Electrical and Electronic Fundamental

DMV10123

ELECTRICAL AND
ELECTRONIC

FUNDAMENTAL



KOLEJ KEMAHIRAN TINGGI MARA KUANTAN

PREFACE
This module is filled with information to give students a ready information for
DMV10123 Electrical and Electronics Fundamental course. It consists of
eight chapters: Concept of electrical quantities, Fundamental Law, Circuit
Analysis, Capacitor and Inductor, Fundamental of solid state principle,
Diodes, Transistors and Operational amplifiers. Each chapter ends with
summary of key points and exercises.

i

TABLE OF CONTENTS

PREFACE i
TABLE OF CONTENTS ii
LIST OF ABBREVIATION/SYMBOL vi

CHAPTER 1 CONCEPT OF ELECTRICAL QUANTITIES 1
Introduction 1
Learning Objectives 1
1.1 Systems of Units 2
1.2 Charge and Current 3
1.3 Voltage 4
1.4 Power and Energy 5
1.5 Circuit Elements 7
8
1.5.1 Passive and Active Elements 9
1.5.2 Independent Source 10
1.5.3 Dependent Source 14
Exercise 15
Summary
17
CHAPTER 2 FUNDAMENTAL LAW 17
Introduction 17
Learning Objectives 18
2.1 Ohm’s Laws 19
23
2.1.1 Resistor 24
2.1.2 Conductance 25
2.2 Nodes, Branches, and Loops 25
2.3 Kirchhoff’s Laws 27
2.3.1 Kirchhoff’s Voltage Law 29
2.3.2 Kirchhoff’s Current Law 29
2.4 Series Resistors and Voltage Division 30
2.4.1 Series Circuit 30
2.5 Parallel Resistors and Current Division 38
2.5.1 Parallel Circuit 39
2.6 Wye-Delta Transformations 39
2.6.1 Delta to Wye Conversion 42
2.6.2 Wye to Delta Conversion 43
Exercise
Summary 45
45
CHAPTER 3 CIRCUIT ANALYSIS 45
Introduction
Learning Objectives

ii

3.1 Nodal Analysis 45
3.2 Node Analysis with Voltage Sources 51
51
3.2.1 Supernode 56
3.3 Mesh Analysis 59
3.4 Mesh Analysis with Current Source 60
64
3.4.1 Supermesh 65
3.5 Nodal Versus Mesh Analysis 68
Exercise
Summary 69
69
CHAPTER 4 CAPACITOR AND INDUCTOR 69
Introduction 70
Learning Objectives 70
4.1 Capacitors 71
72
4.1.1 Capacitance 73
4.1.2 Voltage/Current Relationships 74
4.1.3 Voltage Across Capacitor 74
4.1.4 Current Through Capacitor 75
4.2 Series and Parallel Capacitors 76
4.2.1 Capacitors in Parallel 77
4.2.2 Current Through Capacitor in Parallel 79
4.2.3 Capacitors in Series 79
4.2.4 Voltage Across Capacitor in Series 81
4.3 Inductors 82
4.3.1 Inductance 82
4.3.2 Voltage / Current Relationships 84
4.3.3 Current Through Inductor 84
4.3.4 Voltage Across Inductor 85
4.4 Series and Parallel Inductors 86
4.4.1 Inductors in Parallel 87
4.4.2 Current Flow in Inductor in Parallel 90
4.4.3 Inductors in Series 91
4.4.4 Voltage Across Inductor in Series
Exercise 93
Summary 93
93
CHAPTER 5 FUNDAMENTAL OF SOLID STATE PRINCIPLES 93
Introduction 95
Learning Objectives 96
5.1 Atomic Theory 97
98
5.1.1 Charge and Conduction 101
5.1.2 Covalent Bonding 103
5.1.3 Conduction
5.2 Doping
5.3 The PN Junction
5.4 Bias

