Resistance and resistivity — Paper 2 · A Level Physics 9702

7 A circuit contains three similar lamps A, B and C. The circuit also contains three switches, S1, S2 and S3, as shown in Fig. 7.1. A X S1 S2 S3 B C Y Fig. 7.1 One of the lamps is faulty. In order to detect the fault, an ohm-meter (a meter that measures resistance) is connected between terminals X and Y. When measuring resistance, the ohm- meter causes negligible current in the circuit. Fig. 7.2 shows the readings of the ohm-meter for different switch positions. switch meter reading S1 S2 S3 / Ω open open open ∞ closed open open 15 Ω open closed open 30 Ω open closed closed 15 Ω Fig. 7.2 (a) Identify the faulty lamp, and the nature of the fault. faulty lamp: … nature of fault: … [2] (b) Suggest why it is advisable to test the circuit using an ohm-meter that causes negligible current rather than with a power supply. … … [1] Use (c) Determine the resistance of one of the non-faulty lamps, as measured using the ohm- meter. resistance = …………………… Ω[1] (d) Each lamp is marked 6.0 V, 0.20 A. Calculate, for one of the lamps operating at normal brightness, (i) its resistance, resistance = …………………… Ω[2] (ii) its power dissipation. power = …………………… W [2] (e) Comment on your answers to (c) and (d)(i). … … … [2]

10 marks

Mark scheme: 7 (a) lamp C M1 lamp is shorted A1 [2] (b) shorted lamp A would cause damage to the supply/lamps /blow fuse in supply B1 [1] (c) 15 Ω B1 [1] (d) (i) V = I R C1 R = 30 Ω A1 [2] (ii) P = VI or I2R or V2 / R C1 P = 1.2 W A1 [2] (e) filament is cold when measuring with ohm-meter in (b) B1 resistance of filament rises as temperature rises B1 [2]

This question in 9702/21 May/June 2006

Use 6 A car battery has an internal resistance of 0.060 Ω. It is re-charged using a battery charger having an e.m.f. of 14 V and an internal resistance of 0.10 Ω, as shown in Fig. 6.1. car 0.10 Ω 0.060 Ω battery battery + charger 14 V – E Fig. 6.1 (a) At the beginning of the re-charging process, the current in the circuit is 42 A and the e.m.f. of the battery is E (measured in volts). (i) For the circuit of Fig. 6.1, state 1. the magnitude of the total resistance, resistance = … Ω 2. the total e.m.f. in the circuit. Give your answer in terms of E. e.m.f. = … V [2] (ii) Use your answers to (i) and data from the question to determine the e.m.f. of the car battery at the beginning of the re-charging process. e.m.f. = … V [2] Examiner’s Use (b) For the majority of the charging time of the car battery, the e.m.f. of the car battery is 12 V and the charging current is 12.5 A. The battery is charged at this current for 4.0 hours. Calculate, for this charging time, (i) the charge that passes through the battery, charge = … C [2] (ii) the energy supplied from the battery charger, energy = … J [2] (iii) the total energy dissipated in the internal resistance of the battery charger and the car battery. energy = … J [2] (c) Use your answers in (b) to calculate the percentage efficiency of transfer of energy from the battery charger to stored energy in the car battery. efficiency = … % [2]

12 marks

Mark scheme: 6 (a) (i) 1 total resistance = 0.16 Ω A1 2 e.m.f. = either (14 – E) or (E – 14) A1 [2] (ii) either 14 – E = 42 × 0.16 or (E – 14) = -42 × 0.16 C1 E = 7.3 V A1 [2] (b) (i) charge = It C1 = 12.5 × 4 × 60 × 60 = 1.8 × 105 C A1 [2] (ii) either energy = EQ or energy = Eit C1 either energy = 14 × 1.8 × 105 or energy = 14 × 12.5 × 4 × 3600 = 2.52 × 106 J A1 [2] (iii) energy = I2Rt or Vit and V = IR C1 = 12.52 × 0.16 × 4 × 3600 = 3.6 × 105 J A1 [2] (c) efficiency = (2.52 × 106 – 3.6 × 105)/(2.52 × 106) C1 = 86% A1 [2]

This question in 9702/21 May/June 2007

6 An electric shower unit is to be fitted in a house. The shower is rated as 10.5 kW, 230 V. The shower unit is connected to the 230 V mains supply by a cable of length 16 m, as shown in Fig. 6.1. copper wire cable shower unit 230 V supply 10.5 kW 230 V copper wire 16 m Fig. 6.1 (a) Show that, for normal operation of the shower unit, the current is approximately 46 A. [2] (b) The resistance of the two wires in the cable causes the potential difference across the shower unit to be reduced. The potential difference across the shower unit must not be less than 225 V. The wires in the cable are made of copper of resistivity 1.8 × 10–8 Ω m. Assuming that the current in the wires is 46 A, calculate (i) the maximum resistance of the cable, resistance = … Ω [3] Examiner’s Use (ii) the minimum area of cross-section of each wire in the cable. area = … m2 [3] (c) Connecting the shower unit to the mains supply by means of a cable having wires with too small a cross-sectional area would significantly reduce the power output of the shower unit. (i) Assuming that the shower is operating at 210 V, rather than 230 V, and that its resistance is unchanged, determine the ratio power dissipated by shower unit at 210 V . power dissipated by shower unit at 230 V ratio = … [2] (ii) Suggest and explain one further disadvantage of using wires of small cross-sectional area in the cable. … … … [2]

12 marks

Mark scheme: 6 (a) power = VI C1 current = 10.5 × 103 / 230 M1 = 45.7 A A0 [2] (b) (i) p.d. across cable = 5.0 V C1 R = 5.0 / 46 C1 = 0.11 Ω A1 [3] (ii) R = ρL / A C1 0.11 = (1.8 × 10–8 × 16 × 2) / A C1 A = 5.3 × 10–6 m2 A1 [3] (wires in parallel, not series, allow max 1/3 marks) GCE A/AS LEVEL – October/November 2007 9702 02 (c) (i) either power = V 2 / R or power ∝ V 2 C1 ratio = (210 / 230)2 = 0.83 A1 [2] (ii) resistance of cable is greater M1 greater power loss/fire hazard/insulation may melt wire may melt/cable gets hot A1 [2]

This question in 9702/21 Oct/Nov 2007

1 (a) State the most appropriate instrument, or instruments, for the measurement of the following. (i) the diameter of a wire of diameter about 1 mm … [1] (ii) the resistance of a filament lamp … [1] (iii) the peak value of an alternating voltage … [1] (b) The mass of a cube of aluminium is found to be 580 g with an uncertainty in the measurement of 10 g. Each side of the cube has a length of (6.0 ± 0.1) cm. Calculate the density of aluminium with its uncertainty. Express your answer to an appropriate number of significant figures. density = … ± … g cm–3 [5]

8 marks

Mark scheme: 1 (a) (i) micrometer (screw gauge) / travelling microscope … B1 [1] (ii) either ohm-meter or voltmeter and ammeter or multimeter/avo on ohm setting … B1 [1] (iii) either (calibrated) c.r.o. or a.c. voltmeter and × √2 … B1 [1] (b) density = mass / volume … C1 = 580 / 63 = 2.685 g cm-3 …(allow 2.68, 2.69, 2.7) … A1 % uncertainty in mass = (10 / 580) × 100 = 1.7% … C1 % uncertainty in volume = 3 × (0.1 / 6) × 100 = 5.0% … C1 uncertainty in density = 0.18 g cm-3 density = 2.7 ± 0.2 g cm-3 … A1 [5] (answer 2.69 ± 0.09 g cm-3 scores 4 marks)

This question in 9702/21 May/June 2009

7 A network of resistors, each of resistance R, is shown in Fig. 7.1. For Examiner’s Use A Z R R R B Y R C X Fig. 7.1 (a) Calculate the total resistance, in terms of R, between points (i) A and C, resistance = … [1] (ii) B and X, resistance = … [1] (iii) A and Z. resistance = … [1] (b) Two cells of e.m.f. E1 and E2 and negligible internal resistance are connected into the For network in (a), as shown in Fig. 7.2. Examiner’s Use E1 A Z R R I1 I3 R B Y I2 R C X E2 Fig. 7.2 The currents in the network are as indicated in Fig. 7.2. Use Kirchhoff’s laws to state the relation (i) between currents I1, I2 and I3, … [1] (ii) between E2, R, I2 and I3 in loop BCXYB, … [1] (iii) between E1, E2, R, I1 and I2 in loop ABCXYZA. … [1]

6 marks

Mark scheme: 7 (a) (i) R … B1 [1] (ii) 0.5R … B1 [1] (iii) 2.5R …(allow e.c.f. from (ii)) … B1 [1] (b) (i) I1 + I2 = I3 … B1 [1] (ii) E2 = I3R + I2R … B1 [1] (iii) E1 – E2 = 2I1R – I2R … B1 [1]

This question in 9702/21 May/June 2009

1 The volume of fuel in the tank of a car is monitored using a meter as illustrated in Fig. 1.1. FUEL ½ ¼ ¾ 0 1 Fig. 1.1 The meter has an analogue scale. The meter reading for different volumes of fuel in the tank is shown in Fig. 1.2. 60 50 volume / litre 40 30 20 10 0 0 ¼ ½ ¾ 1 empty full meter reading Fig. 1.2 The meter is calibrated in terms of the fraction of the tank that remains filled with fuel. (a) The car uses 1.0 litre of fuel when travelling 14 km. The car starts a journey with a full For tank of fuel. Examiner’s Use (i) Calculate the volume of fuel remaining in the tank after a journey of 210 km. volume = … litres [2] (ii) Use your answer to (i) and Fig. 1.2 to determine the change in the meter reading during the 210 km journey. from full to … [1] (b) There is a systematic error in the meter. (i) State the feature of Fig. 1.2 that indicates that there is a systematic error. … … [1] (ii) Suggest why, for this meter, it is an advantage to have this systematic error. … … [1]

5 marks

Mark scheme: 1 (a) (i) car uses 210 / 14 = 15 litres of fuel … C1 volume reading = 45 litres . … A1 [2] (ii) from ‘full’ to ‘3/4’ mark … B1 [1] (b) (i) line/graph does not pass through (‘empty, 0) / there is an intercept … B1 [1] (do not allow ‘non-linear’) (ii) (meter shows zero fuel when there is some left in the tank so) acts as a ‘reserve’ … B1 [1] [Total: 5]

This question in 9702/21 Oct/Nov 2009

4 A uniform wire has length L and area of cross-section A. For The wire is fixed at one end so that it hangs vertically with a load attached to its free end, as Examiner’s shown in Fig. 4.1. Use wire load W Fig. 4.1 When the load of magnitude W is attached to the wire, it extends by an amount e. The elastic limit of the wire is not exceeded. The material of the wire has resistivity ρ. (a) (i) Explain what is meant by extends elastically. … … … [2] (ii) Write down expressions, in terms of L, A, W, ρ and e for 1. the resistance R of the unstretched wire, R = … [1] 2. the Young modulus E of the wire. E = … [1] (b) A steel wire has resistance 0.44 Ω. Steel has resistivity 9.2 × 10–8 Ω m. For Examiner’s A load of 34 N hung from the end of the wire causes an extension of 7.7 × 10–4 m. Use Using your answers in (a)(ii), calculate the Young modulus E of steel. E = … Pa [3]

7 marks

Mark scheme: 4 (a) (i) returns to original shape / size / length etc. … B1 when load / distorting forces / weight / strain is removed … B1 [2] (ii) 1 R = ρL / A … B1 [1] 2 E = WL / Ae … B1 [1] (b) E = WR / eρ … C1 = (34 × 0.44) / (7.7 × 10-4 × 9.2 × 10-8) … C1 = 2.1 × 1011 Pa … A1 [3] [Total: 7]

This question in 9702/21 Oct/Nov 2009

6 (a) Two resistors, each of resistance R, are connected first in series and then in parallel. For Examiner’s Show that the ratio Use combined resistance of resistors connected in series combined resistance of resistors connected in parallel is equal to 4. [1] (b) The variation with potential difference V of the current I in a lamp is shown in Fig. 6.1. 0.15 I / A 0.10 0.05 0 0 1.0 2.0 3.0 V / V Fig. 6.1 Calculate the resistance of the lamp for a potential difference across the lamp of 1.5 V. For Examiner’s Use resistance = … [2] (c) Two lamps, each having the I-V characteristic shown in Fig. 6.1, are connected first in series and then in parallel with a battery of e.m.f. 3.0 V and negligible internal resistance. Complete the table of Fig. 6.2 for the lamps connected to the battery. p.d. across resistance of combined resistance each lamp / V each lamp / of lamps / lamps connected in series lamps connected in parallel Fig. 6.2 [4] (d) (i) Use data from the completed Fig. 6.2 to calculate the ratio combined resistance of lamps connected in series . combined resistance of lamps connected in parallel ratio = … [1] (ii) The ratios in (a) and (d)(i) are not equal. By reference to Fig. 6.1, state and explain qualitatively the change in the resistance of a lamp as the potential difference is changed. … … … … [3]

11 marks

Mark scheme: 6 (a) total resistance in series = 2R total resistance in parallel = ½R … M1 ratio is 2R / ½R = 4 … (allow mark if clear numbers in the ratio) … A0 [1] (b) at 1.5 V, current is 0.10 A … C1 1.5 resistance = V/I = 0.1 = 15 Ω … A1 [2] (use of tangent or any other current scores no marks) (c) p.d. across resistance of combined each lamp / V each lamp / Ω resistance / Ω series 1.5 15 30 parallel 3.0 20 10 column 1 … A1 columns 2 and 3: max 3 marks with -1 mark for each error or omission … A3 [4] (d) (i) ratio is 3 … (allow e.c.f.) … A1 [1] (ii) resistance increases as potential difference increases … B1 increasing p.d. increases current … B1 current increases non-linearly so resistance increases … B1 [3] [Total: 11] GCE A/AS LEVEL – October/November 2009 9702 22

This question in 9702/22 Oct/Nov 2009

6 An electric heater is to be made from nichrome wire. Nichrome has a resistivity of For 1.0 × 10–6 Ω m at the operating temperature of the heater. Examiner’s The heater is to have a power dissipation of 60 W when the potential difference across its Use terminals is 12 V. (a) For the heater operating at its designed power, (i) calculate the current, current = … A [2] (ii) show that the resistance of the nichrome wire is 2.4 Ω. [2] (b) Calculate the length of nichrome wire of diameter 0.80 mm required for the heater. length = … m [3] (c) A second heater, also designed to operate from a 12 V supply, is constructed using the For same nichrome wire but using half the length of that calculated in (b). Examiner’s Explain quantitatively the effect of this change in length of wire on the power of the Use heater. … … … … [3]

10 marks

Mark scheme: 6 (a) (i) P = VI …………………………..……………………………….…..……………… C1 60 = 12 × I I = 5.(0) A …………………………………………….…………………………… A1 [2] (ii) either V = IR or P = I 2R or P = V2 / R ….………..………………. C1 either 12 = 5 × R or 60 = 52 × R or 60 = 122/R ….……….…………… M1 R = 2.4 Ω …………………………………………………………………………. A0 [2] (b) R = ρL/A …………………………..…………………………………………………….. C1 A = π × (0.4 × 10–3)2 (= 5.03 × 10–7) .…………..……………………………………… C1 L = (2.4 × 5.03 × 10–7)/(1.0 × 10–6) = 1.2 m …………..……………….……………………………………………………. A1 [3] (c) resistance is halved ……………………………….…………………………………… M1 either current is doubled or power ∝ 1/R ….……… ……………………………… M1 power is doubled …………………….……..…………………………………………… A1 [3] GCE AS/A LEVEL – May/June 2010 9702 21

This question in 9702/21 May/June 2010

6 (a) (i) State what is meant by an electric current. For Examiner’s … Use … [1] (ii) Define electric potential difference. … … [1] (b) The variation with potential difference V of the current I in a component Y and in a resistor R are shown in Fig. 6.1. 0.7 I / A 0.6 componentcomponentcomponent YYY 0.5 resistor R 0.4 0.3 0.2 0.1 0 0 2 4 6 8 10 V / V Fig. 6.1 Use Fig. 6.1 to explain how it can be deduced that resistor R has a constant resistance For of 20 Ω. Examiner’s Use … … … [2] (c) The component Y and the resistor R in (b) are connected in parallel as shown in Fig. 6.2. E Y R 20 Ω Fig. 6.2 A battery of e.m.f. E and negligible internal resistance is connected across the parallel combination. Use data from Fig. 6.1 to determine (i) the current in the battery for an e.m.f. E of 6.0 V, current = … A [1] (ii) the total resistance of the circuit for an e.m.f. of 8.0 V. resistance = … Ω [2] (d) The circuit of Fig. 6.2 is now re-arranged as shown in Fig. 6.3. For Examiner’s Use Y R E Fig. 6.3 The current in the circuit is 0.20 A. (i) Use Fig. 6.1 to determine the e.m.f. E of the battery. E = … V [1] (ii) Calculate the total power dissipated in component Y and resistor R. power = … W [2]

10 marks

Mark scheme: 6 (a) (i) movement/flow of charged particles B1 [1] (ii) work done per unit charge (transferred) B1 [1] (b) straight line through origin B1 resistance = V / I , with values for V and I shown M1 = 20 Ω A0 [2] (using the gradient loses the last mark) (c) (i) 0.5 A A1 [1] (ii) either resistance of each resistor is 20 Ω or total current = 0.8 A C1 either combined resistance = 10 Ω or R = E / I = 10 Ω A1 [2] (d) (i) 10 V A1 [1] (ii) power = EI C1 = 10 × 0.2 = 2.0 W A1 [2] GCE AS/A LEVEL – May/June 2010 9702 23

This question in 9702/23 May/June 2010

6 (a) A lamp is rated as 12 V, 36 W. For Examiner’s (i) Calculate the resistance of the lamp at its working temperature. Use resistance = … Ω [2] (ii) On the axes of Fig. 6.1, sketch a graph to show the current-voltage (I–V) characteristic of the lamp. Mark an appropriate scale for current on the y-axis. I / A 0 6 12 V / V Fig. 6.1 [3] (b) Some heaters are each labelled 230 V, 1.0 kW. The heaters have constant resistance. For Examiner’s Determine the total power dissipation for the heaters connected as shown in each of the Use diagrams shown below. (i) 230 V power = … kW [1] (ii) 230 V power = … kW [1] (iii) 230 V power = … kW [2]

9 marks

Mark scheme: 6 (a) (i) either P = V 2 / R or P = VI and V = IR C1 R = 4.0 Ω A1 [2] (ii) sketch vertical axis labelled appropriately B1 (straight) line from origin then curved in correct direction B1 line passes through 12 V, 3.0 A B1 [3] (b) (i) 2.0 kW A1 [1] (ii) 0.5 kW A1 [1] (iii) total resistance = 3R / 2 C1 power = 0.67 kW A1 [2] GCE AS/A LEVEL – October/November 2010 9702 21

