Q1. (a) A transformer operating on a 230 V mains supply provides a 12 V output. There are 1150 turns on the primary coil.

Transcription

1 Q1. (a) A transformer operating on a 230 V mains supply provides a 12 V output. There are 1150 turns on the primary coil. (i) Calculate the number of turns on the secondary coil. answer =... turns (1) (ii) A number of identical lamps rated at 12 V, 24 W are connected in parallel across the secondary coil. The primary circuit of the transformer includes a 630 ma fuse. Calculate the maximum number of lamps that can be supplied by the transformer if its efficiency is 85%. answer =... lamps (2) (iii) The transformer circuit includes a fuse. Explain why this is necessary. (1) (iv) Why is the fuse placed in the primary circuit rather than in the secondary circuit? (1) Page 1 of 19

2 (b) The figure below shows an experimental arrangement that can be used to demonstrate magnetic levitation. The iron rod is fixed vertically inside a large coil of wire. When the alternating current supply to the coil is switched on, the aluminium ring moves up the rod until it reaches a stable position floating above the coil. (i) By reference to the laws of electromagnetic induction explain why a current will be induced in the ring, why the ring experiences a force that moves it upwards, why the ring reaches a stable position. The quality of your written communication will be assessed in your answer. (6) Page 2 of 19

3 (ii) What would happen to the ring if the alternating current in the coil was increased without changing the frequency? Explain your answer. (2) (Total 13 marks) Q2. (a) (i) Outline the essential features of a step-down transformer when in operation (2) (ii) Describe two causes of the energy losses in a transformer and discuss how these energy losses may be reduced by suitable design and choice of materials. The quality of your written communication will be assessed in this question. (Allow one lined page). (6) Page 3 of 19

4 (b) Electronic equipment, such as a TV set, may usually be left in standby mode so that it is available for instant use when needed. Equipment left in standby mode continues to consume a small amount of power. The internal circuits operate at low voltage, supplied from a transformer. The transformer is disconnected from the mains supply only when the power switch on the equipment is turned off. This arrangement is outlined in the diagram below. When in standby mode, the transformer supplies an output current of 300 ma at 9.0V to the internal circuits of the TV set. (i) Calculate the power wasted in the internal circuits when the TV set is left in standby mode. answer =... W (1) (ii) If the efficiency of the transformer is 0.90, show that the current supplied by the 230 V mains supply under these conditions is 13 ma. (2) (iii) The TV set is left in standby mode for 80% of the time. Calculate the amount of energy, in J, that is wasted in one year through the use of the standby mode. 1 year = s answer =... J (1) Page 4 of 19

5 (iv) Show that the cost of this wasted energy will be about 4, if electrical energy is charged at 20 p per kwh. (2) (c) The power consumption of an inactive desktop computer is typically double that of a TV set in standby mode. This waste of energy may be avoided by switching off the computer every time it is not in use. Discuss one advantage and one disadvantage of doing this (2) (Total 16 marks) Q3. A transformer has 1200 turns on the primary coil and 500 turns on the secondary coil. The primary coil draws a current of 0.25 A from a 240 V ac supply. If the efficiency of the transformer is 83%, what is the current in the secondary coil? A B C D 0.10 A 0.21 A 0.50 A 0.60 A Page 5 of 19

6 Q4. The primary coil of a step-up transformer is connected to a source of alternating pd. The secondary coil is connected to a lamp. Which line, A to D, in the table correctly describes the flux linkage and current through the secondary coil in relation to the primary coil? A >1 <1 B <1 <1 C >1 >1 D <1 >1 (Total 1 mark) Q5. Which one of the following is not a cause of energy loss in a transformer? A B C D good insulation between the primary and secondary coil induced currents in the soft iron core reversal of magnetism in the soft iron core resistances in the primary and secondary coil (Total 1 mark) Page 6 of 19

