EC LAB MANUAL
Experiment No. 1
Objective:
To study the electric circuit lab equipment
Apparatus:
a)
b)
c)
d)
e)
Bread Board
Digital Multimeter
Triple Output Variable DC Power Supply
Dual Channel Digital Oscilloscope
Function Generator
Theory:
a) Bread Board:
A bread board is a construction base for prototyping of electronics. It is a circuit testing
device with a number of tiny perforations. Interconnections can be done with the help of
these holes by wires usually called jump wires.
Fig.1.1: Bread Board
Working:
i.
ii.
On each bread board there are always two sets of two rows on top and bottom. These are
connected horizontally. These are usually used for the ‘+’ and ‘–’ contacts of the battery.
And between these two sets there are holes that are connected in column by conductive
metal strips
Fig.1.2: Using Bread Board
b) Digital Multimeter (DMM):
A digital multimeter is a test tool used to measure two or more electrical valuesprincipally voltage in Volts, current in Amperes, and resistance in Ohms. It is a standard
diagnostic tool for technicians in the electrical or electronic industries.
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A digital multimeter has three parts:
i.
Display
ii. Selection Knob
iii. Port
Fig.1.3: Digital Multimeter(DMM)
Working:
There will be some main connections for the probes. Although only two probes are needed at
one time.
i.
Common: It is use with all measurements and this will take the negative or black lead
probe.
ii.
Volts, Ohms: This connection is used for measurement of voltage and resistance. And
will take the positive or red probe.
iii.
Ampere and milliamperes: This connection is used for the low current measurements
and will again take the red lead and probe.
iv.
High current: This is often a separate connection for high current measurements.
c) Triple Output Variable DC Power Supply:
This power supply has three outputs. Two outputs are continuously adjustable at 0-30V
DC and 0-3A, and the third output is fixed at 5V DC and 3A. It has three channels. It is
used as an output device. It has four voltage and current rotators by which we can vary
Voltage and Current.
Fig.1.4: Triple output variable DC power supply
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Function:
A power supply unit (or PSU) converts mains AC to low-voltage regulated DC power for the
internal components of a computer. ... Some power supplies have a manual switch for
selecting input voltage, while others automatically adapt to the mains voltage.
Working:
i.
ii.
iii.
iv.
v.
vi.
Turn ON power Supply.
Press output button on power supply.
Select one channel (CH-1, CH-2, CH-3).
Select voltage on the channel on Power Supply.
Connect the positive (red) probe of DMM with the red point on power supply and
similarly, connect the negative (black) probe with the black point of selected channel.
DMM shows the voltage that is coming from the Power Supply.
d) Dual Channel Digital Oscilloscope:
It is a device which stores and analyses the signal digitally.
Fig.1.5: Dual Channel Digital Oscilloscope
Function:
i.
Shows and calculate the frequency and amplitude of an oscillating signal.
ii.
Shows the voltage and time of a particular signal. This function is the main used of all the
functions described here.
iii.
Helps to troubleshoot any malfunction components of a project by looking at the expected
output after a particular component.
iv.
Shows the content of the AC voltage or DC voltage in a signal.
Working:
i.
ii.
It gives voltage (V p-p) and Time Period by constructing voltage-time graph. Voltage appear
on Y-axis and Time Period on X-axis.
The wave which is displayed by oscilloscope has small boxes. Each small box is divided into
five parts.
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iii. From graph we can calculate peek to peek voltage and time period manually. And from time
period we can find frequency.
iv. It has two channels. Each channel has two probes one is small probe which is negative and
the is bigger which is positive.
The square waveform displayed by Oscilloscope is shown below;
Fig.1.6: Square Waveforms
e) Function generator:
It is usually a piece of electronic test equipment or used to generate different types of
electrical waveforms over a wide range of frequencies. Some of the most common
waveforms produced by the function generator are the Sine, Square, Triangular and Sawtooth
shapes. We can adjust the frequency range by pressing the button.
Fig.1.7: Different waveforms
We can form these types of waveforms by selecting the buttons that is present on the
function generator. If we want to form square waveform on oscilloscope we first select
square waveform and then select our desired frequency and voltage. And then connect the
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function generator with the oscilloscope. Oscilloscope will show the square waveform in the
form of graph and from graph we can calculate the frequency and voltage manually.
Function generator is shown as below;
Fig.1.8: Function Generator
Working:
i.
ii.
iii.
iv.
Power on the generator and select the desired output signal: square wave, sine wave or
triangle wave.
Connect the output leads to an oscilloscope to visualize the output signal and set its
parameters using the amplitude and frequency controls.
Attach the output leads of the function generator to the input of the circuit you wish to
test.
Attach the output of your circuit to a meter or oscilloscope to visualize the resulting
change in signal.
Procedure:
Task 1: Calculate the peek to peek 5.96V and 187.27Hz from the graph available in
oscilloscope.
i.
First, I turned on the digital oscilloscope.
ii.
Then, I connected the positive terminal of probe with the positive terminal of oscilloscope
and negative terminal of probe with the negative terminal of oscilloscope.
iii. Then I checked the voltage error in oscilloscope which was approximately 1.5%.
iv.
Then I checked the frequency error in oscilloscope which was approximately 3.6%.
v.
Observations:
Task 1:
Fig.1.9: Testing Oscilloscope
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Task 2: Generate square waveform signal from function generator and observe it.
i.
First, we turned ON the function generator and oscilloscope.
ii.
We connected probe with the function generator.
iii. Then we connected probes with digital oscilloscope in channel 1.
iv.
We connected positive terminal of probe (connected with oscilloscope) with the positive
terminal of probe (connected with function generator) and similarly negative terminal
with negative.
v.
Then we set the square waveform from the function generator.
vi.
We set the peek to peek voltage of function generator at 3.76 Vp-p.
vii.
Then We set the frequency of function generator at 150.2 Hz.
viii. We adjusted the size of waveform by voltage division and time division button.
ix. Then we calculated the voltage (p-p) and frequency mathematically by using graph shown
on oscilloscope.
Observations:
Task 2: The readings set by the function generator are:
Fig.1.9: Function generator readings
The graph shown on oscilloscope:
Fig.1.10: Graph shown on oscilloscope
Conclusions:
From this experiment, we have learned how to use oscilloscope and how to take its readings
by connecting it with function generator. Also, we have learned the use of various electric
circuit lab equipment.
Precautions:
i
ii
iii
iv
Plug the switch carefully.
Don’t interchange the probe of Oscilloscope and function generator.
Use all safety measures to prevent yourself from any type of electric shock.
Adjust the waveform position carefully. And Measure the values carefully.
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Experiment No. 2
Objective:
Determine the resistance of a resistor by color code method.
Apparatus:
a) Bread Board
b) Digital Multimeter (DMM)
c) Resistor
Theory:
a) Resistor:
A resistor is a passive two-terminal electrical component that implements electrical
resistance as a circuit element.
Uses of resistor:
In electronic circuits, resistors are used to reduce current flow, adjust signal
levels, to divide voltages, bias active elements, and terminate transmission lines, among other
uses. High-power resistors that can dissipate many watts of electrical power as heat, may be
used as part of motor controls, in power distribution systems, or as test loads for generators.
Fixed resistors have resistances that only change slightly with temperature, time or operating
voltage. Variable resistors can be used to adjust circuit elements (such as a volume control or
a lamp dimmer), or as sensing devices for heat, light, humidity, force, or chemical activity.
