Chap_ter
BASIC DEFINITIONS AND U NITS
t
The primary objective of this chapter is to introduce the basic terms that play the vital role in
electrical engineering. The definition of these terms has been given along with their units.
1.1 CHARGE
It is known that every atom consists of electron(s), proton(s), and neutron(s). The electron has
negative charge, proton has positive charge and neutron has no charge In an atom, the total
number of electrons is always equal to the total number of pr9tons. The atom is neutral only when
the magnitude of total charge of electron(s) is equal to that of proton(s). Therefore, if an atom
loses electron(s), it is positively charged. Similarly reverse case occurs when an atom loses
proton(s). ~ unit of charge is coulomb (C).
1.2 DKIFf VELOCITY
Let us consider a conductor XY as shown in
Figure 1.1. The two ends of the conductor are
connected to a battery. Before connecting to the
battery, the electrons in the conductor were in
the random motion. As soon as the battery is in
action, the electric field is established. The free
electrons at the end Y will now experience a
force -eE in the direction Y to X. The electrons
x.-------------, y1--- -
Figure 1.1
V
Conductor connected to
a battery.
in the conductor are accelerated in this direction. The electrons collide with each other and also
with the positive ions in the conductor. At each collision, the momentum gained by the ~lectrons
are lost. The electron is accelerated afresh after each collision.
A backward force acts on the electrons due to collisions. This force is known as collision
drag. The net effect of these collis~ons is that the electrons slowly drift with a constant average
drift velocity in the direction of - E.
The drift velocity is defined as the vector average velocity of the charge carrier moving
under the effect of electric field, the average being macroscopic, i.e., taken over a large volume as
compared to molecular volume. ·
1
of lltctrrca, ""~ --
ClJRKENT /tar e through a conductor. f flow of charge is constane)
It ls defined as the rate of flow of c g ductor in a time t (rat~ o . d the unit of current is:
If a total charge q passes through a con d it of time is sccon '
. f h o is coulomb an un
then current / = qlt. The umt O c ar0 e
1
= Coulomb = Ampere
. Second value of current is given by
. ti ,. instantaneous
If the rate of flow of charge 1s not constant, le
. dq
l = -
. . . . . cit <l d is a macroscopic quantity.
The current 1s charactenst1c of a particular con uctor an fl vs from higher
As water always falls from hig11er to lowc1 levels, the current alwayds b
0
~ , rr JW and a
potential to lower potential. Current flows through an clement is rcprcsente Y an <l
1
h' . . t
· · · -• , I e represents t e cur ren
value (vanable of fixed). Arrow. represents a reference dJrcct1on anu va u
flow in the reference direction as shown in Figure 1.2. ·
Figure 1.2 Representation of current.
Figure 1.3 shows another example. In Figure l.3(a), 5 A current is flowing from left to right
whereas in Figure l.3(b), -5 A current is flowing from right to left. These two cases are identical
to each other.
+--- -5A
(a) (b)
Figure 1.3 Two representation of same current.
1.4 CURRENT DENSITY
The current density at any point in a conductor ca1 rying current is th~ curn.!nl per unit area of
cross-section of the conductor. The aic;a is in the direction nonnal to tlw L"lllt'L''nt. The current
density J is gi vcn by
Its unit is ampere/metcr2 ( A/m2).
1.5 VOLTAGE
I
J = -
A
Energy ~s required to move a char'ge from one point to another. If W joules of energy is required
for movmg charge Q from point a to b, the voltage V between a and b is given by
wV= -
Q
J w-=- V
C Q
1 volt = 1 joule/I coulomb
unit of voltage is volt (V).
ID Figure l.4(a), terminal A is +8 V above terminal B. In Figure l.4(b), tenninal •
terminal A or in other words +8 V below A. Also it can be written as VM •
l.4(a)] and VBA = -8 V [Figure l.4(b)] by using double subscript notation and in
AB=-VBA.
0
I
0 0
I 0
A B A B
+ 8V 8V +
(a) (b)
Figure 1:4 Two equivalent voltage representations.
ENERGY
...
