Design, Engineering
and Technology
Electronic and Electrical
Fundamentals
Course Tutor Guide
[INTERMEDIATE 2]
David Campbell
James Watt College


2
Acknowledgements
This pack was developed by the Colleges Open Learning Exchange
Group (COLEG) on behalf of Learning and Teaching Scotland.
The author would like to thank the following for their assistance in
developing these materials:
Marian Lever for her advice and patience
SQA for the inclusion of general information from Higher Still
Material.
CONTENTS
Introduction to the course
1
Course structure
Additional time for course
Aims of the course
Core skills
Progression and related study
1
1
1
2
2
Course content
Summary of course content
Unit 1: Electrical Fundamentals (Int 2)
Unit 2: Semiconductor Applications: An Introduction (Int 2)
Unit 3: Combinational Logic (Int 2)
2
2
2
3
3
External assessment information
How students will be assessed
When and where students will be assessed
What students have to achieve
Grade descriptions
External assessment: appeals
4
4
4
4
5
5
Preparation for external assessment
Integration
Learning and teaching approaches
Consolidation
Exemplar materials
6
6
6
6
6
Electrical Fundamentals: exemplar questions and solutions
7
Semiconductor Applications: exemplar questions and solutions
Combinational Logic: exemplar questions and solutions
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INTRODUCTION TO THE COURSE
This course introduces the field of electronic and electrical engineering
at Intermediate 2 level. It achieves this through the study of basic
electrical principles and introductory analogue and digital processing.
The course provides sufficient electrical and electronic knowledge to
allow candidates to progress on to a wide range of technolog ical areas
at Higher level.
Course structure
The course comprises these units:
D132 11
D133 11
D134 11
Electrical Fundamentals (Int 2)
Semiconductor Applications: An
Introduction (Int 2)
Combinational Logic (Int 2)
1 Credit (40 hours)
1 Credit (40 hours)
1 Credit (40 hours)
Additional time for course
In common with all courses, this course includes 40 hours over and
above the 120 hours for the component units. This is for induction,
extending the range for learning and teaching approaches, support ,
consolidation, integration of learning and preparation for external
assessment. This is the responsibility of each individual centre.
Aims of the course
Specifically this course aims to provide the candidate with improved
technological capability in the field of electronic and electrical
engineering through the study of basic electrical principles and
introductory analogue and digital processing.
In a more general context it also aims to contribute to the general
education and personal development of the candidate and in particular
foster a greater awareness of what electricity is, how it may be
generated and how it may be utilised to perform processing tasks.
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INTRODUCTION TO THE COURSE
Core skills
This course contributes to the following core skills unit:
Core skills components for the course Using Number Int 2
Progression and related study
The course serves as a bridge to Higher Electrical Engineering, Higher
Electronics or Higher Mechatronics and will also be an asset in the
wider scientific and engineering fields.
Course content
The course content for Electrical and Electronic Fundamentals is
summarised below.
Summary of course content
Unit l: Electrical Fundamentals (Int 2) – 40 hours
This unit has been designed to introduce candidates to the basic
electrical engineering principles and laws. It covers the relationships
between current, voltage, resistance, power and energy in a dc network.
It also considers the factors relating to the generation of electricity as
an ac (sinusoidal) waveform.
Outcomes
1. Determine the current, voltage and resistance relationships in a
resistive dc network.
2.
Solve problems on power and energy in dc resistive systems.
3.
Determine the relationships between the factors relating to the
force acting on a current-carrying conductor situated in a magnetic
field.
4.
Determine the factors that relate to the generation of a sinusoidal
voltage waveform.
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INTRODUCTION TO THE COURSE
Unit 2: Semiconductor Applications: An Introduction (Int 2) – 40
hours
This unit has been designed to introduce candidates to analogue
electronics. It covers the analysis of diode parameters and circuits,
including power applications using thyristors and triacs. It also
considers the analysis of amplifier circuits using discrete transistors
(bipolar and field effect transistors) as well as operational amplifiers.
Outcomes
1. Interpret the operation of semiconductor diode circuits.
2.
Outline the use of power control devices.
3.
Interpret the operating conditions of a single-stage resistanceloaded small-signal amplifier.
4.
Investigate operational amplifier circuits.
Unit 3: Combinational Logic (Int 2) – 40 hours
This unit has been designed to introduce candidates to digital
electronics. It considers binary representation, basic logic gates and the
analysis and synthesis of simple combinational circuits. It offers an
introduction to binary systems that is suitable for Intermediate 2. It
outlines the function of basic gates and allows for simple design and
construction.
Outcomes
1.
Perform simple binary operations.
2.
Identify the function of logic gates.
3.
Assemble and investigate a combinational logic circuit.
4.
Solve combinational logic system problems.
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INTRODUCTION TO THE COURSE
External assessment information
How students will be assessed
To achieve the National Qualification in:
Electronic and Electrical Fundamentals (Intermediate 2)
Students must pass all the unit assessments within the course and pass
an external assessment.
D132 11
D133 11
D134 11
Electrical Fundamentals (Int 2)
Semiconductor Applications: An
Introduction (Int 2)
Combinational Logic (Int 2)
1 Credit (40 hours)
1 Credit (40 hours)
1 Credit (40 hours)
When and where students will be assessed
Currently there is only one external examination diet for National
Qualifications: in the centre from which they are registered.
For the Open Learner who lives some distance from the centre, an
invigilation system may be set up at a recognised support centre local to
the student.
What students have to achieve
 The external assessment will comprise one written examination
paper.
 The time allocated for the question paper will be 2 hours and 30
minutes. The paper will sample across the Electronic and Electrical
Fundamentals course. The paper will be divided into two parts. The
first part will comprise a mandatory section within which al l
questions must be answered. This section will be worth a total of 50
marks. The questions in this section will cover all areas of the
course but will be short in nature. The second section provides a
choice of two from three questions. These questions are generally
longer and require greater knowledge and analytical skills than the
mandatory section.