A usual carbon resistor is shown below:
Fig2.1: Carbon Resistor
Resistors are common elements of electrical networks and electronic circuits
and are ubiquitous in electronic equipment. Practical resistors as discrete
components can be composed of various compounds and forms. Resistors are
also implemented within integrated circuits.
In circuits resistor is shown by the following symbol:
Fig.2.2: Circuit Symbol of resistor
Color-Code of Resistance:
We can observe the resistance of a resistor by a color code sequence printed in form of bands
on it.
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i.
ii.
iii.
iv.
The first band indicates the first digit of the value of resistance.
The second band gives digit.
The third band is decimal multiplier. It gives the number of zeroes after the first two
digits.
The fourth band gives the tolerance value.
A simple table including method of determining resistance is shown below:
Table No.2.1: Color code of resistance
Tolerance:
The possible variation of resistance of carbon resistor from marked value is called tolerance.



Silver band indicates the tolerance of ±5%
Golden band indicates the tolerance of ±10%
If there is no fourth band then the tolerance value is ±20%
For Example:
Advantages of Carbon Resistors:
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

They are cheap.
They have small size.
Disadvantages of Carbon Resistors:



They are noisy.
They are very sensitive to temperature changes.
The power dissipation capacity is small.
Procedure:
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
I hold the resistor in my right hand with the tolerance band pointing towards the thumb.
I observed the color of first band and wrote its value from above table.
Similarly, I observed the color of 2nd band and its values from table.
Then I observed the color of 3rd band which is also called multiplier band and wrote its
value from table
Then, I observed the tolerance band and wrote its percentage value by using above table.
Then I calculated resistance by using above measures.
Then I calculated resistance of resistor using digital multimeter by inserting it in a bread
board.
Then I took difference of theoretical and observed value.
Observations and Calculations:
Sr. no.
First
band
Second
band
Third
band
Tolerance
Calculatd
resistance
Reading
on DMM
Differnce
1
Brown
1
Black
0
Red
102
Gold
±5%
1000±5%
Ω
996Ω
4Ω
2
Red
2
Red
2
Red
102
Gold
±5%
2200±5%
Ω
2195Ω
5Ω
3
Orange
3
Black
0
Brown
101
Gold
±5%
300±5%Ω
297Ω
3Ω
4
Blue
6
Green
8
Brown
101
Gold
±5%
680±5%Ω
674Ω
6Ω
5
Yellow
4
Violet
7
Brown
101
Gold
±5%
470±5%Ω
472Ω
2Ω
Precautions:
i.
ii.
Before observing color bands make sure you hold tolerance band towards your thumb of
your right hand. And observe color bands carefully.
Insert resistor in bread board carefully. Resistor should be inserted in bread board in two
different columns. And take readings from digital multimeter accurately.
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Experiment No. 3
Objective:
Verification of Ohm’s law.
Apparatus:
a)
b)
c)
d)
e)
Bread Board
Digital Multimeter
Resistors
D.C. Power Supply
Jumper wires
Theory:
Ohm’s Law:
Ohm’s law states that, the current through a conductor between two points is directly
proportional to the voltage across the two points provided the physical properties of the
object do not change. It introduces a constant of proportionality resistance.
V= IR
Where I is the current through the conductor in units of ampere, V is the voltage across
the conductor in units of volts and R is the resistance of the conductor in units of ohms.
Ohm’s law is an empirical relation which accurately describes the conductivity of the vast
majority of electrically conductive materials over many orders of magnitude of current.
The ohm (symbol: Ω) is the SI unit of electrical resistance, named after Georg Simon
Ohm. An ohm is equivalent to a volt per ampere. Since resistors are specified and
manufactured over a very large range of values, the derived units of milliohm (1mΩ =10-3
Ω), kilohm(1kΩ =1000Ω), and megaohm(1 MΩ = 106 Ω) are also in common usage.
Ohmic Conductors: The conductors which obey ohm’s law at every temperature are
called ohmic conductors.
Non-ohmic conductors: The conductors which do not obey ohm’s law are called nonohmic.
 Ohm’s law can be used to solve simple circuits. A complete circuit is one which is
closed. It contains at least one voltage source and at least one load. The sum of
voltages across a complete circuit is zero.
 In circuit analyses three equivalent expressions of ohm’s law are use:
𝑉
𝑉
𝑉 = 𝐼𝑅, 𝑅 = 𝐼 and I=𝑅.
Circuit Diagram:
To verify ohm’s law, a simple circuit is taken including four different resistors and
voltage source of 5V.
V1
5V
R1
R2
10kΩ
470Ω
R3
2200Ω
R4
1kΩ
Fig.3.1: Circuit to verify ohm’s law.
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Procedure:
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
ix.
First of all, I made the circuit using resistors on breadboard as shown in above circuit
diagram.
Then, I used jumper wires to connect circuit with the DC power supply.
I turned ON the DC power supply and set its voltage at 6V.
Then, I measured voltage across each resistor using digital multimeter and wrote all
the values shown by DMM.
To measure current across each resistor, I removed one terminal of resistor from
breadboard and kept the other connected with power supply. I connected positive
probe of DMM with the power supply terminal of resistor and negative probe with the
removed terminal of resistor I noted current readings in the table.
I repeated the same procedure for the other three resistors.
After noting all the readings, I calculated resistance by using ohm’s law relation
i.e. by dividing Voltage with Current.
Then I calculated the difference between the actual resistance and calculated
resistance.
At the end, I calculated the percentage error.
Observations and Calculations:
Fig.3.2: Circuit made on bread board
Fig.3.3: Measuring Voltage Using DMM.
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Fig3.4: Voltage measured on DMM.
Fig.3.5: Current measured by DMM.
Resistance
No.
Resistance
(KΩ)
Voltage
(V)
Current
(mA)
Ohm’s Law
resistance
(KΩ)
Difference
Percenatage
Error
R1
1
0.29
0.29
1
0
0
R2
4.7
3.20
6.86
4.6
0.1
2.1
R3
9.7
3.15
0.327
9.6
0.4
4.1
R4
10
4.751
0.472
10.06
0.06
0.6
Conclusion:
We have verified Ohm’s law using DMM, Breadboard and various resistances.
Precautions:
v Plug the switch carefully.
vi Before taking values make sure your circuit is not short.
vii If your circuit change it before connecting with power supply.
viii Measure the values carefully.
ix While taking current value across resistor make sure your one terminal of resistor
is removed from breadboard.
x Use all safety measures to prevent yourself from any type of electric shock.
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Experiment No. 4
Objective:
To verify Kirchoff’s Voltage Law.
Apparatus:
a)
b)
c)
d)
e)
Digital multimeter
Resistors(2.2K,1K,2.2K,10K )
DC Power Supply
Breadboard
Jumper Wires
Theory:
Kirchoff’s Voltage Law:
This law is also called Kirchhoff's
and Kirchhoff's second rule.
second
law, Kirchhoff's
loop (or mesh) rule,
The principle of conservation of energy implies that, the directed sum of the
electrical potential differences (voltage) around any closed network is zero, or:
More simply, the sum of the emfs in any closed loop is equivalent to the sum of the potential
drops in that loop, or: The algebraic sum of the products of the resistances of the conductors
and the currents in them in a closed loop is equal to the total emf available in that loop.
