observe the result, the voltage is higher
than that of input voltage 6V.
3. Now you can connect any combination of a lower turn primary coil and large turns secondary coil.
4. Record the result in the given table.
5. Now observe the result by changing the core.
No. of turns
Type of Core
AC
O/P voltage
I/P Voltage
Primary Coil Secondary Coil
400 400 6V
400 800 6V
400 1600 6V
400 3200 6V
800 1600 6V
800 3200 6V
1600 3200 6V
200 400 6V
200 800 6V
200 1600 6V
200 3200 6V

Page No.39
Experiment 17
Objective:
Study of the effects of moving I core on a step up transformer
Procedure:
1. Take U core and fit it on the trainer board.
2. Take a 400 turn coil and insert it into the U core as primary coil.
3. Take a 1600 turn coil and insert it into the U core as secondary coil.
4. Now take a U shape small object and I core with the long screw from the accessories box.
5. Now fit the object on U core in such a way that the long screw of I core should be matched with the
given hole on the object. (when I-core is placed on U-core)
6. Tight the bottom screw of object at its extreme and upper screw of object slightly less than its
extreme position.
7. Now if you move long screw, of I-core the I-core will move from its position towards upper or lower
direction.
8. Now connect primary coil to 6V AC power supply and secondary to multimeter. Switch ON the
trainer.
9. If you move the screw, I-core will move and because of that power linkages will be changed and can
be observed in multimeter.
If a light bulb is connected across the secondary the effect of moving the I-core can be
demonstrated to the students.
Figure 17.1

Page No.40
Experiment 18
Objective:
Study the connection of a voltmeter in network and measure voltage through it.
Equipments Needed:
1. Meter Demonstrator
2. Patch chords
3. One 100W bulb as AC load
4. One 6V bulb as DC load
Circuit diagram:
Connection diagram for the experiment is shown in figure 18.1 and figure 18.2.
Figure 18.1

Page No.41
Figure 18.2
Procedure:
1. Connect 230V AC supply to the Meter Demonstrator.
2. Keep the voltage adjust knob OFF.
3. Now connect Variable AC supply to AC/DC supply input i.e. terminals 1 and 2 to terminal 5 and 6
respectively.
4. Connect voltmeter across the load (i.e. V1 and V2 output of the AC voltmeter to terminals 9 and 10
respectively).
5. Connect a 100W bulb to the AC load section.
6. Now connect terminal 9 and 10 to the terminal 14 and 15 respectively.
7. Slightly move the voltage adjust knob so that it become ON.
8. Measure the reading that is pointed by the AC voltmeter.
9. The least count of scale is 1V (0.1V in case of DC Voltmeter).
10. Now move the voltage adjust knob and take various readings of voltages.
11. Similarly perform the experiment for DC Voltmeter with Variable DC supply and DC load. (refer to
figure5)
Result:
As the voltage adjust knob is rotated, the voltmeter shows more voltage on its scale.

Page No.42
Experiment 19
Objective:
Study the connection of Ammeter in network and measure current through it.
Equipments Needed:
2. Patch chords
4. One 6V bulb as DC load
Circuit diagram:
Connection diagram for the experiment is shown in figure 19.1 and figure 19.2.
Figure 19.1
Procedure:
3. Now connect Variable AC supply to AC/DC supply input (i.e. terminals 1 and 2 to terminal 5 and 6
respectively).

Page No.43
Figure 19.2
4. Connect Ammeter in series to the load i.e. A1 and A2 output of the AC voltmeter to terminals 7 and
8 respectively.
5. Connect a 100W bulb to the AC load section.
6. Connect terminal 6 and 8 to the terminal 9 and 10 respectively.
9. Measure the reading that is pointed by the AC Ammeter.
10. The least count of scale is 0.05A.
11. Now move the voltage adjust knob and take various readings of Current.
12. Similarly perform the experiment for DC Ammeter with Variable DC supply and DC load (refer to
figure 19.2).
Result:
As the voltage adjust knob is rotated the Ammeter shows more current on its scale.

Page No.44
Experiment 20
Objective:
Study the connection of a Wattmeter in network and measure of power through it.
Equipments Needed:
2. Patch chords
3. Two 100W bulb as AC load
Circuit diagram: Connection diagram for the experiment is shown in figure 20.1.
Figure 20.1
Procedure:
3. Now connect Variable AC supply to AC/DC supply input i.e. terminals 1 and 2 to terminal 5 and 6
respectively.
4. Connect Wattmeter across the load i.e. W1, W2 and W3V to terminals 11, 13 and 12 respectively.
5. Connect a 100W bulb to the AC load section & Connect terminal 5 and 6 to the terminal 11 and 12
respectively.
8. Measure the reading that is pointed by the AC Wattmeter. The least count of scale is 20W.
9. Now move the voltage adjust knob and take various readings of Power.
10.Now connect another 100W bulb to the AC load section and take readings.
Result:
The power measured by the wattmeter increases with two bulbs as compared with one
bulb, because of the increased current.

