fet first
NATED Series
electrotechnics N4
Student’s Book
Jowaheer Consulting and Technologies
FET FIRST Electrotechnics N4
Student’s Book
© Jowaheer Consulting and Technologies, 2013
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Contents
Module 1: Principles of electricity
Unit 1.1: Basic atomic theory.............................................................1
Unit 1.2: Resistance, potential difference, electric current and
electromotive force.............................................................5
Unit 1.3: Ohm’s Law and Kirchhoff’s laws...........................................8
Unit 1.4: Power, energy, efficiency and Joule’s Law............................ 13
Unit 1.5: Factors influencing the resistance of a material..................... 14
Module 2: Series and parallel networks
Unit 2.1: Series circuits................................................................... 21
Unit 2.2: Parallel circuits................................................................. 24
Unit 2.3: Series-parallel circuits........................................................26
Module 3: Grouping of cells
Unit 3.1: Cells...............................................................................30
Unit 3.2: Series grouping of cells......................................................32
Unit 3.3: Parallel grouping of cells....................................................33
Unit 3.4: Series-parallel grouping of cells...........................................34
Module 4: Magnetism and electromagnetism
Unit 4.1: Magnetism.......................................................................36
Unit 4.2: Electromagnetism.............................................................40
Module 5: Electromagnetic induction
Unit 5.1: The principle of electromagnetic induction............................46
Unit 5.2: Inductance......................................................................49
Module 6: Capacitors
Unit 6.1: Capacitors.......................................................................52
Unit 6.2 Capacitors in series...........................................................55
Unit 6.3: Capacitors in parallel.........................................................56
Unit 6.4: Capacitors in series-parallel................................................57
Module 7: DC machines
Unit 7.1: Overview of DC machines..................................................59
Unit 7.2: Calculating the emf generated in an armature winding............66
Unit 7.3: DC generators..................................................................67
Unit 7.4: DC motors....................................................................... 76
Module 8: AC circuit theory
Unit 8.1: Fundamentals of AC circuit theory.......................................84
Unit 8.2: Complex numbers............................................................90
Unit 8.3: Series AC circuits............................................................. 91
Unit 8.4: Parallel AC circuits.......................................................... 100
Unit 8.5: Power in an AC circuit..................................................... 107
Module 9: Transformers
Unit 9.1: Overview of transformers................................................. 112
Unit 9.2: Single-phase transformer calculations................................ 118
Module 10: AC machines
Unit 10.1: Three-phase induction motors and motor starters.............. 125
Unit 10.2: Single-phase induction motors........................................ 133
Module 11: Generation and supply of AC power
Unit 11.1: Three-phase distribution networks................................... 143
Module 12: Measuring instruments
Unit 12.1: Overview of measuring instruments................................. 148
Unit 12.2: Extending the range of voltmeters and ammeters............... 151
Unit 12.3: Measurement of resistance............................................. 153
Module 13: Switchgear and protection devices
Unit 13.1: Fuses.......................................................................... 156
Unit 13.2: Circuit breakers, contactor and protection devices
on motor starters.......................................................... 161
Module 14: Solid state control
Unit 14.1: The decimal and binary number systems.......................... 165
Unit 14.2: Arithmetic operations using the binary system................... 169
Unit 14.3: Logic gates.................................................................. 171
Module 15: Rectification
Unit 15.1: Rectifiers..................................................................... 177
Syllabus grid......................................................................................... 183
Module 1
Principles of electricity
Units in this module
•
•
•
•
•
Unit 1.1 Basic atomic theory
Unit 1.2 Resistance, potential difference, electric current and electromotive force
Unit 1.3 Ohm’s Law and Kirchhoff’s Laws
Unit 1.4 Power, energy, efficiency and Joule’s Law
Unit 1.5 Factors influencing the resistance of a material
Unit 1.1 Basic atomic theory
of that compound. A molecule contains
atoms of each of the elements that form the
compound. See Figure 1.1.
Electricity is the most useful source of
energy there is. We rely on electricity to
run our homes, businesses and industry.
Can you imagine what life would be like
if there were no electricity? In this unit we
will investigate the basic building block of
electricity, namely the atom.
Atoms and elements
All matter is made up of tiny particles
called atoms. Elements like gold, copper
and carbon are made up of atoms. The
atom is the smallest part of an element
that can exist and still retain the properties
of the element. An element is a substance
which cannot be broken down into a
simpler substance by chemical means, for
example gold, copper, carbon, aluminium
and silver.
Figure 1.1 Water molecule
Atomic structure
Protons, neutrons and electrons
An atom consists of three types of particles,
namely electrons, protons and neutrons.
The centre part of an atom is called the
nucleus and consists of protons and
neutrons. Protons have a positive electrical
charge whereas neutrons are electrically
neutral as they have no electrical charge.