It can be stated as:
𝑛
∑ 𝑉𝑘 = 0
𝑘=1
Here, n is the total number of voltages measured. This law is based on the conservation of
energy whereby voltage is defined as the energy per unit charge. The total amount of energy
gained per unit charge must be equal to the amount of energy lost per unit charge, as energy
and charge are both conserved.
Before applying this rule for the analysis of complex network it is worthwhile to thoroughly
understand the rules for finding the potential charges.
i.
ii.
If a source of emf is traversed from negative to positive terminal, the potential change is
positive, it is negative in opposite direction.
If a resistor is traversed in the direction of current, the change in potential is negative, it is
positive in the opposite direction.
Circuit Diagram:
Circuit to study kirchoff’s law is shown below;
R1
R2
220Ω
100Ω
R3
220Ω
VA
8V
VB
6V
R4
1000Ω
VC
5V
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Procedure:
i.
ii.
iii.
iv.
v.
vi.
vii.
First, I made the circuit on breadboard as shown above in circuit diagram using four
resistors.
Then, I connected two channels of DC power supply according to circuit diagram.
I set one channel of DC power supply at 8V and the other channel at 6V and a third one
5V.
Then, I measured voltage across each resistor with the help of digital multimeter.
Then I applied KVL on loop 1 and loop 2 and the observed that the sum of voltages was
close to zero.
Then I turned the Channel 2 OFF from power supply and again applied KVL on loops
and calculated sum of voltages which was almost zero.
Then I turned the Channel 1 OFF and repeated above steps.
Observations and Calculations:
Fig.4.2: Circuit made on breadboard
R1
R2
220Ω
100Ω
R3
220Ω
VA
8V
VB
6V
R4
1000Ω
VC
5V
If we take the current direction as shown in above figure then we will have ;
For Loop 1:
-VA+VR1+VR3+VB=0
………………(1)
For Loop 2:
VR3+VR2+VR4-VC +VB=0 ………………(2)
For Loop 3:
This is the Outer Loop
-VA+VR1+VR2+VR4 –VC =0 ………………….(3)
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Supply Voltage
Loop 1
Loop 2
Loop 3
VA
VB
VC
VR1
VR3
VR2
VR3
VR4
VR1
VR2
VR4
8.01
6.19
5.15
2.01
0.20
1.01
0.20
10.13
2.01
1.01 10.13
6.84
5.33
5.15
1.77
0.26
0.93
0.26
9.3
1.77
0.93
9.3
7.78
5.15
2.40
0.46
1.13
0.46
11.29
2.40
1.13 11.29
9.3
Putting values in eq (1),(2) and(3). We get
For VA = 8.01V, VB = 6.19V and VC = 5.15V
Loop 1: -8.01+2.01+0.20+6.19 = 0
Loop 2: -5.15+1.01+0.20-6.19+10.13 = 0
Loop 3: -8.01+2.01+1.01+10.13-5.15 = 0
For VA = 6.84V, VB = 5.33V and VC = 5.15V
Loop 1: -6.84+2.01+0.20+5.33 = 0
Loop 2: -5.15+1.01+0.20-5.33+10.13 = 0
Loop 3: -6.84+2.01+1.01+10.13-5.15 = 0
For VA = 9.3V, VB = 7.78V and VC = 5.15V
Loop 1: -9.3+2.01+0.20+7.78 = 0
Loop 2: -5.15+1.01+0.20-7.78+10.13 = 0
Loop 3: -9.3+2.01+1.01+10.13-5.15 = 0
Conclusion:
We verified Kirchoff’s voltage law.
Precautions:






Plug the switch carefully.
Before taking values make sure your circuit is not short.
If your circuit change it before connecting with power supply.
Measure the values carefully.
While taking current value across resistor make sure your one terminal of resistor is
removed from breadboard.
Use all safety measures to prevent yourself from any type of electric shock.
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Experiment No. 5
Objective:
To verify Kirchhoff’s current Law.
Apparatus:
a)
b)
c)
d)
e)
Digital multimeter (DMM)
DC power supply
Bread board
Jumper wires
Resistors
Theory:
Kirchhoff’s Current Law:
Kirchhoff's current law states that:
“The algebraic sum of all currents entering and exiting a node must equal zero.”
This law is used to describe how a charge enters and leaves a wire junction point or node on a
wire. Kirchhoff’s current law (KCL) is Kirchhoff’s first law that deals with the conservation
of charge entering and leaving a junction.
To determine the amount or magnitude of the electrical current flowing around an electrical
or electronic circuit, we need to use certain laws or rules that allows us to write down these
currents in the form of an equation. The network equations used are those according to
Kirchhoff’s laws, and as we are dealing with circuit currents, we will be looking at
Kirchhoff’s current law, (KCL).
Kirchhoff’s Current Law is one of the fundamental laws used for circuit analysis. His current
law states that for a parallel path the total current entering a circuits junction is exactly equal
to the total current leaving the same junction. This is because it has no other place to go as no
charge is lost.
In other words, the algebraic sum of ALL the currents entering and leaving a junction must
be equal to zero as:
Σ IIN = Σ IOUT
This idea by Kirchhoff is commonly known as the Conservation of Charge, as the current is
conserved around the junction with no loss of current. Let’s look at a simple example of
Kirchhoff’s current law (KCL) when applied to a single junction.
An Example including Single Junction:
Fig. No:5.1
Here in this simple single junction example, the current IT leaving the junction is the algebraic
sum of the two currents, I1and I2 entering the same junction. That is: IT = I1 + I2
Note that we could also write this correctly as the algebraic sum of: IT - (I1 + I2) = 0
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So, if I1 equals 5 amperes and I2 is equal to 3 amperes, then the total current IT leaving the
junction will be 5 + 3 = 8 amperes, and we can use this basic law for any number of junctions
or nodes as the sum of the currents both entering and leaving will be the same.
Also, if we reversed the directions of the currents, the resulting equations would still hold
true for I1 or I2. As I1 = IT - I2 = 5 - 3= 2A, and I2 = IT - I1 =8 - 5 = 3A Thus, we can think of
the currents entering the junction as being positive, while the ones leaving the junction as
being negative.
Then we can see that the mathematical sum of the currents either entering or leaving the
junction and in whatever direction will always be equal to zero, and this forms the basis of
Kirchhoff’s Junction Rule, more commonly known as Kirchhoff’s Current Law, or (KCL).
Circuit Diagram:
R1
1kΩ
V1
5V
R2
R7
4.1kΩ
10kΩ
R3
100Ω
R4
4.6Ω
R6
2.2kΩ
R5
220Ω
Procedure:
i.
ii.
iii.
iv.
v.
vi.
vii.
First, I made the circuit shown in the circuit diagram on the bread board.
Then, I connected the circuit with DC power supply and turned it ON.
According to the circuit diagram, I applied 9V using channel 1.
After completing the circuit and applying the voltage I took the digital multimeter and
checked the current across each resistor.
Then, I applied the Kirchoff’s current Law in all loops.
Then, I compared the experimental, theoretical and simulated values.
Then, I took sum of currents at every node to verify Kirchhoff’s current law.
Observation and Calculations:
Voltage = 5.15V
Equivalent resistance = 10.96 Ω
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At node 2:
I1 = I2 + I3
At node 3:
I3 = I4 + I5
At node 4:
I2 = I7 + I6
At all nodes current entering is equal to current leaving.