Page No.45
Experiment 21
Objective:
Plot the voltage v/s current characteristics for a resistive load.
Equipments Needed:
2. Patch chords
3. Two 100W AC bulb as AC load
Circuit diagram:
Connection diagram for the experiment is shown in figure 21.1.
Figure 21.1
Procedure:
2. Keep the voltage adjust knob at its left most position i.e. off position
3. Now connect Variable AC supply to AC/DC supply input i.e. terminal 1 and 2 to terminal 5 and 6
respectively.
4. Connect Ammeter terminals A1 and A2 to the terminal 7 and 8 respectively.
5. Connect Voltmeter terminals V1 and V2 to the terminals 9 and 10 respectively
6. Connect Wattmeter across the load i.e. W1, W2 and W3V to terminals 11, 13 and 12
respectively.

Page No.46
7. Connect a load of 100W (i.e. bulb) to the AC load section.
8. Connect terminal 5 and 6 to the terminal 7 and 10 respectively.
11. Measure the voltage, current and power readings in AC Voltmeter, AC Ammeter and AC
Wattmeter.
12. Now move the voltage adjust knob and take various readings in the observation table.
13. Plot a graph between Voltage and Current, keeping Voltage at X axis and Current at Y axis.
14. The plot will be a straight line.
15. Now estimate the Power factor from the readings using following formula i.e.
Reactive Power = V * I * Cos θ
Observation Table:
Sr. Voltage Current Active Power Wattmeter Reading
No.
(V) (A) V * I (W)
1
2
3
4
5
Result:
1. The Voltage v/s Current Characteristics would be a straight line.
2. The Wattmeter reading differs from Active power by a factor Cos θ, called Power Factor.

Page No.47
Experiment 22
Objective:
Study of the operation of Moving Coil type instruments
Equipments Needed:
2. Patch chords of appropriate length
3. One 6V DC bulb as DC load
Circuit diagram:
Connection diagram for the experiment is shown in figure 22.1
Figure 22.1
Procedure:
2. Connect Variable DC supply to AC/DC supply input.
3. Connect DC voltmeter and DC Ammeter to their appropriate place.
4. Connect circuit properly with the load.
5. Now slightly move the voltage adjust knob and observe the motion in meters.
6. The coil wounded around the soft iron start moving with the changing voltage.
7. A permanent magnet and a hair spring can also be observed.
Result:
The coil wounded to the soft iron deflects with respect to the torque produced due to the
voltage (or current).

Page No.48
Experiment 23
Objective:
Study of the operation of Moving Iron type instruments
Equipments Needed:
2. Patch chords
Circuit diagram:
Figure 23.1
Procedure:
1) Connect 230V AC supply to the Meter Demonstrator.
2) Connect Variable AC supply to AC/DC supply input.
3) Connect AC voltmeter and AC Ammeter to their appropriate place.
4) Connect circuit properly with the AC load.
5) Now slightly move the voltage adjust knob and observe the motion in meters.
6) The Iron in the center attached with the pointer starts moving.
7) A fixed coil and a hair spring can also be observed.
Result:
The soft iron deflect with respect to the torque produced due to the voltage (or current),
while the coil remain stationary.

Page No.49
Experiment 24
Objective:
Study of the operation of Dynamometer type instruments
Equipments Needed:
2. Patch chords of appropriate length
Circuit diagram:
Figure 24.1
Procedure:
2. Connect Variable AC supply to AC/DC supply input.
3. Connect AC Wattmeter its appropriate place.
4. Connect circuit properly with the AC load.
5. Now slightly move the voltage adjust knob and observe the meter.
6. The small coil in the center (pressure coil) starts moving.
7. Two fixed coils (current coils) and a hair spring can also be observed.
Result:
The pressure coil (inner) deflects with respect to the torque produced due to the voltage
& current, while two current coils remain stationary.

Page No.50
Experiment 25
Objective:
To Measure the Power Factor of Resistive (R) Load.
Equipments Needed:
o Patch Cords
o CRO or Digital Oscilloscope. Mains Cord
Circuit Diagram:
Connection diagram shown in below figure 25.1 with black lines
Figure 25.1 : Power Factor Demonstrator
Procedures:
1. Connect the mains cord to the trainer board.
2. Connect terminal 1 to terminal 3 and terminal 2 to terminal 4 as shown in figure1 using patch chords.
3. Connect resistance from Network Component section to Load Section as shown in figure1. Connect
terminal 9 to terminal 5 and terminal 10 to terminal 6.
4. Connect terminal 6 to terminal 8.
Note: In above figure 1 only R1 is connected to Load Section but user can make different value of
resistance using series or parallel combination.
5. Connect BNC cables from Output of panel (V & I) to both channel of Oscilloscope/DSO.
6. Now switch ‘on’ the panel.
7. Note down readings of Voltage, Current, True Power, Apparent Power, and Power Factor from LCD
display.