iii

5.4.1 Forward bias 103
5.4.2 Reverse Bias 105
Exercise 106
Summary 107

CHAPTER 6 DIODES 109
Introduction 109
Learning Objectives 109
6.1 Introduction to PN Junction Diode 110
111
6.1.1 Voltage Current Characteristic of a Diode 114
6.2 Ideal and Practical Diode 117
6.3 Zener Diode 121
6.4 Light Emitting Diode 122
6.5 Series and Parallel Diode 123
6.6 Half wave rectifier 127
6.7 Full wave rectifier 128
130
6.7.1 Center tapped Full Wave 134
6.7.2 Full Wave Bridge rectifier 139
6.8 Power supply 139
6.9 Clipper 142
6.9.1 Limiter without Biasing 145
6.9.2 Limiter with Voltage Biasing 147
Exercise
Summary 149
149
CHAPTER 7 TRANSISTOR 149
Introduction 150
Learning Objectives 150
7.1 Introduction to Bipolar Junction Transistor (BJT) 151
151
7.1.1 BJT Structure and Symbol 153
7.2 BJT Characteristic 154
155
7.2.1 Basic Transistor Operation 162
7.2.2 DC Load Line 165
7.3 DC Biasing 173
7.3.1 Fixed Bias Circuit 176
7.3.2 Emitter Bias
7.3.3 Voltage Divider Bias 177
Exercise 177
Summary 177
178
CHAPTER 8 OPERATIONAL AMPLIFIER 180
Introduction
Learning Objectives
8.1 Introduction to Operational Amplifier

8.1.1 Practical symbol for IC op-amp

iv

8.1.2 Ideal and Practical op-amp characteristics 181
8.1.3 Open Loop and Close Loop Configurations 183
8.2 Comparator 185
8.3 Inverting and Non-Inverting Operational Amplifier 188
8.4 Voltage Follower 192
8.5 Summing Amplifier 192
8.6 Multistage Amplifier 194
Summary 198
REFERENCES 199

v

LIST OF ABBREVIATION/SYMBOL

A- Ampere
Alternating current
ac - Bipolar Junction Transistor
Current- controlled current source
BJT - Current- controlled voltage source
Close loop
CCCS - Direct current
Dual Inline Package
CCVS - Farad
Conductance
CL - Henry
Current
dc - kilo
Kirchoff’s Current Law
DIP - Kirchoff’s Voltage Law
mili
F- Not connected
Open loop
G- Operational Amplifier
Power
H- Charge

I-

k-

KCL -

KVL -

m-

NC -

OL -

Op-amp -

P-

q-

vi

R - Resistor
SI - International System of Units
S - Siemen
W - Watt
V - Volt
VCCS - Voltage- controlled current source
VCVS - voltage- controlled voltage source
Ω - Ohm

vii

CHAPTER 1

CONCEPT OF ELECTRICAL QUANTITIES

Introduction
Electric circuit theory and electromagnetic theory are the two

fundamental theories upon which all branches of electrical engineering are
built. Many branches of electrical engineering, such as power, electric
machines, control, electronics, communications, and instrumentation, are
based on electric circuit theory.

In electrical engineering, we are often interested in communicating or
transferring energy from one point to another. To do this requires an
interconnection of electrical devices. Such interconnection is referred to as
an electrical circuit, and each component of the circuit is known as an
element.

Learning Objectives
The objectives of this chapter are to :
1. be familiar with the International System of Measurement.
2. be familiar with a variety of derived SI units.
3. be familiar with electric charge.

1

4. be capable of analysing simple applications involving current and
voltage.

5. have an understanding of the power and energy associated with
electric circuits and networks.

6. have an understanding of circuit elements.

1.1 Systems of Units

An electric circuit or electric network is a collection of electrical
elements interconnected in some specified way. Standard system of units
used is the International System of Units (abbreviated SI), adopted by the
General Conference on Weights and Measures in 1960.

In this system, there are six basic units in the SI and all other units are
derived from them. They are meter, kilogram, second, coulomb, kelvin and
the candela. The electrical quantities which are usually used are given in
Table 1.1

Table 1.1: Electrical quantities

Abbreviation Electrical Unit
Quantity
I Current Ampere (A)
V Voltage Volt (V)
R Resistance Ohm (Ω)
P Watt (W)
L Power Henry (H)
C Inductance Farad (F)
G Capacitance
Conductance Siemen (S)/mho (Ʊ)

Besides the electrical quantities and units in SI, there has one of the
greatest advantages of the SI which is the decimal system to relate larger

2

and smaller units to the basic unit. The various powers of 10 are denoted by
standard prefixes, some of which are given, along with their abbreviations
in Table 1.2.

Table 1.2: Standard prefixes

Multiple Prefix Symbol

1012 tera T
109 giga G
106 mega M
103 kilo k
10-3 mili m
10-6 mikro μ
10-9 nano n
10-12 piko p

1.2 Charge and Current

Charge, q is an electrical property of the atomic particles of which
matter consists, measured in coulombs (C).Each atom is consisting of
electrons, proton and neutrons. The charge e on an electron is a negative
charge with magnitude of 1.602 x 10-19C, while a proton carries a positive
charge of the same magnitude as the electron.

