Potential difference and power — 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

3 A shopping trolley and its contents have a total mass of 42 kg. The trolley is being pushed For along a horizontal surface at a speed of 1.2 m s–1. When the trolley is released, it travels a Examiner’s distance of 1.9 m before coming to rest. Use (a) Assuming that the total force opposing the motion of the trolley is constant, (i) calculate the deceleration of the trolley, deceleration = … m s–2 [2] (ii) show that the total force opposing the motion of the trolley is 16 N. [1] (b) Using the answer in (a)(ii), calculate the power required to overcome the total force opposing the motion of the trolley at a speed of 1.2 m s–1. power = … W [2] (c) The trolley now moves down a straight slope that is inclined at an angle of 2.8° to the For horizontal, as shown in Fig. 3.1. Examiner’s Use 2.8° Fig. 3.1 The constant force that opposes the motion of the trolley is 16 N. Calculate, for the trolley moving down the slope, (i) the component down the slope of the trolley’s weight, component of weight = … N [2] (ii) the time for the trolley to travel from rest a distance of 3.5 m along the length of the slope. time = … s [4] (d) Use your answer to (c)(ii) to explain why, for safety reasons, the slope is not made any steeper. … … [1]

12 marks

Mark scheme: 3 (a) (i) v2 = 2as 1.22 = 2 × a × 1.9 M1 a = 0.38 m s–2 A1 [2] (ii) F = ma = 42 × 0.38 M1 = 16 N A0 [1] (b) power = Fv C1 = 16 × 1.2 = 19 W A1 [2] (c) (i) component = 42 × 9.8 × sin2.8 C1 = 20.1 N A1 [2] (ii) accelerating force = 20.1 – 16 = 4.1 N C1 acceleration of trolley = 4.1 / 42 = 0.098 m s–2 C1 s = ½at2 3.5 = ½ × 0.098 × t2 C1 t = 8.5 s A1 [4] GCE A/AS LEVEL – May/June 2008 9702 02 (d) either allows plenty of time to stop runaway trolley or speed of trolley increases gradually or trolley will travel faster B1 [1] (answer must be unambiguous when read in conjunction with question)

This question in 9702/21 May/June 2008

6 An electric heater consists of three similar heating elements A, B and C, connected as shown For in Fig. 6.1. Examiner’s Use 240V S1 A B S2 C S3 Fig. 6.1 Each heating element is rated as 1.5 kW, 240 V and may be assumed to have constant resistance. The circuit is connected to a 240 V supply. (a) Calculate the resistance of one heating element. resistance = ……………….……….. Ω [2] (b) The switches S1, S2 and S3 may be either open or closed. For Examiner’s Complete Fig. 6.2 to show the total power dissipation of the heater for the switches in Use the positions indicated. S1 S2 S3 total power / kW open closed closed closed closed open closed closed closed closed open open closed open closed [5] Fig. 6.2

7 marks

Mark scheme: 6 (a) either P = VI and V = IR or P = V2 / R C1 resistance = 38.4 Ω A1 [2] (b) zero B1 1.5 kW B1 3.0 kW B1 0.75 kW B1 2.25 kW B1 [5]

This question in 9702/21 May/June 2008

6 A cell has electromotive force (e.m.f.) E and internal resistance r. It is connected in series For with a variable resistor R, as shown in Fig. 6.1. Examiner’s Use E r R Fig. 6.1 (a) Define electromotive force (e.m.f.). … … … [2] (b) The variable resistor R has resistance X. Show that power dissipated in resistor R X = . power produced in cell X + r [3] (c) The variation with resistance X of the power PR dissipated in R is shown in Fig. 6.2. For Examiner’s Use 2.0 PR / W 1.5 1.0 0.5 0 0 0.5 1.0 1.5 2.0 2.5 3.0 X / Ω Fig. 6.2 (i) Use Fig. 6.2 to state, for maximum power dissipation in resistor R, the magnitude of this power and the resistance of R. maximum power = … W resistance = … Ω [2] (ii) The cell has e.m.f. 1.5 V. Use your answers in (i) to calculate the internal resistance of the cell. internal resistance = … Ω [3] (d) In Fig. 6.2, it can be seen that, for larger values of X, the power dissipation decreases. Use the relationship in (b) to suggest one advantage, despite the lower power output, of using the cell in a circuit where the resistance X is larger than the internal resistance of the cell. … … [1]