Page No.51
8. Now observe lagging or leading between the voltage & current waveform on the oscilloscope and
also measure the peak to peak voltage and current by adjusting the value of voltage/division knob
& time/division knob of Oscilloscope/DSO.
Note: 230V AC signal is not possible to measure on CRO/DSO, for that we have to divide the actual wave
form by the factor of 40 to easily observe the waveform on CRO.
9.Calculate the all values theoretical by using concept & formula, which has been provided in theory
section and compare it with practical result.
Observation Table:
S. No. Voltage(v) Resistance(R) Current(I)
True Apparent Power
Power(P) Power(S) Factor(PF)
1.
2.
3.
4.
5.
Calculation:
Formula used:
Current : I = V/R
Impedance : Z = R + JX, |Z| = √R²+ XL² (Magnitude of impedance)
Apparent Power : S = I2Z,
True Power : P = VI cos (Ø) (watts),
Apparent Power : S = VI
Power Factor : PF = cos (Ø)
where Ø = tan -1(X/R)
Example:
V=230V, R1 = 530E , I = .433 Amp
Z = R + jX, Z = 530 + J0, Z= 530E
Where,
Z = Impedance of Circuit
X = Impedance of reactive element for the purely resistive circuit X=0
So
Z = R
Now
Ø = tan -1 (0/530) = 0 (since X = 0)
PF = cos (Ø) = cos (0) = 1
Note: The resistance are not purely resistive in nature. It will have some internal Capacitive & Inductive
effects. Internal Capacitance & Inductance effects of resistor will depend on its value. Due to this
lagging or leading effect will be shown on the display.

Page No.52
In above figure CH 1 = Voltage waveform, CH 2 = Current waveform
Figure 25.2
Conclusion :
Voltage and Current in purely resistive network in same phase one can see wave form.

Page No.53
Experiment 26
Objective:
To Measure the Power Factor of Resistive & inductive (RL) Load.
Equipments Needed:
Patch Cords
CRO or Digital Oscilloscope. Mains Cord
Circuit Diagram: Connection diagram shown in below figure 26.1 with dark black lines.
Figure 26.1: Power Factor Demonstrator
Procedures:
2. Connect terminal 1 to terminal 3 and terminal 2 to terminal 4 as shown in figure2 (a) using patch
chords.
3. Connect Resistor & Inductor from Network Component section to Load Section as shown in figure
26.1.
4. Connect terminal 9 to terminal 5 and terminal 10 to terminal 6.
5. Connect terminal 6 to terminal 13 and terminal 7 to terminal 15 and terminal 14 to terminal 16
6. Connect terminal 7 to terminal 8.
Note: In above figure only R1 is connected to Load Section but user can make different value of resistance
using series or parallel combination.

Page No.54
display.
10. Now observe lagging or leading between the voltage & current waveform on the oscilloscope/DSO
and also measure the peak to peak voltage and current by adjusting the value of voltage/division
knob & time/division knob of Oscilloscope/DSO.
Note: 230V AC signal is not possible to measure on CRO/DSO, for that we have to divide the actual wave
form by the factor of 40 to easily observe the waveform on CRO.
11. Calculate the all values theoretically by using concept & formula, which has been provided in theory
Observation table:
S. no. Voltage Resistance Inductor Current True Power Apparent Power
(V) (R) (L) (I) (P) Power (S) Factor (PF)
1.
2.
3.
4.
Calculation:
Formula used:
Current : I = V/|Z|
|Z| = √R²+ XL² (Magnitude of impedance)
Z = R + jXL
XL =jωL (XL= Impedance of inductor)
ω = 2πf
Real Power : P = S x cos (Ø)
Power Factor : PF = cos (Ø)
Ø = tan -1(XL/R)
Example:
Theoritically value calculate in ideal case :
R1= 530 ohm, L = 400mH, V= 230V , freq. = 50 Hz By using above mention formula
XL = 125.6 ohm, Z = 655.6, I = .422 Amp, Ø = 13.32 degree |Z| = 544.6
Real power P = 94.43 Watt, Apparent power S = 97.06VA, PF = 0.97
Practically by trainer:
Note: Practical value depends on supply voltage at the time experiment R1= 530 ohm, L = 400mH,
By using above mention formula XL = 125.6 ohm, |Z| =544.6,
V= 219V
F= 50Hz

Page No.55
I = 0.32, Ø = 17.25 degree Real Power = 66.5 Watt
Apparent Power = 68.6 VA P.F. = 0.955
Power triangle diagram shown in below figure which show the relation between real
power, apparent power and reactive power.
Figure 26.2
Conclusion :
There are phase difference between voltage and current in RL circuit. Voltage is
lead with respect current due to inductive effect.
There are some power consumed by reactive load so real power is less than apparent power.
Note: Waveform of current is distorted due to inductor and capacitor because it is energy stored
device and third & fourth harmonics of current is superimposed on original wave form of current.
Figure 26.3

Page No.56
Experiment: 27
Objective:
To Analysis the Improvement of Power Factor of (RL) Network through capacitor.
Equipments Needed:
1. Patch Cords
2. CRO or Digital Oscilloscope. Mains Cord
Circuit Diagram:
Connection diagram shown in below fig.27.1 with black lines
Procedures:
2. Connect terminal 1 to terminal 3 and terminal 2 to terminal 4 as shown in figure2 (b) using patch
chords.
3. Connect Resistor, inductor & capacitor from Network Component section to Load Section as shown
in figure 2b.
5. Connect terminal 6 to terminal 13 and terminal 7 to terminal 15 and terminal 14 to terminal 16
Note: In above figure only R1 is connected to Load Section but user can make different value of
resistance using series or parallel combination