Electrons are particles with a negative
charge that move around the nucleus in
Compounds and molecules
A compound is a combination of two or
more elements. For example:
2 atoms of hydrogen (H) + 1 atom of
oxygen (O) = 1 molecule of the compound
called water (H2O).
A molecule is the smallest particle of
a compound that has all the properties
1
The presence of protons in the nucleus
makes it positively charged. The nucleus
makes up almost the entire mass of an atom
and this is termed the atomic mass of the
element. The more protons and neutrons
an atom has, the greater is its mass. The
total number of protons and neutrons in
an atom is called its mass number (A).
Electrons have such a small mass that they
are left out of the mass calculation.
The atomic number (Z) and mass
number (A) of an element are written
with the symbol of the element in a specific
way. Look at the example of carbon in
Figure 1.4.
orbitals or shells, just like the planets that
circle continuously around the sun. See
Figures 1.2 and 1.3.
electrons
protons
nucleus
electron shells
Atomic number
Chemical symbol
Figure 1.2 Three-dimensional representation of
an atom
Name of element
electron shells
Relative atomic mass
Figure 1.4 Atomic number (Z) and mass number
(A) of carbon
electrons
Valence and free electrons
Electrons orbit the nucleus in orbitals
or shells. The electrons that orbit the
nucleus in the outermost shell are called
valence electrons. The valence of an
atom determines its ability to gain or
lose electrons. This, in turn, determines
the chemical and electrical properties of
the atom. An atom that lacks one or two
electrons in its outer shell will easily gain
electrons to complete its shell.
A large amount of energy is required
to free electrons from an atom because
all the orbiting electrons experience a
force of attraction towards the nucleus.
However, the further an electron is from
the nucleus, the less this force of attraction
is. Therefore, valence electrons can be set
nucleus
Figure 1.3 Two-dimensional representation of an
atom
In a neutral atom the number of protons
is equal to the number of electrons. An
element can be identified by the number
of protons in its nucleus. This is called
the atomic number (Z) of the element.
For example, carbon has six protons in its
nucleus, so its atomic number is 6 (Z = 6).
2
free quite easily. See Figure 1.5. A valence
electron that is removed from its orbit is
called a free electron. See Figure 1.6.
are neutral. An atom usually has an equal
number of positively charged protons
and negatively charged electrons. These
charges cancel each other out and the atom
is then electrically neutral.
However, we saw that atoms try to
become stable by filling their outer shells.
To do this, the atom either has to gain or
lose electrons. If it gains electrons, it has
surplus electrons and becomes negatively
charged (–). If it loses electrons, it has a
deficit or a lack of electrons and becomes
positively charged (+).
Benjamin Franklin named these two
types of charges positive charges and
negative charges. Figure 1.7 shows how
similar poles of a magnet repel each other,
while Figure 1.8 shows how dissimilar
poles of a magnet attract each other. The
same applies to electrical charges. Particles
with opposite positive and negative
charges attract one another, while particles
with the same type of charge, for example
positive and positive charges, push each
other away or repel one another. See Figure
1.9. The force with which atoms attract or
repel each other is called electric force or
electrostatic force. Electricity is the effect
caused by the movement of the charged
particles in atoms.
Figure 1.5 Electrons moving from one energy
level to another
free
electrons
Figure 1.7 Similar poles of a magnet repel each
other.
Figure 1.6 Free electrons
Positive and negative charges
The particles that make up an atom are held
together by electrical charges. Remember
that protons have a positive charge and
electrons have a negative charge. Neutrons
Figure 1.8 Dissimilar poles of a magnet attract
each other.
3
F
F
F
F
F
the negative terminal. This is known as the
conventional current flow theory.
However, we know today that current flow
in a circuit is the movement of electrons
through the conductor. We also know
that electrons have a negative charge and
that unlike charges attract each other. So
electrons move from the negative terminal
of the battery to the positive terminal. This
is called the electron current flow theory.
See Figure 1.10.
The idea of conventional current flow
remains in use today even though we know
that current is the movement of electrons
from negative to positive terminals.
F
Figure 1.9 Movement of charges
Conductors and insulators
Conductors
We saw earlier that in some materials the
electrons in the outer shell are not attracted
strongly to the nucleus and that they can
move between atoms. When this happens,
there is a flow of electrical charge. This flow
of electrons is known as an electric current.
Materials that allow electric current to
pass through them are called conductors.
These materials have a large number
of free electrons and a high electrical
conductivity. Conductivity is a measure of
a material’s ability to carry electric current.
Aluminium, gold, silver and copper are
examples of conductors. Most metals are
good conductors. Carbon is the only nonmetal that is a good conductor.
electron current
flow
conventional
current flow
Figure 1.10 Conventional current flow and
electron current flow
Exercise 1.1
1. State whether the following are
true or false. If false, write the
correct statement.
a) An atom consists of only two
particles.
b) An electron is negatively
charged.
c) The central part of an atom is
called the nucleus.
d) The element aluminium has 13
electrons.
e) The nucleus of an atom is
always negatively charged.