11 marks

Mark scheme: 6 (a) energy transferred from source / changed from some form to electrical … M1 per unit charge (to drive charge round a complete circuit) … A1 [2] (b) and power in R = I 2X … M1 E = I (X + r) … M1 power in cell = EI and algebra clear leading to ratio = X / (X + r) … A1 [3] GCE A/AS LEVEL – October/November 2009 9702 21 (c) (i) 1.4 W … A1 0.40 Ω ………(allow ±0.05 Ω) … A1 [2] (ii) current in circuit = 1.4/0.4 = 1.87 A … C1 1.5 = 1.87 (r + 0.40) … C1 r = 0.40 Ω … … A1 [3] (d) either less power lost / energy wasted / lost or greater efficiency (of energy transfer) … B1 [1] [Total: 11]

This question in 9702/21 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) A metal wire of constant resistance is used in an electric heater. For In order not to overload the circuit for the heater, the supply voltage to the heater is Examiner’s reduced from 230 V to 220 V. Use Determine the percentage reduction in the power output of the heater. reduction = … % [2] (b) A uniform wire AB of length 100 cm is connected between the terminals of a cell of e.m.f. 1.5 V and negligible internal resistance, as shown in Fig. 6.1. 1.5 V 100 cm C A B L 5.0 Ω A Fig. 6.1 An ammeter of internal resistance 5.0 Ω is connected to end A of the wire and to a contact C that can be moved along the wire. Determine the reading on the ammeter for the contact C placed (i) at A, reading = … A [1] (ii) at B. For Examiner’s Use reading = … A [1] (c) Using the circuit in (b), the ammeter reading I is recorded for different distances L of the contact C from end A of the wire. Some data points are shown on Fig. 6.2. 0.4 I / A 0.3 0.2 0.1 0 0 20 40 60 80 100 L / cm Fig. 6.2 (i) Use your answers in (b) to plot data points on Fig. 6.2 corresponding to the contact C placed at end A and at end B of the wire. [1] (ii) Draw a line of best fit for all of the data points and hence determine the ammeter reading for contact C placed at the midpoint of the wire. reading = … A [1] (iii) Use your answer in (ii) to calculate the potential difference between A and the For contact C for the contact placed at the midpoint of AB. Examiner’s Use potential difference = … V [2] (d) Explain why, although the contact C is at the midpoint of wire AB, the answer in (c)(iii) is not numerically equal to one half of the e.m.f. of the cell. … … … [2]

10 marks

Mark scheme: 6 (a) either P ∝ V 2 or P = V 2/R …………………………………………………………. C1 reduction = (2302 – 2202)/2302 = 8.5 % ………………………………………………………………….. A1 [2] (b) (i) zero ……………………………………………………………………………. A1 [1] (ii) 0.3(0) A ……………………………………………………………………….. A1 [1] (c) (i) correct plots to within ± 1 mm ………………………………………………. B1 [1] (ii) reasonable line/curve through points giving current as 0.12 A allow ± 0.005A) ………………………………………………………………. B1 [1] (iii) V = IR …………………………………………………………………………. C1 V = 0.12 × 5.0 = 0.6(0) V …………………………………………………………………... A1 [2] (d) circuit acts as a potential divider/current divides/current in AC not the same as current in BC ………………………………………………………………………. B1 resistance between A and C not equal to resistance between C and B ……. B1 or current in wire AC × R is not equal to current in wire BC × R B1 [2] any 2 statements GCE AS/A LEVEL – May/June 2010 9702 22

This question in 9702/22 May/June 2010

3 A cyclist is moving up a slope that has a constant gradient. The cyclist takes 8.0 s to climb For the slope. Examiner’s The variation with time t of the speed v of the cyclist is shown in Fig. 3.1. Use 8 v / m s–1 6 4 2 0 0 2 4 6 8 t / s Fig. 3.1 (a) Use Fig. 3.1 to determine the total distance moved up the slope. distance = … m [3] (b) The bicycle and cyclist have a combined mass of 92 kg. For The vertical height through which the cyclist moves is 1.3 m. Examiner’s Use (i) For the movement of the bicycle and cyclist between t = 0 and t = 8.0 s, 1. use Fig. 3.1 to calculate the change in kinetic energy, change = … J [2] 2. calculate the change in gravitational potential energy. change = … J [2] (ii) The cyclist pedals continuously so that the useful power delivered to the bicycle is 75 W. Calculate the useful work done by the cyclist climbing up the slope. work done = … J [2] (c) Some energy is used in overcoming frictional forces. For Examiner’s (i) Use your answers in (b) to show that the total energy converted in overcoming Use frictional forces is approximately 670 J. [1] (ii) Determine the average magnitude of the frictional forces. average force = … N [1] (d) Suggest why the magnitude of the total resistive force would not be constant. … … … [2]