Page No.57
6. Now switch ‘on’ the panel
Display.
8. Measure the peak to peak Voltage and Current by adjusting the value of voltage/division knob &
time/division knob of Oscilloscope/DSO.
Calculate the value of power factor theoretical by using concept & formula, which has
been provided, in theory section and match with practical result.
Observation Table:
S.No. Voltage Resistance Inductor Capacitor True Apparent Power Improved
(v) (R) (L) (C) Power(P) Power(S) Factor(PF) Power
Factor(PF)
1.
2.
3.
4.
Calculation:
Formula used
Current : I = V/|Z|
Impedance of Capacitor : Xc =1/jωC, ω = 2πf ,
Impedance of Inductor : XL = jωL, ω = 2πf,
Effective Impedance : X effective = Xc XL / XL + Xc
|Z| = √R²+ (X effective) ²
Z = R+J X effective
Real Power : P = VI cosØ
Ø = tan-1(X effective /R)
Power Factor : PF = cosØ,
Example:

Page No.58
R1= 530ohm, L= 400mH, XL = 125.6ohm, C1 = 12.5 F, Xc = 254.77 ohm, V= 230V, Frequency = 50 Hz
Practically Power Factor of RL network shown in experiment 2(a) = 0.955
If we add a capacitor 12.5 F parallel with inductor
X effective = 84.12V, |Z| = 536.63 ohm, I = .428Amp, Ø =9.01 degree
Real power = 97.16 watt,
Apparent power = 98.44 VA
Power factor = 0.987
Note: Practical value depends on supply voltage at the time experiment.
V = 219 V
Frequency = 50 Hz
I = .33Amp, Ø = 15.63 degree
Real Power = 70.3 watt
Apparent Power = 73.0 VA
We can see that phase angle reduce and power factor improved over only RL load.
Ø1 =17.25 degree, Ø2= 15.63 degree
Figure 27.2
Conclusion :
There are reduction in phase angle between voltage and current circuit.
Voltage is still lead with respect current due to inductive effect but inductive effect is less
over purely RL load.

Page No.59
There are some power consumed by reactive load so real power is less than apparent
power. Reactive power is decreased over purely RL network due to that real power increased.
Note : Waveform of current is distorted due to inductor and capacitor because it is energy stored device
and third and fourth harmonics of current is superimposed on original wave form of current.
Figure 27.3

Page No.60
Experiment 28
Objective:
To Measure the Power Factor of Resistive& Capacitive (RC) Load.
Equipments Needed:
Patch Cords
Circuit Diagram:
Connection diagram shown in below fig.28.1 with black lines.
Procedures:
2. Connect terminal 1 to terminal 3 and terminal 2 to terminal 4 as shown in figure3 (a) using patch
chords.
3. Connect Resistor & Inductor from Network Component section to Load Section as shown in figure
3a.
i. Connect terminal 9 to terminal 5 and terminal 10 to terminal 6.
ii. Connect terminal 7 to terminal 17 and terminal 8 to terminal 19 and terminal 18 to
terminal 20.
iii. Connect terminal 7 to terminal 6.

Page No.61
display.
knob & time/division knob of Oscilloscope/DSO.
8. Note: 230V AC signal is not possible to measure on CRO/DSO, for that we have to divide the actual
wave form by the factor of 40 to easily observe the waveform on CRO.
9. Calculate the value of power factor theoretical by using concept &formula, which has been
provided, in theory section and match with practical result.
Observation table:
S.No. Voltage Resistance Capacitor Current True Apparent Power
(V) (R) (C) (I) Power(P) Power(S) Factor(PF)
1.
2.
3.
4.
Calculation:
Formula used:
Current : I = V/|Z|
Impedance of Capacitor : Xc =1/jωc, ω = 2πf ,
Magnitude of impedance: |Z| = √R²+ XC²
Real Power : P = VI cosØ
Power Factor : PF = cosØ,
Ø = tan-1
(XC/R)
Example:
R1 = 530 ohm, C= 7.962 F, V = 230V, Freq. = 50 Hz
XC = 399.98 ohm, |Z| = 663.9 ohm, I = 0.346Amp, Ø = 37.04 degree
Real Power = 63.5 Watt, Apparent Power = 79.58 VA, Power Factor =0.798
Note: Practical value depends on supply voltage at the time experiment. R = 530 ohm, C= 7.962 F, XC =
399.98 ohm, |Z|= 663.9,
V = 216 V,

Page No.62
Freq. = 50 Hz,
I = 0.33 Amp, Ø = 40.44degree Real Power = 54.8 Watt
Apparent Power = 72.0 VA Power Factor =0.761
Power triangle diagram shown in below figure which show the relation between real
power, apparent power and reactive power.
Figure 28.2
Conclusion :
There are phase difference between voltage and current in RC circuit.
Current is lead with respect current which show capacitive effect in circuit. This effect easily seen
on wave form .
Note : Waveform of current is distorted due to inductor and capacitor because it is energy stored
device and third and fourth harmonics of current is superimposed on original wave form of
current.
Figure 28.3