This question in 9702/21 Oct/Nov 2010

6 The variation with temperature of the resistance RT of a thermistor is shown in Fig. 6.1. For Examiner’s Use 4.0 3.0 RT / kΩ 2.0 1.0 0 0 5 10 15 20 25 temperature / °C Fig. 6.1 The thermistor is connected into the circuit of Fig. 6.2. 1.6 kΩ 1.2 kΩ A B V 9.0 V Fig. 6.2 The battery has e.m.f. 9.0 V and negligible internal resistance. The voltmeter has infinite For resistance. Examiner’s Use (a) For the thermistor at 22.5 °C, calculate (i) the total resistance between points A and B on Fig. 6.2, resistance = … Ω [2] (ii) the reading on the voltmeter. voltmeter reading = … V [2] (b) The temperature of the thermistor is changed. The voltmeter now reads 4.0 V. Determine (i) the total resistance between points A and B on Fig. 6.2, resistance = … Ω [2] (ii) the temperature of the thermistor. For Examiner’s Use temperature = … °C [2] (c) A student suggests that the voltmeter, reading up to 10 V, could be calibrated to measure temperature. Suggest two disadvantages of using the circuit of Fig. 6.2 with this voltmeter for the measurement of temperature in the range 0 °C to 25 °C. 1. … … 2. … … [2]

10 marks

Mark scheme: 6 (a) (i) at 22.5 °C, RT = 1600 Ω or 1.6 kΩ C1 total resistance = 800 Ω A1 [2] (ii) either use of potential divider formula or current = 9 / 2000 (4.5 mA) C1 V = (0.8/2.0) × 9 V = (9/2000) × 800 = 3.6 V = 3.6 V A1 [2] (b) (i) total resistance = 4/5 × 1200 C1 = 960 Ω A1 [2] (ii) for parallel combination, 1/960 = 1/1600 + 1/RT RT = 2400 Ω / 2.4 kΩ C1 temperature = 11 °C A1 [2] GCE A LEVEL – October/November 2010 9702 22 (c) e.g. only small part of scale used / small sensitivity B1 non-linear B1 [2] (any two sensible suggestions, 1 each, max 2)

This question in 9702/22 Oct/Nov 2010

8 An electric heater has a constant resistance and is rated as 1.20 kW, 230 V. For Examiner’s The heater is connected to a 230 V supply by means of a cable that is 9.20 m long, as Use illustrated in Fig. 8.1. copper wires in cable heater, rated 230 V 1.20 kW 230 V 9.20 m Fig. 8.1 The two copper wires that make up the cable each have a circular cross-section of diameter 0.900 mm. The resistivity of copper is 1.70 × 10–8 Ω m. (a) Show that (i) the resistance of the heater is 44.1 Ω, [2] (ii) the total resistance of the cable is 0.492 Ω. [2] (b) The current in the cable and heater is switched on. Determine, to three significant For figures, the power dissipated in the heater. Examiner’s Use power = … W [3] (c) Suggest two disadvantages of connecting the heater to the 230 V supply using a cable consisting of two thinner copper wires. 1. … … 2. … … [2] Please turn over for Question 9.

9 marks

Mark scheme: 8 (a) (i) either P = V2 / R or I = 1200 / 230 or 5.22 C1 R = (230 × 230) / 1200 R = 2302 / 1200 or R = 230 / 5.22 M1 = 44.1 Ω = 44.1 Ω A0 [2] (ii) R = ρL / A C1 = (1.7 × 10–8 × 9.2 × 2) / (π × {0.45 × 10–3}2) M1 = 0.492 Ω A0 [2] (b) current = 230 /44.6 C1 power = (230 /44.6)2 × 44.1 C1 = 1170 W A1 [3] (allow full credit for solution based on potential divider) (c) e.g. less power dissipated in the heater / smaller p.d. across heater / more power loss in cable / current lower B1 cable becomes heated / melts B1 [2] (any two sensible suggestions, 1 each, max 2)

This question in 9702/23 Oct/Nov 2010

1 Measurements made for a sample of metal wire are shown in Fig. 1.1. quantity measurement uncertainty length 1750 mm ± 3 mm diameter 0.38 mm ± 0.01 mm resistance 7.5 Ω ± 0.2 Ω Fig. 1.1 (a) State the appropriate instruments used to make each of these measurements. (i) length … [1] (ii) diameter … [1] (iii) resistance … [1] (b) (i) Show that the resistivity of the metal is calculated to be 4.86 × 10–7 Ω m. [2] (ii) Calculate the uncertainty in the resistivity. uncertainty = ± … Ω m [4] (c) Use the answers in (b) to express the resistivity with its uncertainty to the appropriate For number of significant figures. Examiner’s Use resistivity = … ± … Ω m [1]

10 marks

Mark scheme: 1 (a) (i) metre rule / tape (not ‘rule’) B1 [1] (ii) micrometer (screw gauge) / digital caliper B1 [1] (iii) ammeter and voltmeter / ohmmeter / multimeter on ‘ohm’ setting B1 [1] (b) (i) resistivity = RA / L C1 = [7.5 × π × (0.38 × 10–3)2 / 4] / 1.75 M1 = 4.86 × 10–7 Ω m A0 [2] (ii) (uncertainty in R =) [0.2 / 7.5] × 100 = 2.7% and (uncertainty in L =) [3 / 1750] × 100 = 0.17% C1 (uncertainty in A =) 2 × (0.01 / 0.38) × 100 = 5.3 % C1 total = 8.13% C1 uncertainty = 0.395 × 10–7 (Ω m) A1 [4] (missing 2 factor in uncertainty in A, then allow max 3/4) (c) resistivity = (4.9 × 10–7 ± 0.4 × 10–7) Ω m A1 [1]

This question in 9702/21 May/June 2011

5 (a) For a cell, explain the terms For Examiner’s (i) electromotive force (e.m.f.), Use … … [1] (ii) internal resistance. … … [1] (b) The circuit of Fig. 5.1 shows two batteries A and B and a resistor R connected in series. R 3.0 V 12 V A B 0.10 Ω 0.20 Ω Fig. 5.1 Battery A has an e.m.f. of 3.0 V and an internal resistance of 0.10 Ω. Battery B has an e.m.f. of 12 V and an internal resistance of 0.20 Ω. Resistor R has a resistance of 3.3 Ω. (i) Apply Kirchhoff’s second law to calculate the current in the circuit. current = … A [2] (ii) Calculate the power transformed by battery B. power = … W [2] (iii) Calculate the total energy lost per second in resistor R and the internal For resistances. Examiner’s Use energy lost per second = … J s–1 [2] (c) The circuit of Fig. 5.1 may be used to store energy in battery A. Suggest how your answers in (b) support this statement. … … … [1]

9 marks

Mark scheme: 5 (a) (i) energy converted from chemical to electrical when charge flows through cell or round complete circuit B1 (ii) (resistance of the cell) causing loss of voltage or energy loss in cell B1 [2] (b) (i) EB – EA = I (R + rB + rA) 12 – 3 = I (3.3 + 0.1 + 0.2) C1 I = 2.5 A A1 [2] (ii) Power = E × I = 12 × 2.5 C1 = 30 W A1 [2] (iii) P = I2 × R or P = V2 / R or P = VI = (2.5)2 × 3 = 92 / 3.6 = 9 × 2.5 C1 = 22.5 J s–1 A1 [2] (c) power supplied from cell B is greater than energy lost per second in circuit B1 [1]

This question in 9702/22 May/June 2011

5 (a) (i) On Fig. 5.1, sketch the I – V characteristic for a filament lamp. For Examiner’s Use I / A 0 0 V / V Fig. 5.1 [2] (ii) Explain how the resistance of the lamp may be calculated for any voltage from its I – V characteristic. … … [1] (b) Two identical filament lamps are connected first in series, and then in parallel, to a 12 V power supply that has negligible internal resistance. The circuits are shown in Fig. 5.2 and Fig. 5.3 respectively. 12 V 12 V Fig. 5.2 Fig. 5.3 (i) State and explain why the resistance of each lamp when they are connected in For series is different from the resistance of each lamp when they are connected in Examiner’s parallel. Use … … … … … [3] (ii) Each lamp is marked with a rating ‘12 V, 50 W’. Calculate the total resistance of the circuit for the two lamps connected such that each lamp uses this power. total resistance = … Ω [3]

9 marks

Mark scheme: 5 (a) (i) Start from (0,0) and smooth curve in correct direction B1 Curve correct for end section never horizontal B1 [2] (ii) R = V / I hence take co-ords of V and I from graph and calculate V / I B1 [1] (b) (i) each lamp in parallel has a greater p.d. / greater current M1 lamp hotter M1 resistance of lamps in parallel greater A1 [3] (ii) P = V2 / R or P = VI and V = IR C1 R = 144 / 50 = 2.88 for each lamp C1 total R = 1.44 Ω A1 [3]

This question in 9702/23 May/June 2011

5 (a) Define the ohm. For Examiner’s … [1] Use (b) Determine the SI base units of resistivity. base units of resistivity = … [3] (c) A cell of e.m.f. 2.0 V and negligible internal resistance is connected to a variable resistor R and a metal wire, as shown in Fig. 5.1. 2.0 V R metal wire 900 mm Fig. 5.1 The wire is 900 mm long and has an area of cross-section of 1.3 × 10–7 m2. The resistance of the wire is 3.4 Ω. (i) Calculate the resistivity of the metal wire. resistivity = … [2] (ii) The resistance of R may be varied between 0 and 1500 Ω. For Calculate the maximum potential difference (p.d.) and minimum p.d. possible across Examiner’s the wire. Use maximum p.d. = … V minimum p.d. = … V [2] (iii) Calculate the power transformed in the wire when the potential difference across the wire is 2.0 V. power = … W [2] (d) Resistance R in (c) is now replaced with a different variable resistor Q. State the power transformed in Q, for Q having (i) zero resistance, power = … W [1] (ii) infinite resistance. power = … W [1]

12 marks

Mark scheme: 5 (a) ohm = volt / ampere B1 [1] (b) ρ = RA / l or unit is Ω m C1 units: V A–1 m2 m–1 = N m C–1 A–1 m2 m–1 C1 = kg m2 s–2 A–1 s–1 A–1 m2 m–1 = kg m3 s–3 A–2 A1 [3] (c) (i) ρ = [3.4 × 1.3 × 10–7] / 0.9 C1 = 4.9 × 10–7 (Ω m) A1 [2] (ii) max = 2.(0) V A1 min = 2 × (3.4 /1503.4) = 4.5 × 10–3 V A1 [2] (iii) P = V2 / R or P = VI and V = IR C1 = (2)2 / 3.4 = 1.18 (allow 1.2) W A1 [2] (d) (i) power in Q is zero when R = 0 B1 [1] (ii) power in Q = 0 / tends to zero as R = infinity B1 [1] GCE AS/A LEVEL – October/November 2011 9702 21

This question in 9702/21 Oct/Nov 2011

4 (a) The output of a heater is 2.5 kW when connected to a 220 V supply. For Examiner’s (i) Calculate the resistance of the heater. Use resistance = … Ω [2] (ii) The heater is made from a wire of cross-sectional area 2.0 × 10–7 m2 and resistivity 1.1 × 10–6 Ω m. Use your answer in (i) to calculate the length of the wire. length = … m [3] (b) The supply voltage is changed to 110 V. (i) Calculate the power output of the heater at this voltage, assuming there is no change in the resistance of the wire. power = … W [1] (ii) State and explain quantitatively one way that the wire of the heater could be changed to give the same power as in (a). … … … [2]

8 marks

Mark scheme: 4 (a) (i) R = V2 / P or P = IV and V = IR C1 = (220)2 / 2500 = 19.4 Ω (allow 2 s.f.) A1 [2] (ii) R = ρl / A C1 l = [19.4 × 2.0 × 10–7] / 1.1 × 10–6 C1 = 3.53 m (allow 2 s.f.) A1 [3] (b) (i) P = 625, 620 or 630 W A1 [1] (ii) R needs to be reduced C1 Either length ¼ of original length or area 4× greater or diameter 2× greater A1 [2]

This question in 9702/21 May/June 2012

4 A battery of electromotive force 12 V and negligible internal resistance is connected to two For resistors and a light-dependent resistor (LDR), as shown in Fig. 4.1. Examiner’s Use 8.0 kΩ S 12 V X 12 kΩ A Y Fig. 4.1 An ammeter is connected in series with the battery. The LDR and switch S are connected across the points XY. (a) The switch S is open. Calculate the potential difference (p.d.) across XY. p. d. = … V [3] (b) The switch S is closed. The resistance of the LDR is 4.0 kΩ. Calculate the current in the ammeter. current = … A [3] (c) The switch S remains closed. The intensity of the light on the LDR is increased. State For and explain the change to Examiner’s Use (i) the ammeter reading, … … … [2] (ii) the p.d. across XY. … … … [2]

10 marks

Mark scheme: 4 (a) total resistance = 20 (kΩ) C1 current = 12 / 20 (mA) or potential divider formula C1 p.d. = [12 / 20] × 12 = 7.2 V A1 [3] (b) parallel resistance = 3 (kΩ) C1 total resistance 8 + 3 = 11 (kΩ) C1 current = 12 / 11 × 103 = 1.09 × 10–3 or 1.1 × 10–3 A A1 [3] (c) (i) LDR resistance decreases M1 total resistance (of circuit) is less hence current increases A1 [2] (ii) resistance across XY is less M1 less proportion of 12 V across XY hence p.d. is less A1 [2]

This question in 9702/22 May/June 2012

2 (a) Define electrical resistance. For Examiner’s … Use … [1] (b) A circuit is set up to measure the resistance R of a metal wire. The potential difference (p.d.) V across the wire and the current І in the wire are to be measured. (i) Draw a circuit diagram of the apparatus that could be used to make these measurements. [3] (ii) Readings for p.d. V and the corresponding current І are obtained. These are shown in Fig. 2.1. 0.30 0.25 0.20 I / A 0.15 0.10 0.05 0 0 1.0 2.0 3.0 4.0 5.0 V / V Fig. 2.1 Explain how Fig. 2.1 indicates that the readings are subject to For Examiner’s 1. a systematic uncertainty, Use … … [1] 2. random uncertainties. … … [1] (iii) Use data from Fig. 2.1 to determine R. Explain your working. R = … Ω [3] (c) In another experiment, a value of R is determined from the following data: Current І = 0.64 ± 0.01 A and p.d. V = 6.8 ± 0.1 V. Calculate the value of R, together with its uncertainty. Give your answer to an appropriate number of significant figures. R = … ± … Ω [3]

12 marks

Mark scheme: 2 (a) resistance = potential difference / current B1 [1] (b) (i) metal wire in series with power supply and ammeter B1 voltmeter in parallel with metal wire B1 rheostat in series with power supply or potential divider arrangement or variable power supply B1 [3] (ii) 1. intercept on graph B1 [1] 2. scatter of readings about the best fit line B1 [1] (iii) correction for zero error explained B1 use of V and corrected І values from graph C1 resistance = V / І = 22.(2) Ω [e.g. 4.0 / 0.18] A1 [3] (c) R = 6.8 / 0.64 = 10.625 C1 %R = %V + %І = (0.1 / 6.8) × 100 + (0.01 / 0.64) × 100 C1 = 1.47% + 1.56% ∆R = 0.0303 × 10.625 = 0.32 Ω R = 10.6 ± 0.3 Ω A1 [3] GCE AS/A LEVEL – October/November 2012 9702 21

This question in 9702/21 Oct/Nov 2012

6 (a) Define potential difference (p.d.). For Examiner’s … [1] Use (b) A power supply of e.m.f. 240 V and zero internal resistance is connected to a heater as shown in Fig. 6.1. 240 V Fig. 6.1 The wires used to connect the heater to the power supply each have length 75 m. The wires have a cross-sectional area 2.5 mm2 and resistivity 18 nΩ m. The heater has a constant resistance of 38 Ω. (i) Show that the resistance of each wire is 0.54 Ω. [3] (ii) Calculate the current in the wires. current = … A [3] (iii) Calculate the power loss in the wires. power = … W [3] (c) The wires to the heater are replaced by wires of the same length and material but For having a cross-sectional area of 0.50 mm2. Without further calculation, state and explain Examiner’s the effect on the power loss in the wires. Use … … … [2]

12 marks

Mark scheme: 6 (a) p.d. = work (done) / charge OR energy transferred from (electrical to other forms) / (unit) charge B1 [1] (b) (i) R = ρl / A C1 ρ = 18 × 10–9 C1 R = (18 × 10–9 × 75) / 2.5 × 10–6 = 0.54 Ω A1 [3] (ii) V = IR C1 R = 38 + (2 × 0.54) C1 I = 240 / 39.08 = 6.1 (6.14) A A1 [3] GCE AS/A LEVEL – October/November 2013 9702 21 (iii) P = I 2R or P = VI and V = IR or P = V2/R and V = IR C1 P = (6.14)2 × 2 × 0.54 C1 P = 41 (40.7) W A1 [3] (c) area of wire is less (1/5) hence resistance greater (×5) M1 OR R is ∝ 1/A therefore R is greater p.d. across wires greater so power loss in cables increases A1 [2]

This question in 9702/21 Oct/Nov 2013

6 (a) Define potential difference (p.d.). For Examiner’s … [1] Use (b) A power supply of e.m.f. 240 V and zero internal resistance is connected to a heater as shown in Fig. 6.1. 240 V Fig. 6.1 The wires used to connect the heater to the power supply each have length 75 m. The wires have a cross-sectional area 2.5 mm2 and resistivity 18 nΩ m. The heater has a constant resistance of 38 Ω. (i) Show that the resistance of each wire is 0.54 Ω. [3] (ii) Calculate the current in the wires. current = … A [3] (iii) Calculate the power loss in the wires. power = … W [3] (c) The wires to the heater are replaced by wires of the same length and material but For having a cross-sectional area of 0.50 mm2. Without further calculation, state and explain Examiner’s the effect on the power loss in the wires. Use … … … [2]

12 marks

Mark scheme: 6 (a) p.d. = work (done) / charge OR energy transferred from (electrical to other forms) / (unit) charge B1 [1] (b) (i) R = ρl / A C1 ρ = 18 × 10–9 C1 R = (18 × 10–9 × 75) / 2.5 × 10–6 = 0.54 Ω A1 [3] (ii) V = IR C1 R = 38 + (2 × 0.54) C1 I = 240 / 39.08 = 6.1 (6.14) A A1 [3] GCE AS/A LEVEL – October/November 2013 9702 22 (iii) P = I 2R or P = VI and V = IR or P = V2/R and V = IR C1 P = (6.14)2 × 2 × 0.54 C1 P = 41 (40.7) W A1 [3] (c) area of wire is less (1/5) hence resistance greater (×5) M1 OR R is ∝ 1/A therefore R is greater p.d. across wires greater so power loss in cables increases A1 [2]

This question in 9702/22 Oct/Nov 2013

6 (a) Distinguish between electromotive force (e.m.f.) and potential difference (p.d.). … … … [2] (b) A battery of e.m.f. 12 V and internal resistance 0.50 Ω is connected to two identical lamps, as shown in Fig. 6.1. 12 V 0.50 1 Fig. 6.1 Each lamp has constant resistance. The power rating of each lamp is 48 W when connected across a p.d. of 12 V. (i) Explain why the power dissipated in each lamp is not 48 W when connected as shown in Fig. 6.1. … … … [1] (ii) Calculate the resistance of one lamp. resistance = … Ω [2] (iii) Calculate the current in the battery. current = … A [2] (iv) Calculate the power dissipated in one lamp. power = … W [2] (c) A third identical lamp is placed in parallel with the battery in the circuit of Fig. 6.1. Describe and explain the effect on the terminal p.d. of the battery. … … … … [2] Please turn over for Question 7.