7 Q6. When a mobile phone is being recharged, the charger heats up. The efficiency of the transformer in the charger can be as low as 15% when drawing a current of 50 ma from a 230 V mains supply. If the charging current required is 350 ma, what is the approximate output voltage at this efficiency? A B C D 4.9 V 11 V 28 V 33 V (Total 1 mark) Q7. A transformer, which is not perfectly efficient, is connected to a 230 V rms mains supply and is used to operate a 12 V rms, 60 W lamp at normal brightness. The secondary coil of the transformer has 24 turns. Which line, A to D, in the table is correct? number of turns on primary coil rms current in primary coil A 92 less than 0.26 A B 92 more than 0.26 A C 460 less than 0.26 A D 460 more than 0.26 A (Total 1 mark) Q8. Which one of the following statements concerning power losses in a transformer is incorrect? Power losses can be reduced by A B C D laminating the core. using high resistance windings. using thick wire. using a core made of special iron alloys which are easily magnetised. (Total 1 mark) Page 7 of 19

8 Q9. The primary winding of a perfectly efficient transformer has 200 turns and the secondary has 1000 turns. When a sinusoidal pd of rms value 10 V is applied to the input, there is a primary current of rms value 0.10 A rms. Which line in the following table, A to D, gives correct rms output values obtainable from the secondary when the primary is supplied in this way? rms output emf/v rms output current/a A B C D (Total 1 mark) Q10. Why, when transporting electricity on the National Grid, are high voltages and low currents used? A B C D The energy lost by radiation from electromagnetic waves is reduced. The electrons move more rapidly. The heat losses are reduced. The resistance of the power lines is reduced. (Total 1 mark) Page 8 of 19

9 Q11. Using the circuit shown, and with the switch closed, a small current was passed through the coil X. The current was slowly increased using the variable resistor. The current reached a maximum value and was then switched off. The maximum reading on the microammeter occurred when A B C D the small current flowed at the start. the current was being increased. the current was being switched off. the current in X was zero. (Total 1 mark) Page 9 of 19

10 M1. (a) (i) use of = gives N S = = 60 (turns) 1 (ii) max output power = (= 123 W) max number of lamps = 5 (no mark for non-integer answer) [or efficiency = gives 0.85 = (and max I S = 10.3 (A)) max number of lamps = 5 ] 2 (iii) (iv) fuse prevents transformer from overheating [or prevents transformer from supplying excessive currents] (all of) transformer is disconnected from supply when fuse fails [or fuse in secondary circuit would leave primary circuit live] 1 1 (b) (i) The candidate s writing should be legible and the spelling, punctuation and grammar should be sufficiently accurate for the meaning to be clear. The candidate s answer will be assessed holistically. The answer will be assigned to one of three levels according to the following criteria. High level (good to excellent) 5 or 6 marks The information conveyed by the answer is clearly organised, logical and coherent, using appropriate specialist vocabulary correctly. The form and style of writing is appropriate to answer the question. The candidate states that the ac in the coil produces a constantly changing magnetic field that passes through the ring, causing an emf to be induced according to Faraday s law. The candidate recognises that the induced emf will cause a current to flow in the ring, that the current is likely be large because the coil acts as a single conductor with low resistance, and that this current also produces a magnetic field. The candidate appreciates that Lenz s law indicates that the direction of the induced current is such as to produce a magnetic field that will oppose the existing field, and that the two fields will interact. The candidate refers to the force that acts on a current-carrying conductor when it is in a magnetic field and that this force lifts the ring upwards (into an area where the magnetic field is weaker) until the upwards magnetic force is equal to the downwards weight of the ring. Page 10 of 19