2. Fill in the missing words:
a) The direction of flow of
electrons is from ____ to ____ .
b) The direction of conventional
current is from ____ to ____ .
Insulators
Materials that block the flow of electricity
are called insulators. These materials have
very few free electrons and have a low
conductivity. In other words, they do not
have the ability to allow a current to pass
through them. Plastic, wood, rubber and
porcelain are examples of insulators.
The conventional current flow theory
and the electron current flow theory
In the 1800s scientists believed that a
positive charge represented an increase
in the amount of electricity and that a
negative charge was a reduced amount of
electricity. They assumed that current flows
from the positive terminal of a battery to
4
Unit 1.2 Resistance, potential
difference, electric current
and electromotive force
3. State what will happen to the
charges shown in Figure 1.11.
Resistance
Resistance is the opposition to the flow of
electric current. Electrical resistance (R) is
measured in ohms (Ω). We can compare
resistance to a water tap that regulates the
flow of water.
Potential difference (pd)
Consider the two spheres A and B shown
in Figure 1.12. According to Coulomb’s
Law, they will be attracted to each other.
In other words, there is a potential to do
work. By definition, this potential to do
work is known as energy. If the spheres
are freed, the attraction will move them
together. The energy per unit charge is
called the potential difference (pd). When
the spheres cannot move, the potential
difference represents the energy stored to
do work.
Figure 1.11 Two negative charges
4. Complete the following
statements:
a) A glass rod rubbed with a
silk cloth attracts rubber
rubbed with fur. If the rubber
is negatively charged, then
the glass rod will be ____
charged.
b) The amount of attraction or
repulsion force which acts
between two electrically
charged bodies in free space
depends on ____ and ____ .
c) Negative charge is created in
a neutral body through an
accumulation of excess ____ .
d) Static charges are created by
____ .
e) The space between and around
charged bodies is called an
____ .
potential
difference
A
F
B
F
Figure 1.12 A charge on two bodies represents a
potential to do work.
When the charge is moving as shown in
Figure 1.13, the potential difference
represents the energy lost in moving the
charge from one point to the other, for
example from C to D.
C
potential
difference
D
Figure 1.13 Potential difference between two
points
5
One volt of potential difference exists
between two points when one joule of
work must be done to move one coulomb
of charge from one point to the other.
A larger unit of quantity of electricity is the
ampere-hour (A·h). 1 A·h = 3 600 C.
Example 1
A battery is charged at a rate of 3 A for
2 hours (h). Determine the quantity of
electricity in:
a) coulombs
b) ampere-hours.
In equation form we have:
Potential = work done (W) in joules (J)
difference (pd) charge (Q) in coulombs (C)
Potential difference is also known as closed
circuit voltage.
Given: I = 3 A; t = 2 h
Solution
Electric current
The unit of current is the ampere (A).
The symbol for current is I and current is
measured with an ammeter. The ampere
is defined as the constant current that
will produce a force of 2 × 10 –7 newton
per metre of length between two straight,
parallel conductors of infinite length, of
negligible circular cross-section placed
one metre apart in a vacuum.
In terms of charge, one ampere is the
movement of charge at a rate of one
coulomb or 6,242 × 1018 electrons per
second as shown in Figure 1.14. Note that
the direction of the current is indicated
with an arrow.
Q
[I
a) Q = It
=t]
= 3 × 2 × 60 × 60 [1 h = 60 × 60 s]
= 21 600 C
b) Ampere-hours = I × hours
=3×2
= 6 A·h
Electromotive force (emf)
Electromotive force is the greatest
potential difference that can be produced
by a source of electric current in a
circuit. The unit of emf is the volt (V).
Electromotive force is measured over an
open circuit and the value is always higher
than the potential difference.
A source of electromotive force has two
terminals between which it maintains a
potential difference. Some examples of
sources of emf are a:
• cell or battery
• generator
• thermocouple.
time = 1 s
A
Figure 1.14 Movement of electrons
Electromotive force is calculated using the
formula:
In equation form, the current-charge
relationship is as follows:
Emf = voltage at terminal of source of
supply + voltage drop in source of supply
I = Q
t
or
6,242 × 1018 electrons
E = V + Ir
where:
I = current in amperes (A)
Q = charge in coulombs (C)
t = time in seconds (s).
where:
E = electromotive force in volts (V)
V = potential difference in volts (V)
6
I = current in amperes (A)
r = internal electrical resistance of the
source in ohms (Ω).
potential, just like a stream of water under
high pressure will move towards a point
where the pressure is lower. You cannot
measure the potential of a charged particle,
but you can measure the difference in
potential between two points in a circuit.
Potential difference is calculated using the
formula:
V = E – Ir
Example 2
A cell with an emf of 1,49 V and internal
resistance of 0,5 Ω is connected across
a bulb. The current flowing through the
bulb is 0,25 A. Determine the potential
difference across the bulb.