 The questions will test the candidate’s ability to integrate knowledge
and skills acquired across the component Units.
 The question paper will be worth 100 marks.
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INTRODUCTION TO THE COURSE
Grade descriptions
External assessment will provide the basis for grading attainment in the
course award. Grades are intended to assist candidates,
teachers/lecturers and users of the certificate and to help establish
standards when question papers are being set.
The grade of award A, B or C will be based on the total score obtained
in the examination paper and students are advised to contact you for
full descriptions of the nature of the achievement that is required for the
award of these grades in their course assessment.
External assessment: appeals
Where a student fails to attain the estimated grade, it will be open to a
centre to submit an appeal on his or her behalf. Before doing so, your
centre should establish that it has evidence that will support an
improved award. The evidence may comprise Unit assessments, class
tests and/or the outcomes of a prelim, and the evidence should relate to
as much of the course as possible.
In evaluating evidence, examiners find it useful to know the conditions
under which it has been produced, for example whether the assessed
activity was seen or unseen, open or closed book, carried out under
supervision or elsewhere. Examiners find evidence produced under
supervision more compelling. Centres will therefore be asked to assist
examiners by identifying the conditions under which each item of
evidence was produced. In considering an assessment appeal, the
examiners will scrutinise the evidence or other work produced for
internal assessment and, where a centre has run preliminary
examinations, candidate responses to these.
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INTRODUCTION TO THE COURSE
Preparation for external assessment
Integration
Candidates should be encouraged to link the information from each unit
in order to develop an integrated approach to Electronic and Electrical
Fundamentals. This can be difficult to achieve as students normally
progress from one Unit to the next but it should be a key feature of the
preparation for external assessment.
Learning and teaching approaches
A variety of learning and teaching approaches should be used to sustain
interest and have maximum effect on learning. Extensive use should be
made of simulation. In particular pre-prepared circuits should be
provided which allow students to interact with the circuits they are
studying and develop familiarity and understanding of their operation.
Practical exercises, again using pre-prepared circuits, must also be
undertaken. Tutorial exercises reflecting typical unit and external
examination questions must also be worked through. To this end the
exemplar questions and solutions provided in this package should
provide significant help.
Consolidation
Consolidation of the subject matter can be achieved by the type of
integration highlighted above applied in various contexts, by utilisi ng
part of the additional 40 hours available as part of the course.
Exemplar materials
In order to support preparation for the external examination a set of
exemplar exercises have been developed. These exercises allow
candidates to utilise and apply knowledge in contexts that are similar to
that of a final examination. Worked solutions have also been provided
which, as well as providing the correct answer, give the candidate a
greater insight into how a solution is formulated.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
ELECTRICAL FUNDAMENTALS
Outcome 1
?1
(a)
What is the terminology used to describe how the resistors are
connected in the networks shown in Fig SAQ 1(i)?
(b)
Calculate the total resistance between nodes X and Y in circuits A and B.
1 K
10 
X
X
R1
R1
R2
4 K
40 
R2
R3
Y
Y
5 K
A
B
Fig SAQ 1(i)
(c)
If
Y


X
a dc voltage source of 10 V is applied between the terminals X and
in each circuit, as shown in Fig SAQ 1(ii), calculate in each case:
the current that would flow from the source
the potential difference (voltage) across each resistor.
10 
1 K
X
R1
R1
R2
R2
40 
R3
Y
Y
5 K
A
B
Fig SAQ 1(ii)
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4 K
ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?1
Solution
(a)
The resistors within the two networks given in Fig SAQ 1(i) are
connected in series. When connected in series resistors form a
‘string’ in which there is only one path for current to flow along.
1 K
10 
X
X
R1
R1
4 K
R2
40 
R2
R3
Y
Y
5 K
A
B
Fig SAQ 1(i)
(b)
The total resistance of resistors connected in series is:
R TOTAL = R 1 + R 2 + R 3 + …
Hence for circuit A the total resistance between nodes X and Y is:
R TOTAL = 10 + 40 = 50 
For circuit B the total resistance between nodes X and Y is:
R TOTAL = 1 + 4 + 5 = 10 K
(c)
For circuit A the total resistance between X and Y was calculated
to be 50 . This means that a total of 50  would appear across
the terminals of the 10 V, i.e. as far as the source is concerned the
circuit connected across its terminals is:
I
X
10 V
RTOTAL
50 
Y
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
As we know the voltage between nodes X and Y, and we know the resistance
between nodes X and Y, we can calculate the current that would flow through
the resistor, and hence from the source, simply by applying Ohm ’s law:
I=
V 10
=
= 0.2 A = 200 mA
R 50
This argument can equally be applied to circuit B:
I
X
10 V
10 K
RTOTAL
Y
Again, as we know the voltage between nodes X and Y, and we know
the resistance between nodes X and Y, we can calculate the current that
would flow through the resistor, and hence from the source, by applying
Ohm’s law:
I=
V
10
=
= 1 mA
R 10 K
As the resistors are connected in series the current that flows from the source
will flow through each of the resistors in the path. Applying Ohm ’s law we
can calculate the voltage between the ends of each resistor in circuit A:
V1 = I × R = 200 mA × 10 Ω = 2 V
V2 = I × R = 200 mA × 40 = 8 V
I = 0.2 A
V1 = 2 V
X
R1
10 
10 V
R2
40 
V2 = 8 V
Y
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
Similarly, for circuit B:
V1 = I × R1 = 1 mA × 1 K = 1 × 10 –3 × 1 × 103 = 1 V
V2 = I × R2 = 1 mA × 4 K = 1 × 10 –3 × 4 × 103 = 4 V
V3 = I × R3 = 1 mA × 5 K = 1 × 10 –3 × 5 × 103 = 5 V
1 = 1 mA
V1 = 1 V
X
R1
1 K
10 V
R2
4 K
V2 = 4 V
R3
5 K
Y
V3 = 5 V
In both circuits notice how the voltage applied between nodes X and Y
is distributed across all the resistors in the series path. Also notice the
direction of the voltage arrows; the arrow always points in the direction
of higher potential. For example, in circuit B node X is at 10 V. The
voltage between the ends of R 1 is 1 V, therefore the voltage at the node
connecting R 1 to R 2 is 9 V (10 V – 1 V). The voltage between the ends
of R 2 is 4 V, therefore the voltage at the node connecting R 2 to R 3 is 5 V
(9 V – 4 V). Finally the voltage between the ends of R 3 is 5 V,
therefore the voltage at the node connecting R 3 to the negative terminal
of the supply must be 0 V (5 V – 5 V). This is exactly what we would
expect as we are using a 10 V supply.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?2
(a)
What is the terminology used to describe how the resistors are
connected in the networks shown in Fig SAQ 2(i)?
