24-11-2024

Electrical And
Electronics Technology
Dr. Sathish Shet K
sathisha.shet@pilani.bits-pilani.ac.in
Associate Professor
BITS Pilani Department of Electrical & Electronics Engineering
BITS Pilani
Pilani|Dubai|Goa|Hyderabad

BITS Pilani
Pilani Campus

Electrical And Electronics Technology

ENGG ZC112/ PE ZC112
LECTURE-1

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Course Objectives:
To get familiar with the concepts of electrical circuit analysis employed in various
fields of industry in measurements, production, etc.
To understand the concepts of various electrical machines like motors, alternators
etc. employed in domestic and industrial environments
To learn the applications of digital circuits including semiconductor theory to for
understanding various digital instruments, control logics etc
Text Book(s):

Smith, I McKenzie (Ed), Hughes Electrical & Electronics

Technology, Pearson Education, 10th Edition, 2008.

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Reference Book

R1: Vincent Del Toro, Electrical Engineering

Fundamentals, PHI, 2nd Ed., 1994
R2:John Bird, Electrical and Electronic Principles and
Technology, Elsevier, 3rd Ed.,2007 (available on online
digital library)

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Contact session-1 Details

Contact List of Topic Sub-Topics Refere Prerequi

Hour Title nce site
 D.C Circuits T1- Recorded
Simple DC  Simulation of D.C Ch.1 Lecture
circuits, circuits
1-2
Network  Network Analysis
analysis of D.C circuits
 Numerical & Quiz

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

An Electrical System
A basic electrical system has four constituent parts.
1. The source: The function of the source is to provide the energy for the
electrical system. A source may usually be thought of as a
battery or a generator, although for simplicity we might even
think of a socket outlet as a source.
2. The load: The function of the load is to absorb the electrical
energy supplied by the source. Most domestic electrical
equipment constitutes loads. Common examples include lamps
and heaters, all of which accept energy from the system.
3. The transmission system: This conducts the energy from
the source to the load. Typically the transmission system
consists of insulated wire.
4. The control apparatus : As the name suggests, its function
is to control. The most simple control is a switch which permits
the energy to flow or else interrupts the flow
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An Electrical System

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Electrical and Electronic Systems

 Basic functions include elements concerned with the

manipulation of electrical energy.

 Common functions are:

– generation

– transmission of communication

– control or processing

– utilization

– storage

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Basic Electrical System

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Basic Electrical System

System examples
– electrical and electronic systems often fall within a
range of categories, such as those responsible for:

• power generation and distribution

• monitoring of some equipment or process
• control of some equipment or process
• signal processing
• communications

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System Inputs and Outputs

Systems may often be described simply by their inputs, their output

and the relationship between them

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 Nature of inputs/outputs will depend on where we draw

our system boundaries. For example:

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 By changing the system boundary we change the nature of the inputs

and outputs

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BITS Pilani, Pilani Campus

Current flow in a circuit

1. There must be a complete circuit around which the electrons

may move. If the electrons cannot return to the point of starting,
then eventually they will all congregate together and the flow will
cease.
2. There must be a driving influence to cause the continuous flow.
This influence is provided by the source which causes the current
to leave at a high potential and to move round the circuit until it
returns to the source at a low potential.

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Current flow in a circuit

The driving influence is termed the electromotive force,

hereafter called the e.m.f.
Each time the charge passes through the source, more
energy is provided by the source to permit it to
continue round once more.
This is a continuous process since the current flow is
continuous.
It should be noted that the current is the rate of flow of
charge through a section of the circuit.

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Electromotive force and potential difference

The energy introduced into a circuit is transferred to the load unit by

the transmission system, and the energy transferred due to the
passage of unit charge between two points in a circuit is termed the
potential difference (p.d.).
If all the energy is transferred to the load unit, the p.d. across the load
unit is equal to the source e.m.f.
It will be observed that both e.m.f. and p.d. are similar quantities.
How ever, an e.m.f. is always active in that it tends to produce an
electric current in a circuit whereas a p.d. may be either passive or
active.
A p.d. is passive whenever it has no tendency to create a current in a
circuit.

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Electromotive force and potential difference

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Electrical units

 Current : Symbol: I Unit: ampere (A)

 Quantity of electricity : Q [coulombs] = I [amperes] × t

[seconds]
∴ Q = It Charge Symbol: Q Unit: coulomb (C)

 Potential difference: Electric potential Symbol: V , Unit: volt

(V) V= , P = VI

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Electrical units

 Resistance : Electric resistance Symbol: R Unit: ohm (Ω)

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Ohm’s Law
Defines the relationship between voltage, current, and resistance in an electric
circuit
Ohm’s Law:
Current in a resistor varies in direct proportion to the voltage applied
to it and is inversely proportional to the resistor’s value.
Stated mathematically: V
I=
R
V Where: I is the current (amperes)
+ - V is the potential difference (volts)
I R R is the resistance (ohms)

Note: Direction of currents and polarities of voltages are crucial for applying Ohm’s law.