Look at Figure 1.15.
Given: E = 1,49 V; r = 0,5 Ω; I = 0,25 A
Solution
V = E – Ir
Potential difference (pd):
V= 1,49 – (0,25 × 0,5)
= 1,49 – 0,125
= 1,365 V
Exercise 1.2
1. Fill in the missing words to
complete the sentence:
a) Resistance is the ____ to
current flow.
b) ____ is measured in ohms.
2. List two sources of emf.
3. A battery of 12 V supplies a
current of 1,2 A for five minutes.
How much energy is supplied in
this time?
4. A battery has an emf of 13,5 V.
When a 25 Ω resistor is connected
across the battery, the terminal pd
is 12 V. Calculate the:
a) current through the 25 Ω
resistor
b) internal resistance of the
battery.
Figure 1.15 Emf of a cell
A and B = terminals of the source of
supply
R = load resistor
V = potential difference (voltage)
measured across load R
r = internal resistance of the cell
I = current that passes through the circuit
Ir = voltage drop across the internal
resistance
E = emf of the cell.
A positively charged particle has a higher
potential than a negatively charged particle
or one that has no charge. A positive charge
will tend to move towards a point of lower
7
Given: V = 220 V; R = 25 Ω
Unit 1.3 Ohm’s Law and
Kirchhoff’s Laws
Solution
Ohm’s Law
There is a fundamental relationship
between the value of resistance (R)
measured in ohms (Ω), the current (I)
measured in amperes (A) and the voltage
(V) measured in volts (V).
From Ohm’s Law:
I = V
R
= 220
25
= 8,8 A
Ohm’s Law
Ohm’s Law states that the current
flowing in a circuit is directly
proportional to the applied voltage
across the circuit and inversely
proportional to the resistance of the
circuit, provided that the temperature
remains constant.
Written as a mathematical expression,
Ohm’s Law states:
25 Ω
220 V
heater
element
I=V
R
Figure 1.16 Element in a heater
where:
I = current in amperes (A)
V = potential difference (voltage) in volts (V)
R = resistance in ohms (Ω).
Example 2
Calculate the voltage which must be
applied to a 1 kilo-ohm (kΩ) resistor so
that the current flow is 2 milli-amperes
(mA).
The above equation can be arranged as:
V = IR
or
R=V
I
Given: R = 1 kΩ; I = 2 mA
Solution
From Ohm’s Law:
V = IR
= 2 × 10–³ × 1 × 10³
=2V
Example 1
The following information is available
about a heater (see Figure 1.16):
• Supply voltage = 220 V
• Resistance of heater element = 25 Ω.
Calculate the current drawn from the
supply.
8
physicist, Gustav Robert Kirchhoff. There
are two basic laws of an electric circuit, as
illustrated below.
Exercise 1.3
1. An electric bulb draws 0,5 A when
connected to a 230 V supply.
Determine the resistance of the
bulb filament.
2. A 220 V geyser draws 4 A from the
supply. Determine the resistance
of the geyser element.
3. Determine the potential difference
that must be applied to a 22 Ω
resistor if the current flowing
through it is 500 mA.
4. Refer to Figure 1.17. Calculate the
value of unknown resistor R x.
A
Kirchhoff’s Current Law (KCL)
The sum of the currents entering a
junction is equal to the sum of the
currents leaving the junction.
Alternatively, at any instant the
algebraic sum (Σ) of the currents at a
junction in a network is zero (ΣI = 0).
Consider the junction 0 in Figure 1.18.
I1
I4
I2
Rx
I5
I3
V
Figure 1.18 Kirchhoff’s Current Law (KCL)
I1 + I2 + I3 = I4 + I5
or
Figure 1.17 Circuit with a resistor
I1 + I2 + I3 – I4 – I5 = 0
Meter
Reading
Note
Voltmeter
12 V
Ammeter
0,25 A
Σ is the Greek capital letter sigma and
means ‘sum of’.
Kirchhoff’s Voltage Law (KVL)
The voltage applied to a closed circuit is
equal to the sum of the voltage drops
in that circuit.
Kirchhoff’s Laws
Calculations using Ohm’s Law are not
possible if one of the branches of a parallel
circuit contains a source of emf or if the
current to be calculated is part of a network
in which sources of emf may be present.
In this situation, we use Kirchhoff’s Laws.
These laws are the basic laws of electric
circuits and were formulated by a German
Alternatively, at any instant in a closed
loop, the algebraic sum of the emfs
acting around the loop is equal to the
algebraic sum of the pds around the
loop. You have already used this law in
the series circuit analysis:
VT = V1 + V2 + V3
where VT = total or supply voltage.
