UNESCO-NIGERIA
UNESCO
TECHNICAL &
VOCATIONAL EDUCATION
REVITALISATION PROJECT-PHASE
PROJECT
II
NATIONAL DIPLOMA IN
ELECTRICAL ENGINEERI
ENGINEERING
NG TECHNOLOGY
Low
Low
High
_
I
10
10 Ohms
Ohms
High
_
+
V
+
R
100 volts
ELECTRICAL ENGINEERING
SCIENCE (I)
COURSE CODE:
CODE EEC 115
YEAR II SEMESTER I
THEORY
Version 1: December 2008
TABLE OF CONTENTS
ELECTRICAL ENGINEERING SCIENCE (I)
THEORY
LIST OF CONTENTS
WEEK 1:
1.1 Atom
1.2 Structure and composition of an atom
1.3 Conductors, insulators and semiconductor
WEEK 2:
1.4 Concepts of current and electron flow
1.5 Electric current, potential difference, electromotive force and resistance
1.6 Multiples and sub multiples of electric quantities
WEEK 3:
2.1 Direct current
2.2 Analogy between current flow and water flow
2.3 Basic d.c. circuit
WEEK 4
2.4 Ohm’s law
WEEK 5:
2.4 Ohm’s law
WEEK 6 :
2.5 Resistivity and conductivity
2.6 Series and parallel circuit
WEEK 7:
2.7 Solved problems on resistivity and conductivity
2.8 Equivalent resistance of series and parallel circuits
WEEK8:
2.9 Kirchhoff’s laws
WEEK 9:
2.10 Superposition principle
WEEK 10:
2.11 Temperature coefficient of resistance
WEEK 11:
3.1 Energy and its various types
3.2 Relationship between electrical, mechanical and thermal energy
3.3 Electrical power
WEEK 12:
3.4 Joule’s law
WEEK 13:
4.1 Electric charge
4.2 Coulomb’s law of electrostatics
WEEK 14:
4.3 Definition of terms used in electrostatics
4.4 Capacitance
WEEK 15:
4.5 Capacitor connections
4.6 Energy stored in a capacitor
Concept Of Electric Current Flow
Week 1
At the end of this week, the students should be able to:
•
Define an atom
•
Explain the structure and composition of an atom
•
Differentiate between conductors, insulators and semi conductors
1.1 ATOM
Defination: An atom is the smallest indivisible particle which all matter is made of up.
1.2 STRUCTURE AND COMPOSITION OF AN ATOM
According to the electron theory of matter, an atom is made up of tiny particles. The central
part of the atom is known as the nucleus. The nucleus of the atom consists of particles called
protons and neutrons held together in a compact form by a binding energy. The outer portion
of the atom is made up of orbiting particles called electrons. Electrons normally move about
the centre of an atom in paths which are referred to as shells orbits. Each of these shells can
contain only a certain maximum number of electrons. The central part of the atom shows its
nucleus. We should note that protons are positively charged particles while electrons are
negatively charged particles, since opposite charges attract, the negatively charged electrons
are attracted to the positively charged protons in the nucleus. Consequently, this electric force
of attraction holds the electrons in their orbit.
The number of orbiting electrons normally equals the number of protons in the nucleus. For
that reason an atom is said to be neutrally charged.
1.3 CONDUCTORS, INSULATORS AND SEMICONDUCTOR
Conductor
In a conductor, electric current can flow freely, in an insulator it cannot. Metals such as
copper typify conductors, while most non-metallic solids are said to be good insulators,
having extremely high resistance to the flow of charge through them. "Conductor" implies
that the outer electrons of the atoms are loosely bound and free to move through the material.
Most atoms hold on to their electrons tightly and are insulators. In copper, the valence
electrons are essentially free and strongly repel each other. Any external influence which
5
moves one of them will cause a repulsion of other electrons which propagates, "domino
fashion" through the conductor.
Types of conductors and their properties
Many types of materials can conduct electricity.
Take these materials listed here and put them in order from lightest to heaviest
with respect to their densities, that is, weight per unit volume:
aluminum, copper, gold, iron, lead, mercury, silver, water, zinc
aluminum
copper
Insulator
An Insulator is a material or object which resists the flow of heat (thermal insulators) or
electric charge (electrical insulators).
The term insulator has the same meaning as the term dielectric, but the two terms are
used in different contexts. The opposite of insulators are conductors and semiconductors,
which permit the flow of charge. Semiconductors are strictly speaking also insulators, since
they prevent the flow of electric charge at low temperatures, unless doped with atoms that
release extra charges to carry the current). However, some materials (such as silicon dioxide)
are very nearly perfect electrical insulators, which allow flash memory technology. A much
larger class of materials, (for example rubber and many plastics) are "good enough"
insulators to be used for home and office wiring (into the hundreds of volts) without
noticeable loss of safety or efficiency.
6
wood
Rubber
Semiconductor
A semiconductor is a material with an conductance that is intermediate between that of an
insulator and a conductor.
A semiconductor behaves as an insulator at very low temperature, and has an
appreciable conductance at room temperature
temperature. A semiconductor can be distinguished
inguished from a
conductor by the fact that, at absolute zero,
zero the uppermost filled electron energy band is fully
filled in a semiconductor, but only partially filled in a conductor.
conductor. The distinction between a
semiconductor and an insulator is slightly more arbitrary. A semiconductor has a band gap
which is small enough such that its conduction band is appreciably thermally populated with
electrons at room temperature, whilst an insulator has a band gap which is too wide for there
to be appreciable thermal electrons in its conduction band at room temperature.
Band structure of a semiconductor
Fig 1.1: Band structure of a semiconductor showing a full valence band and an empty
conduction band. The Fermi level lies within the forbidden band gap
In the parlance of solid-state physics, semiconductors (and insulators) are defined as
solids in which at absolute zero (0 K), the uppermost band of occupied electron energy states,
known as the valence band, is completely full. Or, to put it another way, the Fermi energy of
the electrons lies within the forbidden band gap. The Fermi energy, or Fermi level can be
thought of as the energy up to which available electron states are occupied at absolute zero.
At room temperatures, there is some smearing of the energy distribution of the
electrons, such that a small, but not insignificant number have enough energy to cross the
energy band gap into the conduction band. These electrons which have enough energy to be
in the conduction band have broken free of the covalent bonds between neighbouring atoms
in the solid, and are free to move around, and hence conduct charge. The covalent bonds from
which these excited electrons have come now have missing electrons, or holes which are free
to move around as well. (The holes themselves don't actually move, but a neighbouring
electron can move to fill the hole, leaving a hole at the place it has just come from, and in this
way the holes appear to move.)
It is an important distinction between conductors and semiconductors that, in
semiconductors, movement of charge (current) is facilitated by both electrons and holes.
Contrast this to a conductor where the Fermi level lies within the conduction band, such that
the band is only half filled with electrons. In this case, only a small amount of energy is
needed for the electrons to find other unoccupied states to move into, and hence for current to
flow.
