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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 75 133 iv © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 1 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 2 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 3 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 4 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 5 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 6 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 7 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 8 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 9 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 10 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 11 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 12 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 13 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 14 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 15 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 16 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 17 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 18 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 19 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 20 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 21 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 22 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 23 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 24 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 25 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 26 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 27 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 28 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 29 © Learning and Teaching Scotland 2004 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= ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 30 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 31 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 32 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 33 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 34 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 35 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 36 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 37 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 38 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 39 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?6 Solution V 240 = = 48 I 5 (a) R= (b) R = V × I = 240 × 5 = 1200 W ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 40 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 41 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 42 © Learning and Teaching Scotland 2004 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: ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 43 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 44 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 45 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 46 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 47 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 48 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 49 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 50 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 51 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 52 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS (c) A I I B I I I I C ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 53 © Learning and Teaching Scotland 2004 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? ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 54 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 55 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 56 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS SAQ 5(a) could also have been solved by applying Fleming’s left-hand rule. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 57 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 58 © Learning and Teaching Scotland 2004 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 59 © Learning and Teaching Scotland 2004 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 60 © Learning and Teaching Scotland 2004 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). ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 61 © Learning and Teaching Scotland 2004 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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 62 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?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 . ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 63 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?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? ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 64 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS (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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 65 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?3 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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 66 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 67 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 68 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 69 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?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 ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 70 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 71 © Learning and Teaching Scotland 2004 360 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 72 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS ?6 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) ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 73 © Learning and Teaching Scotland 2004 ELECTRICAL FUNDAMENTALS: EXEMPLAR QUESTIONS AND SOLUTIONS (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. ELECTRONIC AND ELECTRICAL FUNDAMENTALS (INT 2 ) 74 © Learning and Teaching Scotland 2004