9
This can also be stated as follows:
voltage applied = sum of voltage drop
or
Figure 1.20 Potential rise
voltage applied – sum of voltage drop = 0
• Refer to Figure 1.21. If the direction
is from positive to negative, assume
the voltage is negative. This is called
potential drop. If the branch is traced
from A to B, then it should be taken
as –IR in the equations. If the branch
is traced from B to A, it is a potential
rise and should be taken as +IR in the
equations.
VT – (V1 + V2 + V3) = 0
ΣV = VT – V1 – V2 – V3 = 0
Consider Figure 1.19.
E1
E2
I
R1
R2
R3
Figure 1.19 Kirchhoff’s Voltage Law (KVL)
Figure 1.21 Potential drop
E1 + E2 = IR1 + IR2 + IR3
Example 3
Use Kirchhoff’s Voltage Law to determine
the voltage drop across R2 as shown in
Figure 1.22.
ΣE – ΣIR drops = 0 (around closed
loop)
ΣE = ΣIR
Σ potential rises = Σ potential drops
An arbitrary current direction for each
branch current is assumed. If the solution
for the current is negative, then the
direction of that current is opposite to the
direction assumed.
R1
R2
R3
2V
V2
3V
6V
Figure 1.22 Resistors in a series circuit
Apply the following rules:
• Refer to Figure 1.20. If the direction
is from negative to positive, assume
the voltage is positive. This is called
potential rise. For example, if the
branch AB is traced from A to B,
then the drop across it must be taken
as a potential rise and as +IR in the
equations. If the branch is traced from
B to A, then it should be taken as –IR in
the equations.
Solution
ΣV = 0
6 – 2 – V2 – 3 = 0
V2 = 1 V
[VT – V1 – V2 – V3]
Example 4
Refer to Figure 1.23. Write an equation by
using Kirchhoff’s Voltage Law around the
loop ABCDA.
10
Equation 1 × 5:
5I1 + 10I2 = 30 ... Equation 3
Equation 3 + Equation 2:
5I1 + 10I2 = 30
11I1 – 10I2 = 12
∴16I1 = 42
∴ I1 = 42 16
= 2,625 A
2,625 + 2I2= 6
2I2 = 6 – 2,625
= 3,375
∴ I2 = 3,375 2
= 1,688 A
Figure 1.23 Loop ABCDA
Solution
–I1R1 + E2 – I2R2 – I3R3 – I4R4 + E1 = 0
∴ E1 + E2 = I1R1 + I2R2 + I3R3 + I4R4
Current through 10 Ω = I1 – I2
= 2,625 – 1,688
= 0,937 A
Example 5
Use Kirchhoff’s Law to calculate the current
in each branch of the network as shown
in Figure 1.24. Use the loops ABEFA and
BCDEB to write down the equations relating
to the current. Solve these equations to find
the value of these currents.
Example 6
Apply Kirchhoff’s Law to Figure 1.25 to
determine the value of the current in each
branch of the circuit.
Figure 1.25 Circuit with loops
Figure 1.24 Network with loops
Solution
Solution
For loop ABEFA:
For loop ABEFA:
(12 – 6) – (I1 + 2I2)= 0
6 – I1 – 2I2 = 0
I1 + 2I2 = 6 ... Equation 1
( 15 – 12) – [I1 + 0,2(I1 – I2)] = 0
3 – [I1 + 0,2I1 – 0,2I2] = 0
3 – 1,2I1 + 0,2I2 = 0
1,2I1 – 0,2I2 = 3 ... Equation 1
For loop BCDEB:
–I1 – 10(I1 – I2) + 12 = 0
–I1 – 10I1 + 10I2 + 12 = 0
11I1 – 10I2 = 12 ... Equation 2
11
For loop ACDFA:
2. Write down the equation for
current I1 in Figure 1.27.
6 – (I1 + 0,2I2) = 0
I1 + 0,2I2 = 6 ... Equation 2
Equation 1 + Equation 2:
I2
I1
1,2I1 – 0,2I2 = 3
I1 + 0,2I2 = 6
∴ 2,2I1 = 9
I1 = 9 2,2
= 4,091 A
I3
Figure 1.27 Current flow through a parallel
connection
3. Two batteries A and B are
connected in parallel to supply a
load resistance of 4 Ω as shown in
Figure 1.28. Use Kirchhoff’s Law
to determine the:
a) value of the current I1 and I2
b) voltage across the load resistor.
Substitute I1 in Equation 2:
4 ,091 + 0,2I2= 6
0,2I2 = 1,909
1,909
I2 =
0,2
= 9,545 A
∴ I1 – I2= 4,091 – 9,545
= –5,45 A
Exercise 1.4
1. Figure 1.26 shows a network
junction. Determine the current I1.
12 A
7A
Figure 1.28 Batteries connected in parallel
I1
4. A generator with an emf of 30 V
and an internal resistance of 2 Ω is
used to charge two batteries A and
B, each with an emf of 24 V and
20 V and with internal resistance
of 0,2 Ω and 1 Ω respectively. Use
Kirchhoff’s Law to determine the
value of the current:
a) supplied by the generator
b) through battery A
c) through battery B.