1C = 6.24 x 1018e

The Law of Conservation of charge states that charges can neither
be created nor destroyed, they can only be transferred. The motion of
charges creates electric current. It is conventional to take the current flow
as the movement of positive charges, that is, opposite to the flow of
negative charges.

Electric current, I is the time rate of change of charge, measured in
Amperes (A). Mathematically the relationship between current, charge
and time is given by:

3

dq (1.1)
I = dt (1.2)
1A = 1C/s

If the current does not change with time, but remains constant, we
call it a direct current (dc). By convention the symbol I is used to represent
such a constant current. An alternating current (ac) is a current that varies
sinusoidally with time.

1.3 Voltage

Voltage (or potential difference), V is the energy required (absorbed
or expended) to move a unit charge through an element. The voltage Vab
between two points a and b in an electric circuit is the energy (or work)
needed to move a unit charge from a to b; mathematically,

dw (1.3)
Vab = dq

Where w is energy in Joules (J) and q is charge in coulombs (C).

1V=1J/C (1.4)

The + and - signs are used to define reference direction or voltage
polarity. The Vab can be interpreted in 2 ways:

1. point a is at the potential of Vab volts higher than point b, or

2. the potential at point a with respect to point b is Vab.

Vab = -Vba (1.5)

4

1.4 Power and Energy

Power, p is the time rate of expending or absorbing energy,
measured in watts (W). For practical purposes, we need to know how much
power an electric device can handle.

p= dw (1.6)
dt (1.7)

p = vi

p is a time varying quantity and is called the instantaneous power.
Thus, the power absorbed or supplied by an element is the product of the
voltage across the element and the current through it.

+p: power is being delivered to element
: power is absorbed by the element

-p: power is being supplied by the element
How to know the power have + or - signs?
CURRENT DIRECTION and VOLTAGE POLARITY! (At the element)

By the Passive Sign Convention:
+p: current enters through the + polarity of the voltage (element) [draw the
element]
-p: current enters through the - polarity of the voltage (element).
Passive Sign Convention is satisfied when the current enters through the
positive terminal of an element and p = +vi, if the current enters through the
negative terminal, p = -vi.
Energy, w is the capacity to do work, measured in Joule (1).

5

The Law of Conservation of Energy: The algebraic sum of power in
the circuit at any instant of time must be zero.

� = 0 (1.8)

Again, the total power supplied to the circuit must balance the total
power absorbed. From equation (1.6), the energy absorbed or supplied by
an element from time t0 to time t is

(1.9)

= � = �



Example 1.1:

Figure 1.1 shows a circuit with five elements. If P1 = -205 W, P2 = 60 W, P4 = 45
W, P5 = 30 W, calculate the power P3 received or delivered by element 3.

24
135

Figure 1.1

Solution:
From equation 1.8, the algebraic sum of power in the circuit at any instant
of time must be zero. Hence,

6

P1 + P2 + P3 + P4 + P5 = 0
or,

-205 + 60 + P3 + 45 + 30 = 0
or,

P3 = 205 - 60 - 45 - 30
Power received by the element 3,

P3 = 70W

1.5 Circuit Elements
The basic ideal electrical element as shown in Figure 1.2 can be described as
below:
1. The element cannot be divided.
2. There are only 2 terminals.
3. The element shall be represented in V and I equations.

A B
Basic Ideal Element

Figure 1.2: Basic ideal electrical element

7

1.5.1 Passive and Active Elements

Passive element in electrical circuit is an element which absorbs or
receives energy from the energy source in whole circuit. The total energy
delivered to it from the rest of the circuit is always nonnegative. That is, with
reference to equation below for all t we have;

tt (1.10)

w(t) = � p(t) dt = � vi dt ≥ 0

-∞ -∞

The polarities of v and i are shown as Figure 1.3. The current direction
shall be flowing through an element from its positive terminal. Therefore, the
potential different at positive terminal is higher than negative terminal.
Examples of passive elements are resistors (R), capacitors (C), and inductors
(L).

i _

Passive Element
+

Figure 1.3: Passive element

While, an active element is an element which supplying energy or
delivering power to the rest 0 an electrical circuit. Figure 1.4 shows the
polarities of v and i for the active element. The elements are voltage
sources or current sources such as generators, batteries, testing devices
and electronic devices (transistor, amplifier etc.) that require power
supplies.

i _

Active Element
+

Figure 1.4: Active element

8

1.5.2 Independent Source
Two major categories of power sources are independent source and

dependent source. There are two types of independent sources such as
Independent Voltage Source and Independent Current Source. Figure 1.5
and Figure 1.6 have shown the symbols and description of the independent
power source.