To transfer a charge through an element, work is being_done. To determine the energy being
supplied to the element or by the element to the rest of the circuit, both the polarity of the
voltage across the element and the direction of the current through the el.ement must be known. If
a positive current enters the positive terminal of an element, an external force must drive_the
current. Therefore, the external force delivering energy to the element. On the other hand, if the
current leaves the positive terminal or enters the negative terminal, the element is delivering
energy to the external citcuit.
In Figures l.5(a) and l.5(b), the element is absorbing energy whereas in Figures l.5(c) and
1.5(d), the element is delivering energy.
4A
(a)
+
8V
4A
(b)
8V
+
4A
(c)
+
8V
4A
av
+
(d)
Figure 1.5 Different voltage-current relatlonshlps.
is v (say) and a small eharge Aq is m
1.-.11Vo.:
of Electrical and Electronic
LW ~
Lt - == V Lt At
t ➔ O ~t 1 ➔ 0 u
dW dq
dt == vdt
Since p = dW/dt, by definition
dW - dq == vi
p = dt - V dt
The current and voltage reference directions O
} l 11.-------0of the element shown in Figure 1.6. If the current L______._
reference direction arrow points the positive end + v
of the voltage reference direction, the current and Figure
1
_6
Current and voltage reference
voltage defined above are said to satisfy the directions of an element.
passive sign convention. . .
If vi> 0 it means element is absorbing power. If vi < 0, it means element is dehvermg power.
' · b" wer
In Figures l.S(a) and l.S(b), p =vi= 8 x 4 = 3'.?, W. Therefore, the element 1s ab~or ~ng po ·
In Figures l.5(c) and 1.5 (d), p =vi= 8 x (-4) =-32 W. Therefore, the element is dehvermg power.
1.8 VOLTAGE DROP AND VOLTAGE KISE
Vxv
X +
From Figure I.7, point X is at a higher potential
as compared to point Y. If we go from X to Y,
there will be voltage drop. If we go from Y to X,
there will be voltage rise. Figure 1.7 Voltage drop or rise.
1.9 BASIC CIRCUIT
It consists of the following: (i) source,
(ii) connecting wires, and (iii) load. Figure 1.8
shows a basic circuit.
1.10 S.I. UNITS
Voltage +
sour-ce
Outgoing conductor
Incoming conductor
Figure 1.8 A basic circuit.
Table 1.1
-0
ro
0
_J
y
Physical quantity Unit Abbreviation Physical quantity Unit Abbreviation
Length metre m Electric current ampere A
Mass kilogramme kg Absolute temperature kelvin K
Time second s, sec. Luminous intensity candela cd
Temperature in kelvin = Temperature in cclsius + 273
The word degrees_ kelvin is not correct unit. Simply kelvin is the correct one.
The S.I. unit of plane angle is radian where 2rr radians indicate one complete revolution.
Also one revolution indicates 360°.
2rc radians = 360 degrees
1 d
. 360
ra 1an = 21C degrees = 57.296 degrees
.Basic Definitions and Units [TI
1.11 MULTIPLES AND SUB-MULTIPLES OF UNITS
exa
peta
. tera
giga
mega
kilo
Abbreviation
E
p
T
G
M
k
Multiplication
1018
1015
1012
109
106
103
Table 1.2
factor Prefix Abbreviation
milli m
micro µ
nano n
pico p
femto f
atto a
Multiplication factor
10-3
10-6
10-9
10-12
10-15
10-18
t Eample 1.1 Convert the following: (i) 124 MW into watt, (ii) 33 kV into volt, (iii) 12 µF into
farad, (iv) 23 mA into ampere.
. Solution
(i) 124 MW= 124 x 106
W Ans.
(ii) 33 kV =33 x 103
A Ans.
.. (iii) 12 µF = 12 x 1o-6
F Ans.
(iv) 23 mA = 23 x 10-3
A Ans.
SOME DERIVED UNITS
Quantity
Area
Volume
Linear velocity
Unit
m2
m3
m/s
Table 1.3
Exercises
.1 Define current and give its definition.