Node 1
Node 2
Node 3
Voltage
I1
(A)
I2
(A)
I3
(A)
I3
(A)
I4
(A)
I5
(A)
I2
(A)
I7
(A)
I6
(A)
5.15
3.75
0.182
3.56
3.56
0.17
3.36
0.182
0.033
0.15
6.1
4.5
0.1
4.27
4.27
0.202
4.03
0.1
0.04
0.18
7.3
5.39
0.26
5.07
5.07
0.24
4.78
0.26
0.05
0.21
Conclusion:
In this experiment, we verified Kirchhoff’s current law.
Precautions:





Make the circuit on the bread board very carefully.
Insert all the electric components in bread board carefully.
Connect the probes carefully.
Measure current across through each resistor with the help of DMM very carefully.
Follow all the safety precautions to avoid yourself from electric shocks or any other
damage.
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Experiment No. 6
Objective:
To verify Voltage divider rule.
Apparatus:
a)
b)
c)
d)
e)
DC power supply
Digital Multimeter (DMM)
Resistors
Breadboard
Jumper Wires
Theory:
Voltage Divider Rule:
The voltage division rule (voltage divider) is a simple rule which can be used in solving circuits to
simplify the solution. Applying the voltage division rule can also solve simple circuits thoroughly.
The statement of the rule is simple:
‘‘The voltage is divided between two series resistors in direct proportion to their resistance.’’
Proof of Voltage Divider Rule:
It is easy to prove Voltage divider in the following circuit:
Fig.6.1: Circuit To prove Voltage Divider Rule
The Ohm's law implies that
V1(t)=R1i(t)….. (I)
V2(t)=R2i(t)….. (II)
Therefore
v(t)=R1i(t)+R2i(t)=(R1+R2)i(t)
Hence
i(t)= V(t)∕R1+R2.
Substituting in I and II
𝑉(𝑡)
v1(t)=R1
𝑅1+𝑅2
v2(t)=R2
V(t)
R1+R2
Consequently,
V1(t)=
𝐑𝟏
𝐑𝟏+𝐑𝟐
𝐕(𝐭)
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V2(t)=
𝐑𝟐
𝐑𝟏+𝐑𝟐
V(t)
Which shows that the voltage is divided between two series resistors in direct proportion to
their resistance.
Example:
The rule can be easily extended to circuits with more than two resistors. For example,
Fig6.2: Example circuit.
Voltage Division among four resistors is:
𝑅1
𝑉1(𝑡) =
𝑅1+𝑅2+𝑅3+𝑅4
𝑉2(𝑡) =
𝑉3(𝑡) =
𝑉4(𝑡) =
𝑉(𝑡)
R2
R1+R2+R3+R4
𝑅3
𝑅1+𝑅2+𝑅3+𝑅4
𝑅4
𝑅1+𝑅2+𝑅3+𝑅4
𝑉(𝑡)
𝑉(𝑡)
𝑉(𝑡)
In above example, we calculated voltage drop each resistor by using voltage divider rule.
Applications of Voltage Divider:
The voltage divider is used only there where the voltage is regulated by dropping a particular
voltage in a circuit. It mainly used in such systems where energy efficiency does not
necessary to be considered seriously.
In our daily life, most commonly the voltage divider is used in potentiometers. The best
examples for the potentiometers are the volume tuning knob attached to our music systems
and radio transistors, etc. The basic design of the potentiometer includes three pins which are
shown above. In that two pins are connected to the resistor which is inside of the
potentiometer and the remaining pin is connected with a wipe contact which slides on the
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resistor. When someone changes the knob on the potentiometer then the voltage will be
appeared across the stable contacts and wiping contact according to the voltage divider rule.
Voltage dividers are used to adjust the signal’s level, for voltage measurement and bias of
active devices in amplifiers. A multimeter and Wheatstone bridge includes voltage dividers.
Voltage dividers can be used to measure the resistance of the sensor. To form a voltage
divider, the sensor is connected in series with a known resistance and known voltage is
applied across the divider. The analogue to digital converter of the microcontroller is
connected to the center tap of the divider so that tap voltage can be measured. By using the
known resistance, measured voltage resistance can be calculated.
Circuit Diagram:
Fig.6.3: Circuit
Procedure:
1. First of all, I made the diagram on the bread board as shown in above circuit diagram.
2. After making the circuit I connected the circuit with the DC power supply with the
help of jumper wires.
3. I applied 15 volts from channel 1 of the DC power supply with the circuit according
to the above circuit.
4. After completing the circuit and applying the voltage I took the digital multimeter and
checked the Voltage across each resistor.
5. Then, I calculated the value of voltages across each resistor by using voltage divider
rule.
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Observations and Calculations:
Sr NO
Resistors
Software
Experimental
Percentage
voltage
voltage
Error
R1
10k
13.72V
13.69V
0.2
R2
1k
1.28V
1.28V
0
R3
15k
1.28V
1.28V
0
R4
20k
212.78mV
212.76mV
0.01
R5
100k
1.06V
1.03V
2.8
Conclusion:
In this experiment, we verified voltage divider rule by making a circuit on
breadboard of resistors.
Precautions:






Make the circuit on the bread board very carefully.
Insert all the electric components in bread board carefully.
Turn on the DC power supply carefully.
Connect the probes carefully.
Measure voltage across through each resistor with the help of DMM very carefully.
Follow all the safety precautions to avoid electric shocks or any other damage.
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Experiment No. 7
Objective:
To verify Current Divider Rule.
Apparatus:
a)
b)
c)
d)
e)
Digital multimeter (DMM)
DC power supply
Bread board
Jumper wires
Resistors
Theory:
Statement: The electrical current entering the node of a parallel
circuit is divided into the branches. Current divider formula is employed to
calculate the magnitude of divided current in the circuits.
Let's understand the basic definitions:
Node: A point where two or more than two components are joined.
Parallel circuit: The circuit in which one end of all components share a
common node, and the other end of all components share the other common
node.
General formula:
A parallel circuit with 'n' number of resistors and an input voltage source is
illustrated below. We are interested to find the current which is flowing through
Rx.
In the above formula:
Ix: The current through Rx.
It: The total current which enters the circuit.
Rx: The resistance of the component whose current value is to be determined
Rt: The equivalent resistance of the parallel circuit
For two resistors:
Let's consider a parallel circuit having two resistors R1 and R2. The current
It enters the node. We are interested to calculate the current that is flowing
through. The general formula and circuit now take the form:
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We can modify the previous equation to obtain an alternative formula:
Circuit Diagram:
R
23OΩ
V1
V
R1
330Ω
R2
2.2KΩ
R3
1000Ω
R4
2.2KΩ
Procedure:
1. First of all I made the circuit shown in the circuit diagram on the bread
board.
2. After making the circuit I connected the circuit with the DC power supply
with the help of jumper wires.
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3. According to the circuit diagram I applied 15 voltage from channel 1 of the
DC power supply with the circuit.
4. After completing the circuit and applying the voltage I took the digital
multimeter and checked the current across each resistor.
5. After checking the current across each resistor I calculate the value of
current across each resistor by current divider rule .
Observation and Calculations:
Sr NO
Resistors
Software
Experimental
Percentage
Current(mA)
current(mA)
Error
R1
230ꭥ
34.69
34.66
0.08
R1
330ꭥ
21.28
21.27
0.04
R1
2.2k
3.19
3.18
0.31
R1
1k
7.02
7.01
014
R1
2.2k
3.19
3.18
0.31
Precautions:
 Make the circuit on the bread board very carefully.