11 marks

Mark scheme: 6 (a) e.m.f.: energy converted from chemical / other forms to electrical per unit charge B1 p.d.: energy converted from electrical to other forms per unit charge B1 [2] (b) (i) the p.d. across the lamp is less than 12 V or there are lost volts / power / energy in the battery / internal resistance B1 [1] (ii) R = V2 / P (or V = RI and P = VI) C1 = 144 / 48 = 3.0 Ω A1 [2] GCE AS/A LEVEL – May/June 2014 9702 21 (iii) I = E / (RT + r) C1 = 12 / 2.0 = 6.0 A A1 [2] (iv) power of each lamp = I 2R = (3.0)2 × 3.0 C1 = 27 W A1 [2] (c) less resistance (in circuit) / more current M1 more lost volts / less p.d. across battery A1 [2]

This question in 9702/21 May/June 2014

6 A battery is connected in series with resistors X and Y, as shown in Fig. 6.1. 24 V I X Y B A C 6.0 1 R Fig. 6.1 The resistance of X is constant. The resistance of Y is 6.0 Ω. The battery has electromotive force (e.m.f.) 24 V and zero internal resistance. A variable resistor of resistance R is connected in parallel with X. The current І from the battery is changed by varying R from 5.0 Ω to 20 Ω. The variation with R of І is shown in Fig. 6.2. 2.5 I / A 2.0 1.5 5 10 15 20 R / 1 Fig. 6.2 (a) Explain why the potential difference (p.d.) between points A and C is 24 V for all values of R. … … [1] (b) Use Fig. 6.2 to state and explain the variation of the p.d. across resistor Y as R is increased. Numerical values are not required. … … … [2] (c) For R = 6.0 Ω, (i) show that the p.d. between points A and B is 9.6 V, [2] (ii) calculate the resistance of X, resistance = … Ω [3] (iii) calculate the power provided by the battery. power = … W [2] (d) State and explain qualitatively how the power provided by the battery changes as the resistance R is increased. … … [1]

11 marks

Mark scheme: 6 (a) there are no lost volts / energy lost in the battery or there are no lost volts / energy lost in the internal resistance B1 [1] (b) the current / I decreases (as R increases) M1 p.d. decreases (as R increases) A1 or the parallel resistance (of X and R) increases M1 p.d. across parallel resistors increases, so p.d. (across Y) decreases A1 [2] GCE AS/A LEVEL – May/June 2014 9702 22 (c) (i) current = 2.4 (A) C1 p.d. across AB = 24 – 2.4 × 6 = 9.6 V M1 or total resistance = 10 Ω (= 24 V / 2.4 A) C1 (parallel resistance = 4 Ω), p.d. = 24 × (4 / 10) = 9.6 V M1 [2] (ii) R (AB) = 9.6 / 2.4 = 4.0 Ω C1 1 / 6 + 1 / X = 1 / 4 [must correctly substitute for R] C1 X = 12 Ω A1 or IR = 9.6 / 6.0 = 1.6 (A) (C1) IX = 2.4 – 1.6 = 0.8 (A) (C1) X (= 9.6 / 0.8) = 12 Ω (A1) [3] (iii) power = VI or EI or V 2 / R or E 2 / R or І 2R C1 = 24 × 2.4 or (24)2 / 10 or (2.4)2 × 10 = 57.6 W (allow 2 or more s.f.) A1 [2] (d) power decreases M0 e.m.f. constant or power = 24 × current, and current decreases or e.m.f. constant or power = 242 / resistance, and resistance increases A1 [1]

This question in 9702/22 May/June 2014

5 (a) Explain why the terminal potential difference (p.d.) of a cell with internal resistance may be less than the electromotive force (e.m.f.) of the cell. … … … [2] (b) A battery of e.m.f. 4.5 V and internal resistance r is connected in series with a resistor of resistance 6.0 Ω, as shown in Fig. 5.1. battery 4.5V r I 6.0 1 Fig. 5.1 The current I in the circuit is 0.65 A. Determine (i) the internal resistance r of the battery, r = … Ω [2] (ii) the terminal p.d. of the battery, p.d. = … V [2] (iii) the power dissipated in the resistor, power = … W [2] (iv) the efficiency of the battery. efficiency = … [2] (c) A second resistor of resistance 20 Ω is connected in parallel with the 6.0 Ω resistor in Fig. 5.1. Describe and explain qualitatively the change in the heating effect within the battery. … … … … [3]

13 marks

Mark scheme: 5 (a) lost volts / energy used within the cell / internal resistance B1 when cell supplies a current B1 [2] GCE AS/A LEVEL – May/June 2014 9702 23 (b) (i) E = І(R + r) C1 4.5 = 0.65 (6.0 + r) r = 0.92 Ω A1 [2] (ii) І = 0.65 (A) and V = ІR C1 V = 0.65 × 6 = 3.9 V A1 [2] (iii) P = V 2 / R or P = І2R and P = ІV C1 = (3.9)2 / 6 = 2.5 W A1 [2] (iv) efficiency = power out / power in C1 = І 2R / І 2(R + r) = R / (R + r) = 6.0 / ( 6.0 + 0.92 ) = 0.87 A1 [2] (c) (circuit) resistance decreases B1 current increases M1 more heating effect A1 [3]

This question in 9702/23 May/June 2014

3 The resistance R of a uniform metal wire is measured for different lengths l of the wire. The variation with l of R is shown in Fig. 3.1. 4.0 3.0 R / 1 2.0 1.0 0 0 0.20 0.40 0.60 0.80 1.00 l / m Fig. 3.1 (a) The points shown in Fig. 3.1 do not lie on the best-fit line. Suggest a reason for this. … … [1] (b) Determine the gradient of the line shown in Fig. 3.1. gradient = … [2] (c) The cross-sectional area of the wire is 0.12 mm2. Use your answer in (b) to determine the resistivity of the metal of the wire. resistivity = … Ω m [3] (d) The resistance R of different wires is measured. The wires are of the same metal and same length but have different cross-sectional areas A. On Fig. 3.2, sketch a graph to show the variation with A of R. R 0 0 A Fig. 3.2 [2]

8 marks

Mark scheme: 3 (a) random error (in the measurements) of the length OR resistance B1 [1] (b) gradient = (3.6 – 1.9 ) / (0.8 – 0.4) C1 = 4.25 A1 [2] (c) R = ρl / A C1 ρ = gradient × area = 4.25 × 0.12 × 10–6 C1 = 5.1(0) × 10–7 Ω m A1 [3] (d) resistance decreasing with increasing area B1 correct shape with curve being asymptote to both axes B1 [2]

This question in 9702/21 Oct/Nov 2014

5 A battery of electromotive force (e.m.f.) 12 V and internal resistance r is connected in series to two resistors, each of constant resistance X, as shown in Fig. 5.1. 12 V r I1 X X Fig. 5.1 The current Ι1 supplied by the battery is 1.2 A. The same battery is now connected to the same two resistors in parallel, as shown in Fig. 5.2. 12 V r I2 X X Fig. 5.2 The current Ι2 supplied by the battery is 3.0 A. (a) (i) Show that the combined resistance of the two resistors, each of resistance X, is four times greater in Fig. 5.1 than in Fig. 5.2. [2] (ii) Explain why Ι2 is not four times greater than Ι1. … … … [2] (iii) Using Kirchhoff’s second law, state equations, in terms of e.m.f., current, X and r, for 1. the circuit of Fig. 5.1, … 2. the circuit of Fig. 5.2. … [2] (iv) Use the equations in (iii) to calculate the resistance X. X = … Ω [1] (b) Calculate the ratio power transformed in one resistor of resistance X in Fig. 5.1 . power transformed in one resistor of resistance X in Fig. 5.2 ratio = … [2] (c) The resistors in Fig. 5.1 and Fig. 5.2 are replaced by identical 12 V filament lamps. Explain why the resistance of each lamp, when connected in series, is not the same as the resistance of each lamp when connected in parallel. … … … … [2]

11 marks

Mark scheme: 5 (a) (i) in series 2X or in parallel X / 2 M1 other relationship given and 4× greater in series (than in parallel) A1 [2] (ii) due to the internal resistance B1 total resistance for series circuit is not four times greater than resistance for parallel circuit B1 [2] (iii) 1. E = I1(2X + r) or 12 = 1.2(2X + r) A1 2. E = I2(X/2 + r) or 12 = 3.0(X/2 + r) A1 [2] (iv) 2X + r = 10 and X/2 + r = 4 X = 4.0 Ω A1 [1] (b) P = I2R or V2 / R or VI C1 ratio = [(1.2)2 × 4] / [(1.5)2 × 4] = 0.64 A1 [2] (c) the resistance (of a lamp) changes with V or I B1 V or I is greater in parallel circuit or circuit 2 or V or I is less in series circuit or circuit 1 B1 [2]

This question in 9702/22 Oct/Nov 2014

6 (a) A wire has length 100 cm and diameter 0.38 mm. The metal of the wire has resistivity 4.5 × 10–7 Ω m. Show that the resistance of the wire is 4.0 Ω. [3] (b) The ends B and D of the wire in (a) are connected to a cell X, as shown in Fig. 6.1. 2.0 V cell X 1.0 Ω l B C D 1.5 V metal wire 0.50 Ω cell Y Fig. 6.1 The cell X has electromotive force (e.m.f.) 2.0 V and internal resistance 1.0 Ω. A cell Y of e.m.f. 1.5 V and internal resistance 0.50 Ω is connected to the wire at points B and C, as shown in Fig. 6.1. The point C is distance l from point B. The current in cell Y is zero. Calculate (i) the current in cell X, current = … A [2] (ii) the potential difference (p.d.) across the wire BD, p.d. = … V [1] (iii) the distance l. l = … cm [2] (c) The connection at C is moved so that l is increased. Explain why the e.m.f. of cell Y is less than its terminal p.d. … … … [2]

10 marks

Mark scheme: 6 (a) R = ρl / A C1 A = [π × (0.38 × 10–3)2] / 4 (= 0.113 × 10–6 m2) C1 R = (4.5 × 10–7 × 1.00) / ( [π × (0.38 × 10–3)2] / 4 ) = 4.0 (3.97) Ω M1 [3] (b) (i) І = V / R C1 = 2.0 / 5.0 = 0.4(0) A A1 [2] (ii) p.d. across BD = 4 × 0.4 = 1.6 V A1 [1] (iii) p.d. across BC (l) = 1.5 (V) C1 BC (l) = (1.5 / 1.6) × 100 = 94 (93.75) cm A1 [2] (c) p.d. across wire not balancing e.m.f. of cell OR cell Y has current B1 energy lost or lost volts due to internal resistance B1 [2]

This question in 9702/23 Oct/Nov 2014

5 The variation with potential difference (p.d.) V of current I for a semiconductor diode is shown in Fig. 5.1. 12.0 10.0 8.0 I / mA 6.0 4.0 2.0 0 – 0.5 0 0.5 1.0 V / V Fig. 5.1 (a) Use Fig. 5.1 to describe the variation of the resistance of the diode between V = −0.5 V and V = 0.8 V. … … … … [2] (b) On Fig. 5.2, sketch the variation with p.d. V of current I for a filament lamp. Numerical values are not required. I 0 0 V Fig. 5.2 [2] (c) Fig. 5.3 shows a power supply of electromotive force (e.m.f.) 12 V and internal resistance 0.50 Ω connected to a filament lamp and switch. 12 V 0.50 1 Fig. 5.3 The filament lamp has a power of 36 W when the p.d. across it is 12 V. (i) Calculate the resistance of the lamp when the p.d. across it is 12 V. resistance = … Ω [1] (ii) The switch is closed and the current in the lamp is 2.8 A. Calculate the resistance of the lamp. resistance = … Ω [3] (d) Explain how the two values of resistance calculated in (c) provide evidence for the shape of the sketch you have drawn in (b). … … [1]

9 marks

Mark scheme: 5 (a) very high/infinite resistance for negative voltages up to about 0.4 V B1 resistance decreases from 0.4 V B1 [2] (b) initial straight line from (0,0) into curve with decreasing gradient but not to horizontal M1 repeated in negative quadrant A1 [2] (c) (i) R = 122 / 36 = 4.0 Ω A1 or I = P / V = 36 / 12 = 3.0 A and R = 12 / 3.0 = 4.0 Ω (A1) [1] (ii) lost volts = 0.5 × 2.8 = 1.4 (V) or E = 12 = 2.8 × (R + r) C1 R = V / I = (12 – 1.4) / 2.8 or (R + r) = 4.29 Ω C1 = 3.8 (3.79) Ω or R = 3.8 Ω A1 [3] (d) resistance of the lamp increases with increase of V or I B1 [1]

This question in 9702/21 May/June 2015

1 (a) Use the definition of work done to show that the SI base units of energy are kg m2 s−2. [2] (b) Define potential difference. … … [1] (c) Determine the SI base units of resistance. Show your working. units … [3]

6 marks

Mark scheme: 1 (a) (work =) force × distance or force × displacement or (W =) F × d M1 units of work: kg m s–2 × m = kg m2 s–2 A1 [2] work (done) or energy (transform ed) (from electrical to other forms) (b) (p.d. = ) B1 [1] charge (c) R = V / I B1 units of V: kg m2 s–2 / A s and units of I: A C1 or R = P / I2 [or P = VI and V = IR] (B1) units of P: kg m2 s–3 and units of I: A (C1) or R = V2 / P (B1) units of V: kg m2 s–2 / A s and units of P: kg m2 s–3 (C1) units of R: (kg m2 s–2 / A2 s =) kg m2 s–3 A–2 A1 [3]

This question in 9702/22 May/June 2015

5 A uniform resistance wire AB has length 50 cm and diameter 0.36 mm. The resistivity of the metal of the wire is 5.1 × 10–7 Ω m. (a) Show that the resistance of the wire AB is 2.5 Ω. [2] (b) The wire AB is connected in series with a power supply E and a resistor R as shown in Fig. 5.1. E R M B A 2.5 1 C N D Fig. 5.1 The electromotive force (e.m.f.) of E is 6.0 V and its internal resistance is negligible. The resistance of R is 2.5 Ω. A second uniform wire CD is connected across the terminals of E. The wire CD has length 100 cm, diameter 0.18 mm and is made of the same metal as wire AB. Calculate (i) the current supplied by E, current = … A [4] (ii) the power transformed in wire AB, power = … W [2] (iii) the potential difference (p.d.) between the midpoint M of wire AB and the midpoint N of wire CD. p.d. = … V [2]

10 marks

Mark scheme: 5 (a) R = ρl / A C1 = (5.1 × 10−7 × 0.50) / π(0.18 × 10−3)2 = 2.5 (2.51) Ω M1 [2] (b) (i) resistance of CD = 8 × resistance of AB = 20 (Ω) C1 circuit resistance = [1 / 5.0 + 1 / 20]−1 = 4.0 (Ω) C1 current = V / R = 6.0 / 4.0 C1 = 1.5 A A1 [4] (ii) power in AB = I 2R or power = V 2 / R C1 = (1.2)2 × 2.5 = 3.6 W = (3.0)2 / 2.5 = 3.6 W A1 [2] (iii) potential drop A to M = 1.25 × 1.2 = 1.5 V M1 potential drop C to N = 3.0 V p.d. MN = 1.5 V A1 [2]

This question in 9702/23 May/June 2015

6 (a) Define electromotive force (e.m.f.) for a battery. … … [1] (b) A battery of e.m.f. 6.0 V and internal resistance 0.50 Ω is connected in series with two resistors X and Y, as shown in Fig. 6.1. 6.0 V 0.50 1 4.0 1 12 1 X Y Fig. 6.1 The resistance of X is 4.0 Ω and the resistance of Y is 12 Ω. Calculate (i) the current in the circuit, current = … A [2] (ii) the terminal potential difference (p.d.) across the battery. p.d. = … V [1] (c) A resistor Z is now connected in parallel with resistor Y in the circuit in (b). The new arrangement is shown in Fig. 6.2. 6.0 V 0.50 1 4.0 1 12 1 X Y Z Fig. 6.2 Resistor Y is made from a wire of length l and diameter d. Resistor Z is a wire made from the same material as Y. The length of the wire for Z is l / 2 and the diameter is d / 2. (i) Calculate the resistance R of the combination of resistors Y and Z. R = … Ω [3] (ii) State and explain the effect on the terminal p.d. across the battery. A numerical value is not required. … … … … [2] (d) For the circuits given in (b) and (c), show that the ratio power developed in the external circuit in Fig. 6.1 power developed in the external circuit in Fig. 6.2 is approximately 0.8. [3]

12 marks

Mark scheme: 6 (a) energy converted from chemical to electrical per unit charge B1 [1] (b) (i) current = E / (R + r) C1 = 6.0 / (16 + 0.5) = 0.36 (0.364) A A1 [2] (ii) terminal p.d. = (0.36 × 16) = 5.8 V or (6 – 0.36 × 0.5) = 5.8 V A1 [1] (c) (i) use of R = ρl / A or proportionality with length and inverse proportionality with area or d 2 C1 d / 2 and l / 2 gives resistance of Z = 2RY = 24 (Ω) C1 R = resistance of parallel combination = [1/24 + 1/12]–1 = 8(.0) (Ω) A1 [3] (ii) resistance of circuit less therefore current larger B1 lost volts greater therefore terminal p.d. less B1 [2] (d) power = I 2 R or VI or V 2 / R C1 current in second circuit (= 6.0 / 12.5) = 0.48 (A) B1 ratio = [(0.36)2 × 16] / [(0.48)2 × 12] = 0.75 [0.77 if full s.f. used] B1 [3]