The ease with which electrons in a semiconductor can be excited from the valence
band to the conduction band depends on the band gap between the bands, and it is the size of
this energy bandgap that serves as an arbitrary dividing line between semiconductors and
insulators. Materials with a bandgap energy of less than about 3 electron volts are generally
considered semiconductors, while those with a greater bandgap energy are considered
insulators..
The current-carrying electrons in the conduction band are known as "free electrons,"
although they are often simply called "electrons" if context allows this usage to be clear. The
holes in the valence band behave very much like positively-charged counterparts of electrons,
and they are usually treated as if they are real charged particles.
8
Concept Of Electric Current Flow
Week 2
At the end of this week, the students should be able to:
•
Explain the concepts of current and electron flow
•
Define electric current, potential difference, electromotive force (e.m.f) and resistance
•
State multiples and submultiples of electric quantities
1.4
CONCEPT OF CURRENT AND ELECTRON FLOW
At any instant in time, the electrons in a conductor are in random motion. However, if a
directional force e.g. electromotive force EMF from a battery is applied to the conductor as
shown in Fig 2.1 below, then end A of the conductor is positive while end B is negative. This
results in directed flow of electrons.
The directional movement of free electrons is referred to as current flow and the conventional
current flow is in the opposite direction of electron flow as can be seen in the diagram.
Normally, conventional current flow is generally used.
B -ve terminal
Power
Electron flow
source
Current flow
A
+ve terminal
Fig 2.1
1.5 ELECTRIC CURRENT, POTENTIAL DIFFERENCE, ELECTROMOTIVE FORCE AND RESISTANCE
Electric Current
Electric current is the rate of flow of electric charges round a circuit. It has a symbol I and is
measured in amperes (A).
Potential Difference (P.d.)
The potential difference between two points is defined as the number of joules of electrical
energy transferred from one side of the points to the other with the passage of one coulomb
from one point to another. It is measured in volts (V).
9
Electromotive Force (e.m.f)
Electromotive force is the force that gives rise to electric current in a circuit; also it’s the
force that makes flow of electric current in a circuit. This force arises from many effects
including chemical (e.g. battery cells) and magnetic, as (e.g. generator). The unit of
electromotive force is the volt, symbol E.
Resistance
The resistance can be defined as an opposing force experience by the flow of charge through
a material. The opposition is due to the collision between electrons and other atoms in the
material, and it converts electrical energy into heat energy. The unit of resistance is Ohm (Ω),
and has a symbol R.
1.6
MULTIPLES AND SUBMULTIPLES
Since it is often necessary to describe quantities that exist in large multiples or submultiples
of a unit, therefore standard prefixes are used to denote powers of 10 of SI units. See below
table;
Table 2.1. SI prefixes
PREFIX
Tera
Giga
Mega
Kilo
Hector
Deka
Deci
Centi
Milli
Micro
Nano
Pico
SYMBOL
T
G
M
K
h
da
d
c
m
µ
n
p
MULTIPLIER
1012
109
106
103
102
101
10-1
10-2
10-3
10-6
10-9
10-12
10
Simple d.c. Circuits
Week 3
At the end of this week, the students should be able to:
•
Define d.c current
•
State the analogy between current flow and water flow
•
Describe basic d.c. circuits
2.1 DIRECT CURRENT
Direct current is that current which flows in only direction at a time. The d.c. circuits
normally contain a battery, which is the source, and this battery has two terminals namely:
Negative and Positive terminals as shown in figure 3.1 below.
+
Fig. 3.1: symbol of a battery
2.2
ANALOGY BETWEEN CURRENT FLOW AND WATER FLOW
The three main concepts in electricity are voltage, current and resistance. Voltage or potential
difference can be thought of as the driving force (although it is not really a "force") behind
the electric current. In the water analogy, in figure (3.2), you can think of voltage as the
pressure difference created by a pump that causes the water to flow through the pipe in the
water system. Also you can think of the electrons flow in the electric circuit as equivalent to
the flow of the water in the pipes that caused by the pump.
Another important concept in electricity is resistance. It’s Property of a material that
opposes the flow of electrons through it. There is always a resistance flow between two
points in a pipe. This is analogous to the resistance (R) in an electric circuit.
11
Figure (3.2) water analogy
2.3
BASIC D. C. CIRCUIT
A basic d.c. (direct current) circuit consist of an electrical connection of a d.c power source
(e.m.f source), connecting wires and a load as shown in Fig 3.3. In this case, the load can be
any electrical component (e.g. a resistor) or a network of resistors. The d.c. power sources
such as dry cells (such as used in flash lights), car batteries and laboratory d.c power supply
unit. The symbol of a d.c source is as shown in Fig. 3.3. It is represented by two vertical
lines, with the longer line marked (+) and the shorter line marked (-).
+
E.m.f, E
source _
R (load)
Connecting wire
Fig 3.3
12
Concept Of Electric Current Flow
Week 4
At the end of this week, the students should be able to:
•
State ohm’s law
•
Explain ohm’s law
•
Draw the graph of current versus voltage
•
Draw the graph of current versus resistance
2.4 OHM’S LAW
In 1826 George Simon Ohm found that the current, voltage, and resistance are
related in a specific way. Ohm expressed this relationship with a formula that is
known today as Ohm‘s law. In this chapter, you will learn Ohm‘s law and how to used
it in solving circuit problems.
Electric circuit can be of two basic forms: series and parallel. In this chapter,
series, and parallel circuits are studied. you will also see how Ohm‘s law is used in
series, and parallel circuits
Ohm‘s law is the most important mathematical relationship between
voltage, current and resistance in electricity. Ohm’s Law is used in three forms
depending on which quantity voltage, current or resistance you need to
determine. In this section, you will learn each of these forms.
2.4.1 Explanation of Ohm‘s Law
In the electric circuit, if the voltage across constant resistor value is increase,
the current through the resistor will also increase; and, if the voltage is decrease, the
current will decrease. For example, if voltage is double, the current will doubled. If
the voltage is halved, the current will also be halved. This relationship is illustrated in
the figure (4.1), with used voltage and current meter.
13
Low
_
I
High
Low
+
_
High
V
10 Ohms
+
R
200 volts
(a) Decrease V, decrease I
(b ) Increase V, increase I
Figure (4.1): Effect of changing the voltage at constant resistance
Ohm‘s law also stated that if the voltage is kept constant, less resistance results
in more current, and, also, more resistance results in less current. For example, if
resistance is halved, the current will doubled. If the resistance is doubled, the current
is halved.
This relationship is illustrated by the meter indications in the figure (4.2).
Where the resistance is increase and the voltage is constant.
(a) Decrease R, increase I
(b ) Increase R, decrease I
Figure (4.2): Effect of changing the resistance at constant voltage
From previous illustrated,
Ohm‘s law states that, current and voltage are linearly
proportional at constant resistance; current and resistance are
inversely related at constant voltage.