Independent sources are usually meant to deliver power to the
external circuit and not to absorb it. Thus, if v is the voltage across the source
and its current i is directed out of the positive terminal, then the source is
delivering power, given by p = vi, to the external circuit. Otherwise it is
absorbing power.

An independent voltage source is a two terminal element, such as a
battery or a generator, which maintains a specified voltage between its
terminals. The voltage is completely independent of the current through the
element.

V

Figure 1.5: Independent Voltage Source

An independent current source is a two terminal element through
which a specified current flow. The current is completely independent of
the voltage across the element.

9

Figure 1.6: Independent Current Source

1.5.3 Dependent Source
Meanwhile, the dependent sources can be divided by four types

which are voltage- controlled voltage source (VCVS), current- controlled
voltage source (CCVS), voltage- controlled current source (VCCS), and
current- controlled current source (CCCS). Figure 1.7 to Figure 1.10 has
shown the symbols of each dependent power sources.

A dependent or control voltage source is one whose terminal voltage
depends on, or is controlled by, a voltage or a current existing at some other
place in the circuit. A voltage- controlled voltage source (VCVS) is a
voltage source controlled by voltage, and a current- controlled voltage
source (CCVS) is a voltage source controlled by current.

+

+
V1 _ V = µV1
_

Figure 1.7: Voltage-controlled voltage source (VCVS)
10

i1

+
_ V = ri1

Figure 1.8: Current-controlled voltage source (CCVS)
A dependent, or controlled current source, is one whose current is
dependent on a voltage or a current existing elsewhere in the circuit. A
voltage-controlled current source (VCCS) is control by a voltage, and a
current-controlled current source (CCCS) is controlled by current.

+
V1 i = gV1
_
Figure 1.9: Voltage-controlled current source (VCCS)

11

i1
i = βi1

Figure 1.10: Current-controlled current source (CCCS)

Example 1.2
Calculate the power supplied or absorbed by each element in Figure 1.11.

I = 10A + 10V _ + 8V _ 4A

P2 14A P4
+ P5
30V + P1 +
_ P3 12V
20V _ 0.4I
_

Figure 1.11

Solution:
By applying the sign convention for passive and active elements

shown in Figure 1.3 and Figure 1.4.

12

For P1, the 5A current is out of the positive terminal. Hence,

P1 = 20(-5) = -100W Supplied power

For P2 and P3, the current flows into the positive terminal of the element in
each case.

P2 = 12(5) = 60W Absorbed power
P3 = 8(6) = 48W Absorbed power

Since P4 is parallel with P3, therefore the voltage across P4 is similar to P3
which is 8V.

When the current flows out of the positive terminal,

P2 = 8(-0.2I) = 8(-0.2 × 5) = -8W Supplied power

13

Exercise
1. Find the power absorbed by each of the elements in Figure 1.12.

I = 10A + 10V _ + 8V _ 4A

P2 14A P4

+ + + P5
30V _ P1 0.4I
20V P3 12V
_ _

Figure 1.12

2. Determine Is in the circuit of Figure 1.13.

+ IS _ 2V + 2A
4V +
5A _ 6V
_
+
4V

_

Figure 1.13

14

3. Find Vo in the circuit of Figure 1.14.

Io = 2A

+ _
28V

6A + 12V _ 1A

P2 3A + _
28V

30V + + _
_ + 5Io
Vo
_ 3A

6A

Figure 1.14

Summary

In this chapter we have studied that:

1. An electric circuit consists of electrical elements connected
together.

2. The International System of Units (SI) is the international measurement
language, which enables engineers to communicate their results.

3. Current is the rate of charge flow, �I = dq�

dt

4. Voltage is the energy required to move 1C of charge through an
element, �v = ddwq�.

15

5. Power is the energy supplied or absorbed per unit time. It is also the

product of voltage and current, �p = dw� = vi

dt

6. According to passive sign convention, power assumes a positives sign
when the current enters the positive polarity of the voltage across an
element.