(b)
Calculate the total resistance between nodes X and Y in circuits A
and B.
X
X
R1
10 
R1
10 
R2
40 
Y
R3
20 
R2
20 
Y
A
B
Fig SAQ 2(i)
(c)
If a dc voltage source of 10 V is applied between terminals X and
Y of circuits A and B, as shown in Fig SAQ 2(ii), calculate:
 the current that would flow from the source
 the current flowing through each resistor.
X
X
10 V
10 V
R1
10 
R1
10 
R2
40 
Y
R3
20 
R2
20 
Y
A
B
Fig SAQ 2(ii)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?2
Solution
(a)
The resistors in Fig SAQ 2(i) are connected in parallel. Resistors
are said to be connected in parallel when two or more resistors a re
individually connected between the same two points. A parallel
circuit provides more than one path for current to flow along.
(b)
The total resistance of resistors connected in parallel is:
1
1
1
1
=
+
+
+ ...
RTOTAL
R1
R2
R3
In the case of two resistors in parallel:
1
RTOTAL
=
RTOTAL =
1
1
R + R2
+
= 1
R1
R2
R1 × R2
R1 × R2
R1 + R2
This is a particularly useful version of the rule for parallel connected
resistors, which we will apply immediately to circuits A and B.
X
R1
10 
R2
40 
Y
In circuit A the ends of R 1 and R 2 are connected to the same two
points (nodes X and Y), so they are in parallel.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
The total resistance between nodes X and Y can be calculated
using the equation for two resistors connected in parallel:
RTOTAL =
R1 × R2
10 × 40
400
=
=
=8
R1 + R2
10 + 40
50
In circuit B the ends of R 1 , R 2 and R 3 are all connected to the same
two points (nodes X and Y) so they are also in parallel:
X
R1
10 
R2
20 
R3
20 
Y
There are two approaches to solving this problem. The first is to
apply the parallel resistance equation:
1
RTOTAL
1
RTOTAL
1
RTOTAL
=
1
1
1
+
+
R1 R2
R3
=
1
1
1
+
+
10
20
20
=
2+1+1
4
=
20
20
RTOTAL =
20
=5Ω
4
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
The second approach is to repeatedly use the two resistors
equation. To illustrate this consider circuit B again. Although all
three resistors are in parallel, we can break the problem into
simpler parts by only considering two of the resistors and working
out their equivalent resistance first.
X
Calculate the equivalent
resistance of two of the
resistors…we have
selected R2 and R3
R1
10 
R2
20 
R3
20 
Y
R4 =
R2 × R3
20 × 20 400
=
=
= 10 
R2 + R3
20 + 20
40
The circuit can now be re-drawn as:
X
R1
10 
R4
10 
Y
Again this circuit is simply two resistors in parallel so we apply
the equation once more:
RTOTAL =
R1 × R4 10 × 10 100
=
=
=5
R1 + R4 10 + 10
20
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
This is the same answer as we obtained before. It really doesn’t
matter which method is chosen as long as we get the correct answer.
There are two important points to note:

when resistors are connected in parallel the total equivalent
resistance is always less than that of the any of the resistors
when two resistors connected in parallel are the same their
equivalent resistance is exactly half: this only works for two
equal resistors.

The total resistance between nodes X and Y in circuit B is 5 .
X
R1
5
Y
(c)
As we know the total resistance between nodes X and Y we can easily
calculate the current that flows from the source in both examples.
In A the total resistance was calculated to be 8 . Hence by
applying Ohm’s law we can calculate the current that flows from
the source:
I =
VSOURCE
10
=
= 1.25 A
RTOTAL
8
I = 1.25 A
I = 1.25 A
X
X
10 V
10 V
R1
10 
Y
R2
40 
R1
8
Y
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
In B the total resistance was calculated to be 5 . Again by
applying Ohm’s law we can calculate the current that flows from
the source:
I =
VSOURCE
10
=
=2A
RTOTAL
5
I=2A
I=2A
X
X
10 V
10 V
R1
10 
R2
20 
R1
5
R3
20 
Y
Y
As we know the voltage between nodes X and Y, and every
resistor is connected between nodes X and Y, we can easily
calculate the current using Ohm’s law.
In circuit A:
I1 =
VXY
10
=
=1A
R1
10
I2 =
VXY
10
=
= 0.25 A
R2
40
I = 1.25 A
X
I1 = 1 A
I2 = 0.25 A
10 V
R1
10 
R2
40 
Y
I = 1.25 A
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
A few points to note here:
 I = I 1 + I 2 , i.e. the current splits and flows through each of the
resistors before rejoining and flowing into the negative
terminal
 notice the way that the current splits; the biggest part goes
through the 10  resistor and the smaller part goes through the
bigger resistor – this is what we would expect.