According this law the following three equivalent formula is derived:
14
Formula for current
I=
V
R
Equation (4.1)
This form of Ohm‘s low is used to determine current if voltage and resistance values
are known
Formula for voltage
V=IR
Equation (4.2)
This form of Ohm‘s low is used to determine voltage if current and resistance values
are known
Formula for resistance
R ==
V
I
Equation (4.3)
This form of Ohm‘s low is used to determine resistance if voltage and current values
are known
2.4.2 Ohm‘s Law triangle
There is an easy way to remember which formula to use. By arranging current,
voltage and resistance in a triangle, one can quickly determine the correct formula.
Ohm‘s Law triangle is shown in figure (4.3)
Figure(4.3) : Ohm‘s Law triangle
15
To use the triangle, cover the value you want to calculate, the remaining letters make
up the formula as shown in figure (4.4).
I=
V
R
V=IR
R ==
V
I
Figure(4.4) : Easy way to remember which formula to use
Remember the following three rules:
Ohm‘s Law can only give the correct answer when the correct values are used.
Current is always expressed in Amperes or Amps (A)
Voltage is always expressed in Volts (V)
Resistance is always expressed in Ohms (Ω)
2.4.3The Relationship of Current, Voltage and Resistance
Ohm‘s law describes how current is related to voltage and resistance. current
and voltage are linearly proportional at constant resistance; current and resistance are
inversely related at constant voltage.
2.4.3.1 The Linear Relationship of Current and Voltage
Current and voltage linearly proportional; that is, if one is increased by a
certain percentage, the other will increase or decrease by the same percentage,
assuming that the resistance is constant value.
To draw a graph of current verses voltage, let‘s take a constant value of
resistance. For example, R=10Ω, and calculate the current for several value of voltage
ranged from 0 to 100 V by used the formula I=V/10 where R=10 Ω. the current values
obtained are shown in the table (4.1):
16
I (A)
V(v) I(A)
6
30
3
5
40
4
4
50
5
60
6
70
7
V
I=
10 Ω
Table (4.1)
10
2
=
20
R
1
Ω
7
10
3
2
1
0
0 10 20 30 40 50 60 70
Figure (4.5) : Graph of current versus voltage for
R= 10 Ω
The graph of the current values versus the voltage values is shown in the Fig.
(4.5). This graph tells us that a change in voltage results in a linearly proportional
change in current. From graph the change in voltage from 20 to 30 is increased by
50% by calculation the current must be increase by the same percentages 50% I = 2 +
2 X 50/100 = 3A.
2.4.3.2 Current and Resistance are Inversely Related
Current varies inversely with resistance as expressed ohms law, I=V/R. when
the resistance is decreased, the current goes up; when the resistance is increased, the
current goes down. For example, let‘s take a constant value of voltage V=10volts, and
calculate the current for several value of resistance ranged from 10 to 100 Ω by used
the formula I=10/R. the current values obtained are shown in the next
table (3-2),
The graph of the current values versus the voltage values is shown in the figure(3-6).
17
R (Ω) I (A)
I (A)
10
1
20
0.500
30
0.333
40
0.250
0.4
50
0.200
0.3
60
0.167
0.2
70
0.143
0.6
0.5
0.1
0
I=
10 v
R
Table ( 4.2)
R (Ω)
0
10
20
30
40 50
60
70
0Figure (4.6) : Graph of current versus resistance for
V= 10 v
18
Simple d.c. Circuits
Week 5
At the end of this week, the students should be able to:
•
Calculate current using ohm’s law
•
Calculate voltage using ohm’s law
•
Calculate the resistance using ohm’s law
2.4.4 Calculating Current
In this section, you will learn to determine the current values when you known
the values of voltage and resistance. As examples, by using the following formula I =
V/R . In order to get current in amperes, you must express the value of voltage in volt
and the value of resistance in ohms.
Example 5.1
The voltage supplied by the battery is 10 volts, and the resistance is 5 Ω
in the circuit of figure (3-7). Calculate the current in amperes?
I
V
10v
R
10Ω
Figure (6.1)
Solution:
use the formula
V
R
I=
I=
10
10
V = 100 volts
R = 10 Ω
I= ?
=
1A
19
Example5-2
If the resistance in figure 5.7 is changed to 0.1 KΩ and the voltage to
50V, what is the new value of current?
V
R
I=
Solution:
V = 500 volts
R = 0.1 KΩ
= 0.1 X 1000 Ω
50
100
I=
=
0.25 A
= 100 Ω
I=?
2.4.5 Calculating Current
In this section, you will learn to determine the current values when you known the
values of voltage and resistance. As examples, by using the following formula I =
V/R . In order to get current in amperes, you must express the value of voltage in volt
and the value of resistance in ohms.
Example 5.1
The voltage supplied by the battery is 10 volts, and the resistance is 5 Ω
in the circuit of figure (3-7). Calculate the current in amperes?
Figure (5.2)
Solution:
I=
use the formula
V = 100 volts
V
R
I=
R = 10 Ω
10
10
I= ?
=
1A
20
Example 5.2
If the resistance in figure 3-7 is changed to 0.1 KΩ and the voltage to
50V, what is the new value of current?
I
I=
Solution:
VV
R
10v
R
V =10
500
Ω volts
R = 0.1 KΩ
= 0.1 X 1000 Ω
50
100
I=
=
= 100 Ω
0.25 A
V
I=?
R
2.4.6 Calculating Resistance
In this section, you will learn to determine the resistance values when you
I=
known the values of voltage and current. As examples, by using the following formula
R = V/I
In order to get resistance by ohms, you must express the value of voltage in
volts and the value of current in amperes.
Example 5.5
In the circuit in figure (3-9), how much resistance is needed to draw 4mA of
current ?
I = 6A
V
24 v
R
Figure (5.3)
V = 24 v
Solution:
use the formula R= V / I
R=?
I = 6A
R=
R=
V
I
24
6
= 4Ω
21
Example 5.6
The circuit in figure (5.3), the voltage is changed to 1.2 KΩ, and the current
is changed to 4 mA, what is the resistance
Solution:
V = 1.2 kV
use the formula R= V / I
R=
V
I
1200
=
= 3,000,000 Ω
4×1000
= 1.2 X 1000
= 1200V
R=?
I = 4 mA
If required Resistance by M Ω
R = 3,000,000 /1,000,000 = 3 MΩ
22
Simple d.c. Circuits
Week 6
At the end of this week, the students should be able to:
•
Define resistivity and conductivity of a conductor
•
State the relationship between series and parallel circuits
•
Differentiate between series and parallel circuits
2.5
RESISTIVITY AND CONDUCTIVITY
Resistivity
Resistivity ρ of a material is the resistance of unit length of material and unit cross sectional
area at a given temperature.