13 marks

Mark scheme: 3 (a) evidence of use of area below the line B1 distance = 39 m (allow ±0.5 m) A2 [3] (if > ±0.5 m but ≤ 1.0 m, then allow 1 mark) (b) (i) 1 EK = ½mv 2 C1 ∆ EK = ½ × 92 × (62 – 32) = 1240 J A1 [2] 2 EP = mgh C1 ∆ EP = 92 × 9.8 × 1.3 = 1170 J A1 [2] (ii) E = Pt C1 E = 75 × 8 = 600 J A1 [2] (c) (i) energy = (1240 + 600) – 1170 M1 = 670 J A0 [1] (ii) force = 670/39 = 17 N A1 [1] (d) frictional forces include air resistance B1 air resistance decreases with decrease of speed B1 [2] GCE AS/A LEVEL – May/June 2010 9702 23

This question in 9702/23 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

5 (a) A variable resistor is used to control the current in a circuit, as shown in Fig. 5.1. For Examiner’s Use 12 V I1 R A 6.0 Ω Fig. 5.1 The variable resistor is connected in series with a 12 V power supply of negligible internal resistance, an ammeter and a 6.0 Ω resistor. The resistance R of the variable resistor can be varied between 0 and 12 Ω. (i) The maximum possible current in the circuit is 2.0 A. Calculate the minimum possible current. minimum current = … A [2] (ii) On Fig. 5.2, sketch the variation with R of current I1 in the circuit. 2.0 I1 / A 1.0 0 0 4 8 12 R / Ω [2] Fig. 5.2 (b) The variable resistor in (a) is now connected as a potential divider, as shown in Fig. 5.3. For Examiner’s Use 12 V I2 A 6.0 Ω Fig. 5.3 Calculate the maximum possible and minimum possible current I2 in the ammeter. maximum I2 = … A minimum I2 = … A [2] (c) (i) Sketch on Fig. 5.4 the I – V characteristic of a filament lamp. I 0 0 V [2] Fig. 5.4 (ii) The resistor of resistance 6.0 Ω is replaced with a filament lamp in the circuits of For Fig. 5.1 and Fig. 5.3. State an advantage of using the circuit of Fig. 5.3, compared Examiner’s to the circuit of Fig 5.1, when using the circuits to vary the brightness of the filament Use lamp. … … … [1]

9 marks

Mark scheme: 5 (a) (i) I = 12 / (6 + 12) C1 minimum current = 0.67 A A1 [2] (ii) correct start and finish points M1 correct shape for curve with decreasing gradient A1 [2] (b) maximum current = 2.0 A A1 minimum current = 0 A1 [2] (c) (i) smooth curve starting at (0,0) with decreasing gradient M1 end section not horizontal A1 [2] (ii) full range of current / p.d. possible or currents / p.d. down to zero or brightness ranging from off to full brightness B1 [1] GCE AS/A LEVEL – May/June 2011 9702 21

This question in 9702/21 May/June 2011

3 (a) (i) Explain what is meant by work done. For Examiner’s … Use … [1] (ii) Define power. … … [1] (b) Fig. 3.1 shows part of a fairground ride with a carriage on rails. 4.1 m 9.5 m s–1 30° Fig. 3.1 The carriage and passengers have a total mass of 600 kg. The carriage is travelling at a speed of 9.5 m s–1 towards a slope inclined at 30° to the horizontal. The carriage comes to rest after travelling up the slope to a vertical height of 4.1 m. (i) Calculate the kinetic energy, in kJ, of the carriage and passengers as they travel towards the slope. kinetic energy = … kJ [3] (ii) Show that the gain in potential energy of the carriage and passengers is 24 kJ. [2] (iii) Calculate the work done against the resistive force as the carriage moves up the For slope. Examiner’s Use work done = … kJ [1] (iv) Use your answer in (iii) to calculate the resistive force acting against the carriage as it moves up the slope. resistive force = … N [2]

