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1 2 A lamp rated at 12 V 60 W is connected to the secondary coil of a step-down transformer and is at full brightness. The primary coil is connected to a supply of 230 V. The transformer is 75% efficient. What is the current in the primary coil? A 0.25 A B 0.35 A C 3.75 A D 5.0 A A transformer has 1150 turns on the primary coil and 500 turns on the secondary coil. The primary coil draws a current of 0.26 A from a 230 V ac supply. The current in the secondary coil is 0.50 A. What is the efficiency of the transformer? A 42% B 50% C 84% D 100% (Total 1 mark) Page 1 of 16 3 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) 4 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 0.10 A B 0.21 A C 0.50 A D 0.60 A Page 2 of 16 5 A hydroelectric power station has a power output of 2.0 MW when water passes through its turbines at a rate of 1.4 m3 s–1 . The water is supplied from a reservoir which is 750 m above the power station turbines, as shown in the diagram below. density of water = 1000 kg m–3 (a) Calculate (i) the mass of water passing through the turbines each second, ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (ii) the loss of potential energy per second of the water flowing between the reservoir and the power station turbines, ............................................................................................................... ............................................................................................................... ............................................................................................................... (iii) the efficiency of the power station. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (6) Page 3 of 16 (b) The turbines drive generators that produce alternating current at an rms potential difference of 25 kV which is then stepped up to an rms potential difference of 275 kV by means of a transformer. (i) Calculate the rms current supplied by the generators to the transformer when the power output of the generators is 2.0 MW. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (ii) The transformer has an efficiency of 95%. Calculate the output current of the transformer. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (4) (Total 10 marks) 6 A transformer is required to produce an r.m.s. output of 2.0 × 103 V when it is connected to the 230 V r.m.s. mains supply. The primary coil has 800 turns. (a) Calculate the number of turns required on the secondary coil, assuming the transformer is ideal. (2) (b) The transformer suffers from eddy current losses. (i) Explain how eddy currents arise. ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... ............................................................................................................... (4) (ii) State the feature of transformers designed to minimise eddy currents. ............................................................................................................... (1) (Total 7 marks) Page 4 of 16 7 (a) Explain what is meant by the term magnetic flux linkage. State its unit. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (2) (b) Explain, in terms of electromagnetic induction, how a transformer may be used to step down voltage. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (4) (c) A minidisc player is provided with a mains adapter. The adapter uses a transformer with a turns ratio of 15:1 to step down the mains voltage from 230 V. (i) Calculate the output voltage of the transformer. (2) (ii) State two reasons why the transformer may be less than 100% efficient. ............................................................................................................... ............................................................................................................... ............................................................................................................... (2) (Total 10 marks) Page 5 of 16 8 A 230 V, 60 W lamp is connected to the output terminals of a transformer which has a 200 turn primary coil and a 2000 turn secondary coil. The primary coil is connected to an ac source with a variable output pd. The lamp lights at its normal brightness when the primary coil is supplied with an alternating current of 2.7 A. What is the percentage efficiency of the transformer? A 3% B 10% C 97% D 100% (Total 1 mark) 9 (a) Calculate the length of copper wire that has a diameter of 1.6 × 10–3 m and a resistance of 25 Ω. resistivity of copper = 1.7 × 10–8 Ω m Length of wire ....................................................... (3) Page 6 of 16 (b) The resistance of copper wire is not zero. Explain why this fact leads to the use of alternating current rather than direct current when transmitting electrical energy. ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ ........................................................................................................................ (3) (Total 6 marks) 10 Which one of the following statements concerning power losses in a transformer is incorrect? Power losses can be reduced by A laminating the core. B using high resistance windings. C using thick wire. D using a core made of special iron alloys which are easily magnetised. (Total 1 mark) Page 7 of 16 11 A transformer with 3000 turns in its primary coil is used to change an alternating pd from an rms value of 240 V to an rms value of 12 V. When a 60 W, 12 V lamp is connected to the secondary coil, the lamp lights at normal brightness and a rms current of 0.26 A passes through the primary coil. Which line, A to D, in the table gives correct values for the number of turns on the secondary coil and for the transformer efficiency? number of turns on the secondary coil efficiency A 150 96% B 60 000 96% C 150 90% D 60 000 90% (Total 1 mark) 12 Which one of the following would not reduce the energy losses in a transformer? A B C D using thinner wire for the windings using a laminated core instead of a solid core using a core made from iron instead of steel using a core that allows all the flux due to the primary coil to be linked to the secondary coil (Total 1 mark) Page 8 of 16 13 Which one of the following is not a cause of energy loss in a transformer? A good insulation between the primary and secondary coil B induced currents in the soft iron core C reversal of magnetism in the soft iron core D resistances in the primary and secondary coil (Total 1 mark) Page 9 of 16 Mark schemes 1 2 3 4 5 B [1] C [1] A [1] C [1] (a) (i) mass per sec ( = density × vol per sec) = 1000 × 1.4 (1) = 1400 kg (s–1) (ii) loss of Ep per sec = 1400 × 9.8 × 750 (1) = 1.0 × 107 J (s–1) (1) (1.03 × 107 J s–1) (allow C.E. for value of mass per sec from (i)) (iii) efficiency = 0.2 (1) (allow C.E. for value (ii)) 6 Page 10 of 16 (b) (i) (use of P = IV gives) Irms = (1) = 80 A (1) (ii) power output = (0.95 × power input) = 0.95 × 2.0 (MW) = 1.9 (MW) (1) = 6.9 A (1) [or I for 100% efficiency = 7.3 (A) (1) I for 95% efficiency = 95% of 7.3 = 6.9 A] 4 [10] 6 (a) C1 7000 (6960) A1 (2) (b) (i) changing magnetic field B1 emf or changing magnetic field is in the core B1 e.m.f. induced (due to changing magnetic field) not back emf B1 current flows as core is made from a conducting material B1 (4) (ii) laminated core B1 (1) [7] 7 (a) product of flux and number of turns B1 Wb or equivalent C1 (2) Page 11 of 16 (b) changing primary magnetic field due to alternating voltage B1 (applied to primary) varying flux links with secondary B1 induced emf ∑ rate of change of flux linkage B1 NS < NP so less voltage on secondary C1 (4) (c) (i) equation or correct substitution C1 15.3 V A1 (2) (ii) <100% flux linkage / flux leakage / copper losses / iron losses / hysterysis losses not just “heating” or “heat loss” B2 (2) [10] 8 9 C [1] (a) R = ρL/A C1 A = 2.0 × 10–6 (m2) or π(0.8 × 10–3)2 seen in equation (condone π(1.6 × 10–3)2 or 8.04 × 10–6 seen) C1 L = 2900 m, 2940 m, 2960 or 3000 m A1 3 Page 12 of 16 (b) resistance leads to loss of heat/energy/power or I2R loss or voltage drop (across cable) B1 lower current lowers loss of heat/energy/power or reduces voltage drop B1 ac can be transformed (to lower transmission current) B1 3 [6] 10 11 12 13 B [1] A [1] A [1] A [1] Page 13 of 16 Examiner reports 1 2 3 4 5 This question was a transformer calculation that caused few problems. Its facility was 78% and it discriminated very well. It is possible that the values chosen for three of the distractors in this question, on transformer efficiency, were so obviously wrong that there was no need for the students to perform any calculation. However, only 54% gave the correct response when the question was pre-tested in 2009. The 2014 A level students made amends for this, because the facility of the question was 87% this time. 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. Transformers were also the subject under test, where 68% of the responses were correct. This was a fairly straightforward calculation involving transformer efficiency. The majority of candidates gained maximum marks in both parts (a)(i) and (a)(ii), but a few candidates, who scored reasonably well elsewhere, lost marks in part (ii) as a result of attempting unnecessarily complicated solutions involving expressions for loss of potential energy and gain of kinetic energy. In part (iii), candidates often knew the correct equation for efficiency but used incorrect data. Candidates often gained both marks in part (b)(i) but gave confused and incorrect answers in part (ii), with many of the weaker candidates attempting to convert from rms to peak values or vice versa. Some candidates failed to recognise that the output current from the generators was the input current to the transformer. 6 (a) Most candidates used the correct equation but a surprisingly large number rearranged it incorrectly and gave an answer indicating that there would be fewer turns on the secondary coil than on the primary coil. Significant figure penalties were common here. (b) (i) This explanation was not well done. Few candidates made it clear that the induction was taking place in the core and very few mentioned that the core was an electrical conductor. Many candidates made weak references to back e.m.f’s and a large number seemed to be trying to explain what happened in the secondary coil. Questions of this type are often asked and candidates would be well advised to practice making sequential explanations. (ii) Many candidates knew about laminations. A few forgot to specify that it was the core that should be laminated. Some wanted to laminate the coils. Page 14 of 16 7 8 9 10 (a) Few of the candidates were able to give a good explanation of magnetic flux linkage. several confusing the term with flux density. Few correctly identified the Weber as the correct unit. (b) Most candidates were unable to give a coherent explanation of how electromagnetic induction underpinned the working of the step down transformer. The most common observation gaining credit was that there are more turns on the primary than the secondary coil. (c) (i) Most candidates calculated the output voltage correctly. (ii) Few candidates’ answers gained credit for this part. It was common for candidates to cite heating or eddy currents as the reason for less than 100% efficiency. Although these answers are not incorrect they are too vague to allow credit at this level. Knowledge of the efficiency of a transformer was tested in this question, which had a facility of 58%. The output power from the transformer must have been 60 W, because the lamp was lit at its normal brightness. The turns ratio indicated that the primary voltage was 23 V, whilst the question stated that the primary current was 2.7 A. Hence the input power could be found using 2.7 × 23 = 62.1 W. 25% of the candidates chose the incorrect distractor B. (a) There were many errors in recall of the resistivity formula and a sizeable proportion of those who recalled it correctly simply substituted the value of the diameter for the area. Of those who appreciated that a calculation of area was necessary, a disappointing number used an incorrect formula or did not distinguish between diameter and radius. (b) This part was not done well. Candidates needed to explain that resistance leads to energy loss in transmission, that lower current reduces this loss and that lower current can be achieved using a transformer with ac. Many misconceptions were, however, apparent in responses. For example, ‘using dc you lose current’; ‘ac goes both ways so you lose energy when it goes one way and get it back when it goes the other’; ‘current makes resistance so if current goes both ways the resistances cancel out’; ‘use a lower current because if current is reduced then resistance is reduced’. 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. 11 This question tested both the turns ratio equation and efficiency; again there were few problems and the facility was 72%. 12 This question which were each correctly answered by just over three-quarters of the students, respectively tested the rotating coil and energy losses in a transformer. Page 15 of 16 13 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. Page 16 of 16