6A
Figure 1.26 Network junction
12
E = Pt
Unit 1.4 Power, energy,
efficiency and Joule’s Law
where:
E = energy in joules (J)
P = power in watts (W)
t = time in seconds (s).
Power
Power is the rate at which work is done
or energy is used. Power is equal to the
product of voltage and current, and is
measured in watt (W). Power is calculated
using the formula:
When a current flows through a resistor,
heat is produced. This leads to Joule’s
Law.
P = VI ... Equation 1
Joule’s Law
where:
P = power in watts (W)
V = voltage in volts (V)
I = current in amperes (A).
Joule’s Law states that the heat
produced by the current flow through
a conductor is directly proportional to
the square of the current, the resistance
and time.
The above equation can be arranged as:
V = P ... Equation 2
I
P
I = ... Equation 3
V
Written as a mathematical expression,
Joule’s Law is as follows:
From Equation 1, substitute V = IR:
P = VI
= IRI
= I²R
where:
Q = amount of heat generated in joules (J)
I = current flowing in the circuit in
amperes (A)
R = resistance in ohms (Ω)
t = time in seconds (s).
Q = I²Rt
From Equation 1, substitute I = V :
R
P = VI
=V×V
R
= V²
R
Example 1
An electric kettle consumes 1,8 megajoules
(MJ) when connected across a 230 V
supply for 30 minutes. Calculate the
current drawn from the supply and the
power rating of the kettle.
Energy
Energy is the ability to do work. When
work is done on an object, it gains energy.
If work is done by an object, it loses energy.
The unit for energy is the joule (J).
Given: E = 1,8 MJ; V = 230 V; t = 30 min.
Solution
[E = Pt]
P =E
t
6
= 1,8 × 10
30 × 60
= 1 000 W or 1 kW
Electric energy is the result of power
developed over a certain period of time.
Energy is calculated using the formula:
13
Since P= VI
I =P
V
= 1 000
230
= 4,348 A
Unit 1.5 Factors influencing
the resistance of a material
Factors that influence the resistance
of a material
Length (l)
The longer a conductor, the greater will be
its resistance. In other words, resistance is
directly proportional to the length (l) of
the conductor. See Figure 1.29. The graph
in Figure 1.30 shows the relationship
between resistance and the length of the
conductor. The length of a wire therefore
has an influence on its ability to conduct
electricity. The longer the wire, the more
difficult it is for the current to flow.
Efficiency
The efficiency rating is a measure of how
effectively a device uses the electricity
required to operate it. Efficiency (ŋ) is
therefore the ratio of the output of an
electric machine or circuit to its input
expressed as a percentage (%).
Example 2
If a motor requires a 150 W input and
delivers only 100 W output, then the
efficiency of the motor is calculated as
follows:
10 m
Efficiency ŋ = output × 100
input
= 100 × 100
150 1
= 66,7%
A
l1 > l2
R1 > R2
Exercise 1.5
5m
B
1. State Joule’s Law and give the
mathematical expression.
2. A geyser with a power rating of 3 kW is connected across a 230 V
supply. Calculate the:
a) current drawn from the supply
b) resistance of the geyser
element
c) energy consumed after 1 hour
in MJ.
3. The input power to an electric
motor is 10 kW. If the efficiency
of the motor is 90%, calculate the
output power.
Resistance (Ω)
Figure 1.29 Conductor A will have more
resistance than conductor B.
Length (m)
Figure 1.30 The longer the wire, the greater its
resistance.
14
across two opposite sides of a cube of
the material at a temperature of 20 °C.
Resistivity (ρ) is measured in ohm·metres
(Ω·m). Different materials have different
values of resistivity or specific resistance.
In other words, resistance is directly
proportional to resistivity, as shown in
Figure 1.33.
Cross-sectional area (A)
The larger the cross-sectional area of a
conductor, the less resistance it will have.
In other words, resistance is inversely
proportional to the area (A) measured
in m². If the cross-sectional area of a
conductor is doubled and the length stays
the same, its resistance will be halved. The
graph in Figure 1.31 shows the relationship
between resistance and cross-sectional
area.
m
10
m
aluminium
A
Resistance (Ω)
m
10
B
m
copper
ρ1 > ρ 2
R1 > R2
Figure 1.33 Conductor A will have more
resistance than conductor B.
Temperature
When the temperature of a material
changes, its resistivity and resistance also
change. See Figure 1.34. The effect of
temperature on different materials is as
follows:
• The resistance of most metals increases
as their temperature increases.
• The resistance of carbon, insulators and
semiconductor materials decreases as
the temperature increases.
• The resistance of alloys increases as
their temperature increases but the rate
of increase is not significant.
Cross-sectional area (m2)
Figure 1.31 The thicker the wire, the less its
resistance.