Again, for circuit B we know the voltage between nodes X and Y.
We also know that every resistor is connected between nodes X
and Y; hence we can easily calculate the current using Ohm ’s law:
I1 =
VXY
10
=
=1A
R1
10
I2 =
VXY
10
=
= 0.5 A
R2
20
I3 =
VXY
10
=
= 0.5 A
R3
20
I=2A
1A
X
I1 = 1 A
I2 = 0.5 A
I3 = 0.5 A
R1
10 
R2
20 
R3
20 
10 V
Y
I=2A
1A
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
Again a few points to note:
 I = I 1 + I 2 + I 3 , i.e. the current splits and flows through each of
the resistors before rejoining and flowing into the negative
terminal
 notice the way that the current splits: 1 A flows through the
10  resistor R 1 . The remaining current divides evenly
between R 2 and R 3 as both resistors have an equal value.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?3
For the circuit shown in Fig SAQ 3(i) calculate:
(a)
the total network resistance
(b)
the current drawn from the supply
(c)
the current flowing through R 1 , R 2 , R 3 and R 4
(d)
the voltage across R 1 , R 2 , R 3 and R 4 .
40 
20 
R1
E = 100 V
R2
60 
10 
R4
R3
Fig SAQ 3(i)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?3
Solution
(a)
In this example we have series and parallel resistors connected in
the same circuit. To more easily identify points in this circu it we
will label the nodes. Remember that a node is simply a continuous
conductor (like a piece of wire) within a network. This circuit has
four nodes, as shown below.
40 
N2
N3
R2
20 
N1
10 
60 
R1
R4
N4
R3
E = 100 V
From the circuit we can see that between nodes N2 and N3 we have
two resistors connected in parallel. Using the equation for two
resistors we can calculate the total resistance between these two nodes:
40 
N2
N3
R2
60 
R3
RN 23 =
R2 × R3
40 × 60
=
= 24 
R2 + R3
40 + 60
N2
24 
N3
RN23
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
The circuit can now be drawn as:
N2
N3
20 
24 
10 
R1
RN23
R4
N1
N4
E = 100 V
This is a simple series circuit. The resistance between N1 and N4
can easily be calculated:
R N14 = R 1 + R N23 + R 4 = 20 + 24 + 10 = 54 
54 
N1
RN14
N4
E = 100 V
The total network resistance is 54 
(b)
The current drawn from the supply can be calculated using Ohm ’s
law:
ISOURCE =
VSOURCE
100
=
= 1.85 A
RN14
54
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
(c), (d) The current through R 1 and R 4 is simply the source current.
This is best illustrated using the following representation:
I = 1.85 A
20 
24 
N2
RN23
R1
N1
10 
N3
R4
N4
E = 100 V
Of course if we know the current through R 1 and R 4 we can calculate
the voltage difference across them by applying Ohm’s law:
VR1 = VN12 = IS × R1 = 1.85 × 20 = 37 V
VR4 = VN34 = I S × R4 = 1.85 × 10 = 18.5 V
The voltage between N2 and N3 can now easily be calculated:
VN2 = 100 – 37 = 63 V
VN3 = 18.5
VN23 = VN2 – VN3 = 63 – 18.5 = 44.5 V
R1
I = 1.85 A
N1
20 
37 V
R4
RN23
N2
24 
44.5 V
N3
10 
18.5 V
N4
E = 100 V
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
As we now know the voltage between N2 and N3 we can calculate
the current through the parallel resistors connected between these
nodes by using Ohm’s law:
I2
40 
R2
N2
I3
N3
60 
R3
44.5 V
I2 =
V23
44.5
=
= 1.11 A
R2
40
I3 =
V23
44.5
=
= 0.74 A
R3
60
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?4
For the circuit shown in Fig SAQ 4(i) calculate:
(a)
the total network resistance
(b)
the current drawn from the supply
(c)
the current flowing through all the resistors
(d)
the potential difference across all the resistors.
E = 100 V
20 
20 
30 
R1
R3
R5
R2
60 
R4
80 
R6
Fig SAQ 4(i)
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50 
ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?4
Solution
(a)
As with SAQ 3 the first step in the solution of this problem is to
allocate nodes, as this helps with the analysis of the circuit. There
are five nodes in this circuit, N1 to N5:
20 
N1
20 
N2
R1
30 
N3
E = 100 V
R2
60 
N4
R5
R3
80 
R4
R6
50 
N5
R 5 and R 6 are connected in series:
R5 + R6 = 30 + 50 = 80 
The circuit may be re-drawn as:
20 
20 
N2
R1
N1
E = 100 V
N3
R3
R2
60 
R4
80 
80 
N5
R 4 and the 80  resistor are in parallel. The equivalent resistance
between N3 and N5 can now be calculated:
RN35 =
80 × 80
= 40 
80 + 80
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
20 
R3
R1
N1
E = 100 V
N3
20 
N2
RN35
60 
R2
40 
N5
R 3 and R N35 are in series, giving a total resistance of 60 
20 
N1
N2
R1
E = 100 V
R2
60 
60 
N5
The resistance between N2 and N5 may now be simplified to:
RN25 =
60 × 60
= 30 
60 + 60
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
20 
N2
R1
N1
E = 100 V
RN35
30 
N5
The total resistance between N1 and N5 is simply the series
connection of R 1 and R N35 , giving a total of:
RN15 = 20 + 30 = 50 
N1
E = 100 V
RN15
50 
N5
(b)
As we know the total circuit resistance, the current drawn from the
supply can be calculated using Ohm’s law:
IS =
VS
100
=
=2A
RN15
50
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
(c), (d) The current drawn from the source will flow through R 1 and so
the voltage across it will be 40 V. This current will split at N2
with 1 A flowing through each branch. Hence the voltage across
R 2 will be 60 V.