7. Voltage and current sources can be dependent or independent. A
dependent source is one whose value depends on some other circuit
variable.

16

CHAPTER 2

FUNDAMENTAL LAW

Introduction
Chapter 1 introduced basic concept such as current, voltage, and

power in an electric circuit. To actually determine the values of these
variables in a given circuit requires that we understand some fundamental
laws that govern electric circuits. These laws, known as ohm’s law and
kirchhoff’s laws, form the foundation upon which electric circuit analysis is
built.

In this chapter, in addition to these laws, we shall discuss some
techniques commonly applied in circuit design and analysis. These
techniques include combining resistors in series or parallel, voltage division,
and current division. The application of these laws and techniques will be
restricted to resistive circuits in this chapter.

Learning Objectives
The objectives of this chapter are to :
1. have an understanding of Ohm’s law.
2. apply Ohm’s law to the analysis of simple circuits.
3. be familiar with kirchhoff’s laws.

17

4. recognize series and parallel connected loads.
5. analyze relatively simple circuits and network.

2.1 Ohm’s Laws

Ohm’s law states that the voltage V across a resistor is directly
proportional to the current I flowing through the resistor. That is,

V∝I (2.1)

Figure 2.1 shows that the voltage V is proportional to the current I.
Proportional means that if say the voltage is doubled, the current will also
double. Changes in the current follow changes in the voltage.

V

I
Figure 2.1: I-V characteristic

Ohm defined the constant of proportionality for a resistor to be the
resistance, R. (The resistance is a material property which can change if the
internal or external conditions of the element are altered, e.g., if there are
changes in the temperature.)

18

Therefore, Equation 2.1 becomes (2.2)
V = IR

The resistance R of an element denotes its ability to resist the flow of
electric current. It is measured in ohms (Ω).

To apply Ohm’s Law:

1. Be careful to the current direction and voltage polarity.

2. The direction of I and the polarity of voltage,v must conform with the
passive sign convention.

2.1.1 Resistor

Circuit element to model the current-resisting behavior of a material
is labeled as resistor. It is the simplest passive element, complying to Ohm’s
law [ideal]. Its resistance value is determined by the ratio of the measured
voltage (V) across the element against the measured current (I) through
the element i.e. R = V/I. Value of R can vary from 0 to infinity;

For R=0 (short circuit)

A short circuit is a circuit element with resistance approaching
zero.Figure 2.2 illustrates an element with R=0.

19

+i i = could be anything
v=0 R=0 v=0
R=0
-

Figure 2.2: Short circuit (R=0)

For R=infinity (open circuit)

An open circuit is a circuit element with resistance approaching
infinity. Figure 2.3 shows an element with R=infinity

+ i=0 i=0
v R=∞ v = could be anything
- R = infinity

Figure 2.3: Open circuit (R=infinity)

Measurement of resistance:
1. Uses Ohmmeter, connected across the element
2. Measures V and I of an element, and then compute ratio of V and I
3. Visual inspection using colour coding. Figure 2.4 shows the

measurement of resistor using colour coding.

20

Ω 01 23 45 67 89

Band 1 2 3 Band 1, 2 digits
Band 3 multiplier
Band 4 5% Gold, 10% Silver 20% Nil
Band 5 1% Brown, 0.1% Red, 0.01% Orange, 0.001% Yellow

Figure 2.4: Visual inspection using colour coding

Example 2.1
An electric iron draws 2A at 120V. Find its resistance.

Solution:
From Ohm’s law (equation 2.2),

R = V / I = 120 / 2 = 60 Ω

Example 2.2
What is the colour code for the following resistance?

a. 39 Ω
b. 27 k Ω
c. 560 Ω

21

Solution:
a. By referring to Figure 2.4, 39 Ω = 39 x 100 Ω , where band 1 is 3 (orange),

band 2 is 9 (white), band 3 is 0 (black). Hence, the answer is Orange,
White, Black

b. By referring to Figure 2.4, 27k Ω = 27 x 103 Ω , where band 1 is 2 (red),
band 2 is 7 (violet), band 3 is 3 (orange). Hence, the answer is Red,
Violet, Orange

c. By referring to Figure 2.4, 560 Ω = 56 x 101 Ω , where band 1 is 5 (green),
band 2 is 6 (blue), band 3 is 1 (brown). Hence, the answer is Green,
Blue, Brown

Example 2.3
What is the resistance value for the following resistor having the code:
a. Yellow, violet, orange
b. Brown, black, yellow
c. Blue, grey, black