This question in 9702/21 Oct/Nov 2015

3 A trolley T moves at speed 1.2 m s–1 along a horizontal frictionless surface. The trolley collides with a stationary block on the end of a fixed spring, as shown in Fig. 3.1. block 1.2 m s–1 fixed end of spring T horizontal frictionless surface Fig. 3.1 The mass of T is 250 g. T compresses the spring by 5.4 cm as it comes to rest. The relationship between the force F applied to the block and the compression x of the spring is shown in Fig. 3.2. 6.0 5.0 x / cm 4.0 3.0 2.0 1.0 0 0 2.0 4.0 6.0 F / N Fig. 3.2 (a) Use Fig. 3.2 to determine (i) the spring constant of the spring, spring constant = … N m–1 [2] (ii) the work done by T compressing the spring by 5.4 cm. work done = … J [2] (b) The spring then expands and causes T to move in a direction opposite to its initial direction. At the time that T loses contact with the block, it is moving at a speed of 0.75 m s–1. From the time that T is in contact with the block, (i) describe the energy changes, … … … … [2] (ii) determine the change in momentum of T. change in momentum = … N s [2]

8 marks

Mark scheme: 3 (a) (i) k = F / x or 1 / gradient C1 (k = 4.4 / (5.4 × 10–2) =) 81 (81.48) N m–1 A1 [2] (ii) work done = area under line or ½Fx or ½kx2 C1 (= 0.5 × 4.4 × 5.4 × 10–2 =) 0.12 (0.119) J A1 [2] (b) (i) kinetic energy/Ek of trolley/T (and block) changes to EPE/strain energy/elastic energy of spring B1 EPE changes to KE of trolley/T and KE of block or to give lower KE to trolley B1 [2] (ii) change in momentum = m(v + u) C1 = 0.25 (0.75 + 1.2) = 0.49 (0.488) N s A1 [2]

This question in 9702/22 Oct/Nov 2015

5 A 240 V power supply S with negligible internal resistance is connected to four resistors, as shown in Fig. 5.1. 0.40 A 240 V S I1 A 550 1 950 1 I2 B 350 1 R Fig. 5.1 Two resistors of resistance 550 Ω and 950 Ω are connected in series across S. Two resistors of resistance 350 Ω and R are also connected in series across S. The current supplied by S is 0.40 A. Currents I1 and I2 in the circuit are shown in Fig. 5.1. (a) Calculate (i) current I1, I1 = … A [2] (ii) resistance R, R = … Ω [2] (iii) the ratio power transformed in resistor of resistance 350 Ω . power transformed in resistor of resistance 550 Ω ratio = … [2] (b) Two points are labelled A and B, as shown in Fig. 5.1. (i) Calculate the potential difference VAB between A and B. VAB = … V [2] (ii) The resistance R is increased. State and explain the effect on VAB. … … … [1]

9 marks

Mark scheme: 5 (a) (i) I = V / R C1 ( = 240 / 1500 =) 0.16 A A1 [2] (ii) I2 = 0.40 – 0.16 (= 0.24) C1 0.24 (350 + R) = 240 R = 650 Ω A1 [2] (iii) power = IV or I2R or V2/R C1 ratio = (84 × 0.24) / (88 × 0.16) or [(0.24)2 × 350] / [(0.16)2 × 550] or (842/350) / (882 / 550) or 20.16 / 14.08 = 1.4(3) A1 [2] (b) (i) p.d. across 350 Ω resistor = 0.24 × 350 or p.d. across 550 Ω resistor = 0.16 × 550 C1 V350 = 84 (V) and V550 = 88 (V) gives VAB = 4.0 V or V950 = 152 (V) and VR = 156 V gives VAB = 4.0 V A1 [2] (ii) p.d. across R increases or potential at B increases or V350 decreases hence VAB increases B1 [1]

This question in 9702/22 Oct/Nov 2015

5 (a) The I-V characteristic of a semiconductor diode is shown in Fig. 5.1. 14.0 12.0 10.0 I / mA 8.0 6.0 4.0 2.0 0 0 0.20 0.40 0.60 0.80 V / V Fig. 5.1 (i) Use Fig. 5.1 to explain the variation of the resistance of the diode as V increases from zero to 0.8 V. … … … … … … [3] (ii) Use Fig. 5.1 to determine the resistance of the diode for a current of 4.4 mA. resistance = … Ω [2] (b) A cell of e.m.f. 1.2 V and negligible internal resistance is connected in series to a semiconductor diode and a resistor R1, as shown in Fig. 5.2. 1.2 V 7.6 mA R1 R2 375 1 Fig. 5.2 A resistor R2 of resistance 375 Ω is connected across the cell. The diode has the characteristic shown in Fig. 5.1. The current supplied by the cell is 7.6 mA. Calculate (i) the current in R2, current = … A [1] (ii) the resistance of R1, resistance = … Ω [2] (iii) the ratio power dissipated in the diode . power dissipated in R2 ratio = … [2]

10 marks

Mark scheme: 5 (a) (i) resistance = V / I B1 very high/infinite resistance at low voltages B1 resistance decreases as V increases B1 [3] (ii) p.d. from graph 0.50 (V) C1 resistance = 0.5 / (4.4 × 10–3) = 110 (114) Ω A1 [2] (b) (i) current (= 1.2 / 375) = 3.2 × 10–3 A A1 [1] (ii) current in diode = 4.4 × 10–3 (A) total resistance = 1.2 / 4.4 × 10–3 = 272.7 (Ω) C1 resistance of R1 = 272.7 – 113.6 = 160 (159) Ω A1 or p.d. across diode = 0.5 V and p.d. across R1 = 0.7 V (C1) resistance of R1 = 0.7 / 4.4 × 10–3 = 160 (159) Ω (A1) [2] (iii) power = IV or I2R or V2/ R C1 ratio = (4.4 × 0.5) / (3.2 × 1.2) or [(4.4)2 × 114] / [(3.2)2 × 375] or [(0.5)2 × 375] / [114 × (1.2)2] = 0.57 A1 [2]

This question in 9702/23 Oct/Nov 2015

7 (a) Electric current is a flow of charge carriers. The charge on the carriers is quantised. Explain what is meant by quantised. … [1] (b) A battery of electromotive force (e.m.f.) 9.0 V and internal resistance 0.25 Ω is connected in series with two identical resistors X and a resistor Y, as shown in Fig. 7.1. battery 9.0 V 0.25 1 X Y X 0.15 1 2.7 1 0.15 1 Fig. 7.1 The resistance of each resistor X is 0.15 Ω and the resistance of resistor Y is 2.7 Ω. (i) Show that the current in the circuit is 2.8 A. [3] (ii) Calculate the potential difference across the battery. potential difference = … V [2] (c) Each resistor X connected in the circuit in (b) is made from a wire with a cross-sectional area of 2.5 mm2. The number of free electrons per unit volume in the wire is 8.5 × 1029 m–3. (i) Calculate the average drift speed of the electrons in X. drift speed = … m s–1 [2] (ii) The two resistors X are replaced by two resistors Z made of the same material and length but with half the diameter. Describe and explain the difference between the average drift speed in Z and that in X. … … … … [2] [Total: 10]

10 marks

Mark scheme: 7 (a) charge exists only in discrete amounts B1 [1] (b) (i) E = I(R + r) or V = IR C1 (total resistance =) 2.7 + 0.30 + 0.25 (= 3.25 Ω) M1 I = 9.0 / (2.7 + 0.30 + 0.25) or 9.0 / 3.25 = 2.8 A A1 [3] (ii) V = IRext C1 = 2.77 × 3.0 or 2.8 × 3.0 or V = E – Ir (C1) = 9.0 – 2.77 × 0.25 or 9.0 – 2.8 × 0.25 V = 8.3 (8.31) V or 8.4 V A1 [2] (c) (i) I = nevA v = 2.77 / (8.5 × 1029 × 1.6 × 10–19 × 2.5 × 10–6) M1 = 8.1 (8.147) × 10–6 m s–1 or 8.2 × 10–6 m s–1 A1 [2] (ii) A reduces by a factor 4 (1/4 less) or resistance of Z goes up by 4× M1 current goes down but by less than a factor of 4 (as total resistance does not go up by a factor of 4) so drift speed goes up A1 [2]

This question in 9702/22 May/June 2016

6 (a) Define the ohm. … [1] (b) A 15 V battery with negligible internal resistance is connected to two resistors P and Q, as shown in Fig. 6.1. 15 V P 12 1 Q Fig. 6.1 The resistors are made of wires of the same material. The wire of P has diameter d and length 2l. The wire of Q has diameter 2d and length l. The resistance of P is 12 Ω. (i) Show that the resistance of Q is 1.5 Ω. [3] (ii) Calculate the total power dissipated in the resistors P and Q. power = … W [3] (iii) Determine the ratio average drift speed of the charge carriers in P . average drift speed of the charge carriers in Q ratio = … [3] [Total: 10]

10 marks

Mark scheme: 6 (a) ohm is volt per ampere or volt / ampere B1 [1] (b) (i) R = ρl / A B1 RP = 4ρ(2l) / πd2 or 8ρl / πd2 or RQ = ρl / πd2 or ratio idea e.g. length is halved hence R halved and diameter is halved hence R is 1/4 C1 RQ (= 4ρl / π4d2) = ρl / πd2 = RP / 8 (= 12 / 8) = 1.5 Ω A1 [3] (ii) power = I 2R or V 2 / R or VI C1 = (1.25)2 × 12 + (10)2 × 1.5 or (15)2/12 + (15)2/1.5 or 15 × 11.25 C1 = (18.75 + 150 =) 170 (168.75) W A1 [3] (iii) IP = (15 / 12 =) 1.25 (A) and IQ = (15 / 1.5 =) 10 (A) C1 vP / vQ = IPnAQe / IQnAPe or (1.25 × πd 2) / (10 × πd 2/4) C1 = 0.5 A1 [3]

This question in 9702/23 May/June 2016

6 (a) Define the ohm. … … [1] (b) A cell X of electromotive force (e.m.f.) 1.5 V and negligible internal resistance is connected in series to three resistors A, B and C, as shown in Fig. 6.1. X 1.5 V A 6.0 Ω C Ω B 4.0 3.0 Ω Fig. 6.1 Resistors A and B have resistances 6.0 Ω and 3.0 Ω respectively and are connected in parallel. Resistor C has resistance 4.0 Ω and is connected in series with the parallel combination. Calculate (i) the current in the circuit, current = … A [3] (ii) the current in resistor B, current = … A [1] (iii) the ratio power dissipated in resistor B . power dissipated in resistor C ratio = … [2] (c) The resistors A, B and C in (b) are wires of the same material and have the same length. (i) Explain how the resistors may be made with different resistance values. … [1] (ii) Calculate the ratio average drift speed of the charge carriers in resistor B . average drift speed of the charge carriers in resistor C ratio = … [2] (d) A cell of e.m.f. 1.5 V and negligible internal resistance is connected in parallel with cell X in Fig. 6.1 with their positive terminals together. State the change, if any, to the current in (i) cell X, … [1] (ii) resistor C. … [1] [Total: 12]

12 marks

Mark scheme: 6(a) volt / ampere B1 6(b)(i) RT = [1 / 3.0 + 1 / 6.0]–1 + 4.0 (= 6.0 Ω) C1 Ι = 1.5 / 6.0 C1 = 0.25 A A1 6(b)(ii) VB = 0.5 V I = 0.5 / 3.0 = 0.17 (0.167) A A1 6(b)(iii) P = I 2R or VI or V 2 / R C1 ratio = (0.1672 × 3.0) / (0.252 × 4.0) = 0.33 A1 6(c)(i) vary/change/different radius/diameter/cross-sectional area (of wire) B1 6(c)(ii) v = I / Ane ( ) ( ) = B C / ratio / A A B C I I or × C B A A B C I I C1 (R ∝ 1 / A so) ratio = × B C R R B C I I = × 0.167 3.0 0.25 4.0 = 0.50 A1 6(d)(i) 0.25 A to 0.13 (0.125) A or halved A1 6(d)(ii) no change A1

This question in 9702/21 May/June 2017

1 (a) State two SI base units other than kilogram, metre and second. 1. … 2. … [1] (b) Determine the SI base units of resistivity. base units … [3] (c) (i) A wire of cross-sectional area 1.5 mm2 and length 2.5 m has a resistance of 0.030 Ω. Calculate the resistivity of the material of the wire in nΩ m. resistivity = … nΩ m [3] (ii) 1. State what is meant by precision. … … 2. Explain why the precision in the value of the resistivity is improved by using a micrometer screw gauge rather than a metre rule to measure the diameter of the wire. … … … [2] [Total: 9]

9 marks

Mark scheme: 1(a) kelvin, mole, ampere, candela any two B1 1(b) use of resistivity = RA / l and V = IR (to give ρ = VA/ Il) C1 units of V: (work done / charge) kg m2 s–2 (A s)–1 C1 units of resistivity: (kg m2 s–3 A–1 A–1 m) = kg m3 s–3 A–2 A1 or use of R = ρL / A and P = I2R (gives ρ = PA / I2L) (C1) units of P: kg m2 s–3 (C1) units of resistivity: (kg m2 s–3 × m2) / (A2 × m) = kg m3 s–3 A–2 (A1) 1(c)(i) ρ = (RA/l) C1 = (0.03 × 1.5 × 10–6) / 2.5 (= 1.8 × 10–8) C1 = 18 nΩ m A1 1(c)(ii) 1. precision is determined by the range in the measurements/values/readings/data/results B1 2. metre rule measures to ± 1 mm and micrometer to ± 0.01 mm (so there is less (percentage) uncertainty/random error) B1

This question in 9702/22 May/June 2017

7 (a) Define electromotive force (e.m.f.) of a cell. … … [1] (b) A cell C of e.m.f. 1.50 V and internal resistance 0.200 Ω is connected in series with resistors X and Y, as shown in Fig. 7.1. C 1.50 V A B 0.200 Ω X Y Fig. 7.1 The resistance of X is constant and the resistance of Y can be varied. (i) The resistance of Y is varied from 0 to 8.00 Ω. State and explain the variation in the potential difference (p.d.) between points A and B (terminal p.d. across C). Numerical values are not required. … … … … [3] (ii) The resistance of Y is set at 6.00 Ω. The current in the circuit is 0.180 A. Calculate 1. the resistance of X, resistance = … Ω [2] 2. the p.d. between points A and B, p.d. = … V [2] 3. the efficiency of the cell. efficiency = … [2] [Total: 10]

10 marks

Mark scheme: 7(a) energy transformed from chemical to electrical / unit charge (driven around a complete circuit) B1 7(b)(i) the current decreases (as resistance of Y increases) M1 lost volts go down (as resistance of Y increases) M1 p.d. AB increases (as resistance of Y increases) A1 7(b)(ii)1. 1.50 = 0.180 × (6.00 + 0.200 + RX) C1 RX = 2.1(3) Ω A1 7(b)(ii)2. p.d. AB = 1.5 − (0.180 × 0.200) or 0.18 × (2.13 + 6.00) C1 = 1.46(4) V A1 7(b)(ii)3. efficiency = (useful) power output / (total) power input or IV / IE C1 ( = 1.46 / 1.5) = 0.97 [0.98 if full figures used] A1

This question in 9702/22 May/June 2017

6 (a) Describe the I–V characteristic of (i) a metallic conductor at constant temperature, … … [1] (ii) a semiconductor diode. … … … [2] (b) Two identical filament lamps are connected in series and then in parallel to a battery of electromotive force (e.m.f.) 12 V and negligible internal resistance, as shown in Fig. 6.1a and Fig. 6.1b. 12 V 12 V Fig. 6.1a Fig. 6.1b The I–V characteristic of each lamp is shown in Fig. 6.2. 6.0 I / A 4.0 2.0 0 0 2.0 4.0 6.0 8.0 10.0 12.0 V / V Fig. 6.2 (i) Use the information shown in Fig. 6.2 to determine the current through the battery in 1. the circuit of Fig. 6.1a, current = … A 2. the circuit of Fig. 6.1b. current = … A [3] (ii) Calculate the total resistance in 1. the circuit of Fig. 6.1a, resistance = … Ω 2. the circuit of Fig. 6.1b. resistance = … Ω [3] (iii) Calculate the ratio power dissipated in a lamp in the circuit of Fig. 6.1a . power dissipated in a lamp in the circuit of Fig. 6.1b ratio = … [2] [Total: 11]

11 marks

Mark scheme: 6(a)(i) straight line through the origin B1 6(a)(ii) zero current for one direction (–ve V) up to zero or a few tenths of volt (+ve V) B1 straight line positive gradient/increasing gradient (+ve V) B1 6(b)(i) 1. current = 2.8 A A1 2. 4(.0) A for each lamp C1 current in circuit = 8(.0) A A1 6(b)(ii) use of R = V / I with correct values of V from graph for each arrangement C1 1. series resistance (= 2.1 + 2.1) = 4.2 or 4.3 Ω or (12 / 2.8) = 4.3 Ω A1 2. parallel resistance 1.5 Ω (each lamp 3.0 Ω) or (12 / 8.0) = 1.5 Ω A1 6(b)(iii) power = IV or V 2 / R or I2R C1 ratio = (2.8 × 6.0) / (4.0 × 12) = 0.35 A1

This question in 9702/23 May/June 2017

7 (a) Define the ohm. … [1] (b) Wires are used to connect a battery of negligible internal resistance to a lamp, as shown in Fig. 7.1. wire wire Fig. 7.1 The lamp is at its normal operating temperature. Some data for the filament wire of the lamp and for the connecting wires of the circuit are shown in Fig. 7.2. filament wire connecting wires diameter d 14 d total length L 7.0 L resistivity of metal ρ 0.028 ρ (at normal operating temperature) Fig. 7.2 (i) Show that resistance of filament wire = 1000. total resistance of connecting wires [2] (ii) Use the information in (i) to explain qualitatively why the power dissipated in the filament wire of the lamp is greater than the total power dissipated in the connecting wires. … … … [1] (iii) The lamp is rated as 12 V, 6.0 W. Use the information in (i) to determine the total resistance of the connecting wires. total resistance of connecting wires = … Ω [3] (iv) The diameter of the connecting wires is decreased. The total length of the connecting wires and the resistivity of the metal of the connecting wires remain the same. State and explain the change, if any, that occurs to the resistance of the filament wire of the lamp. … … … … … [3] [Total: 10]

10 marks

Mark scheme: 7(a) (the ohm is) volt / ampere B1 7(b)(i) R = ρ L / A C1 ratio = [ρ L / (πd 2 / 4)] / [0.028ρ × 7.0L / {π(14d)2/ 4}] = 1000 or ratio = 142 / (0.028 × 7) = 1000 A1 7(b)(ii) same current (in connecting and filament wires) and the lamp/filament (wire) has greater resistance B1 7(b)(iii) P = V 2 / R or P = VI or P = I2R C1 (for filament wire) R = 122 / 6.0 or R = 6.0 / 0.502 or R = 12 / 0.50 C1 (for filament wire) R = 24 Ω (for connecting wire) R = 24 / 1000 = 2.4 × 10–2 Ω A1 7(b)(iv) resistance of connecting wire increases B1 current in circuit/lamp/filament (wire) decreases or potential difference across lamp/filament (wire) decreases M1 (so) resistance of lamp/filament (wire) decreases A1