I1 = 2 A
R1
20 
N2
I3 = 1 A
N1
40 V
R3
20 
20 V
I2 = 1 A
RN35
40 
R2
60 
E = 100 V
60 V
N5
If we now replace R N35 with the remainder of the circuit then we
can see that I 3 will split evenly at N3 as the resistance offered by
each path at the junction is identical. The appropriate voltages can
then be calculated from these currents, as shown below.
I1 = 2 A
N1
R1
20 
N2
40 V
I3 = 1 A
I2 = 1 A
E = 100 V
60 V
R3
20 
20 V
R2
60 
40 V
N3
I5 = 0.5 A
I4 = 0.5 A
R5
30 
N4
15 V
R4
80 
25 V
N5
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I6 = 0.5 A
R6
50 
ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
Outcome 2
?1
Write down the three expressions for power dissipated in a resistor.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?1
Solution
The three expressions for power dissipated in a resistor are:
P=V ×I
V2
R
2
P=I ×R
P=
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?2
The bulb shown in Fig SAQ 2(i) is rated at 40 W. Calculate:
(a)
the amount of current that will flow through the bulb
(b)
the resistance of the bulb
(c)
how much energy will be used if the bulb is turned on for 10
hours.
12 V
40 W
Fig SAQ 2(i)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?2
Solution
(a)
In this question we know that the voltage across the bulb is 12 V
and that the power dissipated is 40 W. Hence we use the first
equation to find the current I:
P=V ×I
P
40
I=
=
= 3.33 A
V
12
(b)
The resistance of the bulb filament can be found from:
R=
(c)
V
12
=
= 3.6 
I
3.333
The power rating is 40 W, which means 40 J of energy per second.
To solve this problem all we have to do is calculate how many
seconds there are in 10 hours, then multiply by 40.
10 hours = 1 × 60 min = 10 × 60 × 60 s = 36 000 s
Hence the amount of energy used is:
40  36 000 = 1.4  10 6 = 1.4 MJ
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?3
The circuit shown in Fig SAQ 3(i) is used to test banks of 10 LEDs at
one time. If the voltage across all the diodes is 2 V and each diode
takes 10 mA, calculate the value of the resistance needed to operate this
circuit and the required power rating of the resistor (0.125 W, 0.25 W,
0.5 W, 1 W, 2 W).
R
5V
2V
Fig SAQ 3(i)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?3
Solution
For the circuit given there are 10 LEDs, each drawing 10 mA therefore
the total current flowing through the resistor is 10  10 mA = 100 mA.
The supply voltage is 5 V, therefore the voltage across the resistor is 5
– 2 = 3 V.
We can now calculate the power dissipation requirements using the
equation P = VI.
P = V × I = 3 × 100 × 10–3 = 300 mW
Hence the minimum power rating of the resistor is 0.5 W = 500 mW.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?4
(a)
Calculate how much energy is used in one kWh.
(b)
If an electricity company charges 7p per kWh calculate the cost of
a 60 W light bulb being switched on for 8 hours.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?4
Solution
(a)
1 kW = 1000 W = 1000 J per second
1 hour = 3600 seconds
hence 1 kWh = 1000  3600 = 3.6  10 6 = 3.6 MJ
(b)
60 W = 60 J per second
8 hours = 8  60  60 = 28 800 seconds
Hence total energy used = 60  28 800 = 1 728 000 = 1.728 MJ
3.6 MJ of energy costs 7p
therefore 1.728 MJ costs
1.728
× 7 = 3.36p
3.6
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?5
A kettle element is rated at 240 V rms, 3 kW.
(a)
(b)
Determine:
(i)
the rated current of the element
(ii)
the resistance of the element
A full kettle of water takes 3 minutes to boil. At 7p per kWh
calculate how much this costs.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?5
Solution
(a)
(i)
Using the equation P = VI we can calculate the rated current:
I rms =
P
3 kW
3 × 103
=
=
= 12.5 A
Vrms
240 V
240
(ii) The resistance of the element can be found using Ohm’s law:
R=
(b)
Vrms
240
=
= 19.2 
I rms
12.5
From previous calculations 1 kWh = 3.6 MJ, which costs 7p.
A 3 kW kettle will use 3000 J in 1 second.
In 3 minutes it will use 3  60  3000 = 540 000 J, which costs
540 000
× 7 = 1.05p
3.6 × 106
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?6
A heating element connected to a 240 V supply takes a current of 5 A.
Calculate:
(a)
the resistance of the element
(b)
the power dissipated.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?6
Solution
V
240
=
= 48 
I
5
(a)
R=
(b)
R = V × I = 240 × 5 = 1200 W
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?7
An electric motor operates from a 240 V supply and takes a current of
40 A. Determine the energy used in 24 hours in:
 joules
 kilowatt hours.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?7
Solution
The power rating of the motor is:
P = V  I = 240  40 = 9600 W = 9600 J s –1
Number of seconds in 24 hours is:
24  60  60 = 86 400 s
Hence total energy used in 24 hours = 86 400  9600 = 829.440 MJ
1 kWh = 3.6 MJ, hence motor will use
829.440
= 230.4 kWh
3.6
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
Outcome 3
?1
(a)
Define the following terms:
 magnetism
 flux
 flux density
(b)
What is the name given to materials that can be used as permanent
magnets?
(c)
Consider the following statements and indicate whether they are
true or false:
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?1
Solution
(a)
 Magnetism is defined as a force that attracts materials; these
materials are known as ferromagnetic materials.