Page No.63
Experiment 29
Objective:
To Analysis the Improvement of Power Factor of (RC) Network through Inductor.
Equipments Needed:
Patch Cords
Circuit Diagram: Connection diagram shown in below fig.29.1 with black lines
Procedures:
2. Connect terminal 1 to terminal 3 and terminal 2 to terminal 4 as shown in figure 3b using patch
chords.
3. Connect Resistor, Inductor and Capacitor from Network Component section to Load Section as
shown in figure 3b.
a. Connect terminal 9 to terminal 5 and terminal 10 to terminal 6.
i. Connect terminal 7 to terminal 17 and terminal 8 to terminal 19 and terminal 18 to
terminal 20.
b. Connect terminal 6 to terminal 7.
c. Connect terminal 15 to terminal 19 and terminal 16 to terminal 17.

Page No.64
display.
7. Now measure the peak to peak voltage and current by adjusting the value of voltage/ division knob
8. Note: 230V AC signal is not possible to measure on CRO/DSO, for that we have to divide the actual
waveform by the factor of 40 to easily observe the waveform on CRO/DSO.
9. Calculate the all values theoretical by using concept & formula, which has been provided in theory
Observation table:
S.No. Voltage Resistance Inductor Capacitor True Apparent Power Improved
(v) (R) (L) (C) Power(P) Power(S) Factor(PF) Power Factor
1.
2.
3.
4.
Calculation:
Formula used Current : I = V/|Z|
Impedance of Capacitor : Xc =1/jωc, ω = 2πf,
Impedance of inductor : XL = jωL, ω = 2πf,
Effective impedance : X effective = Xc XL / XL + Xc
Z = R+J X effective
Z = √R²+ (X effective)²
True Power : (P) = VI cos (Ø)
Ø = tan-1(X effective /R)
Apparent Power : (S) = VI
Power Factor : PF= cosØ,
Example:
R1= 530 ohm, L1= 200mH, XL = 62.8 ohm, C = 7.962 F, Xc = 399.98 ohm V = 230V, frequency = 50 Hz
Practically Power Factor of RC network in experiment no.3 (a) = 0.761
We add a inductor 200mH parallel with capacitor
X effective = 54.27, |Z| = 532.77, I =0.431 Amp, Ø = 6.27 degree
Real Power = 98.76 Watt, Apparent Power = 99.28VA, Power factor = 0.994
Note: Practical value depends on supply voltage at the time experiment V = 211V
Frequency = 50 Hz
I = .344 Amp, Ø = 8.88 degree Real Power = 71.6
Watt Apparent Power = 72.5VA Power Factor = 0.988

Page No.65
Figure 29.2
Ø1= 40.44 degree, Ø2 = 8.88 degree
Conclusion :
There are reduction in phase angle between voltage and current circuit.
Current is still lead with respect to voltage due to capacitive effect but capacitive effect is less over
purely RC network.
reactive power is decreased over purely RL network due to that real power increased.
Note : Waveform of current is distorted due to inductor and capacitor because it is energy storage device
and third and fourth harmonics of current is superimposed on original wave form of current .
Figure 29.3

Page No.66
Experiment 30
Objective:
To Measure the Power Factor of Resistive, Capacitive, Inductive (RLC) Load.
Equipments Needed:
Patch Cords
Circuit Diagram: Connection diagram shown in below fig.30.1 with black lines.
Figure 30.1: Power Factor Demonstrator
Procedures:
2. Connect terminal 1 to terminal 3 and terminal 2 to terminal 4 as shown in figure 4a using patch
chords.
3. Connect resistance, inductor, and capacitor from Network Component section to Load Section as
shown in figure 4a.
Connect terminal 5 to terminal 9 and terminal 6 to terminal 10. Connect terminal 6 to terminal 13
and terminal 7 to terminal 14. Connect terminal 7 to terminal 17and terminal 8 to terminal 18.
display.

Page No.67
knob & time/division knob of CRO/DSO.
Note: 230V AC signal is not possible to measure on CRO/DSO, for that we have to divide the actual
Observation Table:
S.No. Voltage Resistance Inductor Capacitor True Apparent Power
(V) R (L) C Power(P) Power(S) Factor (PF)
1.
2.
3.
4.
Calculation:
Formula used:
Current : I = V/|Z|
Impedance of RLC Circuit : Z = R+J (XL –Xc), XL >XC
Z = R+J (XC –XL), XC >XL
Magnitude of Impedance : |Z|= √R²+ (XC – XL) ²
Ø = tan-1(XL-XC)/ R, XL. >XC
Ø = tan-1(XC-XL )/ R, XC. >XL
Power Factor : PF = cos Ø,
Real Power : P = VI cos Ø
Example:
a) XC >XL
R1= 530ohm, L1 = 200mH, C1 = 12.5 F,V = 230V, Frequency = 50 Hz All the calculation is done by using
above formula.
I= 0.40Amp, Xc = 254.77 ohm, XL = 62.8 ohm, |Z| = 563.69 ohm
Real Power = 86.48 Watt, Apparent Power: = 92 VA, Ø = 19.94, Power Factor =0 .940
b) XC <XL
R1|| R2 =139.5, C1||C2 = 32.5 F, L1+L2 = 400mH and calculate above parameter similar way.