This question in 9702/21 Oct/Nov 2017

6 (a) State what is meant by an electric current. … [1] (b) A metal wire has length L and cross-sectional area A, as shown in Fig. 6.1. A I L Fig. 6.1 I is the current in the wire, n is the number of free electrons per unit volume in the wire, v is the average drift speed of a free electron and e is the charge on an electron. (i) State, in terms of A, e, L and n, an expression for the total charge of the free electrons in the wire. … [1] (ii) Use your answer in (i) to show that the current I is given by the equation I = nAve. [2] (c) A metal wire in a circuit is damaged. The resistivity of the metal is unchanged but the cross- sectional area of the wire is reduced over a length of 3.0 mm, as shown in Fig. 6.2. 3.0 mm damaged length current 0.69 d d 0.50 A cross-section X cross-section Y Fig. 6.2 The wire has diameter d at cross-section X and diameter 0.69 d at cross-section Y. The current in the wire is 0.50 A. (i) Determine the ratio average drift speed of free electrons at cross-section Y . average drift speed of free electrons at cross-section X ratio = … [2] (ii) The main part of the wire with cross-section X has a resistance per unit length of 1.7 × 10–2 Ω m–1. For the damaged length of the wire, calculate 1. the resistance per unit length, resistance per unit length = … Ω m–1 [2] 2. the power dissipated. power = … W [2] (iii) The diameter of the damaged length of the wire is further decreased. Assume that the current in the wire remains constant. State and explain qualitatively the change, if any, to the power dissipated in the damaged length of the wire. … … … [2] [Total: 12]

12 marks

Mark scheme: 6(a) flow of charge carriers B1 6(b)(i) nALe B1 6(b)(ii) (t is time taken for electrons to move length L) I = Q / t B1 I = nALe / t or I = nALe / (L / v) or I = nAvte / t and I = nAve B1 6(c)(i) ratio = area at X / area at Y = [πd 2 / 4] / [π(0.69d)2 / 4] or d 2 / (0.69d)2 or 1 / 0.692 C1 = 2.1 A1 6(c)(ii) 1. R = ρ L / A or R / L ∝ 1 / A C1 resistance per unit length = 1.7 × 10–2 × (area at X / area at Y) = 1.7 × 10–2 × 2.1 = 3.6 × 10–2 Ω m–1 A1 2. P = I 2R or P = V 2 / R C1 R = 3.6 × 10–2 × 3.0 × 10–3 (= 1.08 × 10–4 Ω) P = 0.502 × 1.08 × 10–4 or P = (5.4 × 10–5)2 / 1.08 × 10–4 = 2.7 × 10–5 W A1 Question Answer Marks 6(c)(iii) (cross-sectional area decreases so) resistance increases M1 (P = I 2R, so) power increases A1

This question in 9702/22 Oct/Nov 2017

6 A filament lamp is rated as 30 W, 120 V. A potential difference of 120 V is applied across the lamp. (a) For the filament wire of the lamp, calculate (i) the current, current = … A [2] (ii) the number of electrons passing a point in 3.0 hours. number = … [2] (b) Show that the resistance of the filament wire is 480 Ω. [2] (c) The filament wire has an uncoiled length of 580 mm and is made of metal. The metal has resistivity 6.1 × 10–7 Ω m at the operating temperature of the lamp. Calculate the diameter of the wire. diameter = … m [3] (d) The potential difference across the lamp is now reduced. State and explain the effect, if any, on the resistance of the filament wire. … … [1] [Total: 10]

10 marks

Mark scheme: 6(a)(i) C1 I = 30 / 120 = 0.25 A A1 6(a)(ii) Q = 0.25 × 3.0 × 3600 (= 2700) C1 number = (0.25 × 3.0 × 3600) / 1.60 × 10–19 = 1.7 × 1022 A1 6(b) R = V / I or R = P / I 2 or R = V 2/ P C1 = 120 / 0.25 or = 30 / 0.252 or = 1202/ 30 = 480 Ω A1 Question Answer Marks 6(c) R = ρl / A C1 A = (6.1 × 10–7 × 580 × 10–3) / 480 (= 7.37 × 10–10) C1 d = [(4 × 7.37 × 10–10) / π]1/2 = 3.1 × 10–5 m A1 6(d) temperature decreases and so resistance decreases B1

This question in 9702/23 Oct/Nov 2017

6 A wire X has a constant resistance per unit length of 3.0 Ω m–1 and a diameter of 0.48 mm. (a) Calculate the resistivity of the metal of wire X. resistivity = … Ω m [3] (b) The wire X is connected into the circuit shown in Fig. 6.1. 5.0 V 2.0 Ω 1.6 A wire X 4.5 Ω R Fig. 6.1 The battery has an electromotive force (e.m.f.) of 5.0 V and an internal resistance of 2.0 Ω. The wire X and a resistor R of resistance 4.5 Ω are connected in parallel. The current in the battery is 1.6 A. (i) Calculate the potential difference across resistor R. potential difference = … V [1] (ii) Determine, for wire X, 1. its resistance, resistance = … Ω [3] 2. its length. length = … m [1] [Total: 8] Please turn over for Question 7.

8 marks

Mark scheme: 6(a) C1 3.0 = ρ / [π × (0.48 × 10–3 / 2)2] C1 ρ = 5.4 × 10–7 Ω m A1 6(b)(i) p.d. = 5.0 – (2.0 × 1.6) = 1.8 V A1 6(b)(ii)1. current in resistor = 1.8 / 4.5 (= 0.40 A) C1 current in wire = 1.6 – 0.40 (= 1.2 A) C1 RX = 1.8 / 1.2 = 1.5 Ω A1 or RT = 1.8 / 1.6 or (5.0 / 1.6) – 2.0 (= 1.125 Ω) (C1) (1 / 1.125) = (1 / 4.5) + (1 / RX) (C1) RX = 1.5 Ω (A1) 6(b)(ii)2. length = 1.5 / 3.0 or 1.5 × 1.8 × 10–7 / (5.4 × 10–7) = 0.50 m A1

This question in 9702/23 May/June 2018

6 (a) Define the volt. … … [1] (b) A battery of electromotive force (e.m.f.) 7.0 V and negligible internal resistance is connected in series with three components, as shown in Fig. 6.1. 7.0 V Z 1.4 V X Y 5.2 Ω 6.0 Ω Fig. 6.1 Resistor X has a resistance of 5.2 Ω. The resistance of the filament wire of lamp Y is 6.0 Ω. The potential difference across resistor Z is 1.4 V. (i) Calculate the current in the circuit. current = … A [2] (ii) Determine the resistance of resistor Z. resistance = … Ω [1] (iii) Calculate the percentage efficiency with which the battery supplies power to the lamp. efficiency = … % [3] (iv) The filament wire of the lamp is made of metal of resistivity 3.7 × 10–7 Ω m at its operating temperature in the circuit. Determine, for the filament wire, the value of α where cross-sectional area α = . length α = … m [2] [Total: 9]

9 marks

Mark scheme: 6(a) joule / coulomb B1 6(b)(i) 7.0 = (I × 5.2) + (I × 6.0) + 1.4 C1 I = 0.50 A A1 6(b)(ii) R = 1.4 / 0.50 = 2.8 Ω A1 6(b)(iii) P = EI or P = VI or P = I2R or P = V2 / R C1 efficiency = [(0.502 × 6.0) / (7.0 × 0.50)] (×100) or efficiency = [(0.50 × 3.0) / (7.0 × 0.50)] (×100) or efficiency = [(3.02 / 6.0) / (7.0 × 0.50)] (×100) C1 efficiency = 43% A1 6(b)(iv) R = ρl / A C1 α = ρ / R = 3.7 × 10–7 / 6.0 = 6.2 × 10–8 m A1

This question in 9702/22 Oct/Nov 2018

6 A battery of electromotive force (e.m.f.) E and internal resistance r is connected to a variable resistor of resistance R, as shown in Fig. 6.1. r E I R V Fig. 6.1 The current in the circuit is I and the potential difference across the variable resistor is V. (a) Explain, in terms of energy, why V is less than E. … … [1] (b) State an equation relating E, I, r and V. … [1] (c) The resistance R of the variable resistor is varied. The variation with I of V is shown in Fig. 6.2. 3.0 V /V 2.0 1.0 0 0 0.5 1.0 1.5 2.0 I /A Fig. 6.2 Use Fig. 6.2 to: (i) explain how it may be deduced that the e.m.f. of the battery is 2.8 V … … [1] (ii) calculate the internal resistance r. r = … Ω [2] (d) The battery stores 9.2 kJ of energy. The variable resistor is adjusted so that V = 2.1 V. Use Fig. 6.2 to: (i) calculate resistance R R = … Ω [1] (ii) calculate the number of conduction electrons moving through the battery in a time of 1.0 s number = … [1] (iii) determine the time taken for the energy in the battery to become equal to 1.6 kJ. (Assume that the e.m.f. of the battery and the current in the battery remain constant.) time taken = … s [3] [Total: 10]

10 marks

Mark scheme: 6(a) energy is dissipated in the internal resistance B1 6(b) E = V + Ir B1 6(c)(i) (graph shows) maximum value of potential difference is 2.8 (V) or (graph shows) when current/I (from battery) is zero, V is 2.8 (V) / E B1 6(c)(ii) r = (–)gradient or r = (E – V) / I or substituted values from the graph for E, V and I C1 r = 1.4 Ω A1 6(d)(i) R = 2.1 / 0.50 = 4.2 Ω A1 6(d)(ii) number = 0.50 / 1.60 × 10–19 = 3.1 × 1018 A1 6(d)(iii) energy = EIt or P = EI and P = W / t C1 (9.2 – 1.6) × 103 = 2.8 × 0.50 × t C1 t = 5.4 × 103 s A1

This question in 9702/21 May/June 2019

1 (a) (i) Define resistance. … … [1] (ii) A potential difference of 0.60 V is applied across a resistor of resistance 4.0 GΩ. Calculate the current, in pA, in the resistor. current = … pA [2] (b) The energy E transferred when charge Q moves through an electrical component is given by the equation E = QV where V is the potential difference across the component. Use the equation to determine the SI base units of potential difference. SI base units … [3] [Total: 6]

6 marks

Mark scheme: 1(a)(i) potential difference / current B1 1(a)(ii) R = 4.0 × 109 (Ω) C1 I = 0.60 / 4.0 × 109 = 1.5 × 10–10 (A) I = 150 pA A1 1(b) units of energy: kg m2 s–2 C1 units of charge: A s C1 units of potential difference: (kg m2 s–2 / A s =) kg m2 A–1 s–3 A1

This question in 9702/23 May/June 2019

6 (a) Define the ohm. … [1] (b) A battery of electromotive force (e.m.f.) E and internal resistance 1.5 Ω is connected to a network of resistors, as shown in Fig. 6.1. 1.5 E I 2.0 RZ 1.8 A Y Z 8.0 0.60 A X Fig. 6.1 Resistor X has a resistance of 8.0 Ω. Resistor Y has a resistance of 2.0 Ω. Resistor Z has a resistance of RZ. The current in X is 0.60 A and the current in Y is 1.8 A. (i) Calculate: 1. the current I in the battery I = … A [1] 2. resistance RZ Ω [2] RZ = … 3. e.m.f. E. E = … V [2] (ii) Resistors X and Y are each made of wire. The two wires have the same length and are made of the same metal. Determine the ratio: cross-sectional area of wire X 1. cross-sectional area of wire Y ratio = … [2] average drift speed of free electrons in X 2. . average drift speed of free electrons in Y ratio = … [2] [Total: 10] Please turn over for Question 7.

10 marks

Mark scheme: 6(a) volt / ampere B1 6(b)(i) 1. I = 1.8 + 0.60 = 2.4 A A1 2. (8.0 × 0.60) = 1.8 × (2.0 + RZ) C1 RZ = 0.67 Ω A1 3. E – (2.4 × 1.5) = (0.60 × 8.0) or E – (2.4 × 1.5) = 1.8 × (2.0 + 0.67) or E = 2.4 × [1.5 + (8.0 × 2.67) / (8.0 + 2.67)] C1 E = 8.4 V A1 6(b)(ii) 1. R = ρL / A or R ∝ 1 / A C1 ratio = RY / RX = 2.0 / 8.0 = 0.25 A1 2. I ∝ Av or IX / IY = AXvX / AYvY C1 ratio = (0.60 / 1.8) × (1 / 0.25) = 1.3 A1

This question in 9702/23 May/June 2019

6 (a) Define electric potential difference (p.d.). … … [1] (b) The variation with potential difference V of the current I in a semiconductor diode is shown in Fig. 6.1. 30 25 I / mA 20 15 10 5 0 0 0.5 1.0 V / V Fig. 6.1 Use Fig. 6.1 to describe qualitatively the variation of the resistance of the diode as V increases from 0 to 1.0 V. … … … … [2] (c) The diode in (b) is part of the circuit shown in Fig. 6.2. 2.0 V 15 mA 60 Ω X Y Fig. 6.2 The cell of electromotive force (e.m.f.) 2.0 V and negligible internal resistance is connected in series with the diode and resistors X and Y. The resistance of Y is 60 Ω. The current in the cell is 15 mA. (i) Use Fig. 6.1 to determine the resistance of the diode. resistance = … Ω [3] (ii) Calculate: 1. the resistance of X resistance = … Ω [3] 2. the ratio power dissipated in resistor Y total power produced by the cell. ratio = … [2]

11 marks

Mark scheme: 6(a) work done / charge or energy (transferred from electrical to other forms) / charge B1 6(b) for V < 0.25 V resistance is infinite/very high (as current is zero) B1 for V > 0.25 V resistance decreases (as V increases) B1 6(c)(i) R = V / I C1 = 0.75 / (15 × 10–3) C1 = 50 Ω A1 Question Answer Marks 6(c)(ii) 1. VY = 15 × 10–3 × 60 (= 0.90 V) C1 VX = 2.0 – 0.90 – 0.75 (= 0.35 V) C1 RX = 0.35 / (15 × 10–3) = 23 Ω A1 or total R = 60 + 50 + RX (C1) 60 + 50 + RX = 2.0 / (15 × 10–3) (C1) RX = 23 Ω (A1) 2. P = VI or P = EI or P = I2R or P = V2 / R C1 ratio = ( ) 2 3 3 15 10 60 2.0 15 10 − − × × × × or 3 3 0.90 15 10 2.0 15 10 − − × × × × or ( ) 2 3 0.90 / 60 2.0 15 10− × × = 0.45 A1

This question in 9702/21 Oct/Nov 2019

6 (a) State Kirchhoff’s first law. … … [1] (b) The variations with potential difference V of the current I for a resistor X and for a semiconductor diode are shown in Fig. 6.1. 15.0 12.5 I / mA resistor X 10.0 7.5 diode 5.0 2.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 V / V Fig. 6.1 (i) Determine the resistance of the diode for a potential difference V of 0.60 V. resistance = … Ω [3] (ii) Describe, qualitatively, the variation of the resistance of the diode as V increases from 0.60 V to 0.75 V. … [1] (c) The diode and the resistor X in (b) are connected into the circuit shown in Fig. 6.2. E 9.3 mA X 7.5 mA Y Fig. 6.2 The cell has electromotive force (e.m.f.) E and negligible internal resistance. Resistor Y is connected in parallel with resistor X and the diode. The current in the cell is 9.3 mA and the current in the diode is 7.5 mA. (i) Use Fig. 6.1 to determine E. E = … V [1] (ii) Determine the resistance of resistor Y. resistance = … Ω [2] (iii) Calculate the power dissipated in the diode. power = … W [2] (iv) The cell is now replaced by a new cell of e.m.f. 0.50 V and negligible internal resistance. Use Fig. 6.1 to determine the new current in the diode. current = … mA [1]

11 marks

Mark scheme: 6(a) sum of current(s) into junction = sum of current(s) out of junction or (algebraic) sum of current(s) at a junction is zero B1 6(b)(i) R = V / I C1 R = 0.60 / 7.5 × 10–3 C1 R = 80 Ω A1 6(b)(ii) resistance decreases B1 6(c)(i) E = 0.60 + 0.30 E = 0.90 V A1 6(c)(ii) (I =) 9.3 – 7.5 C1 I = 1.8 (mA) or 1.8 × 10–3 (A) R = 0.90 / 1.8 × 10–3 = 500 Ω A1 or total resistance = 0.90 / 9.3 × 10–3 = 96.8 (Ω) total resistance of diode and X = 0.90 / 7.5 × 10–3 = 120 (Ω) 1 / 96.8 = 1 / R + 1 / 120 (C1) R = 500 Ω (A1) Question Answer Marks 6(c)(iii) P = VI or I2R or V2 / R C1 P = 0.60 × 7.5 × 10–3 or (7.5 × 10–3)2 × 80 or 0.602 / 80 = 4.5 × 10–3 W A1 6(c)(iv) current = 2.5 mA A1

This question in 9702/22 Oct/Nov 2019

6 A battery of electromotive force (e.m.f.) 12 V and negligible internal resistance is connected to a network of two lamps and two resistors, as shown in Fig. 6.1. 0.50 A R 0.20 A 12 V X Y 28 Ω Fig. 6.1 The two lamps in the circuit have equal resistances. The two resistors have resistances R and 28 Ω. The lamps are connected at junction X and the resistors are connected at junction Y. The current in the battery is 0.50 A and the current in the lamps is 0.20 A. (a) Calculate: (i) the resistance of each lamp resistance = … Ω [2] (ii) resistance R. R = … Ω [2] (b) Determine the potential difference VXY between points X and Y. (c) Calculate the ratio total power dissipated by the lamps . total power produced by the battery ratio = … [2] (d) The resistor of resistance R is now replaced by another resistor of lower resistance. State and explain the effect, if any, of this change on the ratio in (c). … … … … … [2] [Total: 11]

11 marks

Mark scheme: 6(a)(i) C1 resistance = (12 / 0.20) / 2 or 6 / 0.20 = 30 Ω A1 6(a)(ii) I = 0.50 – 0.20 (= 0.30 A) C1 R + 28 = 12 / 0.30 (= 40 Ω) R = 12 Ω A1 Question Answer Marks 6(b) p.d. across lamp = 0.20 × 30 (= 6.0 V) C1 p.d. across R = 0.30 × 12 (= 3.6 V) C1 VXY = 6.0 – 3.6 = 2.4 V A1 or p.d. across lamp = 0.20 × 30 (= 6.0 V) (C1) p.d. across 28 Ω resistor = 0.30 × 28 (= 8.4 V) (C1) VXY = 8.4 – 6.0 = 2.4 V (A1) 6(c) P = VI or P = EI or P = I2R or P = V2 / R C1 ratio = (6.0 × 0.20) × 2 / (12 × 0.50) or 0.20 / 0.50 = 0.40 A1 6(d) no change to V across lamps, so power in lamps unchanged or current in battery/total current increases (and e.m.f. the same) so power produced by battery increases B1 both the above statements and so the ratio decreases B1