 As the force acting on a magnetic material is invisible, it is
necessary to employ some means by which it may be expressed and
visualised. This force can be visualised in terms of imaginary lines
of force known as lines of flux.
 Flux density is the measurement of the number of lines of flux in a
given area.
(b)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?2
(a)
A magnetic field of cross-sectional area 500 cm 2 has a total
magnetic flux of 25 mWb. Calculate the flux density.
(b)
A magnetic circuit of CSA (cross sectional area) 80 mm 2 has a
total flux of 120 Wb. Determine the flux density of the circuit.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?2
Solution
(a)
When calculating flux density care has to be taken that the correct
units are being used.
Area given = 500 cm 2
This needs to be converted into m 2 .
1 m 2 = 100  100 = 10 000 cm 2
(b)
Hence 500 cm 2 =
500
= 0.05 m 2
10 000
Flux density  =
flux
25 mWb
25 × 10 –3
=
=
= 0.5  Wb m –2
2
area
0.05
0.05 m
To solve this problem we must convert mm 2 into m 2 :
1 m 2 = 1000 mm × 1000 mm = 1 × 106 mm 2
80 mm 2 =
 =
80 mm 2
= 80 × 10 –6 m 2
6
2
1 × 10 mm
flux
120  Wb
120 × 10 –6
=
=
= 1.5 Wb m –2
–6
2
–6
area
80 × 10 m
80 × 10
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?3
(a)
Sketch the direction of the magnetic flux around the current carrying conductors shown in Fig SAQ 3(i).
Fig SAQ 3(i)
(b)
For the parallel current-carrying conductors given in Fig SAQ 3(ii)
show the direction of the resulting force between them. For the
parallel current carrying conductors given in Fig SAQ 3(ii) show
the direction of the resulting force between them.
A
B
C
Fig SAQ 3(ii)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?3
Solution
(a)
A simple way to remember the direction of lines of magnetic flux
around a current-carrying conductor is to curl the fingers of your
right hand and point your thumb outward. The direction of your
thumb represents the direction of ‘conventional’ current and the
curl of your fingers represents the direction of flux. Remember
that a dot is like the tip of an arrow coming towards you and a
cross is like the tail feather of an arrow moving away from you,
i.e. a dot means current is flowing out of the paper and a cross
means current is flowing into the paper.
(b) A
The lines of force between the two conductors are all in the
same direction, hence the field strength is at its strongest
here. The net result of this is that the conductors experience
a force that tries to push them apart.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
B
This time the lines of force between the two conductors are in
the opposite direction. The net effect of this is that they
cancel each other out, resulting in the field being weaker
here. Hence the conductors experience a force that tries to
push them together.
C
The argument for the final pair of current-carrying
conductors is identical to that given previously.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?4
(a)
Fig SAQ 4(i) shows a current-carrying coil. Sketch the magnetic
field around the coil and explain why this field occurs.
I
Fig SAQ 4(i)
(b)
For the current-carrying coils shown in Fig SAQ 4(ii) indicate
which end is North and which end is South.
A
I
B
I
C
I
Fig SAQ 4(ii)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
(c)
Consider the answers given in (a) and (b), and delete one
statement to give the correct force between the electromagnets
shown in Fig SAQ 4(iii).
A
I
I
B
I
I
I
I
C
Fig SAQ 4(iii)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?4
Solution
(a)
The diagram below shows the cross-section of the wire as it loops
over the top and reappears underneath. The crosses and dots
indicate the direction of current. Notice that the field is
concentrated inside the ‘winding’ and is moving in the same
direction, entering the winding on the right-hand side and leaving
the winding on the left-hand side. This is identical to a permanent
magnet and is known as an electromagnet.
(b) A
(b) B
(b) C
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
(c)
A
I
I
B
I
I
I
I
C
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?5
(a)
Fig SAQ 5(i) shows two examples of a current-carrying conductor
within a magnetic field. Show the direction of the force the
current-carrying conductors would have exerted on them. Explain
why you have arrived at your decision in each example.
Fig SAQ 5(i)
(b)
The equation that determines the magnitude of the force acting on
a current-carrying conductor in a magnetic field is given by:
F =  lI
Use this equation to determine the solution to the following
problems:
(i)
A conductor lying in and at right angles to a magnetic field has a
force of 0.1 N exerted on it when a current of 5 A is passed
through it. What force will be exerted on the conductor if the
current through it is increased to 12 A?
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
(ii)
A conductor lying in and at right angles to a magnetic field has a
force of 2 N exerted on it when a current of 20 A is passed
through it. What force will be exerted on the conductor if the
current through it is decreased to 8 A?
(iii)
A conductor of length 0.4 m, lying in and at right angles to a
magnetic field, has a force of 3 N exerted on it. If the effective
length of the conductor in the field is increased to 1 m, what will
be the new value of force exerted on the conductor?
(iv)
A 1.2-m conductor, lying in and at right angles to a magnetic
field, has a force of 8 N exerted on it. If the effective length of
the conductor in the field is reduced by 25%, what will be the
new value of force exerted on the conductor?
(v)
The flux density of a magnetic field is 0.75 T. A current carrying conductor placed in and at right angles to the field has a
force of 1.5 N exerted on it. What will be the new value of the
force if the field flux density is increased to 1.25 T?
(vi)
A magnetic field has a flux density of 0.8 T. A current -carrying
conductor placed in and at right angles to the field has a force of
4 N exerted on it. What will be the new value of the force if the
field flux density is reduced by 30%?
(vii)
A copper conductor, carrying a current of 50 mA, lies within,
and at right angles to, a uniform field measuring 10 cm wide in
the direction of the wire. Find the force produced on the
conductor when the flux density is 0.8 T.
(viii) A conductor 0.8 m long carries a current of 20 A and lies at right
angles to a magnetic field of flux density 0.4 T. Determine the
force exerted on the conductor.