 Insert all the electric components in bread board carefully.
 Turn on the DC power supply carefully.
 Connect the probes carefully.
 Measure current across through each resistor with the help of DMM very
carefully.
 Follow all the safety precautions to avoid electric shocks or any other
damage.
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Experiment No. 8
Objective:
To verify Superposition principle.
Apparatus:
a)
b)
c)
d)
e)
Digital Multimeter
Resistors
Bread Board
Jumper Wires
DC power supply
Theory:
Superposition Principle:
Statement: The superposition principle states that the voltage across (or current through) an
element in a linear circuit is the algebraic sum of the voltages across (or currents through)
that element due to each independent source acting alone.
Steps to Apply Superposition Principle:
1.Turn off all independent sources except one source. Find the output (voltage or current) due
to that active source.
2. Repeat step 1 for each of the other independent sources.
3. Find the total contribution by adding algebraically all the contributions due to the
independent sources.
Uses of Superposition principle:
The principle of superposition helps us to analyze a linear circuit with more than one
independent source by calculating the contribution of each independent source separately.
However, to apply the superposition principle, we must keep two things in mind:
1)We consider one independent source at a time while all other independent sources are
turned off. This implies that we replace every voltage source by 0 V (or a short circuit), and
every current source by 0 A (or an open circuit). This way we obtain a simpler and more
manageable circuit.
2) Dependent sources are left intact because they are controlled by circuit variables.
Example:
Use the superposition theorem to find v in the circuit.
Fig.8.1: Circuit Example
Since there are two sources, let
V= V1+V2
where V1 and V2 are the contributions due to the 6-V voltage source and the 3-A current
source, respectively. To obtain V1 we set the current source to zero, as shown in Fig.8.2.
Applying KVL to the loop in Fig. 8.2 gives:
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Fig.8.2: Calculating V 1
12i1 – 6 = 0
i1 = 0.5 A
Thus,
V1 = 4 i1 = 2V
To get we set the voltage source to zero, as in Fig. 8.3. Using current division,
Fig.8.3: Calculating V2
i1 =
8
(6) = 2A
4+8
Hence,
V2 = 4i3 = 8A
And we find
V= V1 + V2
V = 2 + 8 = 10V
Circuit Diagram:
Procedure:
i.
ii.
iii.
iv.
v.
vi.
vii.
First of all I made the circuit as shown in the circuit above.
Then I supplied 15V from channel 2 and 12V from channel 1.
Then I took DMM and measured voltage across resistor.
Then by using DMM I measured current of each resistor.
Then I theoretically calculated current and voltage values.
Then I took difference between theoretical and experimental values.
Then at the last I calculated percentage error.
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Observations and calculations:
Sr No.
Resistors
Both voltage
source connected
𝑽,
𝑽,,
V=𝑽, + 𝑽,,
𝑅1
1k
1.63V|
3.57V
-1.93V
1.64V
𝑅2
10k
13.37V
11.43V
1.95V
13.38V
𝑅3
4.7k
1.37V
11.43V
-10.05V
1.38V
𝑅4
1k
12V
24.31V
-12V
12.31V
𝑅5
4.7k
-117.08nV
24.311V
-141.39V
-
-117.08V
Conclusion:
I verified superposition theorem.
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Experiment No. 9
Objective:
To verify Thevenin’s Theorem.
Apparatus:
a)
b)
c)
d)
e)
Digital multimeter (DMM)
DC power supply
Bread board
Jumper wires
Resistors
Theory:
Thevenin Theorem:
Thevenin’s theorem states that a linear two-terminal circuit can be replaced by an equivalent
circuit consisting of a voltage source VTH in series with a resistor RTH, where VTH is the
open-circuit voltage at the terminals and RTH is the input or equivalent resistance at the
terminals when the independent sources are turned off.
Any combination of batteries and resistances with two terminals can be replaced by a single
voltage source e and a single series resistor r. The value of e is the open circuit voltage at the
terminals, and the value of r is e divided by the current with the terminals short circuited.
To apply this idea in finding the Thevenin resistance, we need to consider two cases.
■ CASE 1: If the network has no dependent sources, we turn off all independent sources. is
the input resistance of the network looking between terminals.
■ CASE 2: If the network has dependent sources, we turn off all independent sources. As
with superposition, dependent sources are not to be turned off because they are controlled by
circuit variables. We apply a voltage source Vo at terminals a and b and determine the
resulting current. Then RTH = Vo/Io, Alternatively, we may insert a current source io at
terminals a-b as shown in fi and find the terminal voltage Vo .Either of the two approaches
will give the same result. In either approach we may assume any value of Vo and io. For
example, we may use Vo= 1V or io=1A, or even use unspecified values of Vo or io.
Fig 9.1
It often occurs that takes a negative value. In this case, the negative resistance
R) implies that the circuit is supplying power.
(V = -I
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As mentioned earlier, a linear circuit with a variable load can be replaced by the Thevenin
equivalent, exclusive of the load. The equivalent network behaves the same way externally as
the original circuit. Consider a linear circuit terminated by a load RL, The current through the
load and the voltage across the load are easily determined once the Thevenin equivalent of
the circuit at the load’s terminals is btained, as shown in Fig.
Fig 9.2
We obtain.
Circuit Diagram:
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Procedure:
1. Here In Thevenin's Theorem, These Steps are applied to Solve the circuit.
2. First we have to remove the load resistance R L from the circuit and Find Open circuit
voltage VTh. Here we connect the two cordes of DMM at the two ends from where we
remove Load Resistance.
R2
10Ω
XMM1
V1
V2
6V
12V
Vth
?
R3
R1
15Ω
25Ω
3. Secondly, we have to find RTh at Load resistance position in such a way that by shortcircuiting the Voltage Sources and Open-circuiting the current sources. Then
Applying Series or parallel combinations to sought out RTh. Here we also use DMM at
the two ends from where we remove load resistance.
R2
10Ω
XMM1
R3
25Ω
R1
15Ω
4. Finally, Now make a Thevenin's circuit and Find Load current IL at Load Resistance.
Moreover, Load voltage can also be found in this step.
5. For
Finding
Voltage across the load resistance, See the figure below:
Rth
Vth
XMM2
12.5Ω
RL
5Ω
9V
6. For Finding IL across the load resistance, See the Figure Below:
Rth
XMM2
12.5Ω
Vth
9V
RL
5Ω
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Observation and Calculations:
For Vth
For Rth
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For IL and VL
-
R1
2.055V
0.137A
15Ὠ
Volts
Amperes
Ohms
Parameters
R2
1.370V
0.137A
10Ὠ
R3
9.425V
0.377A
25Ὠ
RL
2.571V
0.5142A
5Ὠ
Theoretical
Results
Experimental
Results
Simulations
9V
9V
0%
12.5Ὠ
12.5Ὠ
0%
0.5142A
0.5142A
0%
VTH
RTH
IL
Precautions:






Make the circuit on the bread board very carefully.
Insert all the electric components in bread board carefully.
Turn on the DC power supply carefully.
Connect the probes carefully.
Measure current across through each resistor with the help of DMM very carefully.
Follow all the safety precautions to avoid electric shocks or any other damage.
33 | P a g e
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Experiment No. 10
Objective:
To verify Norton’s Theorem.