This question in 9702/23 Oct/Nov 2019

5 (a) Define the ohm. … … … [1] (b) A wire has a resistance of 1.8 Ω. The wire has a uniform cross-sectional area of 0.38 mm2 and is made of metal of resistivity 9.6 × 10–7 Ω m. Calculate the length of the wire. length = … m [3] (c) A resistor X of resistance 1.8 Ω is connected to a resistor Y of resistance 0.60 Ω and a battery P, as shown in Fig. 5.1. 1.2 V P 1.8 Ω 0.60 Ω X Y Fig. 5.1 The battery P has an electromotive force (e.m.f.) of 1.2 V and negligible internal resistance. (i) Explain, in terms of energy, why the potential difference (p.d.) across resistor X is less than the e.m.f. of the battery. … … … [1] (ii) Calculate the potential difference across resistor X. potential difference = … V [2] (d) Another battery Q of e.m.f. 1.2 V and negligible internal resistance is now connected into the circuit of Fig. 5.1 to produce the new circuit shown in Fig. 5.2. 1.2 V Q 1.2 V P 1.8 Ω 0.60 Ω X Y Fig. 5.2 State whether the addition of battery Q causes the current to decrease, increase or remain the same in: (i) resistor X … [1] (ii) battery P. … [1] (e) The circuit shown in Fig. 5.2 is modified to produce the new circuit shown in Fig. 5.3. 1.2 V P 3.6 Ω 1.8 Ω 0.60 Ω X Y Fig. 5.3 Calculate: (i) the total resistance of the two resistors connected in parallel resistance = … Ω [1] (ii) the current in resistor Y. current = … A [2] [Total: 12]

12 marks

Mark scheme: 5(a) volt / ampere B1 5(b) R = ρ L / A C1 L = (1.8 × 0.38 × 10–6) / 9.6 × 10–7 C1 = 0.71 m A1 5(c)(i) thermal energy is dissipated in resistor Y B1 5(c)(ii) V / 1.2 = 1.8 / (1.8 + 0.6) C1 V = 0.90 V A1 or I = 1.2 / (1.8 + 0.6) (= 0.50) (C1) V = 0.50 × 1.8 = 0.90 V (A1) 5(d)(i) remain the same B1 5(d)(ii) decrease B1 5(e)(i) 1 / R = 1 / 1.8 + 1 / 3.6 R = 1.2 Ω A1 Question Answer Marks 5(e)(ii) I = 1.2 / (1.2 + 0.60) C1 = 0.67 A A1 or VY = 1.2 × 0.60 / (1.2 + 0.60) (= 0.40) (C1) I = 0.40 / 0.60 = 0.67 A (A1)

This question in 9702/22 Feb/March 2020

5 (a) Metal wire is used to connect a power supply to a lamp. The wire has a total resistance of 3.4 Ω and the metal has a resistivity of 2.6 × 10–8 Ω m. The total length of the wire is 59 m. (i) Show that the wire has a cross-sectional area of 4.5 × 10–7 m2. [2] (ii) The potential difference across the total length of wire is 1.8 V. Calculate the current in the wire. current = … A [1] (iii) The number density of the free electrons in the wire is 6.1 × 1028 m–3. Calculate the average drift speed of the free electrons in the wire. average drift speed = … m s–1 [2] (b) A different wire carries a current. This wire has a part that is thinner than the rest of the wire, as shown in Fig. 5.1. wire thinner part Fig. 5.1 (i) State and explain qualitatively how the average drift speed of the free electrons in the thinner part compares with that in the rest of the wire. … … … [2] (ii) State and explain whether the power dissipated in the thinner part is the same, less or more than the power dissipated in an equal length of the rest of the wire. … … … [2] (c) Three resistors have resistances of 180 Ω, 90 Ω and 30 Ω. (i) Sketch a diagram showing how two of these three resistors may be connected together to give a combined resistance of 60 Ω between the terminals shown. Ensure you label the values of the resistances in your diagram. [1] (ii) A potential divider circuit is produced by connecting the three resistors to a battery of electromotive force (e.m.f.) 12 V and negligible internal resistance. The potential divider circuit provides an output potential difference VOUT of 8.0 V. Fig. 5.2 shows the circuit diagram. 12 V Fig. 5.2 On Fig. 5.2, label the resistances of all three resistors and the potential difference VOUT. [2] [Total: 12]

12 marks

Mark scheme: 5(a)(i) R = ρL / A A = (2.6 × 10–8 × 59) / 3.4 = 4.5 × 10–7 m2 A1 5(a)(ii) I = 1.8 / 3.4 = 0.53 A A1 5(a)(iii) I = Anvq v = 0.53 / (4.5 × 10–7 × 6.1 × 1028 × 1.60 × 10–19) C1 = 1.2 × 10–4 m s–1 A1 5(b)(i) (cross-sectional) area/A is less M1 (I, n, e the same so) average drift speed is greater A1 5(b)(ii) (area is less so) more resistance/R M1 (I is the same, so) more power/P A1 or (P = I2ρL / A so) P ∝ 1 / A (M1) (A is less so) more P (A1) 5(c)(i) 180 Ω and 90 Ω resistors shown connected in parallel B1 5(c)(ii) resistors connected in parallel labelled as 180 Ω and 90 Ω and the other resistor labelled as 30 Ω M1 VOUT or 8.0 V labelled across the two resistors in parallel A1

This question in 9702/21 May/June 2020

5 One end of a wire is attached to a fixed point. A force F is applied to the wire to cause extension x. The variation with F of x is shown in Fig. 5.1. 0.6 0.5 x / mm 0.4 0.3 0.2 0.1 0 0 5 10 15 20 25 30 35 40 45 F / N Fig. 5.1 The wire has a cross-sectional area of 4.1 × 10–7 m2 and is made of metal of Young modulus 1.7 × 1011 Pa. Assume that the cross-sectional area of the wire remains constant as the wire extends. (a) State the name of the law that describes the relationship between F and x shown in Fig. 5.1. … [1] (b) The wire has an extension of 0.48 mm. Determine: (i) the stress stress = … Pa [2] (ii) the strain. strain = … [2] (c) The resistivity of the metal of the wire is 3.7 × 10–7 Ω m. Determine the change in resistance of the wire when the extension x of the wire changes from x = 0.48 mm to x = 0.60 mm. change in resistance = … Ω [3] (d) A force of greater than 45 N is now applied to the wire. Describe how it may be checked that the elastic limit of the wire has not been exceeded. … … [1] [Total: 9]

9 marks

Mark scheme: 5(a) Hooke’s (law) B1 5(b)(i) σ = F / A C1 = 36 / (4.1 × 10–7) = 8.8 × 107 Pa A1 5(b)(ii) Young modulus = σ / ε or F / Aε C1 ε = 8.8 × 107 / (1.7 × 1011) = 5.2 × 10–4 A1 5(c) R = ρL / A C1 ΔR = ρΔx / A = 3.7 × 10–7 × 0.12 × 10–3 / (4.1 × 10–7) C1 = 1.1 × 10–4 Ω A1 5(d) remove the force/F and wire returns to original length B1

This question in 9702/22 May/June 2020

7 (a) Define the ohm. … … [1] (b) A uniform wire has resistance 3.2 Ω. The wire has length 2.5 m and is made from metal of resistivity 460 nΩ m. Calculate the cross-sectional area of the wire. cross-sectional area = … m2 [3] (c) A cell of electromotive force (e.m.f.) E and internal resistance r is connected to a variable resistor of resistance R, as shown in Fig. 7.1. E r I R Fig. 7.1 The current in the circuit is I. (i) State, in terms of energy, why the potential difference across the variable resistor is less than the e.m.f. of the cell. … … [1] (ii) State an expression for E in terms of I, R and r. E = … [1] (iii) The resistance R of the variable resistor is changed so that it is equal to r. Determine an expression, in terms of only E and r, for the power P dissipated in the variable resistor. P = … [2] [Total: 8]

8 marks

Mark scheme: 7(a) volt / ampere B1 7(b) R = ρL / A C1 A = 460 × 10–9 × 2.5 / 3.2 C1 = 3.6 × 10–7 m2 A1 7(c)(i) energy is dissipated in the internal resistance/r B1 7(c)(ii) E = IR + Ir or E = I (R + r) B1 7(c)(iii) P = I2R or P = I2r C1 I = E / 2r (so) P = E2 / 4r A1

This question in 9702/21 Oct/Nov 2020

3 (a) A spring is fixed at one end and is compressed by applying a force to the other end. The variation of the force F acting on the spring with its compression x is shown in Fig. 3.1. F1 F 0 0 0.045 x / m Fig. 3.1 A compression of 0.045 m is produced when a force F1 acts on the spring. The spring has a spring constant of 800 N m−1. (i) Determine F1. F1 = … N [2] (ii) Use Fig. 3.1 to show that, for a compression of 0.045 m, the elastic potential energy of the spring is 0.81 J. [2] (b) A child’s toy uses the spring in (a) to launch a ball of mass 0.020 kg vertically into the air. The ball is initially held against one end of the spring which has a compression of 0.045 m. The spring is then released to launch the ball. The kinetic energy of the ball as it leaves the toy is 0.72 J. (i) The toy converts the elastic potential energy of the spring into the kinetic energy of the ball. Use the information in (a)(ii) to calculate the percentage efficiency of this conversion. efficiency = … % [1] (ii) Determine the initial momentum of the ball as it leaves the toy. momentum = … N s [3] (c) The ball in (b) leaves the toy at point A and moves vertically upwards through the air. Point B is the position of the ball when it is at maximum height h above point A, as illustrated in Fig. 3.2. B ball reaches maximum height at point B h ball at point A kinetic energy 0.72 J A mass 0.020 kg Fig. 3.2 (not to scale) The gravitational potential energy of the ball increases by 0.60 J as it moves from A to B. (i) Calculate h. h = … m [2] (ii) Determine the average force due to air resistance acting on the ball for its movement from A to B. average force = … N [2] (iii) When there is air resistance, the ball takes time T to move from A to B. State and explain whether the time taken for the ball to move from A to its maximum height will be more than, less than or equal to time T if there is no air resistance. … … [1] [Total: 13]

13 marks

Mark scheme: 3(a)(i) F = kx C1 F1 = 800 × 0.045 = 36 N A1 3(a)(ii) (E =) ½kx2 or ½Fx or area under graph C1 ½ × 800 × (0.045)2 or ½ × 36 × 0.045 = 0.81 (J) A1 3(b)(i) efficiency = (0.72 / 0.81) × 100 = 89% A1 3(b)(ii) E = ½mv 2 C1 p = mv C1 0.72 = ½ × 0.020 × v 2 and p = 0.020 × v p = 0.17 N s A1 3(c)(i) (Δ)E = mg(Δ)h C1 h = 0.60 / (0.020 × 9.81) = 3.1 m A1 3(c)(ii) F = (0.72 – 0.60) / 3.1 C1 = 0.039 N A1 3(c)(iii) resultant force on ball is less (than that with air resistance) so time (taken) is more (than T) B1

This question in 9702/22 Oct/Nov 2020

6 (a) A network of three resistors of resistances R1, R2 and R3 is shown in Fig. 6.1. V I1 R1 I R2 I2 R3 I3 Fig. 6.1 The individual currents in the resistors are I1, I2 and I3. The total current in the combination of resistors is I and the potential difference across the combination is V. Show that the combined resistance R of the network is given by 1 1 1 1 = + + . R R1 R2 R3 [2] (b) A battery of electromotive force (e.m.f.) 8.0 V and internal resistance r is connected to three resistors X, Y and Z, as shown in Fig. 6.2. 8.0 V r X Z 0.49 A Y 0.45 A 16 Ω Fig. 6.2 Resistor Y has a resistance of 16 Ω. The current in resistor X is 0.49 A and the current in resistor Y is 0.45 A. Calculate: (i) the current in the battery current = … A [1] (ii) the internal resistance r of the battery. r = … Ω [2] (c) Resistors X and Y in Fig. 6.2 are made from wires of the same material and cross-sectional area. The average drift speed of the free electrons in X is 2.1 × 10−4 m s−1. Calculate the average drift speed v of the free electrons in Y. v = … m s−1 [2] (d) Resistor Z in Fig. 6.2 is replaced by a new resistor of smaller resistance. State and explain the effect, if any, on the terminal potential difference of the battery. … … … [2] [Total: 9]

9 marks

Mark scheme: 6(a) B1 (V / R) = (V / R1) + (V / R2) + (V / R3) or (I / V) = (I1 / V) + (I2 / V) + (I3 / V) and 1 / R = 1 / R1 + 1 / R2 + 1 / R3 B1 6(b)(i) current = 0.49 + 0.45 = 0.94 A A1 6(b)(ii) 8.0 = (0.94 × r) + (0.45 × 16) C1 r = 0.85 Ω A1 6(c) I = Anvq v = (0.45 / 0.49) × 2.1 × 10–4 C1 = 1.9 × 10–4 m s–1 A1 6(d) total/combined resistance decreases B1 (current in battery increases so terminal) potential difference decreases B1

This question in 9702/22 Oct/Nov 2020

6 (a) Define electric potential difference (p.d.). … … [1] (b) A wire of cross-sectional area A is made from metal of resistivity ρ. The wire is extended. Assume that the volume V of the wire remains constant as it extends. Show that the resistance R of the extending wire is inversely proportional to A2. [2] (c) A battery of electromotive force (e.m.f.) E and internal resistance r is connected to a variable resistor of resistance R, as shown in Fig. 6.1. r E A I R Fig. 6.1 The current in the circuit is I. Use Kirchhoff’s second law to show that R = – r. (EI) [1] (d) An ammeter is used in the circuit in (c) to measure the current I as resistance R is varied. 1 Fig. 6.2 is a graph of R against I. 6 R / Ω 4 2 0 0 0.1 0.2 0.3 0.4 0.5 1 / A–1 I –2 Fig. 6.2 (i) Use Fig. 6.2 to determine the power dissipated in the variable resistor when there is a current of 2.0 A in the circuit. power = … W [3] (ii) Use Fig. 6.2 and the equation in (c) to: 1. state the internal resistance r of the battery r = … Ω 2. determine the e.m.f. E of the battery. E = … V [3] [Total: 10]

10 marks

Mark scheme: 6(a) ( ) ( ) work done /energy transferred from electrical to other forms charge B1 6(b) R = ρL / A B1 V = LA and (so) R = ρV / A2 (with ρ and V constant) B1 6(c) E = IR + Ir or E = I(R + r) or E – Ir = IR and R = (E / I) – r A1 6(d)(i) P = I 2R or P = IV or P = V2 / R C1 R = 5.4 (Ω) or V = 10.8 (V) C1 P = 2.02 × 5.4 = 22 W A1 6(d)(ii) 1. r = 0.60 Ω A1 2. E = gradient C1 = e.g. 5.4 / 0.45 = 12 V A1

This question in 9702/23 Oct/Nov 2020

5 (a) Define the ohm. … … [1] (b) A wire is made of metal of resistivity ρ. The length L of the wire is gradually increased. Assume that the volume V of the wire remains constant as its length is increased. Show that the resistance R of the extending wire is proportional to L2. [2] (c) A battery of electromotive force (e.m.f.) E and internal resistance r is connected to a variable resistor of resistance R, as shown in Fig. 5.1. E r I A R VV Fig. 5.1 An ammeter measures the current I in the circuit. A voltmeter measures the potential difference V across the variable resistor. The resistance R is now varied to change the values of I and V. The variation with I of V is shown in Fig. 5.2. 3 V / V 2 1 0 0 2 4 6 I / A Fig. 5.2 (i) Use Fig. 5.2 to state the e.m.f. E of the battery. E = … V [1] (ii) Use Fig. 5.2 to determine the power dissipated in the variable resistor when there is a current of 5.0 A. power = … W [3] (iii) State what is represented by the value of the gradient of the graph. … [1] [Total: 8]

8 marks

Mark scheme: 5(a) volt / ampere B1 5(b) R = ρL / A B1 (A = V / L) (so) R = ρL2 / V (with ρ and V constant so R ∝ L2) B1 5(c)(i) E = 2.4 V A1 5(c)(ii) P = VI or I2R or V2 / R C1 = 1.3 × 5.0 or 5.02 × 0.26 or 1.32 / 0.26 C1 = 6.5 W A1 5(c)(iii) (–) internal resistance or (–) r B1

This question in 9702/22 May/June 2021

5 (a) State Kirchhoff’s first law. … … … [2] (b) The circuit shown in Fig. 5.1 contains a battery of electromotive force (e.m.f.) E and negligible internal resistance connected to four resistors R1, R2, R3 and R4, each of resistance R. E R1 R4 2.4 V R2 R3 0.30 A Fig. 5.1 The current in R3 is 0.30 A and the potential difference (p.d.) across R4 is 2.4 V. (i) Show that R is equal to 4.0 Ω. [2] (ii) Determine the e.m.f. E of the battery. E = … V [2] (c) The battery in (b) is replaced with another battery of the same e.m.f. E but with an internal resistance that is not negligible. State and explain the change, if any, in the total power produced by the battery. … … … [2] (d) The resistors in the circuit of Fig. 5.1 are made from nichrome wire of uniform radius 240 μm. The length of this wire needed to make each resistor is 0.67 m. Calculate the resistivity of nichrome. resistivity = … Ω m [3] [Total: 11]

11 marks

Mark scheme: 5(a) sum of current(s) in = sum of current(s) out or (algebraic) sum of current(s) is zero M1 at a junction (in a circuit) A1 5(b)(i) (current in R4 or R1 =) 0.30 + 0.30 (= 0.60 A) B1 (R =) 2.4 / 0.60 = 4.0 (Ω) A1 or (p.d. across R3 or R2 =) 2.4 / 2 (= 1.2 V) (B1) (R =) 1.2 / 0.30 = 4.0 (Ω) (A1) 5(b)(ii) E = 2.4 + 2.4 + 1.2 C1 = 6.0 V A1 or total resistance = 10 (Ω) (C1) E = 10 × 0.60 = 6.0 V (A1) 5(c) total resistance increases B1 current decreases (in battery) so total power decreases B1 Question Answer Marks 5(d) resistivity = RA / L C1 = 4.0 × π × (240 × 10–6)2 / 0.67 C1 = 1.1 × 10–6 Ω m A1