Page No.68
Practically by Trainer:
Note: Practical value depends on supply voltage at the time experiment.
R1= 530ohm, L1= 200 mH, C1= 12.5 F, Xc = 254.77 ohm, XL = 62.8 ohm, |Z| = 563.69 ohm V = 215 V
Frequency = 50Hz
I= 0.36Amp, Ø = 16.26 degree
Real Power = 75.0Watt
Apparent Power = 78.3 VA
Power Factor = .960
Figure 30.2
Conclusion :
There are phase difference between voltage and current in RLC circuit.
Current is lead with respect voltage which show capacitive effect in circuit. This effect easily seen
on wave form.
Figure 30.3
Note : Waveform of current is distorted due to inductor and capacitor because it is energy stored
device and third and fourth harmonics of current is superimposed on original wave form of
current .

Page No.69
Experiment 31
Objective:
To analysis improvement of the Power Factor of Resistive, Capacitive & Inductive (RLC)
Load with Inductor.
Equipments Needed:
Patch Cords
Circuit Diagram: Connection diagram shown in below fig.31.1 with black lines
Procedures:
2. Connect terminal 1 to terminal 3 and terminal 2 to terminal 4 as shown in figure 4a using
patch chords.
3. Connect Resistor, Inductor, and Capacitor from Network Component section to Load Section
as shown in figure 4a.
Connect terminal 5 to terminal 9 and terminal 6 to terminal 10.
Connect terminal 7 to terminal 17and terminal 8 to terminal 18.
display.

Page No.70
7. Now observe lagging or leading between the voltage & current waveform on the oscilloscope and
also measure the peak to peak voltage and current by adjusting the value of voltage/division knob
Note: 230V AC signal is not possible to measure on CRO/DSO, for that we have to divide the actual
Observation table:
S.No.
Voltage Resistance Inductor Capacitor True Apparent Power Improved
(V) (R) (L) (C) Power(P) Power(S) Factor(PF) Power Factor
1.
2.
3.
4.
Calculation:
Formula used:
Current : I = V/|Z|
Effective impedance : Xeffective = Xc XL / XL + Xc
Impedance of RLC Circuit : Z = R+J (XL – Xeffective)
Magnitude of impedance : |Z| = √R²+ (XL–Xeffective) ²
Real Power : (P) = VI cos Ø,
Apparent Power : (S) = VI
Power Factor : PF = cos Ø,
Ø = tan-1(XL- Xeffective) / R, XL. >XC
Example :
Theoretically value calculate in ideal case :
Power Factor can be improved over experiment on 4(a) Power Factor = .940 by using a
inductor 200mH parallel with 12.5uf capacitor.
R1= 530ohm , L1 = 200 mH, C1 = 12.5 uf ,V = 230V, Frequency = 50 Hz All the
calculation is done by using above formula.
Xc = 254.77 ohm, XL = 62.8 ohm, X 50.38 Ohm, |Z| = 530.1ohm

Page No.71
I= 0.433Amp, Real Power =99.49 Watt, Apparent Power = 99.59 VA
Power Factor = 0.999, Ø = 1.34 degree
Note: Practical value depends on supply voltage at the time experiment
R= 530ohm, L = 200mH, C = 12.5 F, Xc = 254.77 ohm, XL = 62.8 ohm, |Z| = 530.1 ohm V = 211V
Frequency = 50 Hz
I = .33Amp, Ø = 12.04 degree Real power
= 67.8Watt Apparent Power = 69.3VA
Power Triangle:
Ø1 = 16.26 degree, Ø2 = 12.04 degree
Figure 31.2
Conclusion:
There are phase difference between voltage and current in RC circuit.
Voltage is lead with respect current which show inductive effect in circuit. This effect
easily seen on waveform, but this circuit inductive effect is dominate above experiment 4(a) capacitive
effect.
There are some power consumed by reactive load so real power is less than apparent
power. reactive power is decreased over purely series RLC network due to that real power increased.
Note : Waveform of current is distorted due to inductor and capacitor because it is energy stored device
and third and fourth harmonics of current is superimposed on original wave form of current.

Page No.72
Figure 31.3

Page No.73
Experiment 32
EMT
Study of Speed-Torque Characteristics of Separately Excited DC Shunt Motor
Unit Objective:
On completion of this unit you will understand Speed-Torque Characteristics of Separately
Excited DC Shunt Motor.
Discussions On Fundamentals :
For DC motors basic theory explains that torque developed by motor
T on Flux ф x Armature Current Ia
Separately Excited DC motor is studied in this Unit. In this category air gap flux is
practically independent of Armature current Ia because Field winding producing flux is fed from
independent DC source of power. This indicates that torque
T α Armature Current Ia
Further the relation between Supply voltage V, Back Emf E developed by motor and
Armature drop Ia Ra may be stated as
V = E + IaRa.
Here IaRa drop is negligible compared to other terms in the expression and as such if N is
speed of the motor and <1) is flux present, then as per theory,
E=KN ф = or N α K1 V/ф
Thus this type of DC motor exhibits near to constant speed characteristics over wide load
range (If armature supply voltage and Field voltage are maintained constant.). Further speed can be varied
by varying Armature voltage (Speed Increases with voltage.). Field weakening can also achieve speed
variation. (Speed increases if field current is reduced.)
In above discussions, we neglected laRa drop. In practice actual speed drops only because
of this as this drop increases with increase in load or Ia. However for a well designed motor this drop in
speed from no load to full load is around 5%. In case this drop is compensated using a close loop system,
then practically constant speed can also be achieved from no load to full load.
The explanation that follows assumes the loading of DC motor will be done by coupling 3
phase AC generator to it and further to resistive load. In case you have received any other combination,
refer respective armature.