At a given temperature, the resistance R, of a given material is:
•
directly proportional to the length of the material, l, and
•
inversely proportional to the cross sectional area of the material, A.
mathematically, these two statements can be combined and stated as
R=α l
(5.1)
A
where ρ is a constant of proportionality known as resistivity of the material, (ρ) is Greek
letter pronounced Rho.)
from equation (5.1), we can deduce that
ρ= RXA
l
Ohm X meter2
meter
= Ohm-meter
We should note that when R is in ohms, A in m2 and l in m, then the unit of ρ is ohmmeter (Ω - m)
Conductivity
The inverse of resistivity is called conductivity and is denoted by the symbol σ. (σ is a Greek
letter pronounced sigma) its unit of measurement is (Ω - m)-1, or 1 .
Ω-m
Mathematically, the two terms can be related as
σ = 1 , (Ω - m)-1
(5.2)
ρ
This relationship can be expressed as follows:
the higher the resistivity of a material, the lower its conductivity, and vice versa.
23
2.6 SERIES AND PARALLEL CIRCUITS
Series circuits
A series circuit is formed when any numbers of resistors are connected end-tostart so that there is only one path for current to flow. The resistors can be actual
resistors or other devices that have resistance like lamps.
Figure (6.1) shows two resistors connected in series (end -to –start) between
points A and B. There is only way for current to get from point A to point B. a series
circuit provides only one path for current between two points in a circuit so that the
current in the same through R1and R2.
Figure (6.1) : Resistor in series
In an actual circuits diagram, a series circuit may be easy to identify as those in
figure (6.1) For example. Figure (6.2) shows series resistors drawn in other ways.
Figure (6.2) : Some examples of series circuits. Notice that the current is the
same at all points
Parallel Circuits
A parallel circuit is formed when any number of resistors are connected starts-to-starts
and ends-to-ends so that there is different paths for current to flow. The resistors can be
actual resistors or other devices that have resistance like lamps.
24
Figure (6.3) shows two resistors connected in parallel (start-to-start and end-to-end)
between points A and B. There is two ways for current to get from point A to point B. a
parallel circuit provides more than one path for current between two points in a circuit so that
the current flow in two paths current flow in R1 is called I1 and current flow in R2 is called I2.
Figure (6.3) : Resistor in series
In an actual circuits diagram, a parallel circuit may be easy to identify as those in
figure (6.3). For example. Figure (6.4) shows Some examples of parallel resistors drawn in
other ways.
Figure (6.4) : Some examples of parallel circuits. Notice that the current is
the same at all points
25
Simple d.c. Circuits
Week 7
At the end of this week, the students should be able to:
•
Solve problems involving resistivity and conductivity
•
Deduce the equivalent resistance of series and parallel circuits
•
Solve problems on series and parallel resistors
2.7
SOLVE PROBLEMS INVOLVING RESISTIVITY AND
CONDUCTIVITY
Example 7.1
A cylinder wire of 0.5m in length, 0.5mm in diameter, has a resistance of 2.5Ohms. calculate
the
resistivity of the wire.
Solution
length, l = 0.5m
radius r = 0.5mm = 0.25 mm = 2.5 X 10-4m
2
Cross sectional area, A = πr2 = 22 X (2.5 X 10-4)2 m2
7
ρ = RA = 2.5 X (22/7) X (2.5 X 10-4)2
l
0.5
∴
ρ = 98.19 X 10-8 Ohm – meter.
Example 7.2
A wire of diameter 0.6mm, resistivity 1.1 X 10-6Ω - m has resistance of 44Ω. Calculate the
length of
the wire.
Solution
Here, we are given:
radius, r = 0.6 mm = 3 X 10-4m
2
To find l we deduce it from ρ = RA (i.e. l = RA )
l
ρ
∴
l = 44 X (22/7) X (3 X 10-4)2
1.1 X 10-6
= 11.3m
26
Example 7.3
A 30m conductor of cross sectional area 2mm2 has a resistance of 10Ω. Calculate the
conductivity
of the conductor.
Solution
ρ= l .
RA
=
2.8
30
= 1.5 X 106 (Ω - m)-1
10 X 2(10-3)2
EQUIVALENT RESISTANCE OF SERIES AND PARALLEL CIRCUIT
Equivalent Resistance Of A series Circuit
I1
R1
I2
V(R1)
R2
I3
V(R2)
R3
V(R3)
I
Vs
Figure 7.1 Series circuit
When resistors of resistance R1, R2 and R3 are joint together (fig 7.1), they are said to
be connected in series. It should be noted that:
(i)
The current through a series circuit is the same all over.
(ii)
The voltage drop across each resistor is different depending on its resistances.
(iii)
The applied voltage (Vs) is equal to the sum of the P.d’s across each resistor.
i.e. I=I1=I2=I3
Vs= V(R1) + V(R2) +V(R3)
(7.1)
where Vs = IR; V(R1)= IR1; V(R2) = IR2; V(R3) = IR3
(7.2)
Putting eqtn (7.2) in (7.1) gives
IR = IR1 + IR2 + IR3
=> Req = R1 + R2 +R3
For n resistors connected in series, the equivalent resistance
27
Req. is given by Req. = R1 + R2 +R3 + …. + Rn
(7.3)
Equivalent Resistance Of A Parallel Circuit
I
I1
Vs
I2
R1
I3
R2
R3
Figure 7.2 Parallel circuit
When resistors of resistance R1, R2 and R3 are connected in parallel, the following
statements hold good:
(i)
The p.d across each resistor is the same
(ii)
The current flowing through each resistor is different
(iii)
The supply current (I) is equal to the sum of current flowing through each
resistor
i.e. Vs = V1 = V2 = V3
I = I1 + I2 + I3
(7.4)
where I = Vs/R; I1 = Vs/R1 ; I2 = Vs/R2; I3 = Vs/R3
(7.5)
Inserting eqtn (7.5) in (7.4) gives
Vs/R = Vs/R1 + Vs/R2 + Vs/R3
⇒ 1/R = 1/Req = 1/R1 + 1/R2 + 1/R3
For n resistors connected in parallel,
1/Req = 1/R1 + 1/R2 + 1/R3 + - - - + 1/Rn
(7.6)
28
Example 7.4
In the circuit in figure (7.3), Determine the two voltages drops across R1 and
R2?
+
Figure (7.3)
Solution:
R1= 2.2 Ω
1) First solve for total resistance.
RT = R1 + R2
= 2.2 + 1.8 = 4 Ω
2) Second, solve for current:
V
12
I=
=
= 3 Amp
RT
4
3) Finally, solve for voltage drop across any resistor:
R2= 1.8 Ω
V = 12 v
I=?
VR1= ?
VR2= ?
Voltage drop across R1
VR1= I X R1 = 3 X 2.2 = 6.6 v
Voltage drop across R2
VR2= I X R2 = 3 X 1.8 = 5.4 v
The sum of all voltage drops ( 6.6v + 5.4v ) equals the source voltage (12v).
29
Example 7.5
In the circuit in figure (7.4), Determine the current in each branch
+
Figure (7.4)
Solution:
1) First solve for total resistance.