Apparatus:
a)
b)
c)
d)
e)
Digital multimeter (DMM)
DC power supply
Bread board
Jumper wires
Resistors
Theory:
Nortons theorem is an analytical method used to change a complex circuit into a simple
equivalent circuit consisting of a single resistance in parallel with a current source
Norton on the other hand reduces his circuit down to a single resistance in parallel with a
constant current source.
Nortons Theorem states that “Any linear circuit containing several energy sources and
resistances can be replaced by a single Constant Current generator in parallel with a Single
Resistor“.
As far as the load resistance, RL is concerned this single resistance, RS is the value of the
resistance looking back into the network with all the current sources open circuited and ISis
the short circuit current at the output terminals as shown below.
Nortons equivalent circuit
The value of this “constant current” is one which would flow if the two output terminals
where shorted together while the source resistance would be measured looking back into the
terminals, (the same as Thevenin).
For example, consider our now familiar circuit from the previous section.
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To find the Nortons equivalent of the above circuit we firstly have to remove the
centre 40Ω load resistor and short out the terminals A and B to give us the following circuit.
When the terminals A and B are shorted together the two resistors are connected in parallel
across their two respective voltage sources and the currents flowing through each resistor as
well as the total short circuit current can now be calculated as:
with A-B Shorted Out
If we short-out the two voltage sources and open circuit terminals A and B, the two resistors
are now effectively connected together in parallel. The value of the internal resistor Rs is
found by calculating the total resistance at the terminals A and B giving us the following
circuit.
Find the Equivalent Resistance (Rs)
Having found both the short circuit current, Is and equivalent internal resistance, Rs this then
gives us the following Nortons equivalent circuit.
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Nortons equivalent circuit
Ok, so far so good, but we now have to solve with the original 40Ω load resistor connected
across terminals A and B as shown below.
Again, the two resistors are connected in parallel across the terminals A and B which gives us
a total resistance of:
The voltage across the terminals A and B with the load resistor connected is given as:
Then the current flowing in the 40Ω load resistor can be found as:
which again, is the same value of 0.286 amps, we found using Kirchhoff´s circuit law in the
previous tutorials.
Nortons Theorem Summary




The basic procedure for solving a circuit using Nortons Theorem is as follows:
1. Remove the load resistor RL or component concerned.
2. Find RS by shorting all voltage sources or by open circuiting all the current sources.
3. Find IS by placing a shorting link on the output terminals A and B.
4. Find the current flowing through the load resistor RL.
In a circuit, power supplied to the load is at its maximum when the load resistance is equal to
the source resistance. In the next tutorial we will look at Maximum Power Transfer. The
application of the maximum power transfer theorem can be applied to either simple and
complicated linear circuits having a variable load and is used to find the load resistance that
leads to transfer of maximum power to the load.
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Circuit Diagram:
R1
R3
4Ω
1Ω
XMM1
RL
2Ω
V1
28V
V2
7V
Procedure:
1. First, we have to find RN at Load resistance RL position in such a way that by shortcircuiting the Voltage Sources and Open-circuiting the current sources. Then
Applying Series or parallel combinations to sought out RN. Here we also use DMM at
the two ends from where we remove load resistance.
R4
R2
4Ω
1Ω
XMM2
2.
3. Now in second step ,we have to find Norton’s current which is also known as Short
circuit current.Here we remove the load resistance and short circuit current denoted as
ISc can be find out by mesh analysis or nodal analysis across it .
R1
R3
4Ω
1Ω
XMM1
V2
V1
28V
7V
4.
5. Now Finally we make a Norton’s equivalent circuit to find out the Load current IL
across Load resistance.
XMM2
IN
ISc
14A
RN
0.8Ω
RL
2Ω
6. Formulae: IL=IN ×RN÷RN+RL
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Observations and Calculations:
For RTh :
For ISc :
For Load Current IL :
38 | P a g e
EC LAB MANUAL
Volts
Amperes
Ohms
Parameters
RN
IN
IL
R1
20V
5A
4Ὠ
Theoretical Value
0.8Ὠ
14A
4A
RN
8V
4A
2Ὠ
Simulation Value
0.8Ὠ
14A
4A
R3
1V
1A
1Ὠ
Error%
0%
0%
0%
Precautions:






Make the circuit on the bread board very carefully.
Insert all the electric components in bread board carefully.
Turn on the DC power supply carefully.
Connect the probes carefully.
Measure current across through each resistor with the help of DMM very carefully.
Follow all the safety precautions to avoid electric shocks or any other damage.
39 | P a g e
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Experiment No. 11
Objectives:
1. To plot output waveform of half-wave Rectifier (with and without capacitor).
2. To find the ripple factor for half-wave Rectifier.
3. To find Efficiency using formula.
Apparatus:
a) We have a AC voltage source of 5 V having frequency of 5o Hz.
b) Resistors (for multiple experiments).
c) Digital Multimeter with testing cordes.
d) A Bread Board
e) Cathode ray Oscilloscope (For graphical waveform).
f) A single 1N400G virtual Diode.
g) 100µF Capacitor.
Theory:
A rectifier is a device that converts alternating current (AC) to Direct Current (DC). It is done
by using a diode or a group of diodes. Half wave rectifiers use one diode, while a Full wave
rectifier uses multiple diodes.
A half wave rectifier is defined as “A type of rectifier that only allows one half-cycle of an
AC voltage waveform to pass, blocking the other half-cycle.” Half-wave rectifiers are used to
convert AC voltage to DC voltage, and only require a single diode to construct.The working
of a half wave rectifier takes advantage of the fact that diodes only allow current to flow in
one direction.
First, a high AC voltage is applied to the Primary load(resistance), the voltage which is alternating first
goes through the diode, Thus During the positive half cycle of the AC voltage, the diode will be forward
biased and the current flows through the diode. During the negative half cycle of the AC voltage, the
diode will be reverse biased and the flow of current will be blocked. The half wave rectifier does not
completely block the negative half cycles. It allows a small portion of negative half cycles or small
negative current. This current is produced by the minority carriers in the diode.
The current produced by the minority carriers is very small. So it is neglected. We can’t visually see the
small portion of negative half cycles at the output.
In an ideal diode, the negative half cycles or negative current is zero.
Graphically It will look like
40 | P a g e
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The output DC current or DC signal produced by a positive half wave rectifier is a series of positive half
cycles or positive sinusoidal pulses.
The output waveform we have obtained from the theory above is a pulsating DC waveform.
This is what is obtained when using a half wave rectifier without a filter.
Half Wave Rectification with Capacitor Filter:
Filters are used to smoothen pulsating DC waveforms into constant DC waveforms. They
achieve
this
by
suppressing
the
DC
ripples
in
the
waveform.
Although half-wave rectifiers without filters are also possible, but they can’t be used for any
practical applications. As DC equipment requires a constant waveform, we need to ‘smooth
out’ this pulsating waveform for it to be any use in the real world. A capacitor or
an Inductor can be used as a filter – but half wave rectifier with capacitor filter is most
commonly used. For this purpose the capacitor will be placed parallel to the load.
Ripple Factor of Half Wave Rectifier:
‘Ripple’ is the unwanted AC component remaining when converting the AC voltage
waveform into a DC waveform. Even though we try out best to remove all AC components,
there is still some small amount left on the output side which pulsates the DC waveform. This
undesirable AC component is called ‘ripple’.