This question in 9702/21 Oct/Nov 2021

6 (a) A resistance wire of uniform cross-sectional area 3.3 × 10–7 m2 and length 2.0 m is made of metal of resistivity 5.0 × 10–7 Ω m. Show that the resistance of the wire is 3.0 Ω. [2] (b) The ends of the resistance wire in (a) are connected to the terminals X and Y in the circuit shown in Fig. 6.1. 1.50 V r X Y uniform metal wire, resistance 3.0 Ω Fig. 6.1 The cell has an electromotive force (e.m.f.) of 1.50 V and internal resistance r. The potential difference between X and Y is 1.20 V. Calculate: (i) the current in the circuit current = … A [1] (ii) the internal resistance r. r = … Ω [2] (c) A galvanometer and a cell of e.m.f. E with negligible internal resistance are connected to the circuit in (b), as shown in Fig. 6.2. 1.50 V r P X Y E Fig. 6.2 The resistance wire between X and Y has a length of 2.0 m. The galvanometer has a reading of zero when the connection P is adjusted so that the length XP is 1.4 m. Determine the e.m.f. E of the cell. E = … V [2] (d) The circuit in Fig. 6.2 is modified by replacing the original resistance wire with a second resistance wire. The second wire has the same length as the original wire and is made of the same metal. The second wire has a smaller cross-sectional area than the original wire. Connection P is adjusted on the second wire so that the galvanometer has a reading of zero. State and explain whether length XP for the second wire is shorter than, longer than or the same as length XP for the original wire when the galvanometer reading is zero. … … … … … [3]

10 marks

Mark scheme: 6(a) R = ρL / A C1 (R =) 5(.0) × 10–7 × 2(.0) / 3.3 × 10–7 = 3.0 Ω A1 6(b)(i) I = 1.2 / 3.0 = 0.40 A A1 6(b)(ii) r = (1.50 – 1.20) / 0.40 or 1.50 / 0.40 – 3.0 C1 = 0.75 Ω A1 6(c) E / 1.20 = 1.4 / 2.0 C1 E = 0.84 V A1 or RXP = (1.4 / 2.0) × 3.0 (= 2.1 Ω) E = 2.1 × 0.40 (C1) E = 0.84 V (A1) 6(d) (second wire has) larger resistance/resistance increases M1 p.d. across XY is larger/increases (for second wire) or p.d. across the (second) wire is larger/increases M1 (so) length XP (for second wire) is shorter A1

This question in 9702/23 Oct/Nov 2021

6 The ends of a metal resistance wire are connected to a battery of electromotive force (e.m.f.) 8.0 V and negligible internal resistance, as shown in Fig. 6.1. 8.0 V resistance wire Fig. 6.1 The power dissipated by the resistance wire is 36 W. (a) Calculate: (i) the current in the resistance wire current = … A [2] (ii) the number of free electrons that pass through the resistance wire in a time of 50 s number = … [2] (iii) the resistance of the wire. resistance = … Ω [2] (b) The metal of the resistance wire in the circuit has a resistivity of 1.4 × 10–6 Ω m. The cross-sectional area of the wire is 0.25 mm2. Determine the length of the wire. length = … m [2] (c) The circuit shown in Fig. 6.1 is modified by replacing the original resistance wire with a second resistance wire. The second wire has a greater diameter than the original wire. There are no other differences between the second wire and the original wire. By reference to resistance, state and explain whether the power dissipated by the second wire is more than, less than or the same as the power dissipated by the original wire. … … … … [2] (d) The circuit shown in Fig. 6.1 is modified by connecting a second battery, of e.m.f. 8.0 V and negligible internal resistance, in parallel with the original battery and the original resistance wire, as shown in Fig. 6.2. 8.0 V 8.0 V original resistance wire Fig. 6.2 By reference to the current in the resistance wire, state and explain whether the addition of the second battery causes the power in the original resistance wire to decrease, increase or stay the same. … … … … [2] [Total: 12]

12 marks

Mark scheme: 6(a)(i) C1 I = 36 / 8.0 = 4.5 A A1 6(a)(ii) charge = 4.5 × 50 = 225 C1 number = 225 / 1.6 × 10–19 = 1.4 × 1021 A1 6(a)(iii) R = V 2 / P or R = V / I or R = P / I 2 = 8.02 / 36 or = 8.0 / 4.5 or = 36 / 4.52 C1 = 1.8 Ω A1 Question Answer Marks 6(b) R = ρL / A C1 L = (1.8 × 0.25 × 10–6) / 1.4 × 10–6 = 0.32 m A1 6(c) (larger cross-sectional area, same length, same resistivity and so) less resistance M1 (same p.d. and more current so) more power (dissipated) A1 6(d) current (in wire) is the same M1 (same p.d. across wire so) power stays the same A1

This question in 9702/22 Feb/March 2022

6 (a) State Kirchhoff’s first law. … … [1] (b) A battery is connected to two resistors X and Y, as shown in Fig. 6.1. X Y Fig. 6.1 The resistance of resistor X is greater than the resistance of resistor Y. (i) State and explain which resistor dissipates more power. … … … … … [3] (ii) The two resistors are made of wires that have the same length. Both wires are made from metal of the same resistivity. State and explain which resistor is made of wire with the larger cross-sectional area. … … … [2] (c) A battery of electromotive force (e.m.f.) 9.0 V and negligible internal resistance is connected in series with a light-dependent resistor (LDR) and a fixed resistor of resistance 1800 Ω, as shown in Fig. 6.2. 9.0 V 1800 Ω RL V Fig. 6.2 A voltmeter is connected across the fixed resistor. The reading on the voltmeter is 5.4 V. (i) Calculate the current in the circuit. current = … A [1] (ii) Calculate the resistance RL of the LDR. RL = … Ω [2] (iii) The intensity of the light illuminating the LDR increases. By reference to the current in the circuit, state and explain the change, if any, to the voltmeter reading. … … … … [2] [Total: 11]

11 marks

Mark scheme: 6(a) sum of current(s) into junction = sum of current(s) out of junction or (algebraic) sum of current(s) at a junction is zero B1 6(b)(i) same potential difference (across X and Y as in parallel) B1 power = V2 / R (and RX  RY) or power = VI and IX  IY M1 (so) Y (dissipates more power) A1 6(b)(ii) R = L / A (and RX  RY) M1 (so) Y (has the larger (cross-sectional) area) A1 6(c)(i) current = 5.4 / 1800 = 3.0  10–3 A A1 6(c)(ii) 5.4 / 9.0 = 1800 / (1800 + RL) or RL = (9.0 – 5.4) / 3.0  10–3 C1 RL = 1200  A1 6(c)(iii) resistance of LDR / RL decreases B1 current (in the circuit) increases (so) voltmeter reading increases B1

This question in 9702/21 May/June 2022

6 (a) A network of three resistors of resistances R1, R2 and R3 is shown in Fig. 6.1. R1 R2 R3 I V1 V2 V3 V Fig. 6.1 The individual potential differences across the resistors are V1, V2 and V3. The current in the combination of resistors is I and the total potential difference across the combination is V. Show that the combined resistance R of the network is given by R = R1 + R2 + R3. [2] (b) A battery of electromotive force (e.m.f.) 8.0 V and negligible internal resistance is connected to a thermistor, a switch X and two fixed resistors, as shown in Fig. 6.2. 6.0 kΩ R1 X 8.0 V 4.0 kΩ R2 Fig. 6.2 Resistor R1 has resistance 6.0 kΩ and resistor R2 has resistance 4.0 kΩ. (i) Switch X is open. Calculate the potential difference across R1. potential difference = … V [2] (ii) Switch X is now closed. The resistance of the thermistor is 12.0 kΩ. Calculate the current in the battery. current = … A [2] (c) The switch X in the circuit in (b) remains closed. The temperature of the thermistor decreases. By reference to the current in the battery, state and explain the effect, if any, of the decrease in temperature on the power produced by the battery. … … … … … [3] [Total: 9]

9 marks

Mark scheme: 6(a) V = V1 + V2 + V3 B1 IR = IR1 + IR2 + IR3 or (V / I) = (V1 / I) + (V2 / I) + (V3 / I) and R = R1 + R2 + R3 B1 6(b)(i) V / 8.0 = 6.0  103 / (4.0  103 + 6.0  103) or I = 8.0 / (4.0  103 + 6.0  103) = 8.0  10–4 V = 8.0  10–4  6.0  103 C1 V = 4.8 V A1 6(b)(ii) total resistance in parallel = 3.0  103 () or 3.0 (k) C1 current = 8.0 / (3.0  103 + 6.0  103) = 8.9  10–4 A A1 6(c) thermistor resistance increases B1 (thermistor resistance increases so total resistance increases so) current decreases (in battery) M1 (P = EI and E constant so) power decreases A1

This question in 9702/22 May/June 2022

6 (a) (i) On Fig. 6.1, sketch the I–V characteristic of a filament lamp. I 0 V 0 Fig. 6.1 [2] (ii) Explain the shape of the line in (a)(i). … … … … [3] (b) A conducting wire has length 5.8 m and cross‑sectional area 3.4 × 10–8 m2. The resistivity of the metal of the wire is 5.6 × 10–8 Ω m. Calculate the resistance of the wire. resistance = … Ω [2] (c) A resistor of resistance R is placed in a circuit with a cell of negligible internal resistance, two switches S1 and S2, a second resistor of resistance 2R and three ammeters X, Y and Z. The circuit is shown in Fig. 6.2. X A S1 R Y A S2 Z 2R A Fig. 6.2 The reading on X is 1.0 A when S1 is open and S2 is closed. Complete Table 6.1. Table 6.1 position of switches ammeter readings S1 S2 reading on X / A reading on Y / A reading on Z / A open open 0 0 0 open closed 1.0 closed open closed closed [4] [Total: 11]

11 marks

Mark scheme: 6(a)(i) line passes through (0,0) and is in first and third quadrants only M1 gradient of line becoming less steep in both quadrants and roughly symmetrical A1 6(a)(ii) (as I increases) the temperature (of the filament wire/lamp) increases B1 (as I / temperature / V increases) the resistance (of wire/lamp) increases B1 (as I / temperature / V increases the graph curves because) ratio V / I increases or ratio I / V decreases B1 6(b) R = L / A C1 = (5.6  10−8  5.8) / 3.4  10−8 = 9.6  A1 Question Answer Marks 6(c) position of switches ammeter readings S1 S2 X / A Y / A Z / A open open 0 0 0 open closed 1.0 0 1.0 closed open 2.0 2.0 0 closed closed 3.0 2.0 1.0 second row: both values correct (B1) third row: all three values correct (B1) fourth row: X = Y + Z (any values) (B1) all three values correct (B1) B4

This question in 9702/23 May/June 2022

5 (a) State Ohm’s law. … … … [2] (b) The variation of current I with potential difference V for a filament lamp is shown in Fig. 5.1. 2.0 I / A 1.5 1.0 0.5 0 0 2 4 6 8 10 12 V / V Fig. 5.1 The resistance of the filament lamp increases with potential difference. (i) State how Fig. 5.1 shows this. … … [1] (ii) Explain why the resistance varies in this way. … … [1] (c) Fig. 5.2 shows a circuit with a battery of electromotive force (e.m.f.) 12.0 V connected to a linear potentiometer AB and two identical filament lamps P and Q. 12.0 V A B P Q Fig. 5.2 The battery has negligible internal resistance and the lamps each have the same I–V characteristic shown in Fig. 5.1. When the slider of the potentiometer is at its midpoint, as shown in Fig. 5.2, the current I in the battery is 1.78 A. Determine: (i) the current in lamp P current = … A [1] (ii) the total power dissipated in lamps P and Q total power = … W [2] (iii) the resistance of the potentiometer between its ends A and B. resistance = … Ω [2] (d) The slider of the potentiometer in (c) is moved to end A. State and explain the effect on the brightness of lamps P and Q. lamp P: … … lamp Q: … … [2] [Total: 11]

11 marks

Mark scheme: 5(a) current (through a conductor is directly) proportional to potential difference (across the conductor) M1 (provided that) temperature (of conductor remains) constant A1 5(b)(i) (ratio of) V / I increases (as p.d. increases) B1 5(b)(ii) (as p.d. increases, current increases so) temperature increases B1 5(c)(i) I = 1.55 A A1 5(c)(ii) P = VI or P = I 2R or P = V 2 / R C1 = 6.0  1.55  2 or 1.552  3.87  2 or (6.02 / 3.87)  2 A1 = 19 W 5(c)(iii) I = 1.78 – 1.55 C1 ( = 0.23 A) R = 12.0 / 0.23 A1 = 52  5(d) lamp P: p.d. across lamp decreases to zero so goes ‘out’ B1 lamp Q: p.d. across lamp increases to 12 V so gets brighter B1

This question in 9702/21 Oct/Nov 2022

6 (a) Define the potential difference across a component. … … [1] (b) The variation with potential difference V of the current I in a semiconductor diode is shown in Fig. 6.1. I 0 0 0.5 1.0 V / V Fig. 6.1 Use Fig. 6.1 to describe qualitatively: (i) the resistance of the diode in the range V = 0 to V = 0.25 V … [1] (ii) the variation, if any, in the resistance of the diode as V changes from V = 0.75 V to V = 1.0 V. … [1] (c) A battery of electromotive force (e.m.f.) 12 V and negligible internal resistance is connected to a uniform resistance wire XY, a fixed resistor and a variable resistor, as shown in Fig. 6.2. 12 V 2.7 A resistance wire Z X Y 1.6 m 2.0 m 1.5 A 5.0 Ω W Fig. 6.2 (not to scale) The fixed resistor has a resistance of 5.0 Ω. The current in the battery is 2.7 A and the current in the fixed resistor is 1.5 A. (i) Calculate the current in the resistance wire. current = … A [1] (ii) Determine the resistance of the variable resistor. resistance = … Ω [2] (iii) Wire XY has a length of 2.0 m. Point Z on the wire is a distance of 1.6 m from point X. The fixed resistor is connected to the variable resistor at point W. Determine the potential difference between points W and Z. potential difference = … V [3] (iv) The resistance of the variable resistor is now increased. By considering the currents in every part of the circuit, state and explain whether the total power produced by the battery decreases, increases or stays the same. … … … … … … [3] [Total: 12]

12 marks

Mark scheme: 6(a) energy (transferred from electrical to other forms) per unit charge B1 6(b)(i) (resistance is) infinite / very high B1 6(b)(ii) (resistance) decreases (as V increases) B1 6(c)(i) current = 2.7 – 1.5 A1 = 1.2 A 6(c)(ii) 12 = (1.5  5.0) + (1.5  R) or R = (12 / 1.5) – 5.0 C1 R = 3.0  A1 6(c)(iii) V(XZ) = (1.6 / 2.0)  12 (= 9.6 V) C1 V(XW) = 1.5  5.0 (= 7.5 V) C1 potential difference = 9.6 – 7.5 A1 = 2.1 V or V(ZY) = (0.4 / 2.0)  12 (= 2.4 V) (C1) V(WY) = 1.5  3.0 (= 4.5 V) (C1) potential difference = 4.5 – 2.4 (A1) = 2.1 V 6(c)(iv) current in (fixed / variable) resistor decreases B1 current in (resistance) wire is unchanged B1 (so) current in battery decreases, (same e.m.f. so) power decreases B1

This question in 9702/22 Feb/March 2023

7 A battery of electromotive force (e.m.f.) 9.6 V and negligible internal resistance is connected in series with two fixed resistors and a thermistor, as shown in Fig. 7.1. 9.6 V 5800 Ω V 3400 Ω Fig. 7.1 The fixed resistors have resistances of 3400 Ω and 5800 Ω. The reading on the voltmeter in the circuit is 6.0 V. (a) Calculate the current in the resistor of resistance 5800 Ω. current = … A [2] (b) Calculate the resistance of the thermistor. resistance = … Ω [2] (c) The initial energy stored in the battery is 2.6 × 104 J. Assume that the e.m.f. of the battery is constant. Determine the final energy stored in the battery after a charge of 330 C has moved through it. final stored energy = … J [2] (d) The environmental conditions change causing an increase in the resistance of the thermistor. State whether there is a decrease, increase or no change to: (i) the temperature of the thermistor … [1] (ii) the current in the thermistor … [1] (iii) the potential difference across the thermistor. … [1] [Total: 9]

9 marks

Mark scheme: 7(a) C1 I = 6.2  10–4 A A1 7(b) 9.6 = 6.2  10–4  (3400 + 5800 + R) or 6.0 = 6.2  10–4  (3400 + R) C1 R = 6.3  103  A1 7(c) (∆E =) 9.6  330 (= 3170 J) C1 final stored energy = 2.6  104 – 3170 = 2.3  104 J A1 7(d)(i) decrease B1 7(d)(ii) decrease B1 7(d)(iii) increase B1

This question in 9702/21 May/June 2023

6 (a) The current in a filament lamp decreases. State and explain how the resistance of the lamp changes. … … [1] (b) A cylindrical wire has length L and resistance R. The total number of free electrons (charge carriers) contained in the volume of the wire is N. Each free electron has charge e. The potential difference between the ends of the wire is V. Determine expressions, in terms of some or all of the symbols e, L, N, R and V for: (i) the current in the wire current = … [1] (ii) the average drift speed of the free electrons average drift speed = … [2] (iii) the average time taken for a free electron to move along the full length of the wire. time taken = … [1] [Total: 5]

5 marks

Mark scheme: 6(a) temperature decreases (so) resistance decreases B1 6(b)(i) current = V / R A1 6(b)(ii) I = Anvq n = N / V or n = N / AL C1 v = (V / R) / [(V / L) (N / V) e] or (V / R) / [A (N / AL) e] = VL / RNe A1 or v = L / t = L / (Q / I) (C1) = LI / Q = L(V / R) / Ne = VL / RNe (A1) 6(b)(iii) time = distance / speed or Q / I = L / (VL / RNe) or Ne / (V / R) time = RNe / V A1

This question in 9702/22 May/June 2023

5 A student sets up a circuit with a battery, an ammeter, a heater and a light-dependent resistor (LDR) all in series. The battery has negligible internal resistance. A voltmeter is connected across (in parallel with) the heater. (a) On Fig. 5.1, complete the circuit diagram of this arrangement. Fig. 5.1 [3] (b) The heater is a wire made of metal of resistivity 1.1 × 10−6 Ω m. The wire has length 2.0 m and cross-sectional area 3.8 × 10−7 m2. The reading on the voltmeter is 4.8 V. Calculate: (i) the resistance of the heater resistance = … Ω [2] (ii) the reading on the ammeter. reading on ammeter = … A [1] (c) The heater is replaced by a new wire. The new wire is made of the same metal as the wire in (b) and has the same length but a larger diameter. The resistance of the LDR remains constant. (i) State and explain whether the new wire has a resistance that is greater than, less than or the same as that of the wire in (b). … … … [2] (ii) State and explain whether the new reading on the voltmeter is greater than, less than or equal to 4.8 V. … … … [2] [Total: 10]