11 Intermediate level (modest to adequate) 3 or 4 marks The information conveyed by the answer may be less well organised and not fully coherent. There is less use of specialist vocabulary, or specialist vocabulary may be used incorrectly. The form and style of writing is less appropriate. The candidate is familiar with either or both Faraday s and Lenz s laws but only applies one of them to explain what happens in this demonstration. There are correct references to the two forces that act on the ring, and a reasonable explanation of why the ring reaches a stable position. Low level (poor to limited) 1 or 2 marks The information conveyed by the answer is poorly organised and may not be relevant or coherent. There is little correct use of specialist vocabulary. The form and style of writing may be only partly appropriate. The candidate refers much more superficially to either Faraday s or Lenz s law (or to both of them) but shows some understanding of why the forces acting on the ring cause it to reach equilibrium. The explanation expected in a competent answer should include a coherent selection of the following points concerning the physical principles involved and their consequences in this case. Faraday s law An emf is induced whenever there is a change in the magnetic flux passing through a conductor. The magnitude of the emf is proportional to the rate of change of magnetic flux linkage. The induced emf will cause a current to flow in any complete circuit, such as a single conducting ring. Because the ring is made from aluminium, which is a good conductor, a large initial current will be induced in it. Lenz s law The induced current flows in such a direction as to oppose the increase in magnetic flux when the current is switched on in the coil. The current produces a magnetic field in the opposite direction to that produced by the coil. These two (alternating) fields interact like the fields between two facing like magnetic poles, giving repulsion. Page 11 of 19

12 Forces The ring is a current-carrying conductor in a magnetic field, and consequently it experiences a force. This magnetic force acts upwards, in the opposite direction to the weight of the ring. As the ring rises, the magnetic field to which it is exposed becomes weaker as it moves away from the coil. This reduces the induced current, reducing also the magnetic force on the ring. The ring reaches a stable height when the magnetic force has decreased to the point where it is equal to the weight of the ring. 6 (ii) ring would float higher [or be expelled upwards] because (initial) current or emf (induced) in ring is greater or ring moves into weaker field until magnetic force balances weight [or (initially) magnetic force exceeds weight] 2 [13] M2. (a) (i) primary coil with more turns than secondary coil (1) (wound around) a core or input is ac (1) 2 Page 12 of 19

13 (ii) the mark scheme for this part of the question includes an overall assessment for the Quality of Written Communication QWC goodexcellent descriptor Two causes of energy losses are clearly identified, correct measures to indicate how these two losses may be reduced are stated and a detailed physical explanation of why these measures are effective is given. eg any two from the following four 1 When a transformer is in operation, there are ac currents in the primary and secondary coils. The coils have some resistance and the currents cause heating of the coils, causing some energy to be lost. This loss may be reduced by using low resistance wire for the coils. This is most important for the high current winding (the secondary coil of a step-down transformer). Thick copper wire is used for this winding, because thick wire of low resistivity has a low resistance. 2 The ac current in the primary coil magnetises, demagnetises and re-magnetises the core continuously in opposite directions. Energy is required both to magnetise and to demagnetise the core and this energy is wasted because it simply heats the core. The energy wasted may be reduced by choosing a material for the core which is easily magnetised and demagnetised, ie a magnetically soft material such as iron, or a special alloy, rather than steel. 3 The magnetic flux passing through the core is changing continuously. The metallic core is being cut by this flux and the continuous change of flux induces emfs in the core. In a continuous core these induced emfs cause currents known as eddy currents, which heat the core and cause energy to be wasted. The eddy current effect may be reduced by laminating the core instead of having a continuous solid core; the laminations are separated by very thin layers of insulator. Currents cannot flow in a conductor which is discontinuous (or which has a very high resistance). 4 If a transformer is to be efficient, as much as possible of the magnetic flux created by the primary current must pass through the secondary coil. This will not happen if these coils are widely separated from each other on the core. Magnetic losses may be reduced by adopting a design which has the two coils close together, eg by better core design, such as winding them on top of each other around the same part of a common core which also surrounds them. mark range 5-6 Page 13 of 19