Used the formula when there are two unequal value resistors
RT =
R1 X R2
R1 + R2
=
4X6
4+ 6
=
24
10
= 2.4 Ω
2) Second, solve for voltage across any resistor:
VR1 = VR1 = VT = 12 v
3) Finally, solve for current in each branch :
VR1
12
I1 =
=
= 3 Amp
R1
4
VR1
12
I2 =
=
= 2 Amp
R1
6
VT
12
IT =
=
= 5 Amp
RT
2.4
R1 = 4 Ω
R2 = 6 Ω
V = 12 v
I1 = ?
I2 = ?
The sum of current in each branch (3 + 2) equal the total current 5A
30
Simple d.c. Circuits
Week 8
At the end of this week, the students should be able to:
•
State Kirchhoff’’s laws
•
Explain Kirchhoff’s laws
•
Solve simple problems on Kirchhoff’s laws
2.10 KIRCHHOFF’S LAWS
Kirchhoff’s laws were developed in 1847 by the German physicist G.R. Kirchhoff. These
laws are two in numbers and they are formally known as Kirchhoff’s current law (KCL) and
Kirchhoff’s voltage law (KVL).
Kirchhoff’s current law
Kirchhoff’s current law state that the algebraic sum of currents entering a node is zero.
By this law for purposes of explanation, we may regard currents entering a node as
positive, where currents leaving the node as negative. Considering the node in Fig 8.1 and
applying KCL, we get
i1 + (-i2) + (-i3) + i4 + i5 = 0
(8.1)
or
i1 + i4 + i5 = i2 + i3
(8.2)
From equation (8.2) we notice that the sum of current entering a node is equal to the sum of
the current leaving the node. This is an alternative way of stating KCL.
i1
i5
i2
i4
i3
Fig 8.1
31
Example 8.1
Find the value of I in Fig 8.2
I
6A
5A
P
4A
8A
Fig 8.2
Solution
In order to find the value of I, the following equation can be formed.
Currents towards P = currents flowing from P.
∴
6 + 4 + 5 = I + 8, ⇒ I = 7A
Kirchhoff’s voltage law
Kirchhoff’s voltage law state that in a closed circuit, the algebraic sum of the e.m.f.s is equal
to the algebraic sum of the voltage drops.
Mathematically this can be stated as:
± E1 + ±E2 = ± I1R1 ± I2R2 ± I3R3,
if there are two sources of e.m.f and three resistors R1, R2 and R3
In general, ∑ E = ∑ IR
Example 8.2
Determine the value of Vab in the circuit shown in Fig 8.3 (a) and (b) using KVL.
6V
3V
+
_ 6V+
a
Vab
3V
+
a
Vab
_b
_b
2V
2V
(b)
(a)
Fig 8.3
32
Solution
For Fig 8.3(a), applying KVL, we get
-2V + 3V – 6V – Vab = 0
∴
Vab = -5V
For Fig 8.3 (b) applying KVL, we get
-2V + 3V + 6V – Vab = 0
∴
Vab = 7V
Example 8.3
Applying KCL and KVL to the circuit of Fig. 8.4, write down the equation pertaining to loop
I and II respectively,
I1
I3
I2
R1
R3
I
R2
E1
II
E2
Fig 8.4
Solution
N.B. we are working under the assumption of fixed direction of I1, I2 and I3 as shown
in Fig 8.4
Applying KVL to loop I, we get
-E1 = R1I1 + R2I2
Applying KVL to loop II, we get
E2 = -R2I2 + R3I3
From KCL we known that at node A
I1 = I2 + I3
or
I3 = I1 – I2
Putting this value of I3 into the equation involving E2, we get
E2 = -R2I2 + R3 (I1 – I2)
E2 = R2I2 – (R2 + R3) I2
33
Simple d.c. Circuits
Week 9
At the end of this week, the students should be able to:
•
State superposition principle
•
Explain superposition principle
•
Solve simple problems on superposition principle
2.11 SUPERPOSITION PRINCIPLE
The superposition principle is a method that allows us to determine the current
through or the voltage across any resistor or branch in a network. The advantage of
using this approach instead of Kirchhoff’s laws is that it is not necessary to use
determinant or matrix algebra to analyze a given circuit. The theorem state the
following:
In a circuit containing several sources of e.m.f., the resultant current in any
branch is the algebraic sum of the currents in that branch that would be produced by
each e.m.f, acting alone, all other sources of e.m.f being replaced meanwhile by their
respective interval resistances.
In order to apply the superposition principle it is necessary to remove all
sources other than the one being examined. In order to ‘’Zero’’ a voltage source, we
replace it with a short circuits, since the voltage across a short circuit is zero volts. A
current source is zeroed by replacing it with an open circuit, since the current
through an open circuit is zero amps.
Example 8.1
Solve the same problem in example 3.5 to determine all the branch currents using the
superposition principles.
Solution
To solve this problems we shall follow the steps.
Step I, Re-draw the given circuit diagram with E1 = 15V and E2 = OV, as shown in
Fig 9.1 below
34
I1
I3
A
I2
4Ω
I1”
I2’
2Ω
3Ω
E1 = 15V
E2
10V
I3”
A
4Ω
2Ω
3Ω
E1 = 15V
B
B
(b) circuit diagram with E2 =OV
(a) Original circuit diagram
Fig 9.1
Step II: Next, determine all the branch currents I1’, I2’ and I3’ indicated in Fig 9.1 (b),
With E1 = 15V and E2 = OV.
Step III: we find the equivalent resistance of 3Ω resistor and 2Ω resistor connected in
parallel.
i.e.
R3Ω//R2Ω = 3 X 2 = 6 = 1.2Ω
3+2
5
Step IV: The circuit diagram of Fig 9.1(b) reduces to the form shown in Fig 9.1
4Ω
R3Ω//R2Ω = 1.2Ω
E1 = 15V
Fig 9.2
Step V: we can apply the Kirchhoff’s Voltage law to the circuit diagram of Fig 9.2 to obtain,
I’1 =
E1
.
(4 + R3Ω//R2Ω)
I’1 =
15
= 2.885A
( 4 + 1.2)
Step VI: Turn back to fig 9.1(b) and bear in mind that I1 is the total current that flows into
the 3Ω resistor branch and 2Ω resistor branch which are in parallel. Applying the principle of
current division rule, we get
I’2 =
R2Ω
X I’1 =
(R2Ω) + R3Ω)
2 X 2.885 = 1.154A
(2 + 3)
35
Similarly, I’3 =
R3Ω
X I’1 =
3
X 2.885 = 1.731A
(R2Ω + R3Ω)
(2 + 3)
Step VII: Draw the corresponding circuit diagram if the original circuit diagram of Fig 9.1
(a) has E1 = OV and E2 = 10V, to produce the circuit diagram shown in Fig. 9.3(b)
I1
I3
A
I1”
I2
4Ω
A I3”
2Ω
3Ω
E2
10V
E1 = 15V
B
I2’
E2 = 10V
4Ω
3Ω
2Ω
B
(b) circuit diagram with E1 = OV
(a) Original circuit diagram
Fig 9.3
Following all the stages in step II above, we obtain the following results:
(i)
R4Ω//R3Ω = 4 X 3 = 12 Ω = 1.714Ω
4+3
7
(ii)
The resulting equivalent circuit diagram after combing R4Ω and R3Ω in parallel gives
Fig 9.4.