10 m
A
10 m
diameter
= 1 mm
B
diameter
= 10 mm
Figure 1.32 Conductor A will have more
resistance than conductor B.
Resistivity (ρ)
The resistivity of a material is the resistance
per unit length. The resistance is measured
Figure 1.34 Effect of temperature change of
different materials
15
Figure 1.35 shows a graph that is obtained
experimentally. Of interest on this graph
is the part near –273 °C or absolute zero.
Absolute zero is the temperature at which
a substance has no heat and the movement
of molecules stops. At this temperature,
the resistance of a conductor approaches
0 Ω. This condition of 0 Ω is called
superconductivity.
Formula for the relationship between
resistance and the dimensions of a
material
We can summarise factors that influence
the resistance of a material as follows:
• R is directly proportional to length (R ∝ l).
• R is directly proportional to resistivity
(R ∝ ρ).
• R is inversely proportional to the area
1 R∝ .
A
non-linear
Resistance (Ω)
R2
The formula that expresses the relationship
between resistance and the dimensions of
a material is therefore:
R1
non-linear
–273°
T (absolute zero)
t1
R = ρl
A
t1
Temperature (°C)
Figure 1.35 The resistance of most metals
increases with temperature and is 0 Ω at
absolute zero.
where:
R = electrical resistance of a uniform
sample of the material in ohms (Ω)
ρ = static resistivity of the material in ohm·metres (Ω·m)
l = length of the sample of material in
metres (m)
A = cross-sectional area of the sample of
material in square metres (m2).
Table 1.1 shows some typical resistivity
values measured at room temperature.
Good conductors have low resistivity
whereas good insulators have a high
resistivity.
Type of material
Resistivity in Ω·m
Copper
1,7 × 10
Aluminium
2,6 × 10 –8
Carbon (graphite)
10 × 10 –8
Glass
1 × 1010
Mica
1 × 1013
Gold
2,36 × 10 –8
Silver
1,58 × 10 –8
Lead
22 × 10 –8
Example 1
Calculate the resistance of a copper
conductor with a length of 1 km and a
cross-sectional area of 10 × 10–6 m². Take
the resistivity of copper to be 0,0173 µΩ·m.
–8
Given: l = 1 km = 1 000 m; A = 10 × 10–6 m²;
ρ = 0,0173 × 10–6 Ω·m²
Solution
R = ρl
A
–6
× 1 000
= 0,0173 × 10 –6
10 × 10
= 1,73 Ω
Table 1.1 Typical resistivity values at room
temperature
16
Sometimes the diameter or radius is given
and the area must then be calculated.
Area if diameter is given:
A = πd² where d is the diameter in m.
4
Note
1 km = 1 000 m
Voltage drop across copper conductor
= voltage drop across aluminium
conductor: = 77,945 V
Voltage drop across copper conductor:
= Icu Rcu
77,945 = 55 × Rcu
R = 77,945
cu
55
= 1,417 Ω
1 mm² = 1 × 10–6 m²
Area if radius is given:
A = πr² where r is the radius in m.
Example 2
A 1 km copper conductor is connected
in parallel with an aluminium conductor
of the same length. When a current of
90 A is passed through the combination,
it is found that the current through the
copper conductor is 55 A. If the radius
of the aluminium conductor is 2 mm,
calculate the:
a) voltage drop across the conductors
b) diameter of the copper conductor.
ρ l
b) Rcu = cu cu
A
cu
1,417 =0,018 × 10–6 × 1 000
a
∴ a = 0,018 × 10–6 × 1 000
1,417
= 1,27 × 10–5 m²
But:
A = πd²
4
πd² = 1,27 × 10–5
4
–5
∴ d² =1,27 × 10 × 4
π
–5
∴ d = √1,27 × 10 × 4
π
= 4,02 × 10–3 m
Take the resistivity of copper and
aluminium as 0,018 µΩ·m and 0,028 µΩ·m
respectively.
Given: IT = 90 A; Icu = 55 A; Ial = (90 – 55)
= 35 A; ral = 2 mm = 2 × 10–3 m; l = 1 km
Solution
a)
Voltage drop across aluminium
conductor:
= Ial Ral
= 35 × 2,227
= 77,945 V
Exercise 1.6
Aal = πr²
= π(2 × 10–3)²
= π(22 × 10–6)
= 1,257 × 10–5 m²
1. List the four factors that influence
the resistance of a material.
2. Two wires, A and B, are of the
same length and are made of the
same material. However, wire A
has four times the resistance of
wire B. How many times greater is the diameter of wire B than that
of wire A?
Let the resistance of aluminium = Ral
ρ l
Ral = al al
Aal
= 0,028 × 10–6 × 1 000
1,257 × 10–5
Ral = 2,227 Ω [not rounded up here]
17
Silver, aluminium and copper all have a
positive temperature coefficient. In these
cases, when the temperature increases, the
resistance increases.