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Ohm’s Law Triangle

V V
I= (amperes,A)
I R R

V V
R= (ohms,Ω)
I R I

V
V = IR (volts,V)
I R

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SIMPLE DC CIRCUITS

Volt drops in a series circuit: V = V1 + V2 + V3

In general, V = IR, then V1 = IR1, V2 = IR2

and V3 = IR3, the current I being the same in
each resistor. Substituting in equation
V = IR1 + IR2 + IR3
For the complete circuit, the effective
resistance of the load R represents the ratio
of the supply voltage to the circuit current
hence V = IR
but V = IR1 + IR2 + IR3
hence IR = IR1 + IR2 + IR3
and R = R1 + R2 + R3
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Series-connected resistors

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Example-1
Calculate for each of the circuits shown in Fig. the current flowing
in the circuit given that R == 3 kΩ

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Example-2

Calculate the voltage across each of the resistors shown in Fig.

and hence calculate the supply voltage V.

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Example-3

For the circuit shown in Fig. calculate the circuit current,

given that the supply is 100 V.

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Voltage division between two resistors

Given the supply voltage V, it is required to determine the volt drop across
R1. The total resistance of the circuit is

The ratio of the voltages therefore depends on the ratio of the resistances.
This permits a rapid determination of the division of volt drops in a simple series circuit and
the arrangement is called a voltage divider.

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Example-4

A voltage divider is to give an output voltage of 10 V from an

input voltage of 30 V as indicated in Fig. Given that R2 ==
100 Ω, calculate the resistance of R1.

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Parallel Networks

Currents in a parallel network

the voltage across each branch being the same. Substituting

in equation we get

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Parallel Networks

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Example-5

Calculate the supply current to the network shown in Fig.

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Example-6

For the network shown in Fig. calculate the effective resistance

and hence the supply current.

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Circuit with two resistors in parallel

It may also be convenient to determine the manner in which two parallel

resistors share a supply current. With reference again to Fig.

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Example-7

A current of 8 A is shared between two resistors in the network

shown in Fig. 3.22. Calculate the current in the 2 Ω resistor,
given that
(a) R1 = 2 Ω;
(b) R1 = 4 Ω.

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Comparison Between Series and Parallel Networks

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Kirchhoff’s Laws:
Kirchhoff’s laws: (German physicist Gustav Kirchhoff )
1. Kirchhoff’s current law (KCL)
2. Kirchhoff’s voltage law (KVL)

Kirchhoff’s current law

1st form: At any node of a circuit, at every instant of time, the sum of the currents into
the node is equal to the sum of the currents out of the node.

For the given node, applying 1st form of KCL: 𝑖 + 𝑖 = 𝑖 + 𝑖

2nd form: At any node of a circuit, the currents algebraically sum to zero.
Applying above 2nd form of KCL to the above node.𝑖 + 𝑖 − 𝑖 − 𝑖 = 0.

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Kirchhoff’s current Law: Example

Problems-1
Find current 𝑖 at the node Find current 𝑖 𝑎𝑛𝑑 𝑖 at the nodes
shown below shown below

Ans: Prob1 : i3 = 7A
Prob2: i3 = 4A(toward left), i4 = 6A (towards Up)

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Example on Kirchhoff's Law-2

For the network junction shown in Fig. calculate the current I3,
given that I1 =3 A, I2 = 4 A and I4 = 2 A.

I1 − I2 + I3 − I4 = 0
I3 = −I1 + I2 + I4 = −3 + 4 + 2 = 3 A

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Example on Kirchhoff's Law-3

With reference to the network shown in Fig. , determine the

relationship between the currents I1, I2, I4 and I5
.

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Example on Kirchhoff's Law-4

For the network shown in Fig. I1 = 2.5 A and I2 = −1.5

A.Calculate the current I3.

By Kirchhoff’s law:
I1 + I2 + I3 = 0
I3 = −I1 − I2 = −2.5 + 1.5 = −1.0 A
Kirchhoff’s first law may be applied at any point within
a network

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Power and Energy

By consideration of the problems of our wiring, it has been seen that

a reduction in the voltage and in the current to the lamp bulbs
causes their light out-put to be reduced. The light output is the
rate at which the light energy is given out, i.e. the power of the
lamp. It can be shown that
P ∝ V and P ∝ I
where
P ∝ VI
But it may be recalled that the volt is that potential difference across
a conductor when passing a current of 1 A and dissipating energy
at the rate of 1 W. It follows that P = VI

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Consider again relation, In a simple load V = IR, and

substituting
in equation P = VI P = (IR)I = I2R
∴ P = I2R
It should be noted that the expression I2R represents a
power, i.e. the rate at which energy is transferred or
dissipated, and not the energy itself.
W = Pt = VIt
and W = I2Rt

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Example Problem

A 230 V lamp is rated to pass a current of 0.26 A. Calculate its

power output. If a second similar lamp is connected in parallel
to the lamp, calculate the supply current required to give the
same power output in each lamp

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Example Problem

A current of 3 A flows through a 10 Ω resistor. Find:

(a) the power developed by the resistor;
(b) the energy dissipated in 5 min.