22

Solution:

a. By referring to Figure 2.4, yellow equivalent to 4, violet equivalent to
7, orange equivalent to 3. Hence, the answer is 47 x 103 = 47 k Ω

b. By referring to Figure 2.4, brown equivalent to 1, black equivalent to
0, yellow equivalent to 4. Hence, the answer is 10 x 104 = 100 k Ω

c. By referring to Figure 2.4, blue equivalent to 6, grey equivalent to 8,
black equivalent to 0. Hence, the answer is 68 x 100 = 68 Ω

2.1.2 Conductance

A useful quantity in circuit analysis is the reciprocal of resistance R,
known as conductance and denoted by G.

1i (2.3)
G= R = v (2.4)

G= A = σA (2.5)
ρl l

From Equation 2.3,

i = Gv

Conductance is the ability of an element to conduct electric current.
It is measured in mhos (℧) or siemens (S).

23

2.2 Nodes, Branches, and Loops

Network is an interconnection of elements or devices. Circuit is a
network providing one or more closed paths. A branch represents a single
element such as a voltage source or a resistor. A node is the point of
connection between two or more branches. A loop is any closed path in a
circuit.

The circuit in Figure 2.5 has five branches, namely, the 10V voltage
source, the 2A current source, and the three resistors and has three nodes
a, b, c. Notice that the three points that form node b are connected by
perfectly conducting wires and therefore constitute a single point. The
same is true of the four points forming node c. We demonstrate that the
circuit in Figure 2.5 has only three nodes by redrawing the circuit in Figure
2.6. The two circuits in Figure 2.5 and Figure 2.6 are identical.

a 5Ω b

+ 2Ω 3Ω 2A
10V _

c
Figure 2.5: Branches, nodes and loops

24

a 5Ω b
+ 2A
10V _ 3Ω
2Ω

c

Figure 2.6:Three node circuit of Figure 2.5 is redrawn

A loop is a closed path formed by starting at a node, passing through
a set of nodes, and returning to the starting node without passing through
any node more than once. The loop is said to be independent if it contains
at least one branch which is not a part of any other independent loop.

For example, the closed path abca containing the 2Ω resistor in
Figure 2.6 is a loop. Another loop is the closed path bcb containing the 3Ω
resistor and the current source. Although one can identify six loops in Figure
2.6, only three of them are independent.

A network with b branches, n nodes, and l independent loops will
satisfy the fundamental theorem of network topology:

b=1+n-1 (2.6)

2.3 Kirchhoff’s Laws

2.3.1 Kirchhoff’s Voltage Law

Kirchhoff’s Voltage Law (KVL) states that the algebraic sum of the
potential rises and drops around a closed loop (or path) is zero. Total

25

voltage across active element is equal to the total voltage across passive
elements.

Alternatively, KVL states the sum of voltage rise around a closed loop
is equal to the sum of voltage drops around the same loop.

M (2.7)

� vm = 0

m=1

To illustrate KVL, consider the circuit in Figure 2.7. The sign on each
voltage is the polarity of the terminal encountered first as we travel around
the loop.

I

+ _ + _
V1 V2

+ +
VS _ V3

_

Figure 2.7: A single loop circuit illustrating KVL
From Figure 2.7,

Hence, Vs - V1 - V2 - V3 = 0 (2.8)
(2.9)
Vs = V1 + V2 + V3
26

2.3.2 Kirchhoff’s Current Law

Kirchhoff’s Current Law (KCL) states that the algebraic sum of the
current entering and leaving a junction is zero. Or, the sum of the current
entering a junction must equal the sum of the currents leaving the junction.

Alternatively, KCL states that the sum of currents entering a node is
equal to the sum of currents leaving the node.

N (2.10)

� in = 0

n=1

Consider the node in Figure 2.8

I1 I5

I4
I2

I3

Figure 2.8: Current at a node illustrating KCL (2.11)
(2.12)
Applying KCL gives
I1 - I2 + I3 + I4 - I5 = 0

By rearranging the terms, we get

I1 + I3 + I4 = I2 + I5

27

Example 2.4

For the circuit in Figure 2.9(a), find voltages v1 and v2.