2Ω
R4Ω//R3Ω
E2 = 10V
Fig 9.4
(iii)
Applying Ohm’s law to the circuit diagram of Fig 9.4, we get
I”3 =
(iv)
E2
=
10
≅ 2.693A
2 + R4sΩ // R3Ω
(2 + 1.714)
with reference to Fig 9.3, we see that if I”3 is to serve as the total current then
according to KCL then I”1 and I”2 must be seen to flow into node A. For
(v)
this to happen the direction I”2 must be opposite to what is indicated in the diagram of
Fig 9.3. This means that I”2 must bear a negative value. Only I”1 flows into node A.
Therefore, I”1 =
3
X 2.693 = 1.154A
(4 + 3)
36
Similarly, I”2 = _
4
X 2.693 = - 1.539A
(4 + 3)
Step VIII: The total branch currents can be obtained by adding up algebraic as follows:
I1 = I”1 + I’1 = 2.885 + 1.154 = 4.039A
I2 = I’2 + I”2 = 1.154A – 1.538A = - 0.384A
I3 = I’3 + I”3 = 1.731A + 2.693A = 4.424A
37
Simple d.c. Circuits
Week
10
At the end of this week, the students should be able to:
•
Define temperature coefficient of resistance
•
Draw the graph of resistance against temperature
•
Solve simple problems on temperature coefficient of resistance
2.12 TEMPERATURE COEFFICIENT OF RESISTANCE
If the resistance of a coil of insulated copper wire is measured at various temperatures up to
say 2000C, it is found to vary as shown in the figure 10.1 below.
Resistance (Ω)
1.426
0
–234.5
-200
-100
0
100
200
Temp 0C
Fig 10.1 Variation of resistance of copper with temperature
The resistance at 00C been taking as 1Ω. The resistance increases until it reaches 1 426Ω for
an increase of 1000 C in the temperature.
We may define temperature coefficient of resistance as “the increase in resistance per
original resistance at 00C per temperature change”
Mathematically α =
(10.1)
R1 –Ro
Ro (θ 1- θ 0)
The unit of temperature coefficient of resistance is oC-1
Graph of resistance against temperature
The graph of resistance against temperature is shown in Fig 10.2. below.
38
Resistance (Ω)
R2
R1
R0
θ1
θ2
Fig 10.2
Temperature (0C)
Slope = R2 – R1, also slope = R1 – Ro
θ2 – θ1
θ 1 - θo
Temperature coefficient of resistance = slope of resistance/ temperature
Resistance
i.e. α1 =
R2 – R1
(10.2)
R1 (θ 2 .. θ 1)
α0 =
R1 – R2
(10.3)
Ro (θ 1 – θo)
From (10.3), θo =0
=> α0 = R1 –R2
R0 θ1
= >α0 = Ro (1 + α θ 1), where α0 =α
(10.4)
Also, R2 = R0(1+α θ 2)
(10.5)
Generally Rθ =R0 (1+α θ)
(10.6)
From (10.4) and (10.5)
R1= R0 (1+α θ 1), => R1= 1 +a θ1
(10.7)
R2 R0 (1+α θ 2)
R2 1+a θ2
where R1=Resistance at temperature θ 1
R2=
“
“
“
θ2
α = Temperature coefficient of resistance
39
Example 10.1
A coil has a resistance of 50Ω at 00C, if the temperature coefficient of resistance for the coil
is 0.0043Ω/Ω0C, determine the resistance of the coil at 250C.
Solution
Rθ = Ro (1+αθ) = 50(1+0.0043X25) =55.38Ω
Example 10.2
The resistance of a coil of a copper wire at the beginning of a heat test is 173Ω, the
temperature being 160C. At end of the test, the resistance is 212Ω. Calculate the temperature
raise of the coil. Assuming the temperature coefficient of resistance of copper to be
0.00426/0C at 00C.
Solution
R1=173Ω,R2=212Ω,θ1=160C, θ2=?,α=0.00426/0C
R1=1+αθ1
R2 1+αθ2
θ2 = 0.30896= 72.520C
0.00426
Raise in temperature = θ2 - θ1=72.5-16 = 56.50C
Example 10.3
A copper wire has a resistance of 85Ω at 150C. Calculate its resistance at 900C. Assuming the
temperature coefficient resistance of copper to be 0.0043/0C,
Solution
θ1 = 150C, θ2 = 900, R1 = 85Ω and α0 = 0.00-43/0C,
R15 = 1 + α0 θ1 = 1 + 0.0043 X 15
R90 1 + α0 θ2
1 + 0.0043 X 90
i.e.
R90 = R1
1 + 0.0043 X 90
1 + 0.0043 X 15
∴
R90 = 110.75Ω
40
Electric Energy and Power
Week 11
At the end of this week, the students should be able to:
•
Define and state various forms of energy.
•
Explain the relationship between Electrical, Mechanical and thermal energy.
•
Define Electrical energy.
3.1
ENERGY AND ITS VARIOUS TYPES:
Energy
Energy is the ability to do work. The SI unit of energy is the Joules. The Joules can be
defined as the work done when a force F of 1N (Newton) acts through a distance d of 1m in
the direction of the force.
Work done = F (Newton) x d (meters)
There are many forms of energy, namely;
Electrical Energy
Mechanical Energy
Thermal Energy
Light Energy
Chemical Energy
Wind Energy
Solar Energy
Atomic Energy, e.t.c.
3.2
RELATIONSHIP BETWEEN ELECTRICAL,
MECHANICAL
AND THERMAL ENERGY
One form of energy can be converted into another by means of a suitable process. For
example, an electric motor converts electric energy into mechanical energy. The various
forms of energies are related by the law of conversation of energy.
The Law of Conversation of Energy states that Energy can neither be created nor
destroyed. This means that when needs to be converted; the total energy in a system is equal
to the total energy output from the system. For example a car in motion has mechanical
(Kinetic) energy and when the brake are applied for it to come to rest, this mechanical energy
is converted into heat energy and sound energy.
41
Consider an electric fan. For an electric fan to rotate it takes voltage supplies from the
main supply and causes the electric motor to rotate. This action in turn causes the blades of
the fan to rotate and produce wind energy (a form of mechanical energy) and sound energy.
In this case, electric energy was converted to mechanical energy and sound energy.
3.3
ELECTRICAL POWER
Power is the rate of doing work. The unit of Power is joules/second or watt.
Lager units of power could be kilowatt. Also in dealing with large amount of energy, it is
sometimes more convenient to express energy in kilowatt hour than in joules.