Page No.74
Wiring Sequence
I) DC machine Field connections
Sr. From To Sr. From To
1 EMT8(L1) EMT9L97) 2 EMT8(N1) EMT9L(8)
3 EMT9L(9) DC m/c (Z) 4 EMT9L(13) DC m/c (ZZ)
II] DC machine Armature connections
1 EMT8(L2) EMT9C(7) 2 EMT8(N2) EMT9C(8)
3 EMT9C(9) EMT6B(1) 4 EMT9C(13) EMT6B(3)
5 EMT6B(2) EMT6B(5) 6 EMT6B(6) EMT6A(8)
7 EMT6A(7) DC m/c (A) 8 EMT6B(4) DC m/c (AA)
III] AC generator rotor connections
1 EMT8(L3) EMT9R(7) 2 EMT8(N3) EMT9R(8)
3 EMT9R(9) EMT6A(1) 4 EMT9R(13) EMT6A(3)
5 EMT6A(2) EMT6A(5) 6 EMT6A(6) EMT5B(1)
7 EMT6A(4) EMT5B(3) 8 EMT5B(9) AC/C(A)
9 EMT5B(10) AC m/c (B) 10 EMT5B(11) AC m/c (C)
IV] AC generator Resistor load connections
1 AC m/c (U2) AC m/c (V1) 2 AC m/c (V2) AC m/c (W1)
3 AC m/c (W2) AC m/c (U1) 4 AC m/c (U1) EMT14A(2)
5 AC m/c (V1) EMT14A(6) 6 AC m/c (W1) EMT14A(10)
7 EMT14A(3) EMT14A(5) 8 EMT14A(7) EMT14A(9)
9 EMT14A(11) EMT14A(1)
V] Sensor connections
1 EMT9L(14) EMT8(8) 2 EMT9L(15) EMT8(9)
3 EMT9L(16) EMT8(6) 4 EMT9L(19) EMT8(7)
Table: 32.1 Study of Speed-Torque Characteristics of Separately excited DC Shunt Motor coupled to AC
Generator

Page No.75
Figure 32.1
OR
In case you are supplied with trunnion mounted DC shunt m/c as Dynamometer use
following schematic and wiring schedule.

Page No.76
Fig 32.2 Study of speed- Torque Characteristics of Separately Exci9ted DC Shunt Motor coupled to DC
Generator
Figure 32.2
Replace at above wiring schedule in place of AC generator rotor connection & AC
generator Resistor load connection by DC generator / Resistor load connection as below.
Wiring sequence
I) DC self excited generator connection
1 DC m/c (A) DC m/c (Z) 2 DC m/c (ZZ) DC m/c (AA)
3 DC m/c (A) EMT14B(13) 4 DC m/c (AA) EMT14B(15)
OR
I) DC separately excited generator connection
1 DC m/c (A) EMT14B(13) 2 DC m/c (AA) EMT14B(15)
5 EMT9R(9) Dc M/C(Z) 6 EMT9R(13) DC M/C (ZZ)
7 DMM(+VE) DC M/C (Z) 8 DMM (-VE) DC M/C(ZZ)
 DMM (V) from Lab stock
Procedure:
As shown in the diagram the SCR half bridge EMT 9/1 is used for field supply, EMT 9/2 is
used for armature supply while EMT 9/3 is used for Field excitation of generator (Optional for self excited
generator if used).
Speed sensing signal and load cell signal is processed in EMT 9/ 1 and is fed to EMT 8 for
display. The motor drives generator as load. The load can be increased by increasing load on generator.
Further loading resolution may be achieved by controlling excitation of generator.