2Ω 2Ω

+ _ + _
v1 v1

+ _ 3Ω + i _ 3Ω
20V _ v2 20V _ v2
+ +

(a) (b)
Figure 2.9

Solution:
To find v1 and v2, we apply Ohm’s law and Kirchhoff’s voltage law. Assume
that current, i flows through the loop as shown in Figure 2.9(b).
From Ohm’s law,

v1= 2i and v2 = -3i  (1)

Applying KVL around the loop gives

-20 + v1 - v2 = 0  (2)

Substituting (1) into (2), yields

-20 + 2i - (-3i) = 0

5i = 20

28

Thus,
i = 4A

Substituting in (1) finally gives
v1 = 8V and v2 = -12V

2.4 Series Resistors and Voltage Division

2.4.1 Series Circuit

Figure 2.10 shows three resistors R1, R2 and R3 connected end to end,
i.e. in series, with a battery source of V volts.

I R1

+ + _
VS v1

_ +

v2 R2
_

_ v3 +
R3

Figure 2.10

In a series circuit:
a. the current, is the same in all parts of the circuit.

b. the sum of the voltages, V1, V2 and V3 is equal to the total applied
voltage, V.

29

V = V1 + V2 + V3 (2.13)
From Ohm’s law:
(2.14)
V1 = IR1 (2.15)
V2 = IR2 (2.16)
V = IR where R = R1 + R2 + R3

Dissipated power for each resistor

PR1 = IVR1 (2.17)

PR2 = IVR2 (2.18)

Supplied power from the source = Total dissipated power by the
resistors

Ps = IV = PR1+ PR2 + … + PRN (2.19)

We can calculate the value of voltage across any resistor without
knowing the value of current across that resistor using Voltage Divider Rule.

Vx = Rx × Vs (2.20)
Rtotal

2.5 Parallel Resistors and Current Division

2.5.1 Parallel Circuit

Figure 2.11 shows three resistors, R1,R2 and R3 connected across each
other, i.e. in parallel, across a battery source of volts.

30

IS

I1 I2 I3

+ R1 R2 R3
VS _

Figure 2.11

In a parallel circuit:

a. the sum of the currents I1,I2 and I3 is equal to the total circuit current I

I = I1 + I2 + I3 (2.21)

b. the source V volts, is the same across each of the resistors

Vs = VR1 = VR2 = VR3 (2.22)

from Ohm’s law:

I1 = V , I2 = V I3 = V , ∴I = V (2.23)
R1 R2 , R3 R

Power calculation in parallel circuit: (2.24)
Ps = VsIs = V1I1 + V2I2 + … + VNIN
31

Total resistance in parallel circuit:

1 = 1 + 1 = R1+R2 �i.e. prosudmuct� (2.25)
RT R1 R2 R1R2

RT = R1R2 (2.26)
R1+R2

We can calculate the value of current flowing any resistor without knowing
the value of voltage across that resistor using Current Divider Rule.

Ix = RT × Is (2.27)
Rx

or, in simplified way:

IS

I1 I2
R2
+ R1
VS _

Figure 2.12

I1 = R2 × Is (2.28)
R1+R2 (2.29)

I2 = R1 × Is
R1+R2

32

Example 2.5 1Ω
Find Req for the circuit shown in Figure 2.13.

4Ω

2Ω 5Ω
Req

6Ω 3Ω
8Ω

Figure 2.13

Solution:
To get Req, we combine resistors in series and in parallel. The 6 Ω and 3 Ω
resistors are in parallel, so their equivalent resistance is

6 Ω || 3 Ω = (6 x 3) / (6 + 3) = 2 Ω
(The symbol || is used to indicate a parallel combination.) Also, the 1Ω and
5Ω resistors are in series; hence their equivalent resistance is

1Ω + 5Ω = 6Ω

33

4Ω

Req 2Ω 6Ω
8Ω 2Ω

(a)

4Ω

Req 2.4Ω
8Ω

(b)

Figure 2.14: Equivalent circuit for Example 2.5

Thus the circuit in Figure 2.13 is reduced to that in Figure 2.14(a). In Figure
2.14(a), we notice that the two 2 Ω resistors are in series, so that equivalent
resistance is

2Ω + 2Ω = 4Ω

This 4 Ω resistor is now in parallel with the 6 Ω resistor in Figure 2.14(a); their
equivalent resistance is

4 Ω || 6 Ω =(4 x 6) / (4 + 6) = 2.4 Ω

34

The circuit in Figure 2.14(a) is now replaced with that in Figure 2.14(b). In
Figure 2.14(b), The three resistors are in series. Hence, the equivalent
resistance for the circuit is

Req = 4 Ω + 2.4 Ω +8 Ω = 14.4 Ω

Example 2.6
Reduce each of the circuits in Figure 2.15 to a single resistor at terminals a-
b.