1KWh =1000 watt hours
=1000 x 3600 watt seconds or joules
=3,600,000J = 3.6 Mega joules
From its definition the expression for Electric Power can be written as;
Power = work done /time taken
Therefore Electric Power can be written as;
P = IV, joules
Or
P = I2R, joules
Or
P = V2, joules
R
42
Electric Energy and Power
Week 12
At the end of this week, the students should be able to:
•
State Joule’s law
•
Solve problems involving Electrical energy and power
•
Solve simple problems on Joule’s law
3.4
JOULE’S LAW
Electrical Energy E is the energy required for V volts of electricity to make an electric current
of I amperes to flow through a conductor of resistance R Ohms for a time T seconds.
E = IVt, joules
Or
E = I2Rt, joules
Or
E = V2 t, joules
R
The above expressions are also known as Joules Law, which states that;
The amount of work required to maintain a current of I amperes through a resistance of R
Ohms, for time t seconds is;I2R
Example 12.1
Calculate the work done in moving a through a distance of 25m by a force of 20N which acts in the
direction of the motion of the body.
Solution
Since work done = F X d
Work done = 20 x 25 = 500J
Example 12.2
A force of 40N IS applied to a body to move it at a uniform velocity through a distance of 15m in 10s
in the direction of the force. Calculate the power produced.
Solution
Work done = force x distance
= 40 x 15 =600J
Power
= work done
time taken
= 600 = 60W
10
43
Example 12.3
An electric motor develops 5KW at the speed of 100rev/min. calculate (a) the work done in 30mins
(i) in kilowatt hour (ii) in megajoules (b) the torque in N-m.
Solution
(a) (i), Work done in KWh = power(in kw) x time(in hrs)
= 5 x 30/60 = 2.5KWh
ii, work done in mega joules . but 1kwh = 3.6MJ
Therefore w.d. = 2.5 x 3.6 =9MJ
(b) power = 2π x T x n
5000 = 2πT X 1000/60
T = 47.74N-m
Example 12.4
An electric kettle takes 2KW at 240V. Calculate (a) the current and (b) the resistance of the
heating element.
Solution
(a)
I = P/V = 2000/240
∴
I = 8.33A
(b)
From P = V2/R
R = V2/P = 2402/2000 = 28.8Ω
Example 12.5
The power expended in a certain resistor is given by I
2
R. If the power expended in the
resistor is 175W when the current is 5A, Calculate the power in the resistor when,
(i)
Both current & resistance are double.
(ii)
Current is half and resistance double.
(iii)
When current is double and resistance half.
Solution
Given that: P = 175W, I = 5A
P = I2R,
=> R = P
I2
44
= 175
(5)2
(i)
= 7Ω.
When both current and resistance are double, I=2I; R=2R
Therefore P= (2I) 2 X R=4 X 25X2X7 =1400W
(ii) Half current = I/2; double resistance = 2R
Therefore P= (I/2)2 X 2R = (5/2)2 X 2 X 7 = 87.5W
(iii)
Double current = 2I, half resistance = R/2
Therefore P = 4 X 25 X 7/2 = 350W
45
Concepts of Electrostatics and Capacitance
WEEK 13
At the end of this week, the students should be able to:
•
Explain electric charge
•
State Coulomb’s laws
•
Solve problems involving coulomb’s law
3.5
ELECTRIC CHARGE
Electrostatics is the branch of science, which deals with the phenomena associated with
electricity at rest.
Under normal circumstances an atom is electrically neutral which means that the
aggregate of positive charge of protons is exactly equal to the aggregate of negative charge of
electrons.
But if somehow some electrons are removed from atoms of body, then it is left with a
preponderance of positive charge and the body is said to be pensively charge. If on the other
hand, some electrons are added to the atoms of a body, then negative charge out-balance thee
positive charge and the body is said to be negatively charged.
Therefore the deficiency of electrons of the atoms in a body makes it to be positively
charged while the excess of electrons of the atoms in a body makes it negatively charged, so
the charge is known as the total deficiency or excess of electrons in a body and the unit is in
Coulomb (abbreviated C).
3.6
COULOMBS LAWS OF ELECTROSTATICS
1.
First Law: like charges of electricity repels each other, whereas unlike charges attract
each other.
2.
Second Law: the force exerted between two point charges is;
(i) directly proportional to the product of their strength.
(ii) inversely proportional to the square of the distance between them
.This is known as the coulombs laws of electrostatics and can be expressed
mathematically as;
F α Q1Q2
d2
or
F = K Q1Q2
d2
Q1Q2 =
4πεd
2
Q1Q2
4πε0εrd2
46
where F = magnitude of the force of attraction or repulsion between two charges Q1 and Q2,
(in Newtons, N)
d = the distance between the charges Q1 and Q2, meters (m)
Q1,Q2 = charges, Coulombs (C)
K = 1/4πε, (ε = ε0εr), ε = permittivity (farad/meter).
Now we should know that:
Now
1
4πεo
=
1
.
4π x 8.85 x 1012
which is approximately = 9 x 109
Hence coulombs law can be written as;
F = 9 x 109 X Q1 Q2
εrd2
(in a medium)
F = 9 x 109 X Q1 Q2
d2
(in a air or vacuum)
Example 13.1
Calculate the electrostatic force of repulsion between two alpha particles when at a distance
of
10-13meters from each other. The charge of an alpha particle is 3.2 X 10-19C.
Solution:
Hence Q1 = Q2 = 3.2 X 10–19C
d = 10– 13m
But
F = 9 X 10 9 X Q1 Q2
d2
F = 9 X 109 X 3.2 X 10-19 x 3.2 X 10-19
(10–13)2
= 9.2 X 10–2N
Example 13.2
Calculate the distance of separation between two elections for which the electric force
between them is O.1N. Charge of election is 16 X 10 -19 C.
47
Solution
Given that: d =?, F =0.1N, Q1 = Q2 =Q 1.6 X 10-19C
F=
9 X 109 Q2 ,
d2
∴ d =
9 X 10 9Q2 =
F
=>
d2 =
9 X 10 9Q2
F
9 X 10 9 (1.6 X 10-19)2
0.1
= 4.8 X 10-14m
48
Concepts of Electrostatics and Capacitance Week 14
At the end of this week, the students should be able to:
•
Define electric field, electric field strength, electric flux density, permittivity,
permittivity of free space and relative permittivity
•
Define capacitance and state the expression for the capacitance of parallel plate
capacitors in terms of area, the distance between plate and permittivity of the dielectric
•
Solve problems on electric field strength, electric flux density, permittivity and
capacitance
3.7
DEFINATION OF TERMS USED IN ELECTROSTATICS
3.7.1 Electric fields
Electric fields are region through which electric force is being experienced or felt.
3.7.2 Electric field strength
When a P.d of V volts is applied between two metals plates or electrodes, which are d meters
apart (shown in fig 14.1) an electric field is established between the plates. The arrangement
in fig 14.1 is referred to as parallel–plate capacitor. The medium separating the plates is an
insulator, and it is known as the dielectric. A feature of the capacitor is that they are capable
of storing energy, and this energy is stored in the dielectric material of the capacitor.
The electric field strength is the total voltage acting on a field per length of the line
(field).