E
Figure 1
Quantity
Angular velocity
Linear acceleration
Angular acceleration
"O
co
0
__J
.2 .for the foHowing Figure 1, show the direction of current.
Unit
rad/s
m/s2
rad/s2

Chapter 01

  • 1.
    Chap_ter BASIC DEFINITIONS ANDU NITS t The primary objective of this chapter is to introduce the basic terms that play the vital role in electrical engineering. The definition of these terms has been given along with their units. 1.1 CHARGE It is known that every atom consists of electron(s), proton(s), and neutron(s). The electron has negative charge, proton has positive charge and neutron has no charge In an atom, the total number of electrons is always equal to the total number of pr9tons. The atom is neutral only when the magnitude of total charge of electron(s) is equal to that of proton(s). Therefore, if an atom loses electron(s), it is positively charged. Similarly reverse case occurs when an atom loses proton(s). ~ unit of charge is coulomb (C). 1.2 DKIFf VELOCITY Let us consider a conductor XY as shown in Figure 1.1. The two ends of the conductor are connected to a battery. Before connecting to the battery, the electrons in the conductor were in the random motion. As soon as the battery is in action, the electric field is established. The free electrons at the end Y will now experience a force -eE in the direction Y to X. The electrons x.-------------, y1--- - Figure 1.1 V Conductor connected to a battery. in the conductor are accelerated in this direction. The electrons collide with each other and also with the positive ions in the conductor. At each collision, the momentum gained by the ~lectrons are lost. The electron is accelerated afresh after each collision. A backward force acts on the electrons due to collisions. This force is known as collision drag. The net effect of these collis~ons is that the electrons slowly drift with a constant average drift velocity in the direction of - E. The drift velocity is defined as the vector average velocity of the charge carrier moving under the effect of electric field, the average being macroscopic, i.e., taken over a large volume as compared to molecular volume. · 1
  • 2.
    of lltctrrca, ""~-- ClJRKENT /tar e through a conductor. f flow of charge is constane) It ls defined as the rate of flow of c g ductor in a time t (rat~ o . d the unit of current is: If a total charge q passes through a con d it of time is sccon ' . f h o is coulomb an un then current / = qlt. The umt O c ar0 e 1 = Coulomb = Ampere . Second value of current is given by . ti ,. instantaneous If the rate of flow of charge 1s not constant, le . dq l = - . . . . . cit <l d is a macroscopic quantity. The current 1s charactenst1c of a particular con uctor an fl vs from higher As water always falls from hig11er to lowc1 levels, the current alwayds b 0 ~ , rr JW and a potential to lower potential. Current flows through an clement is rcprcsente Y an <l 1 h' . . t · · · -• , I e represents t e cur ren value (vanable of fixed). Arrow. represents a reference dJrcct1on anu va u flow in the reference direction as shown in Figure 1.2. · Figure 1.2 Representation of current. Figure 1.3 shows another example. In Figure l.3(a), 5 A current is flowing from left to right whereas in Figure l.3(b), -5 A current is flowing from right to left. These two cases are identical to each other. +--- -5A (a) (b) Figure 1.3 Two representation of same current. 1.4 CURRENT DENSITY The current density at any point in a conductor ca1 rying current is th~ curn.!nl per unit area of cross-section of the conductor. The aic;a is in the direction nonnal to tlw L"lllt'L''nt. The current density J is gi vcn by Its unit is ampere/metcr2 ( A/m2). 1.5 VOLTAGE I J = - A Energy ~s required to move a char'ge from one point to another. If W joules of energy is required for movmg charge Q from point a to b, the voltage V between a and b is given by wV= - Q J w-=- V C Q
  • 3.