(ix)
Determine the current required in a 0.3-m long conductor,
situated in and at right angles to a magnetic field of flux density
0.5 T, if a force of 3 N is to be exerted on the conductor.
(x)
A conductor 0.3 m long is situated in and at right angles to a
magnetic field. Determine the strength of the magnetic field if a
current of 30 A in the conductor produces a force on it of 12 N.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?5
Solution
(a)
Direction in
which conductor
will move
Field is weaker at
this point due to
lines of flux from
magnet and
conductor moving
in opposite
directions and
hence cancelling
out
Field is stronger at
this point due to
lines of flux from
magnet and
conductor moving
in same direction
Field is stronger at
this point due to
lines of flux from
magnet and
conductor moving
in same direction
Direction in
which conductor
will move
Field is weaker at
this point due to
lines of flux from
magnet and
conductor moving
in opposite
directions and
hence cancelling
out
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
SAQ 5(a) could also have been solved by applying Fleming’s left-hand
rule.
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
(b) (i)
Force is directly proportional to current, i.e. F I, therefore,
all other factors being the same, if 0.1 N is produced by 5 A
then 12 A will give:
F=
12
× 0.1 = 0.24 N
5
(ii) F  I, therefore, all other factors being the same, if 2 N is
produced by 20 A then 8 A will give:
F=
8
× 2 = 0.8 N
20
(iii) Force is directly proportional to length, i.e. F  l, therefore,
all other factors being the same, if 3 N of force is exerted on
a conductor of 0.4 m then 1 m will give:
1
F=
× 3 = 7.5 N
0.4
(iv) F  l.
A 25% reduction in the length of a 1.2 m conductor gives:
25
× 1.2 = 0.3 m , hence the new length is 1.2 – 0.3 = 0.9 m.
100
Therefore, all other factors being the same, if 8 N of force is
exerted on a conductor of 1.2 m then 0.9 m will give:
0.9
×8=6N
1.2
(v)
Force is directly proportional to flux density, i.e. F   ,
therefore, all other factors being the same, if 1.5 N of force is
exerted on a conductor at right angles to a field of 0.75 T
then an increase in the flux density to 1.25 T will give:
F=
1.25
× 1.5 = 2.5 N
0.75
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
(vi) F  
If the flux density of 0.8 T is reduced by 30% then the new
30
× 0.8 = 0.24 , i.e. the new value of flux
value will be
100
density will be 0.8 – 0.24 = 0.56 T.
Therefore, all other factors being the same, if 4 N of force is
exerted on a conductor at right angles to a field of 0.8 T then
a decrease in the flux density to 0.56 T will give:
F=
0.56
× 4 = 2.8 N
0.8
(vii) This is simply an application of the equation:
F =  lI
F = 0.8  0.1  50  10 –3 = 4 mN
(viii) This is simply an application of the equation:
F =  lI
F = 0.4  0.8  20 = 6.4 N
(ix)
(x)
F =  lI
I =
F
l
I =
3
= 20 A
0.5 × 0.3
F =  lI
F
12
 =
=
= 1.33 T
lI
0.3 × 0.3
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
Outcome 4
?1
When a conductor is moved in a magnetic field an emf is induced. For
the conductor shown in Fig SAQ 1(i), will the end that we can see be at
a higher or lower potential than the opposite end? Explain your answer.
Fig SAQ 1(i)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?1
Solution
To solve this problem we make use of Fleming’s right-hand rule:
Applying this rule to the above:

the flux is from left to right

the direction of motion is downward.
Therefore the direction of current is ‘out’ of the paper. This means that
the higher potential is on the end we can see (remember this is a
generator).
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
?2
Write down the equation that describes the magnitude of the emf, E,
induced in a length of conductor cutting a magnetic field at right
angles.
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?2
Solution
The equation that describes the magnitude of the emf induced in a
length of conductor cutting a magnetic field at right angles is:
E =  lu
where  is the flux density of the magnetic field, l is the effective
length of the conductor in metres and u is the velocity of the conductor
through the magnetic field in m s –1 .
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?3
Use the equation stated in SAQ 2 to determine the following:
(a)
A conductor, moving at uniform velocity through and at right
angles to a magnetic field of flux density 0.8 T, has a voltage of 10
V induced in it. If the flux density is increased to 1.6 T, what will
be the new value of the induced emf?
(b)
A copper conductor, moving at a uniform velocity of 20 m s –1 and
at right angles to a magnetic field of constant flux density, has a
voltage of 120 V induced in it. If the velocity is reduced to
10 m s –1 , what will be the new value of the induced emf?
(c)
A conductor of length 0.5 m, moving at uniform velocity through
and at right angles to a constant magnetic field, has a voltage of 15
V induced in it. If the length of the conductor in the field is
doubled, what will be the new value of the induced emf?
(d)
The flux density of a magnetic field is 0.6 T. A conductor,
moving at uniform velocity through and at right angles to the field,
has a voltage of 8 V induced in it. What will be the new value of
induced emf if the field flux density is increased to 1.5 T?
(e)
A conductor of length 0.6 m, moving at uniform velocity through
and at right angles to a constant magnetic field, has a voltage of 12
V induced in it. If the length of the conductor in the field is
reduced to 0.4 m, what will be the new value of the induced emf?
(f)
A magnetic field has a flux density of 0.9 T. A conductor, moving
at uniform velocity through and at right angles to the field, has a
voltage of 18 V induced in it. What will be the new value of
induced emf if the field flux density is reduced to 0.5 T?
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(g)
A conductor of length 0.8 m is situated in and at right angles to a
uniform magnetic field of flux density 1.25 T and moves at a
velocity of 20 m s –1 . Calculate the emf induced in the conductor
when it is moving at right angles through the field.