Efficiency of Half Wave Rectifier:
Rectifier efficiency (η) is the ratio between the output DC power and the input AC power.
The formula for the efficiency is equal to:
Where; PAC = V2rms / RL
PDC = V2dc / RL
Advantages of Half Wave Rectifier:
The main advantage of half-wave rectifiers is in their simplicity. As they don’t require as
many components, they are simpler and cheaper to setup and construct.
As such, the main advantages of half-wave rectifiers are:

Simple (lower number of components)

Cheaper up front cost (as their is less equipment. Although there is a higher cost over
time due to increased power losses)
Disadvantages of Half Wave Rectifier:
The disadvantages of half-wave rectifiers are:



They only allow a half-cycle through per sinewave, and the other half-cycle is wasted.
This leads to power loss.
They produce a low output voltage.
The output current we obtain is not purely DC, and it still contains a lot of ripple (i.e.
it has a high ripple factor).
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Circuit Diagram:
With Capacitor
D1
1N4001G
V1
5Vpk
50Hz
0°
C1
100µF
R1
1kΩ
Without Capacitor
D1
1N4001G
V1
5Vpk
50Hz
0°
R1
1kΩ
Procedure:
First of all I made the circuit as given above.
Then i set the values as we mentioned earlier and then place the DMM across the components
where we want to determine the RMS values of Voltage in the circuit.
XMM3
XMM1
D1
1N4001G
V1
5Vpk
50Hz
0°
XMM2
R1
1kΩ
Then I connected the oscilloscope to resistor to check the graphical waveform.
XMM3
XSC1
XMM1
Ex t Tri g
+
_
B
A
+
_
+
_
D1
1N4001G
V1
5Vpk
50Hz
0°
R1
1kΩ
XMM2
4: Then I find the software values by running the circuit.
42 | P a g e
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Observations & Calculations:
(With capacitor)
Now we will place a 1µF capacitor in parallel to the resistor and determined the values as
before.
D1
1N4001G
V1
5Vpk
50Hz
0°
C1
100µF
R1
2000Ω
XMM2
(Without Capacitor):
For Vrms at the Input supply:
For Vdc Output at Load:
For Vac Rms at load :
43 | P a g e
EC LAB MANUAL
Result:
With Capacitor:
Sr no.
RL
1
2
3
4
5
1Kꭥ
500ꭥ
1300ꭥ
1800ꭥ
2Kꭥ
VAC (Rms)
(Input)
3.53V
3.53V
3.53V
3.53V
3.53V
VAC (Rms)
(Output)
0.211V
0.3798V
0.168V
0.123V
0.112V
VDC
(Output)
4.01V
3.72V
4.085V
4.17V
4.19V
Ripple factor
Vac (Rms / VDC )
0.0526
0.1021
0.0411
0.0295
0.0267
VAC (Rms)
(Input)
3.53V
3.53V
3.53V
3.53V
3.53V
VAC (Rms)
(Output)
1.675V
1.66V
1.68V
1.686V
1.688V
VDC
(Output)
1.314V
1.3V
1.319V
1.327V
1.329V
Ripple factor
VAC (Rms / VDC)
1.275
1.277
1.274
1.271
1.270
Without capacitor:
Sr no.
RL
1
2
3
4
5
1000ꭥ
500ꭥ
1300ꭥ
1800ꭥ
2000ꭥ
Input Ac Supply Wave Form:
Output Waveform Without Capacitor:
Dc output with Capacitor:
Graphically:
Graph between ripple factor and load resistance
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With capacitor
Without capacitor
Precautions:
 Make the circuit on the bread board very carefully.
 Insert all the electric components in bread board carefully.
 Turn on the DC power supply carefully.
 Connect the probes carefully.
 Measure current across through each resistor with the help of DMM very carefully.
 Follow all the safety precautions to avoid electric shocks or any other damage.
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EXPERIMENT No. 12
Objective:
1. To plot output waveform of full-wave Bridge Rectifier (with and without
capacitor).
2. To find the ripple factor for full-wave Bridge Rectifier.
3. To find Efficiency.
Apparatus:
a)
b)
c)
d)
e)
f)
AC voltage source of 6 V having frequency of 5o Hz.
Resistors
Digital Multimeter (DMM)
A Bread Board
Oscilloscope
1N400G virtual Diode.
g) 100µF Capacitor.
Theory:
Full Wave Rectification:
The circuits which convert alternating current (AC) into direct current (DC) are known as
rectifiers. If such rectifiers rectify both the positive and negative half cycles of an input
alternating waveform, the rectifiers are referred as full wave rectifiers. Alternatively, we can
say, a rectifier is a device that converts alternating current (AC) to direct current (DC).
The half-wave rectifier is not more efficient and we cannot use it for applications which need
a smooth and steady DC output. For more efficient and steady DC, we will use a full wave
rectifier.
Full Wave Bridge Rectifier:
Construction of Full Wave Bridge Rectifier:
A full wave bridge rectifier is a type of rectifier which will use four diodes in a bridge
formation. A full wave bridge rectifier system consists of
1. Four Diodes
2. Resistive Load
We use the diodes namely D1, D2, D3 and D4 which form a bridge circuit. The circuit
diagram is as follows:
The four diodes labelled D1 to D4 are arranged in “series pairs” with only two diodes conducting current
during each half cycle. The Positive Half-cycle:
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During the positive half cycle of the supply, diodes D1 and D2 conduct in series as they are forward
biased while diodes D3 and D4 are reverse biased and the current flows through the load.
The Negative Half-cycle:
During the negative half cycle of the supply, diodes D3 and D4 conduct in series as they are forward
biased, but diodes D1 and D2 switch “OFF” as they are now reverse biased. The current flowing
through the load is the same direction as before.
As we can assume Diodes D1 as A, D2 as C, D3 as B, D4 as D So Our Resultant Wave Form
be like with the time interval as;
Full-wave Rectifier with Smoothing Capacitor:
We get a pulsating DC voltage with a lot of ripples as the output of the full wave bridge
rectifier. We cannot use this voltage for practical applications. So, to convert the pulsating
DC voltage to pure DC voltage, we use a filter (capacitor) in the circuit. So by this method We
can improve the average DC output of the rectifier while at the same time reducing the AC variation of
the rectified output by using smoothing capacitors to filter the output waveform. Smoothing or reservoir
capacitors connected in parallel with the load across the output of the full wave bridge rectifier circuit
increases the average DC output level even higher as the capacitor acts like a storage device.
During the half-wave, the diodes D1 and D2 conduct. It charges the capacitor immediately to
the maximum value of the input voltage. When the rectified pulsating voltage starts
decreasing and less than the capacitor voltage, the capacitor starts discharging and supplies
current to the load. This discharging is slower when compared to the charging of the
capacitor, and it does not get enough time to discharge entirely and the charging starts again
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in next pulse of the rectified voltage waveform. So around half of the charge present in the
capacitor gets discharged. During the negative cycle, the diodes D3 and D4 start conducting,
and the above process happens again. This causes, the current continues to flow through the
same direction across the load.
Ripple Factor:
The output we will get from the rectifier will consist of both AC and DC components. The
AC components are undesirable to us and will cause pulsations in the output. These unwanted
AC components are called Ripple. The ripple factor is the ratio between the RMS value of the
AC voltage and the DC voltage (on the output side) of the rectifier (diode). Note that for us to
construct a good rectifier, we want to keep the ripple factor as low as possible. This is why
we use capacitors as filters to reduce the ripples in the circuit in order to keep the output form
as smooth as possible.