The electric field strength, symbol E, in the dielectric is given by the equation
E = V/d, (V/m)
(14.1)
C
V
Fig 14.1
3.7.3 Electric flux density
One line of electric flux is assumed to emanate from a positive charge of one coulomb, hence
Q lines of flux emanate from a charge of Q coulomb. Therefore,
49
Electric flux = Q, it is equal to the stored charge. If the flux passes through a dielectric of
area A (the area being measured perpendicular to the direction of the flux), then the electric
flux density is the amount of electric flux passing through a small area A at right angle to the
flux per unit area, it has symbol D. Mathematically
D= Q/A = Ψ/A (c/m2)
(14.2)
3.7.4 Permittivity
The permittivity, symbol ε, of a dielectric material is a measure of the ability of the material
to allow an electric flux to be established in it. It has the dimensions of farad per meters’
(f/m). For a given value, a material with a high value of permittivity produces a greater
electrostatic flux than does a material with a lower value of permittivity for a simple parallelplate capacitor.
Permittivity, ε = Electric flux density = D/E
Electric field strength
∴
ε = D/E
(14.3)
Permittivity is also given by
ε = ε0εr
(14.4)
3.7.5 Permittivity of free space
In electrostatics, the ratio of the electric flux density in a vacuum to the electric field strength
is termed the permittivity of free space and is represented by the symbol ε0 Hence,
ε0 = Electric flux density in vacuum
Electric field strength
From equation (14.4), ε0 = ε/εr
(14.5)
From a carefully conducted tests it has been found that the value of ε0 = 8.85 x 10-12F/m
3.7.6 Relative permittivity
Relative permittivity (εr) is the ratio of the absolute permittivity to the permittivity of free
space.
Note that εr has no dimension because ε and ε0 have the same units
50
3.8
CAPACITANCE
A capacitor is an arrangement which has the ability to store electricity as excess of positive
charges
on one plate and deficiency of it on the other. Therefore the property of a capacitor to store
electric charge when its plates are at different potentials is reflected to as its capacitance.
The unit of capacitance is termed the farad (symbol f) and farad can be defined as the
capacitance of a capacitor between the plates of which there appears a difference of potential
of 1 volt when it is charged by 1 coulomb of electricity.
Capacitance (farads) =
Charge (coulombs)
Applied voltage (volts)
Q = CV
(14.6)
3.8.1 Capacitance of Parallel Plate Capacitors interms of Area, the distance between
Plates and permittivity of the dielectric
Recall, C = Q/V. using equations (14.1), (14.2) and (14.6),
C = Q/V = Ψ/V = DA/Ed = D/E(A/d) = εA/d
For parallel capacitors with n numbers of plates
C = ε(n – 1)A
d
(14.7)
where n is the number of plates.
Example 14.1
Two conducting plates are arranged parallel to each other and spaced 1.5mm apart. If the
plates are charged until the potential difference between them is 150V, what is the electric
field strength between the plates?
Solution
E = V = 150 .
d 1.5 X 10-3
∴
E = 10KV/m
Example 14.2
Two plates, each of area 5cm2, are placed parallel to each other and very close together, and a
charge of 15 X 10-8C is stored on the plates. Calculate the electric flux density in the space
between the plates.
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Solution
D = Ψ = (15 X 10-8) = 3 X 10-4C/m2
A
5 X 10-4
Example 14.3
Calculate the charge expected on a 400µF capacitor which is connected to a 100V source.
Solution
Q = CV = 400 X 10-6 X 100 = 4 X 10-2C
i.e.
Q = 40 X 10-3C = 40mC
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Concepts of Electrostatics and Capacitance
Week 15
At the end of this week, the students should be able to:
•
State the expression for the equivalent capacitors in series and in parallel
•
Derive an expression for the energy stored in a capacitor
•
Solve problems involving capacitors in series and parallel, and energy stored in a
capacitor
3.9
CAPACITOR CONNECTIONS
Parallel connection
Capacitors in a parallel configuration each have the same potential difference (voltage). The
reason for putting capacitors in parallel is to increase the total amount of charge stored. In
other words, increasing the capacitance we also increase the amount of energy that can be
stored.
The total capacitance (Ceq) is given by:
(15.1)
Fig 15.1
Series connection
The current through capacitors in series stays the same, but the voltage across each
capacitor can be different. The sum of the potential differences (voltage) is equal to
the total voltage. The reason for putting capacitors in series we get less capacitance
and less charge storage than with either
either alone (the total voltage is divided between the
number of capacitors).
Fig 15.2
In parallel, the total charge stored is the sum of the charge in each capacitor.
While in series, the charge on each capacitor is the same.
the total capacitance is given by:
(15.2)
3.10 ENERGY STORE IN A CAPACITOR
If a capacitor having capacitance C farads is charged at a constant current of I amperes for t
seconds, as in the figure below, the charge Q is It coulombs.
Current
I
v
V
Fig 15.3 Charging of a capacitor
If the final p.d across the capacitor is V volts, the charge is also CV coulombs.
Therefore: CV = It, and V = It
C
The p.d across C is therefore proportional to the duration of the charge and is represented by
the thick
ck line in the figure. It follows that:
average p.d across C during charging = 1/2 V (volts)
and average power of C during charging = I x 1/2 V (watt)
Therefore: energy supplied to C during charging
= average power x time
= 1/2 IVt (Joules)
= 1/2 V x CV (Joules)
i.e. electrostatic energy stored in C = 1/2 CV2 Joules.
⇒ W = ½ CV2 Joules
(15.3)
Example 15.1
Two capacitors, A and B, of capacitance 2µF and 4µF respectively, are connected in series to
a dc supply; the charge store by each capacitor is 0.16mc. Determine:
(a) the p.d across each capacitor
(b) the energy stored in each capacitor
(c) the effective capacitance of the series combination.
Solution
Data: CA = 2µF; CB = 4µF; Q = 0.10m; C = 0.16 x 10-3C
(a)
The p.d across each capacitor:
VA = Q = 0.16 x 10-3 =80V
CA
2 x 10-6
VB = Q = 0.16 x 10-3 = 40V
CB
4 x 10-6
(b)
The energy stored in each capacitor:
W = 1/2 CV2
WA = 1/2 x 2 x 10-6 x 802
= 0.0064J
WB
= 1/2 x 4 x 10-6 x 402
= 0.0032J
(c)
The effective capacitance of the series combination:
C = CA CB
(CA CB)
=2x4
2+4
µF
= 1.333µF
Example 15.2
Three capacitors C1, C2 and C3 are connected in parallel to give an equivalent capacitance of
2µF. If the d.c. supply to the parallel combination of the capacitors is 20V, the capacitance of
C1 is 0.5µF and the charge stored by C2 is 4µC, determine the capacitance of C2 and C3.
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Solution
CP = C1 + C2 + C3
but
CP = 2µF (given)
C1 = 0.5µF (given)
However, C2 = Q2 = 4 = 0.2µF
Vp 20
Finally, C3 = Cp - C1 – C2 = 2 – 0.5 – 0.2 = 1.3µF
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