(a) P = I2R = 3 2 × 10 = 90 W
(b) W = Pt = 90 × (5 × 60) = 2700 J

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Example Problem

For the network shown in Fig. calculate the power

developed by each resistor.

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Resistivity

Certain materials permit the reasonably free passage of electric charge

and are termed conductors, while others oppose such a free passage
and are termed insulators. These abilities are simply taken relative to
one another and depend on the material considered. However, other
factors also have to be taken into account.
Consider a conductor made of a wire which has a resistance of 1 Ω for
every 10 cm of its length. If the wire is made 20 cm long then
effectively there are two sections each of 10 cm connected in series.
This being the case, the resistance will be 2 Ω. This form of argument
may be continued so that 30 cm of wire will have a resistance of 3 Ω
and so on, resulting in the length/resistance characteristic shown in
Fig. Since the graph has the form of a straight line, then the
resistance is proportional to the length of wire, i.e. R ∝ l.

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Example Problem

A cable consists of two conductors which, for the purposes of a

test, are connected together at one end of the cable. The
combined loop resistance measured from the other end is
found to be 100 Ω when the cable is 700 m long. Calculate the
resistance of 8 km of similar cable.

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Area/resistance characteristic of a conductor

Again considering 10 cm pieces of conductor of resistance 1

Ω, if two such pieces are connected in parallel then the
resistance is 0.5 Ω. Equally if three such pieces are connected
in parallel, the total resistance is 0.33 Ω, and so on. The
addition of each piece of wire increases the area of conductor
available to the passage of current and hence the area/
resistance characteristic of Fig. is obtained. The form of this
characteristic is such that the resistance is inversely
proportional to the area available, i.e.

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Example

A conductor of 0.5 mm diameter wire has a resistance of 300 Ω.

Find the resistance of the same length of wire if its diameter
were doubled.

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Combining relations [3.14] and [3.15] we get

Rather than deal in proportionality, it is better to insert a constant into this

relation, thereby taking into account the type of material being used. This
constant is termed the resistivity of the material.
Resistivity is measured in ohm metres and is given the symbol ρ (ρ is the
Greek letter rho)
Resistivity Symbol: ρ Unit: ohm metre (Ω m)

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Typical values of resistivity

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Example

A coil is wound from a 10 m length of copper wire having a

cross-sectional area of 1.0 mm 2.Calculate the
resistance of the coil.

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Temperature coefficient of resistance

The resistance of all pure metals increases with increase of temperature,

whereas the resistance of carbon, electrolytes and insulating materials
decreases with increase of temperature.
Certain alloys, such as manganin, show practically no change of
resistance for a considerable variation of temperature. For a
moderate range of temperature, such as 100 °C, the change of
resistance is usually proportional to the change of temperature,
The ratio of the change of resistance per degree change of temperature
to the resistance at some definite temperature, adopted as standard,
is termed the temperature coefficient of resistance and is represented
by the Greek letter α
Temperature coefficient of resistance Symbol: α Unit: reciprocal degree
centigrade (/°C)

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Variation of resistance of copper with temperature

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Kirchhoff’s Voltage Law

1st form: In traversing any loop in any circuit at every instant of time, the sum of the
voltages having one polarity equals the sum of the voltages having the opposite
polarity.

Apply above KVL: we can deduce that

𝑉3 + 𝑉2 + V6 = V1 + V5

2nd form: Around any loop in a circuit, the voltages algebraically sum to zero. Apply
above form of KVL: we can deduce that
𝑉3 + 𝑉2 + V6 − V1 − V5 = 0

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Electric Circuit Terms

Circuit – a circuit is a closed loop conducting path in which an
electrical current flows.
Path – a single line of connecting elements or sources.
Node – a node is a junction, connection or terminal within a circuit
were two or more circuit elements are connected or joined together
giving a connection point between two or more branches. A node is
indicated by a dot.
Branch – a branch is a single or group of components such as resistors
or a source which are connected between two nodes.
Loop – a loop is a simple closed path in a circuit in which no circuit
element or node is encountered more than once.
Mesh – a mesh is a single closed loop series path that does not contain
any other paths. There are no loops inside a mesh.

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Nodes, Mesh, loop and Branches in a circuit

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