14 modestadequate poorlimited incorrect, inappropriate or no response Up to two sources of energy losses are stated and there is an indication of how these may be minimised by suitable features or materials. There is no clear appreciation of an understanding of the physical principles to explain why these measures are effective. Up to two sources of energy losses are given, but the answer shows no clear understanding of the measures required to minimise them. There is no answer or the answer presented is irrelevant or incorrect Answers which address only one acceptable energy loss should be marked using the same principles, but to max 3. (b) (i) power wasted internally (= I V) = = 2.7 (W) (1) 6 1 (ii) input power = 3.0 (W) (1) mains current (1) (= A) 2 (iii) energy wasted per year (= P t) = = 7.5(6) 10 7 (J) (1) 1 (iv) energy wasted = = 21.0 (kwh) (1) cost of wasted energy = = 420p ( 4.20) (1) 2 (c) answers should refer to: an advantage of switching off (1) cost saving, saving essential fuel resources, reduced global warming etc a disadvantage of switching off (1) inconvenience of waiting, time taken for computer to reboot etc risk of computer failure increased by repeated switching on and off energy required to reboot may exceed energy saved by switching off 2 [16] Page 14 of 19

15 M3. C [1] M4. A [1] M5. A [1] M6. A [1] M7. D [1] M8. B [1] M9. B [1] M10. C [1] M11. C [1] Page 15 of 19

16 E1. The transformer turns ratio equation was familiar territory for most in part (a) (i), but correct application of the efficiency formula proved to be a greater challenge in part (a) (ii). Many correct answers were seen, and almost all students knew that the number of lamps has to be an integer. Most difficulties arose from mixing up data for the secondary coil with that for the primary (for example, multiplying the primary current by the secondary voltage). Parts (a) (iii) and (iv) proved to be an exacting test of whether students could think through to the real reasons or had enough practical experience of transformers to know these reasons. Many attempts at part (iii) were general answers about the reason for fitting a fuse in any circuit rather than specifically in a transformer s circuit. Very few students stated that transformer coils can overheat and become damaged when they handle excessive currents and that they therefore need to be protected. Similarly, it was only a small minority of answers to part (iv) that were properly valid; that stopping the primary current would isolate the whole transformer from the mains or that a failed fuse in the secondary circuit would leave the primary live. Most students find that electromagnetic induction is one of the most demanding topics in the specification. In these circumstances perhaps it should not be surprising that many of the attempts to answer part (b) (i) were very disappointing. Even when pointed at a logical and sequential structured answer by three bullet points, many students could not construct a coherent, ordered response. In assessing the Quality of Written Communication, one aspect that must be taken into consideration is the appropriate use of technical terminology. This was often absent from the responses seen. The term induction has a very special meaning in physics; magnetic induction involves magnetising a material by applying a magnetic field, electrostatic induction involves charge separation by applying an electric field, electromagnetic induction involves producing an emf by applying a changing magnetic field. In all cases, direct contact is unnecessary. Many answers contained the word induced used much more carelessly than its technical meaning in physics; a current was stated to be induced in the coil because it was connected to the ac supply, for example. This current was then said to induce a magnetic field. Many students seemed obsessed by effects in the iron rod, rather than in the aluminium ring. The aluminium ring was confused with the coil for example the coil is pushed upwards by the magnetic field. It was evident that a large proportion of students were familiar with statements of the laws of electromagnetic induction but could not apply them meaningfully to explain what happens in this demonstration. The field produced was regularly referred to as an electric field. The repulsion of the ring was sometimes attributed to Coulomb s law and the repulsion between charges. Fleming s left hand rule was confused with his right hand rule, or with the right hand grasp rule. Broadly, an outline plan of an appropriate answer to this question would be along the following lines. The ac current in the coil produces an alternating magnetic field, which is concentrated in the iron rod and passes through the ring. This changing magnetic field induces an emf in the ring. Because the ring is aluminium it is a good conductor and the emf causes a large current in it. A current-carrying conductor in a magnetic field experiences a force so this current produces a magnetic field whose direction opposes the applied field. Interaction between these fields gives a net upwards repulsion of the ring. As the ring moves upwards the magnetic field becomes weaker and the force on the ring decreases. The ring s position becomes stable when the upwards magnetic force balances its weight. Answers written in this fashion were rare but not difficult to identify and they were rewarded well. Answers to part (b) (ii) suffered from the same lack of general understanding as the previous part. It was often realised that the ring would move to a higher position, or be expelled upwards from the rod. Reasons were less well presented. Reference to a larger induced current (or emf) in the ring was considered a prerequisite for an acceptable explanation. Page 16 of 19