10 marks

Mark scheme: 3 (a) (i) work done equals force × distance moved / displacement in the direction of the force B1 [1] (ii) power is the rate of doing work / work done per unit time B1 [1] (b) (i) kinetic energy = ½ mv2 C1 = 0.5 × 600 (9.5)2 C1 = 27075 (J) = 27 kJ A1 [3] (ii) potential energy = mgh = 600 × 9.81 × 4.1 M1 = 24132 (J) A1 = 24 kJ A0 [2] (iii) work done = 27 – 24 = 3.0 kJ A1 [1] (iv) resistive force = 3000 / 8.2 (distance along slope = 4.1 / sin 30°) C1 = 366 N A1 [2]

This question in 9702/22 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

3 (a) Define the terms For Examiner’s (i) power, Use … [1] (ii) the Young modulus. … … … [1] (b) A crane is used to lift heavy objects, as shown in Fig. 3.1. motor cable object of mass 1800 kg ground Fig. 3.1 The motor in the crane lifts a total mass of 1800 kg from rest on the ground. The cable supporting the mass is made of steel of Young modulus 2.4 × 1011 Pa. The cross-sectional area of the cable is 1.3 × 10– 4 m2. As the mass leaves the ground, the strain in the cable is 0.0010. Assume the weight of the cable to be negligible. (i) 1. Use the Young Modulus of the steel to show that the tension in the cable is 3.1 × 104 N. [2] 2. Calculate the acceleration of the mass as it is lifted from the ground. acceleration = … m s–2 [3] (ii) The motor now lifts the mass through a height of 15 m at a constant speed. For Examiner’s Calculate Use 1. the tension in the lifting cable, tension = … N [1] 2. the gain in potential energy of the mass. gain in potential energy = … J [2] (iii) The motor of the crane is 30% efficient. Calculate the input power to the motor required to lift the mass at a constant speed of 0.55 m s–1. input power = … W [3]

13 marks

Mark scheme: 3 (a) (i) power = work done per unit time / energy transferred per unit time / rate of work done B1 [1] (ii) Young modulus = stress / strain B1 [1] (b) (i) 1. E = T / (A × strain) (allow strain = ε) C1 T = E × A × strain = 2.4 × 1011 × 1.3 × 10–4 × 0.001 M1 = 3.12 × 104 N A0 [2] 2. T – W = ma C1 [3.12 × 104 – 1800 × 9.81] = 1800a C1 a = 7.52 m s–2 A1 [3] (ii) 1. T = 1800 × 9.81 = 1.8 × 104 N A1 [1] 2. potential energy gain = mgh C1 = 1800 × 9.81 × 15 = 2.7 × 105 J A1 [2] (iii) P = Fv C1 = 1800 × 9.81 × 0.55 C1 input power = 9712 × (100/30) = 32.4 × 103 W A1 [3]

This question in 9702/23 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

3 (a) Define power. For Examiner’s … Use … [1] (b) A cyclist travels along a horizontal road. The variation with time t of speed v is shown in Fig. 3.1. 12.0 10.0 8.0 v / m s–1 6.0 4.0 2.0 0 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 t / s Fig. 3.1 The cyclist maintains a constant power and after some time reaches a constant speed of 12 m s–1. (i) Describe and explain the motion of the cyclist. … … … … … [3] (ii) When the cyclist is moving at a constant speed of 12 m s–1 the resistive force is For 48 N. Show that the power of the cyclist is about 600 W. Explain your working. Examiner’s Use [2] (iii) Use Fig. 3.1 to show that the acceleration of the cyclist when his speed is 8.0 m s–1 is about 0.5 m s–2. [2] (iv) The total mass of the cyclist and bicycle is 80 kg. Calculate the resistive force R acting on the cyclist when his speed is 8.0 m s–1. Use the value for the acceleration given in (iii). R = … N [3] (v) Use the information given in (ii) and your answer to (iv) to show that, in this situation, the resistive force R is proportional to the speed v of the cyclist. [1]

12 marks

Mark scheme: 3 (a) power is the rate of doing work or power = work done / time (taken) or power = energy transferred / time (taken) B1 [1] (b) (i) as the speed increases drag / air resistance increases B1 resultant force reduces hence acceleration is less B1 constant speed when resultant force is zero B1 [3] (allow one mark for speed increases and acceleration decreases) GCE AS/A LEVEL – October/November 2012 9702 23 (ii) force from cyclist = drag force / resistive force B1 P = 12 × 48 M1 P = 576 W A0 [2] (iii) tangent drawn at speed = 8.0 m s–1 M1 gradient values that show acceleration between 0.44 to 0.48 m s–2 A1 [2] (iv) F – R = ma C1 600 / 8 – R = 80 × 0.5 [using P = 576] 576 / 8 – R = 80 × 0.5 C1 R = 75 – 40 = 35 N R = 72 – 40 = 32 N A1 [3] (v) at 12 m s–1 drag is 48 N, at 8 m s–1 drag is 35 or 32 N R / v calculated as 4 and 4 or 4.4 and consistent response for whether R is proportional to v or not B1 [1]