3. Calculate the resistance of a
copper conductor with a length of
5,75 km and a cross-sectional area
of 10 × 10–6 m². Take the resistivity
of copper to be 0,0173 µΩ·m.
4. A 500 m conductor has a
resistance of 10 Ω. Calculate the
resistance if the length is 2 km.
5. Calculate the resistance of 1 km of
copper cable which has a diameter
of 10 mm. Take the resistivity of
copper to be 1,7 × 10–8 Ω·m.
6. An aluminium conductor, 2 km
long, is connected in parallel with
a copper conductor of the same
length. When a current of 150 A is
passed through the combination,
it is found that the current
through the copper conductor
is 100 A. If the diameter of the
aluminium conductor is 2 mm,
calculate the:
a) voltage drop across the
aluminium and copper
conductors
b) diameter of the copper
conductor if the resistivity of
copper is 0,017 µΩ·m and that
of aluminium is 0,0027 µΩ·m.
Negative temperature coefficient of
resistance
With some materials, when the temperature increases, the resistance decreases.
Insulating materials and carbon have a
negative temperature coefficient, because
when their temperature increases, the
resistance decreases.
Low coefficient of resistance
In the case of alloys such as nichrome, the
variation in resistance with a change in
temperature is so small that it is negligible.
These materials are used as elements for
geysers, kettles and stove plates.
Table 1.2 below shows some typical values
of temperature coefficients of resistance
measured at 0 degrees Celsius (°C).
Temperature coefficient of resistance
The temperature coefficient of resistance
is the effect that the temperature has
on the resistance of a material. The
temperature coefficient is defined as the
increase in resistance in ohm per degree
rise in temperature per ohm of original
resistance.
Material
Temperature
coefficient per °C
Copper
0,0043
Aluminium
0,0038
Nickel
0,0062
Constantan
0
Eureka
0,00001
Carbon
–0,00001
Tungsten
0,0045
Molybdenum
0,0046
Table 1.2 Some typical values of temperature
coefficients of resistance measured at 0 °C
Note
Positive temperature coefficient of
resistance
With most metals, when the temperature
increases, the resistance also increases.
Such materials are said to have a positive
temperature coefficient of resistance.
The negative sign for the temperature
coefficient of carbon means that the
resistance of carbon decreases when
temperature increases.
18
Therefore, for a change in temperature
between any two values, the formula is:
Calculating final resistance
The formula to calculate the final resistance
of a material after a rise in temperature
from 0 °C is given by:
R (1 + α0T2)
R2 = 1
(1 + α0T1)
Rt = R0(1 + α0 T)
where:
R1 = initial resistance
R2 = final resistance
T1 = initial temperature
T2 = final temperature.
where:
Rt = final resistance in ohms (Ω)
R0 = initial resistance at 0 °C in ohms
α0 = temperature coefficient at 0 °C
T = final temperature in °C.
Example 4
A piece of tungsten filament wire has a
resistance of 175 Ω at 20 °C. Calculate its
resistance at 350 °C. Take α0 = 4,5 × 10 –3
per °C.
Example 3
The resistance of a coil of copper wire at
0 °C is 50 Ω. Calculate the resistance of
the coil at 60 °C. Take the temperature
coefficient of resistance of copper wire at
0 °C as 4,3 × 10–3 per °C.
Given: R1 = 175 Ω; T1 = 20 °C; T2 = 350 °C;
α0 = 4,5 × 10–3
Given: R0 = 50 Ω; α0 = 4,3 × 10–3; T = 60 °C
Solution
Solution
In case R0 is not given, the relationship
between the known resistance R1 at T1 °C
and the unknown resistance R2 at T2 °C
can be found as follows:
R (1 + α0T2)
R2 = 1
(1 + α0T1)
–3
× 350])
= 175(1 + [4,5 × 10
–3
(1 + [4,5 × 10 × 20])
= 175(1 + 1,575)
(1 + 0,09)
= 175(2,575)
1,09
= 413,42 Ω
R2 = R0(1 + α0 T2) ... Equation 1
R1 = R0(1 + α0 T1) ... Equation 2
When the temperature coefficient is not
given at 0 °C, the formula below is used:
where:
R1 = initial resistance
R2 = final resistance
T1 = initial temperature
T2 = final temperature.
Rt = Rθ[1 + αθ (T – θ)]
Rt = R0(1 + α0 T)
= 50(1 + [4,3 × 10–3 × 60])
= 62,9 Ω
Example 5
The field coils of a generator have a
resistance of 100 Ω at 25 °C. After running
at full load the resistance increases to
150 Ω. Find the temperature of the coils if
the temperature coefficient of resistance is
0,004 per °C at 25 °C.
Divide Equation 2 by Equation 1:
R (1 + α0 T2)
R2
= 0
R1R0(1 + α0 T1)
Given: Rt = 150 Ω; Rθ = 100 Ω; αθ = 0,004;
θ = 25 °C
19