17 E2. In part (a) (i), the requirement that N p > N s was regarded as fundamental for the first mark; the second mark was awarded for either a core linking the coils or an ac supply. Some candidates missed the point of the question completely by writing about how transformers change voltages rather than concentrating on their essential features. Most attempts to answer part (a) (ii) fell well short of examiners expectations. In many cases the principal cause of this was a lack of detailed knowledge or confused understanding. When this kind of question is employed to assess the quality of candidates written communication, it will be expected that a good answer will be structured: well organised, and coherent. This was another general error, because candidates often presented answers that rambled on without general direction. Successful answers were primarily expected to address two (and only two) causes of energy loss; four causes are commonly identified by the standard sources that deal with this topic, so asking for two was not unduly demanding. For each cause that was identified, there was a requirement to discuss how the loss could be reduced by suitable features and materials. Answers that could be placed in the good to excellent category (five or six marks) should have backed up this factual knowledge with some physical reasoning. Very few answers addressed these requirements sufficiently successfully to deserve the award of full marks. It seemed that most of the candidates had heard about eddy currents, but not all of them knew exactly what they are or where they occur. Many answers showed considerable confusion; eddy currents caused by the coils can be stopped by using a soft iron core, heat losses from the currents can be reduced by using smaller currents, and energy losses from re-magnetising the core can be cut down by laminating it were typical of the confused responses seen. A proportion of the candidates evidently mixed up the principles of energy loss reduction in transformers with the principles involved in reducing power losses from transmission cables, because there were frequent references to using higher voltages in the transformers in order to reduce the currents causing the heating. The calculations in parts (b) (i) and (ii) were usually correct, with P = I V and the transformer efficiency equation being successfully applied. Almost all of the answers to part (b) (iii) were incorrect, because candidates did not realise that when on standby the transformer, as well as the load, continues to waste energy. Consequently, the power wasted on standby was 3.0 W, not 2.7 W. This error did not prevent candidates from accessing both available marks in part (b) (iv), provided they correctly applied the physical principles there. In this part, relatively few completely correct answers were seen, largely because the candidates were unable to convert an energy value from J to kwh. In part (c) it was usual to award both marks; most candidates knew that it takes an appreciable time for a computer to boot up and this would therefore be a disadvantage of switching off. For the advantage, a more specific point than saves energy was being looked for, because this response does little more than re-state the question. E3. Transformers were also the subject under test, where 68% of the responses were correct. This was a fairly straightforward calculation involving transformer efficiency. Page 17 of 19

18 E4. A fairly demanding test of candidates knowledge of transformers; slightly fewer than half of them selected the correct answer. Among the incorrect responses, distractor C was a common choice (23%), showing that the flux linkage ratio was better understood than the current ratio. E5. The facts about energy losses in transformers were well known in this question, where 73% of candidates gave the correct answer. The strongest incorrect distractor was C, suggesting that the energy losses caused by magnetic hysteresis are not always recognised. E6. This question amounted to a traditional transformer efficiency question, but it was set in the context of a mobile phone charger circuit with low efficiency. The facility of the question was 71%. There was no strong distractor, and the question discriminated much better than it had done when pre-tested. E7. The question showed that most of the candidates were able to apply the turns ratio equation correctly, because over 80% of them selected N p = 460. A simple conservation of power calculation would also show them that the primary current would be 0.26 A if the transformer was perfectly efficient. Since this transformer has an efficiency of less than 100%, the better candidates (51%) realised that the primary current had to be greater than 0.26 A. E8. This qestion was about transformers. Causes of power loss were well known in this question, where three quarters of the candidates evidently saw that using windings with higher resistance would have a detrimental effect. E11. This question needed students to spot that the most rapid change of flux in a transformer circuit occurs when a current is suddenly interrupted, leading to a maximum emf and (in this case) the largest current in the secondary circuit. A conventional car ignition system, now increasingly rare, illustrates this principle most effectively. The facility of the question was 43%, with 25% of the students opting for distractor B (when the primary current is steadily increased). Page 18 of 19

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