Rectifier efficiency (η) is the ratio between the output DC power and the input AC power.
The formula for the efficiency is equal to:
Where; PAC = V2rms / RL
PDC = V2dc / RL
Advantages of Full Wave Rectifiers:
Full wave rectifiers have higher rectifying efficiency than half wave rectifiers This means
that they convert AC to DC more efficiently.
They have low power loss because no voltage signal is wasted in the rectification process.
The output voltage of full wave rectifier has lower ripples than half wave rectifiers.
Disadvantages of Full Wave Rectifiers:
The Full wave rectifier is more expensive than half-wave rectifier and tends to occupy a lot of
space.
Circuit Diagram: (With Capacitor)
1N 4001G
D3
V1
6Vpk
50H z
0°
D1
1N 4001G
D2
1N 4001G
D4
1N 4001G
C1
100µF
R1
1.0kΩ
(Without Capacitor)
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1N4001G
D3
V1
6Vpk
50Hz
0°
D1
1N4001G
D2
1N4001G
D4
1N4001G
R1
1.0kΩ
Procedure:
First of all we will design the circuit as above we make.
Then we will set the values as we mentioned earlier and then place the DMM across the
components where we want to determine the RMS values of Voltage in the circuit. As below;
XMM2
1N4001G
D3
V1
6Vpk
50Hz
0°
D1
1N4001G
D2
1N4001G
XMM1
D4
1N4001G
R1
1.0kΩ
After that for graphical waveform so we connect CRO with the resistor follows;
XMM2
1N4001G
D3
V1
6Vpk
50Hz
0°
XSC1
Ex t Tri g
+
_
D1
1N4001G
D2
1N4001G
B
A
+
_
+
_
XMM1
D4
1N4001G
R1
1.0kΩ
Then We will run the circuit so then we will determine the respective Rms values at the
components for our Required Data.
Observations & Calculations (Without Capacitor):
For Vrms at the Input supply:
For Vdc Output at Load:
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For Vac Rms at load :
Sr no.
RL
VAC (Rms)
(Input)
VAC (Rms)
(Output)
VDC (Output)
Ripple factor
VAC (Rms) / VDC )
1
2
3
4
5
1Kꭥ
500ꭥ
1300ꭥ
1800ꭥ
2Kꭥ
4.242V
4.242V
4.242V
4.242V
4.242V
1.666V
1.654V
1.671V
1.676V
1.678V
2.733V
2.676V
2.754V
2.781V
2.79V
0.6096
0.6181
0.6068
0.6027
0.6014
(With Capacitor):
D3
1N4001G
V1
6Vpk
50Hz
0°
D2
1N4001G
D1
1N4001G
D4
1N4001G
C1
100µF
R1
2000Ω
For Vdc Output at Load:
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For Vrms at the Input supply:
For Vrms at the Output Load:
Sr no.
R_L
1
2
3
4
5
1KὨ
500Ὠ
1300Ὠ
1800Ὠ
2KὨ
V_Ac
(Rms)
(Input)
4.242V
4.242V
4.242V
4.242V
4.242V
V_Ac
(Rms)
(Output)
0.1077V
0.2034V
0.0844V
0.0618V
0.0559V
V_dc(Output)
Ripple factor
Vac (Rms) / Vdc
4.515V
4.348V
4.558V
4.602V
4.615V
0.0239
0.0468
0.0185
0.0134
0.0121
Input Ac Supply Wave Form:
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Output Waveform Without Capacitor:
Dc output with Capacitor:
Graphically:
Graph between ripple factor and load resistance
With capacitor
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Without capacitor
Precautions:






Make the circuit on the bread board very carefully.
Insert all the electric components in bread board carefully.
Turn on the DC power supply carefully.
Connect the probes carefully.
Measure current across through each resistor with the help of DMM very carefully.
Follow all the safety precautions to avoid electric shocks or any other damage.
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Experiment No.13
Objective:
Verification of maximum power theorem.
Apparatus:




Bread board.
DC power supply.
Jumper wires.
Resistors.
 Various load resistors.
Theory:
Statement:
“Maximum Power Transfer occurs when the resistive value of the load is equal to
the value of voltage sources internal resistance; allowing maximum power to be supplied.”
Also when RL = RTh
However, when we connect a load resistance, RL across the output terminals of the power
source, the impedance of the load will vary from an open-circuit state to a short-circuit state
resulting in the power being absorbed by the load becoming dependent on the impedance of
the actual power source. Then for the load resistance to absorb the maximum power possible
it has to be “Matched” to the impedance of the power source and this forms the basis
of Maximum Power Transfer.
The Maximum Power Transfer Theorem can be defined as, a resistive load is connected to
a DC-network, when the load resistance (RL) is equivalent to the internal resistance then it
receives highest power is known as Thevenin’s equivalent resistance of the source network.
RL is the resistor to whom we want to supply maximum power.
Example:
A variable resistance RL is connected to a DC source network as shown in the circuit diagram
in figure A below and the figure B represents the Thevenin’s voltage VTH and Thevenin’s
resistance RTH of the source network. The aim of the Maximum Power Transfer theorem is to
determine the value of load resistance RL, such that it receives maximum power from the DC
source.
Considering figure B the value of current will be calculated by the equation shown below
While the power delivered to the resistive load is given by the equation
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Putting the value of I from the equation (1) in the equation (2) we will get
PL can be maximized by varying RL and hence, maximum power can be delivered when
(dPL/dRL) = 0
Circuit diagram:
R1
R2
1.0kΩ
2.2kΩ
R3
2kΩ
V1
15V
R4
330Ω
R5
100Ω
RL
300Ω
Procedure:
i.
ii.
iii.
iv.
First of all, I made the circuit on the bread board.
Then, I connected the channel of the DC power supply to the circuit.
Then, I removed the load resistor.
Then, I find the RTH as
R1
R2
1.0kΩ
2.2kΩ
R3
2kΩ
R4
330Ω
R5
100Ω
v.
R Th
Then, I calculated the Vth in the circuit.
R1
R2
1.0kΩ
2.2kΩ
R3
2kΩ
V1
15V
R5
100Ω
vi.
vii.
R4
330Ω
V Th
Then, I connected the various provided resistors and measured their resistance.
Then I calculated the maximum power through the following
Pmax = VL/ RL+ RTh
formula
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Observations and calculations:
RTH:
VTH:
Hence RTH = 296.05ohm and VTH = 1.05V
Load Voltage=
Sr. No.
RL
VL
1
360Ω
0.573V
2
340 Ω
0.559V
3
320 Ω
0.543V
4
296.047Ω
0.523V
5
270 Ω
0.498V
6
250 Ω
0.478V
7
230 Ω
0.457V
Load current= IL
(mA)
1.594
1.644
1.697
1.766
1.846
1.914
1.987
PL=POut
(mW)
0.913
0.918
0.921
0.922
0.920
0.917
0.908
Graphically
Conclusion:
After performing this experiment I noted that the maximum power transfer when
RL = RTh.
Precautions:
•
•
•
•
Connect the circuit with DC supply carefully.
Make circuit carefully and check its connectivity by DMM.
Insert all the electric components in bread board carefully.
Check the current carefully across resistors.
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