(h)
A 0.7-m long conductor moves at right angles to a magnetic field
of flux density 0.7 T at a velocity of 25 m s –1 . Calculate the emf
induced in the conductor.
(i)
Determine the speed at which a conductor 0.8 m long must m ove
at right angles to a uniform magnetic field of intensity 0.5 T in
order to induce 12 V in the conductor.
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Solution
(a)
E is directly proportional to the flux density  .  has doubled
from 0.8 to 1.6, therefore the voltage would double also from 1 0 V
to 20 V.
(b)
E is directly proportional to the speed of the conductor through the
magnetic field. If the speed is halved from 20 m s –1 to 10 m s –1
then the voltage is also halved from 120 V to 60 V.
(c)
E is directly proportional to the length of the conductor within the
magnetic field. If the length of the conductor is doubled, then the
voltage generated will also double from 15 V to 30 V.
(d)
The flux density is increased by a factor of
1.5
= 2.5 .
0.6
The induced emf will also increase in value by this factor,
i.e. 2.5  8 = 20 V.
(e)
The length of the conductor has been reduced by a factor of
0.4
= 0.667.
0.6
The induced emf will also be reduced by this factor,
i.e. 0.667  12 = 8 V.
(f)
The flux density has been reduced by a factor of
0.5
= 0.556.
0.9
The induced emf will also be reduced by this factor,
i.e. 0.556  18 = 10 V.
(g)
The induced emf in a conductor is E =  lu = 1.25  0.8  20 = 21 V.
(h)
The induced emf in a conductor is E =  lu = 0.7  0.7  25 = 12.25 V.
(i)
E =  lu
u=
E
12
=
= 30 m s –1
 l 0.5 × 0.8
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?4
(a)
If a conductor were to move in the direction shown in Fig SAQ
4(i), explain why there would be no induced emf between the ends
of the conductor.
Fig SAQ 4(i)
(b)
If the conductor were to move in the direction shown in Fig SAQ
4(ii), explain why the magnitude of the emf induced across the
ends of the conductor could be expressed as:
E =  lu sin 
Fig SAQ 4(ii)
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?4
Solution
(a)
If the conductor were to move in the direction shown then it would be
moving along the direction of magnetic flux and not cutting any lines
of flux. To generate an emf the conductor must be cutting the flux.
(b)
The velocity is made up of two parts:


a horizontal part that is in parallel with the lines of flu x
a vertical part that is at right angles to the lines of flux.
It is only the vertical part that contributes to the generation of a
voltage. As these vectors form a right-angled triangle, the
expression for the vertical component of the velocity may be
derived from the equation:
sin =
opposite
hypotenuse
opposite = hypotenuse × sin
vertical component of velocity = velocity × sin
The flux density  does not change and the length of the conductor
does not change, hence the expression for the voltage generated by a
length of conductor moving at an angle within a magnetic field is:
E =  lu sin 
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?5
Use the equation in SAQ 4 to solve the following questions:
(a)
A conductor 0.8 m long moves with a velocity of 15 m s –1 at an
angle of 40 to a magnetic field of flux density 0.5 T. Calculate
the induced emf in the conductor.
(b)
An aluminum conductor 0.7 m long moves with a velocity of 10 m
s –1 at an angle of 50 to a magnetic field of flux density 1 T.
Calculate the induced emf in the conductor.
(c)
Determine the speed at which a conductor 0.5 m long must move
at an angle of 30 to a uniform magnetic field of intensity 1.4 T in
order to induce an emf of 10 V in the conductor.
(d)
An emf of 2 V is induced in a straight conductor that moves with a
velocity of 5 m s –1 at an angle of 45 to a uniform magnetic field
of density 0.6 T. Calculate the effective length of the conductor.
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?5
Solution
(a)
E =  lu sin 
= 0.5  0.8  15  sin 40
= 3.857 V
(b)
E =  lu sin 
= 1  0.7  10  sin 50
= 5.362 V
(c)
E =  lu sin
u=
(d)
E
10
=
= 28.57 m s –1
 l sin 1.4 × 0.5 × sin 30
E =  lu sin
l=
E
2
=
= 0.943 m
 u sin
0.6 × 5 × sin 45
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?6
A single turn is placed into a magnetic field instead of a single
conductor, as shown in Fig SAQ (i).
Fig SAQ 5(i)
If the coil is moving at 35 m s –1 through a magnetic field of flux density
2.8 T:
(i)
determine the induced emf between the ends of the coil when it
cuts the field at 90
(ii) complete the table given in Fig SAQ 5(ii) showing the
instantaneous value of emf generated from 0 to 360
angle 
0
30
60
90
120
150
180
210
240
270
300
330
emf (E)
Fig SAQ 5(ii)
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ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS
(iii) sketch the waveform obtained on the graph given in Fig SAQ 5(iii)
(iv) determine the rms of the voltage waveform. What is the relevance
of rms?
Fig SAQ 5(ii)
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Solution
The effective length of coil is 2  1.2 = 2.4 m (note we don’t count the
‘sides’ of the coil as they don’t cut the lines of flux).
(i)
E =  lu sin
= 2.8  2.4  35  sin90
= 235.2 V
(ii)
angle 
0
30
60
90
120
150
180
210
240
270
300
330
360
emf (E)
0
117.6
203.7
235.2
203.7
117.6
0
–117.6
–203.7
–235.2
–203.7
–117.6
0
(iii)
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(iv) E max = 235.2 V
E rms = 0.707  E max = 0.707  235.2 = 166.3 V
The rms value of an alternating quantity is that indicated by ac
instruments (in ac circuits values of voltage and current are
usually given in rms values). The rms value of voltage is defined
as that value of direct voltage that produces the same heating
effect as an alternating voltage for a pure sine wave of voltage.
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