    1 volt =1 joule/I coulomb unit of voltage is volt (V). ID Figure l.4(a), terminal A is +8 V above terminal B. In Figure l.4(b), tenninal • terminal A or in other words +8 V below A. Also it can be written as VM • l.4(a)] and VBA = -8 V [Figure l.4(b)] by using double subscript notation and in AB=-VBA. 0 I 0 0 I 0 A B A B + 8V 8V + (a) (b) Figure 1:4 Two equivalent voltage representations. ENERGY ... To transfer a charge through an element, work is being_done. To determine the energy being supplied to the element or by the element to the rest of the circuit, both the polarity of the voltage across the element and the direction of the current through the el.ement must be known. If a positive current enters the positive terminal of an element, an external force must drive_the current. Therefore, the external force delivering energy to the element. On the other hand, if the current leaves the positive terminal or enters the negative terminal, the element is delivering energy to the external citcuit. In Figures l.5(a) and l.5(b), the element is absorbing energy whereas in Figures l.5(c) and 1.5(d), the element is delivering energy. 4A (a) + 8V 4A (b) 8V + 4A (c) + 8V 4A av + (d) Figure 1.5 Different voltage-current relatlonshlps. is v (say) and a small eharge Aq is m 1.-.11Vo.:
  • 4.
    of Electrical andElectronic LW ~ Lt - == V Lt At t ➔ O ~t 1 ➔ 0 u dW dq dt == vdt Since p = dW/dt, by definition dW - dq == vi p = dt - V dt The current and voltage reference directions O } l 11.-------0of the element shown in Figure 1.6. If the current L______._ reference direction arrow points the positive end + v of the voltage reference direction, the current and Figure 1 _6 Current and voltage reference voltage defined above are said to satisfy the directions of an element. passive sign convention. . . If vi> 0 it means element is absorbing power. If vi < 0, it means element is dehvermg power. ' · b" wer In Figures l.S(a) and l.S(b), p =vi= 8 x 4 = 3'.?, W. Therefore, the element 1s ab~or ~ng po · In Figures l.5(c) and 1.5 (d), p =vi= 8 x (-4) =-32 W. Therefore, the element is dehvermg power. 1.8 VOLTAGE DROP AND VOLTAGE KISE Vxv X + From Figure I.7, point X is at a higher potential as compared to point Y. If we go from X to Y, there will be voltage drop. If we go from Y to X, there will be voltage rise. Figure 1.7 Voltage drop or rise. 1.9 BASIC CIRCUIT It consists of the following: (i) source, (ii) connecting wires, and (iii) load. Figure 1.8 shows a basic circuit. 1.10 S.I. UNITS Voltage + sour-ce Outgoing conductor Incoming conductor Figure 1.8 A basic circuit. Table 1.1 -0 ro 0 _J y Physical quantity Unit Abbreviation Physical quantity Unit Abbreviation Length metre m Electric current ampere A Mass kilogramme kg Absolute temperature kelvin K Time second s, sec. Luminous intensity candela cd Temperature in kelvin = Temperature in cclsius + 273 The word degrees_ kelvin is not correct unit. Simply kelvin is the correct one. The S.I. unit of plane angle is radian where 2rr radians indicate one complete revolution. Also one revolution indicates 360°. 2rc radians = 360 degrees 1 d . 360 ra 1an = 21C degrees = 57.296 degrees
  • 5.
    .Basic Definitions andUnits [TI 1.11 MULTIPLES AND SUB-MULTIPLES OF UNITS exa peta . tera giga mega kilo Abbreviation E p T G M k Multiplication 1018 1015 1012 109 106 103 Table 1.2 factor Prefix Abbreviation milli m micro µ nano n pico p femto f atto a Multiplication factor 10-3 10-6 10-9 10-12 10-15 10-18 t Eample 1.1 Convert the following: (i) 124 MW into watt, (ii) 33 kV into volt, (iii) 12 µF into farad, (iv) 23 mA into ampere. . Solution (i) 124 MW= 124 x 106 W Ans. (ii) 33 kV =33 x 103 A Ans. .. (iii) 12 µF = 12 x 1o-6 F Ans. (iv) 23 mA = 23 x 10-3 A Ans. SOME DERIVED UNITS Quantity Area Volume Linear velocity Unit m2 m3 m/s Table 1.3 Exercises .1 Define current and give its definition. E Figure 1 Quantity Angular velocity Linear acceleration Angular acceleration "O co 0 __J .2 .for the foHowing Figure 1, show the direction of current. Unit rad/s m/s2 rad/s2