Page No.77
The DC motor while working, also works as swinging dynamometer. The dynamometer
reading as indicated by load cell measurement in kg., gives force developed at motor shaft. Knowing the
torque arm, shaft torque can be calculated.
1) Make the connections as per the wiring schedule above.
2) Keep EMT 9/3 input disconnected. Connect bulb load to EMT 9/1 and 9/2 instead of DC motor. Put ON
Input to EMT 8. Put on the power supply ON / OFF switch on EMT8. Ensure phase matching and smooth
voltage control over outputs of EMT 9/1, 9/2. Keep the output set pot of EMT9/l in max output position
while that of EMT9/2 in minimum output position. This will ensure that full DC voltage is applied to
motor field. Switch off the set up and reconnect the motor connections.
3) Now switch on power supply input and increase voltage setting on EMT9/2 so that armature voltage
increases. DC Motor will start rotating. The motor supplied with the trainer is rated for 20OV/ 3.5Amps,
as armature rating while maximum field voltage is 200V. Observe the speed variation by varying
armature voltage.
4) (Skip this step if DC generator used) Now switch off power supply and connect input AC to EMT 9/3
through dimmer as shown. Connect it's output to bulb load. Check the output voltage control as
explained in 2 above. Adjust the output voltage to about 50 volts using dimmer, when voltage set pot
on EMT9/3 is at it’s maximum position. This bridge will supply excitation current to AC generator, as
such full voltage will never be required. More over full voltage may damage the winding. This is
avoided by keeping maximum AC at reduced value using the dimmer. Now The excitation can be
controlled by using voltage set pot on EMT9/3.
5) Now switch on all the power supplies and adjust the generator excitation for full output voltage. The
Motor speed on no load can be set at 1500 RPM using slight field weakening if required. The motor at
present supports only no load mechanical losses of generator.
6) Now load the generator by putting on 100W, 2 nos lamps per phase or by using resistance selector
switch on EMTl4A/14B and keeping it to minimum resistance position. This will be maximum load
allowed on generator. Now you can reduce the load by reducing generator excitation. (Reducing
excitation will reduce generated voltage for AC & increase for DC generator.)
7) Take down the readings as per following table
Table No.: 32.1 Speed -Torque characteristics for separately excited DC shunt motor.
Sr. Va Volts DC Ia Amps
DC
Load cell – Kg.
L
Torque
Nm.
Speed
RPM
1 180 3.5
2 180 3
3 180 2.5
4 180 2
5 180 1.5
6 180 1
7 180 0.5
[Torque = L x 9.81 x 0.07 N. Meters]
Take another set of readings by repeating above procedure for Armature voltage to be
l40Volts.Plot the graph with Torque on X axis and speed on Y axis.
Equipment Required
EMT DC machine assembly, EMT 3 phase AC machine assembly. EMT tabletop structure.
Conclusion
Separately excited DC shunt motor exhibits good speed regulation over entire load range.
The drop in speed is only due to IaRa drop.

Page No.78
Experiment 33
Study of Speed —Torque Characteristics of DC Series Motor
Unit Objective :
On completion of this unit you will understand performance of DC Series Motor in respect
of it's Speed —Torque Characteristics.
Discussions On Fundamentals :
DC Series motors exhibit unique characteristics, which is most suited for drive
applications, which require very heavy starting torque. Such applications include Railway - Traction,
Hoists, Heavy-duty cranes, Battery operated vehicles etc. For DC motors basic theory explains that
torque developed by motor
T on Flux ф >< Armature Current Ia
In DC series Motor, field winding and armature windings are in series as such they carry
same current and above relation reduces to
T α Ia2
Thus here the Flux increases very rapidly during starting, and then it tapers off due to
magnetic saturation.
V = E + IaRa.
Here IaRa drop is negligible compared to other terms in the expression and as such if N is
speed of the motor and <1) is flux present, then at constant V
E=KN ф = or N α K 1/ ф =K 1/Ia2
Thus in this type of DC motor speed is inversely proportional to square of load current,
and as such decreases with increase in load. Further the motor can deliver very high starting torque.
This motor cannot be started on No Load as the speed can become dangerously high. Variable
armature voltage will give required speed range.

Page No.79
Fig 33.1 : Study of Speed-torque Characteristics of DC Series Motor
Figure 33.1

Page No.80
Wiring Sequence :
I) DC machine Supply connections
1 EMT8(L1) EMT9L(7) 2 EMT8(N1) EMT9L(8)
3 EMT9L(9) EMT6A(1) 4 EMT9L(13) EMT6A(3)
5 EMT6A(2) EMT6A(5) 6 EMT6A(6) DC M/C(YY)
7 DC M/C(Y) DC M/C(A) 8 CD M/C (AA) EMT6A(4)
II) AC generator rotor connections
3 EMT9R(9) EMT5B(1) 4 EMT9R(13) EMT5B(3)
5 EMT5B(9) AC m/c(A) 6 EMT5B(10) DC M/C(B)
7 EMT5B(11) AC M/C(C)
III) AC generator Resistor load connections
1 AC M/C(U2) AC M/C(V1) 2 AC m/c(V2) DC M/C(W1)
3 AC M/C(W2) AC M/C(U1) 4 AC m/c(U1) EM14A(2)
5 AC M/C(V1) EMT14A(6) 6 AC m/c(W1) EMT14A(10)
7 EMT14A(3) EMT14A(5) 8 EMT14A(7) EMT14A(9)
9 EMT14A(11) EMT14A(1)
IV) Sensor connections
1 EMT9L(14) EMT8(8) 2 EMT9L(15) EMT8(9)
3 EMT9L(16) EMT8(6) 4 EMT9L(19) EMT8(7)
OR
In case you are supplied with trunnion mounted DC shunt m/c as Dynamometer use
following schematic & wiring schedule
Fig 33.2 Study of Speed-Torque Characteristics of DC Series Motor coupled to DC Generator
Figure 33.2

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