10 marks

Mark scheme: 5(a) correct symbol for the heater or for the LDR M1 all correct symbols in series (ignore voltmeter) and no extra symbols A1 correct symbol for voltmeter and in parallel with the heater B1 5(b)(i) R = L / A C1 = (1.1  10−6  2.0) / 3.8  10−7 = 5.8  A1 5(b)(ii) I = 4.8 / 5.8 = 0.83 A A1 5(c)(i) A larger (for new wire) or A  d 2 (and d larger for new wire) or R  1 / d 2 (and d larger for new wire) M1 so R is less (than that of first wire) A1 5(c)(ii) (heater / total resistance decreases so) current (in circuit) increases (so p.d. across LDR increases) or heater resistance decreases so it has a smaller share/proportion/fraction of the (total) voltage / e.m.f. M1 (so voltmeter) reading is less (than 4.8 V) A1

This question in 9702/23 May/June 2023

6 (a) A metal wire has a resistance per unit length of 0.92 Ω m–1. The wire has a uniform cross‑sectional area of 5.3 × 10–7 m2. Calculate the resistivity of the metal of the wire. resistivity = … Ω m [2] (b) A battery of electromotive force (e.m.f.) E and negligible internal resistance is connected in series with a fixed resistor and a light‑dependent resistor (LDR), as shown in Fig. 6.1. E 1400 Ω V 1600 Ω Fig. 6.1 The resistance of the fixed resistor is 1400 Ω. The intensity of the light illuminating the LDR causes it to have a resistance of 1600 Ω. A voltmeter connected across the LDR reads 6.4 V. (i) Show that the current in the LDR is 4.0 × 10–3 A. [1] (ii) Calculate the number of free electrons passing through the LDR in a time of 3.2 minutes. number of free electrons = … [2] (iii) Calculate the e.m.f. E. E = … V [2] (iv) Determine the ratio power dissipated in LDR . power dissipated in fixed resistor ratio = … [2] (c) The environmental conditions change causing a decrease in the resistance of the LDR in (b). The temperature of the environment does not change. State whether there is a decrease, increase or no change to: (i) the intensity of the light illuminating the LDR … [1] (ii) the current in the battery … [1] (iii) the reading of the voltmeter. … [1] [Total: 12]

12 marks

Mark scheme: 6(a) R = L / A C1  = 0.92  5.3  10–7 A1 = 4.9  10–7  m 6(b)(i) (current =) 6.4 / 1600 = 4.0 × 10–3 (A) A1 6(b)(ii) charge = 4.0  10–3  3.2  60 C1 ( = 0.768 C) number = 0.768 / 1.6  10–19 A1 = 4.8  1018 6(b)(iii) 6.4 / E = 1600 / (1400 + 1600) C1 or E = 6.4 + (4.0  10–3  1400) or E = 4.0  10–3 (1600 + 1400) E = 12 V A1 6(b)(iv) P = I2R or P = VI or P = V2 / R C1 ratio = (6.42 / 1600) / [(4.0  10–3)2  1400] A1 = 1.1 6(c)(i) increase B1 6(c)(ii) increase B1 6(c)(iii) decrease B1

This question in 9702/22 Oct/Nov 2023

7 (a) State Ohm’s law. … … … [2] (b) A battery of electromotive force (e.m.f.) 6.2 V and negligible internal resistance is connected in a circuit to a uniform resistance wire, a voltmeter, a fixed resistor and a switch, as shown in Fig. 7.1. 6.2 V S wire of resistance 18 Ω V Fig. 7.1 The resistance wire has resistance 18 Ω, length 0.94 m and cross‑sectional area 7.2 × 10–8 m2. The slider S is positioned half‑way along the length of the wire. (i) Calculate the resistivity ρ of the material of the resistance wire. ρ = … Ω m [2] (ii) The switch is open. State the reading on the voltmeter. voltmeter reading = … V [1] (iii) The switch is now closed. State whether there is an increase, decrease or no change to: ● the current in the battery … ● the voltmeter reading. … [2] (iv) The switch remains closed. The slider S is moved along the resistance wire so that the voltmeter reading is 3.1 V. On Fig. 7.1, draw a cross (×) on the resistance wire to show a possible new position of the slider. [1] (c) The circuit in (b) is altered by changing the battery for one of a different e.m.f. The switch is open. A student records the following data for the resistance wire: current in the wire 0.93 A mean drift speed of charge carriers 1.3 × 10–3 m s–1 number density of charge carriers 9.0 × 1028 m–3. (i) Determine the charge q of a charge carrier in the wire suggested by this data. q = … C [2] (ii) With reference to the value of q, explain why the data recorded by the student cannot be correct. … … [1] [Total: 11]

11 marks

Mark scheme: 7(a) current (through a conductor is directly) proportional to potential difference (across the conductor) or vice versa M1 (provided that) temperature (of conductor remains) constant A1 7(b)(i) R = L / A C1  = (18  7.2  10–8) / 0.94 A1 = 1.4  10–6  m 7(b)(ii) voltmeter reading = 3.1 V A1 7(b)(iii) current in the battery: increase B1 voltmeter reading: decrease B1 7(b)(iv) cross marked on the resistance wire to right of the arrowhead of S, but not touching the right-hand end of the resistance B1 wire 7(c)(i) I = Anvq C1 q = 0.93 / [(7.2  10–8)  (9.0  1028)  (1.3  10–3)] q = 1.1  10–19 C A1 7(c)(ii) charge / q (value) is below 1.6  10–19 (C) B1 or charge cannot be below 1.6  10–19 (C) or (the charge carriers / q) should have a charge of 1.6  10–19 (C)

This question in 9702/23 Oct/Nov 2023

3 Lightning occurs when charge builds up in the atmosphere, creating a potential difference between the ground and the atmosphere. During a lightning strike there is an average current of 3.3 × 10 4 A for a time of 2.6 × 10 –5 s. (a) Calculate the charge transferred during the lightning strike. charge = … C [2] (b) The potential difference between the ground and the atmosphere is 3.0 × 107 V. Calculate the average power, in GW, transferred during the lightning strike. power = … GW [2] (c) A lightning rod is attached to a tall building to conduct charge safely to the ground. The lightning rod is modelled as a uniform cylindrical copper cable of total length 95 m that runs from the ground to the top of the building, as shown in Fig. 3.1. lightning rod building ground Fig. 3.1 (i) The resistance of the lightning rod is 9.6 Ω. The resistivity of copper is 1.7 × 10 –8 Ω m. Determine the radius of the lightning rod. radius = … m [3] (ii) The radius of the copper lightning rod is doubled with no change to its length. State the effect of this change on the resistance of the lightning rod. … [1] (d) A section of the lightning rod of length 0.12 m is removed for testing. A tensile stress of 1.9 × 106 Pa is applied, as shown in Fig. 3.2. lightning rod fixed support tensile stress 1.9 × 106 Pa 0.12 m Fig. 3.2 (not to scale) The section of the rod obeys Hooke’s law. The Young modulus of copper is 1.3 × 1011 Pa. Calculate the extension of the section. extension = … m [3] [Total: 11]

11 marks

Mark scheme: 3(a) C1 = 3.3  104  2.6  10–5 = 0.86 C A1 3(b) P = IV or P = VQ / t or V = IR and P = V 2/R or P = I 2R C1 P = 3.3  104  3.0  107 or P = (3.0  107  0.86) / (2.6  10–5) or P = (3.0  107)2 / 910 or P = (3.3  104)2  910 P = 9.9  1011 (W) = 990 GW A1 3(c)(i) R = L / A C1 9.6 = 1.7  10–8  95 / r 2 C1 r = 2.3  10–4 m A1 3(c)(ii) (resistance) decreases by a factor of four A1 Question Answer Marks 3(d) E =  /  C1 x = L / E = 1.9  106  0.12 / (1.3  1011) C1 = 1.8  10–6 m A1

This question in 9702/22 May/June 2024

7 (a) Fig. 7.1 shows two resistors connected in series with a cell of electromotive force (e.m.f.) 1.50 V and internal resistance 0.28 Ω. 1.50 V 0.28 Ω I 1.0 Ω R Fig. 7.1 One of the resistors has resistance 1.0 Ω. The other resistor has resistance R. The terminal potential difference (p.d.) across the cell is 1.36 V. (i) Show that the current I in the circuit is 0.50 A. [2] (ii) Calculate the combined resistance of the two resistors. resistance = … Ω [2] (iii) Use your answer in (a)(ii) to determine resistance R. R = … Ω [1] (b) The circuit in Fig. 7.1 is disconnected and the two resistors are reconnected to the cell, now in parallel with each other. (i) On Fig. 7.2, complete the circuit diagram to show this arrangement. 1.50 V 0.28 Ω Fig. 7.2 [1] (ii) Explain, without calculation, whether the terminal p.d. across the cell is now less than, equal to or greater than 1.36 V. … … … [2] [Total: 8]

8 marks

Mark scheme: 7(a)(i) (V =) 1.50 – 1.36 B1 (= 0.14 V) current = V / R A1 = 0.14 / 0.28 = 0.50 A 7(a)(ii) resistance = V / I C1 = 1.36 / 0.50 A1 = 2.7  7(a)(iii) R = 2.7 – 1.0 A1 = 1.7  7(b)(i) two resistors correctly shown in parallel with cell and no other components B1 7(b)(ii) (external) resistance is now smaller B1 and (so) current (in cell) is greater / (external) resistance smaller fraction of total resistance / internal resistance larger fraction of total resistance (greater p.d. across internal resistance so) terminal p.d. is less B1

This question in 9702/21 Oct/Nov 2024

6 (a) Define resistance. … … [1] (b) A cylindrical metal wire of length 2.4 m and cross-sectional area 8.0 × 10–6 m2 has a resistance of 0.33 Ω. There is a current in the wire of 4.7 A. (i) Determine the resistivity of the metal from which the wire is made. resistivity = … Ω m [2] (ii) Calculate the charge that passes through the wire in a time of 5.0 minutes. charge = … C [2] (iii) The free electrons (charge carriers) in the wire have an average drift speed of 0.16 mm s–1. Determine the number density of charge carriers in the metal. number density = … m–3 [2] (c) The wire in (b) may be considered to be a fixed resistor. It is connected in series with a thermistor to a battery that has negligible internal resistance. (i) Use circuit symbols to complete Fig. 6.1 to show the circuit diagram of this arrangement. Fig. 6.1 [1] (ii) Explain, without calculation, how the power dissipated in the wire changes as the temperature of the thermistor is increased. … … … … [2] [Total: 10]

10 marks

Mark scheme: 6(a) potential difference per unit current B1 6(b)(i)  = RA / L C1 = (0.33  8.0  10–6) / 2.4 A1 = 1.1  10–6  m 6(b)(ii) Q = It C1 = 4.7  5.0  60 A1 = 1400 C 6(b)(iii) I = nAvq C1 n = 4.7 / (8.0  10–6  0.16  10–3  1.60  10–19) = 2.3  1028 m–3 A1 6(c)(i) correct symbols for resistor and thermistor, shown correctly connected in series with the battery B1 6(c)(ii) (as temperature increases) resistance of thermistor decreases M1 (total resistance decreases so) greater current (in circuit/wire) so power (dissipated in the wire) increases A1 or (total resistance decreases so) greater (share of) p.d. across wire so power (dissipated in the wire) increases

This question in 9702/23 Oct/Nov 2024

6 A cylindrical copper wire P of length 0.24 m is shown in Fig. 6.1. 0.24 m 0.85 A Fig. 6.1 (not to scale) The current in the wire is 0.85 A. The resistance of the wire is 3.3 mΩ. The total number of charge carriers N in the wire is 2.6 × 1022. The resistivity of copper is 1.8 × 10–8 Ω m. (a) Calculate the potential difference between the two ends of the wire. potential difference = … V [2] (b) (i) Show that the cross-sectional area of the wire is 1.3 × 10–6 m2. [2] (ii) Show that the number density of charge carriers in the wire is 8.3 × 1028 m–3. [1] (iii) Calculate the average drift speed of the charge carriers (electrons) in the wire. average drift speed = … m s–1 [2] (c) A different copper wire Q has the same volume as wire P, but non-uniform radius, as shown in Fig. 6.2. X r1 r2 Fig. 6.2 (not to scale) The radius r1 at end X of wire Q is the same as the radius of wire P. Radius r2 is less than r1. (i) State and explain how the resistance of wire Q compares with the resistance of wire P. … … … … … … … [4] (ii) On Fig. 6.3, sketch a graph of the variation of the average drift speed of the charge carriers with distance from end X of wire Q. average drift speed 0 0 distance from X Fig. 6.3 [2] [Total: 13]

13 marks

Mark scheme: 6(a) V = IR C1 = 0.85  3.3  10–3 = 2.8 10–3 V A1 6(b)(i) (A =) L / R C1 = 1.8  10–8  0.24 / 3.3  10–3 = 1.3  10–6 (m2) A1 6(b)(ii) (n =) 2.6  1022 / (1.3  10–6  0.24) = 8.3  1028 (m–3) A1 6(b)(iii) v = I / nAq C1 = 0.85 / (8.3  1028  1.3  10–6  1.6  10–19) = 4.9  10–5 m s–1 A1 OR (C1) v = IL / Nq = 0.85  0.24 / (2.6  1022  1.6  10–19) = 4.9  10–5 m s–1 (A1) 6(c)(i) Length (of Q) is greater (than P) B1 (Average cross-sectional) area (of Q) is less (than P) B1 Resistance is proportional to length / (cross-sectional) area M1 (so) the resistance (of Q) is greater (than P) A1 6(c)(ii) A line starting from a non-zero value of drift speed at distance = 0 B1 A line with an increasing positive gradient B1

This question in 9702/22 Feb/March 2025

7 A nichrome resistance wire has length 150 cm, cross-sectional area 2.45 × 10−7 m2 and resistivity 1.12 × 10−6 Ω m. (a) Calculate, to three significant figures, the resistance of the wire. resistance = … Ω [3] (b) The nichrome wire forms part of a potentiometer circuit together with a cell of electromotive force (e.m.f.) 1.2 V and negligible internal resistance, as shown in Fig. 7.1. 1.2 V 150 cm 64 cm nichrome wire cell X Fig. 7.1 (not to scale) The circuit is used to determine the e.m.f. of cell X. The galvanometer is used in a null method to find the null point 64 cm from the left-hand end of the nichrome wire. (i) Explain what is meant by a null method. … … [1] (ii) Calculate the e.m.f. of cell X. e.m.f. = … V [2] (iii) The cell of e.m.f. 1.2 V is replaced by a new cell with the same e.m.f. but with an internal resistance that is not negligible. State and explain the effect, if any, of the internal resistance of the new cell on the position of the null point. … … … … [2] [Total: 8]

8 marks

Mark scheme: 7(a) R = L / A C1 = (1.12  10−6  1.5) / 2.45  10−7 C1 = 6.86  A1 7(b)(i) (A method where the) reading (on the galvanometer) is zero. B1 7(b)(ii) e.m.f. / 1.2 = 64 / 150 C1 e.m.f. = (64 / 150)  1.2 A1 = 0.51 V 7(b)(iii) (the internal resistance will cause a) drop in p.d. across the wire / the terminal p.d. is lower B1 So the null point will move to the right B1

This question in 9702/23 May/June 2025

3 A wire has length L and cross-sectional area A. The wire is made from a metal that has Young modulus E and resistivity ρ. (a) Define the Young modulus of a material. … … [1] (b) (i) State an expression, in terms of some or all of L, A, E and ρ, for the resistance R0 of the wire. R0 = … [1] (ii) Show that the spring constant k0 of the wire is given by EA k0 = L . [2] (c) The wire is stretched, within the limit of proportionality, by a tensile force F. Assume that any changes in the cross-sectional area of the wire are negligible. (i) On Fig. 3.1, sketch the variation with F of the resistance R of the wire. R R0 0 0 F Fig. 3.1 [1] (ii) On Fig. 3.2, sketch the variation with F of the spring constant k of the wire. k k0 0 0 F Fig. 3.2 [1] (d) Copper has a resistivity of 1.8 × 10–8 Ω m and a Young modulus of 1.3 × 1011 Pa. A copper wire of diameter 1.6 mm has a resistance of 0.034 Ω. (i) Show that the length of the wire is 3.8 m. [1] (ii) Use the equation in (b)(ii) to determine the spring constant of the wire. spring constant = … N m–1 [2] [Total: 9]

9 marks

Mark scheme: 3(a) ratio of stress to strain B1 3(b)(i) R0 = L / A A1 3(b)(ii) k = F / x C1 E = FL / Ax = ((F / x)  (L / A)) = kL / A leading to k0 = EA / L A1 3(c)(i) straight line with positive gradient starting at (0, R0) B1 3(c)(ii) straight horizontal line starting at (0, k0) B1 3(d)(i) L = [0.034    (0.80  10–3)2] / 1.8  10–8 = 3.8 (m) A1 3(d)(ii) k = [1.3  1011    (0.80  10–3)2] / 3.8 C1 or k = EA / [RA / ] = E / R = [1.3  1011  1.8  10–8] / 0.034 k = 6.9  104 N m–1 A1

This question in 9702/23 Oct/Nov 2025

5 A student uses a circuit containing an ammeter, a voltmeter and a cell to take measurements to determine the resistance of a length of nichrome wire. (a) (i) Define resistance. … … [1] (ii) Draw a circuit diagram to show how the components should be connected. Use the symbol for a resistor to represent the nichrome wire. [2] (b) The student also measures the length and the diameter of the wire. Table 5.1 shows the measurements recorded for each quantity. Table 5.1 quantity measurement length (0.864 ± 0.001) m diameter (0.496 ± 0.002) mm voltmeter reading (1.38 ± 0.02) V ammeter reading (0.276 ± 0.001) A (i) Show that the resistance of the wire is 5.00 Ω. [1] (ii) Calculate, to three significant figures, the resistivity ρ of the nichrome. ρ = … Ω m [3] (iii) Calculate the percentage uncertainty in ρ. percentage uncertainty = … % [2] (iv) Determine the absolute uncertainty in ρ. absolute uncertainty = … Ω m [1] [Total: 10]

10 marks

Mark scheme: 5(a)(i) potential difference per unit current B1 5(a)(ii) resistor connected to cell in a closed loop and correct circuit symbols used for all components B1 ammeter connected in series with resistor and voltmeter connected in parallel with resistor B1 5(b)(i) (resistance) = 1.38 / 0.276 = 5.00 () A1 5(b)(ii)  = RA / L C1  = 5.00    (0.496  10–3)2 / (4  0.864) C1 = 1.12  10–6  m A1 5(b)(iii) calculation of a fractional or percentage uncertainty in one quantity C1 0.001 / 0.864 or 0.002 / 0.496 or 0.02 / 1.38 or 0.001 / 0.276 percentage uncertainty = [(0.001 / 0.864) + (2  0.002 / 0.496) + (0.02 / 1.38) + (0.001 / 0.276)]  100 A1 = 2.7% 5(b)(iv)  = 0.03  1.12  10–6 A1 = 3  10–8  m

This question in 9702/24 Oct/Nov 2025