5Ω

a b
8Ω 20Ω

30Ω

2Ω (a) 5Ω b
a 4Ω 10Ω
5Ω 3Ω 4Ω
8Ω

(b)
Figure 2.15

35

Solution:

a. Rab = 5 // (8 + 20 // 30) = 5 // (8 + 12) = 5 × 20 = 4Ω
25

b. Rab = 2+4 // (5 + 3 )// 8+ 5 // 10 // 4 = 2 + 4 // 4 + 5 // 2.857

= 2 + 2 + 1.8181 = 5.818Ω

Example 2.7
Find io and vo in the circuit shown in Figure 2.16. Calculate the power
dissipated in the 3 Ω resistor.

i 4Ω a io

+ +
12V _ 6Ω vo 3Ω

-

b

Figure 2.16

Solution:
The 6 Ω and 3 Ω resistors are in parallel, so their combined resistance is

6×3
6Ω // 3Ω = 6+3 = 2Ω
Thus, our circuit reduces to that shown in Figure 2.17. Notice that vo is
not affected by the combination of the resistors because the resistors are in

36

parallel and therefore have the same voltage vo in the ways. One way is
to apply Ohm’s law to get

i= 12 = 2A
4+2

i 4Ω a

+ +
12V _ vo 2Ω

-

b
Figure 2.17: Equivalent circuit for Figure 2.16

And hence, vo= 2i = 2 x 2 = 4V. Another way is to apply voltage division,
since the 12V in Figure 2.17 is divided between the 4 Ω and 2 Ω resistors.
Hence,

2
vo = 2+4 × 12 = 4V

Similarly, io can be obtained in two ways. One approach is to apply Ohm’s
law to the 3 Ω resistor in Figure 2.16 now that we know vo; thus,

vo = 3io = 4V ⇒ io = 4 V
3

Another approach is to apply current division to the circuit in Figure 2.16
now that we know i, by writing

62 4
io = 6 + 3 i = 3 × 2 = 3 V

37

The power dissipated in the 3 Ω resistor is

po = voio = 4 × 4 = 5.33
3

2.6 Wye-Delta Transformations
Figure 2.18 shows the superposition of Y and ∆ networks as an aid in

transforming one to the other.
Rc

ab

R1 R2

n

Rb Ra
R3

c

Figure 2.18: Superposition of Y and ∆ networks

38

2.6.1 Delta to Wye Conversion

Each resistor in the Y network is the product of the resistors in the two
adjacent ∆ branches, divided by the sum of the three ∆ resistors.

R1 = Ra Rb Rc (2.30)
+ Rb + Rc (2.31)
(2.32)
R2 = Ra Rc Ra Rc
+ Rb +

R3 = Ra Ra Rb Rc
+ Rb +

2.6.2 Wye to Delta Conversion

Each resistor in the ∆ network is the sum of all possible products of Y
resistors taken two at a time, divided by the opposite Y resistor.

Ra = R1R2 + R2R3 + R3R1 (2.33)
R1 (2.34)
(2.35)
Rb = R1R2 + R2R3 + R3R1
R2

Rc = R1R2 + R2R3 + R3R1
R3

39

Example 2.8

What value of R in the circuit of Figure 2.19 would cause the current source
to deliver 800 mW to the resistors.

R R
R

30 mA

RR

Figure 2.19

Solution:
Using R ∆ = 3RY = 3R, we obtain the equivalent circuit shown below:

30 mA 3R R 3R 3R/2

⇒3R 30 mA

3R R

3R // R= 3R × R = 3 R
4R 4

3 R + 3 R = 3 R
4 4 2

40

3 R // 3R = �23 R� (3R) = 9R2 2 =R
2 2 × 9R
3 R + 3R
2

P = I2R 800 x 10-3 = (30 x 10-3)2 R

R = 889 Ω

41

Exercise

1. The voltage across a 5kΩ resistor is 16 V. Find the current through the
resistor.

2. Determine Vo in the circuit in Figure 2.20.

+ + +
9V _ vo _ 3V
-

Figure 2.20

3. Find i1 through i4 in the circuit in Figure 2.21

10Ω i4 i2 20Ω

40Ω i3 i1 30Ω
20A

Figure 2.21

42