This question in 9702/23 Oct/Nov 2012

4 A circuit used to measure the power transfer from a battery is shown in Fig. 4.1. The power is For transferred to a variable resistor of resistance R. Examiner’s Use E r A I R V Fig. 4.1 The battery has an electromotive force (e.m.f.) E and an internal resistance r. There is a potential difference (p.d.) V across R. The current in the circuit is І. (a) By reference to the circuit shown in Fig. 4.1, distinguish between the definitions of e.m.f. and p.d. … … … … [3] (b) Using Kirchhoff’s second law, determine an expression for the current І in the circuit. [1] (c) The variation with current І of the p.d. V across R is shown in Fig. 4.2. For Examiner’s 6.0 Use 4.0 V / V 2.0 0 0 1.0 2.0 3.0 4.0 I /A Fig. 4.2 Use Fig. 4.2 to determine (i) the e.m.f. E, E = … V [1] (ii) the internal resistance r. r = … Ω [2] (d) (i) Using data from Fig. 4.2, calculate the power transferred to R for a current of 1.6 A. power = … W [2] (ii) Use your answers from (c)(i) and (d)(i) to calculate the efficiency of the battery for a current of 1.6 A. efficiency = … % [2]

11 marks

Mark scheme: 4 (a) e.m.f. = chemical energy to electrical energy M1 p.d. = electrical energy to thermal energy M1 idea of per unit charge A1 [3] (b) E = I (R +r) or I = E / (R +r) (any subject) B1 [1] (c) (i) E = 5.8 V B1 [1] (ii) evidence of gradient calculation or calculation with values from graph e.g. 5.8 = 4 + 1.0 × r C1 r = 1.8 Ω A1 [2] (d) (i) P = VI C1 P = 2.9 × 1.6 = 4.6 (4.64) W A1 [2] (ii) power from battery = 1.6 × 5.8 = 9.28 or efficiency = VI / EI C1 efficiency = (4.64 / 9.28) × 100 = 50 % or (2.9 / 5.8) × 100 = 50% A1 [2]

This question in 9702/23 Oct/Nov 2012

6 Two resistors A and B have resistances R1 and R2 respectively. The resistors are connected For in series with a battery, as shown in Fig. 6.1. Examiner’s Use E A B R1 R2 Fig. 6.1 The battery has electromotive force (e.m.f.) E and zero internal resistance. (a) State the energy transformation that occurs in (i) the battery, … … [1] (ii) the resistors. … … [1] (b) The current in the circuit is Ι. State the rate of energy transformation in (i) the battery, … [1] (ii) the resistor A. … [1] (c) The resistors are made from metal wires. Data for the resistors are given in Fig. 6.2. For Examiner’s Use resistor A B resistivity of metal ρ ρ / 2 length of wire l l diameter of wire d 2d Fig. 6.2 Use information from Fig. 6.2 to determine the ratio power dissipated in A . power dissipated in B ratio = … [3] (d) The resistors A and B are connected in parallel across the same battery of e.m.f. E. Determine the ratio power dissipated in A . power dissipated in B ratio = … [2]

9 marks

Mark scheme: 6 (a) (i) chemical to electrical B1 [1] (ii) electrical to thermal / heat or heat and light B1 [1] (b) (i) (PB =) EI or I2(R1 + R2) A1 [1] (ii) (PR =) I2R1 A1 [1] (c) R = ρl / A or clear from the following equation B1 ρl / πd 2 ratio = I2R1 / I2R2 = 2 or R1 has 8 × resistance of R 2 C1 ρ (2l ) / π (2d ) = 8 or 8:1 A1 [3] (d) P = V2 / R or E2 / R C1 (V or E the same) hence ratio is 1/8 or 1:8 = 0.125 (allow ecf from (c)) A1 [2]

This question in 9702/21 May/June 2013

6 (a) Define charge. For Examiner’s … [1] Use (b) A heater is made from a wire of resistance 18.0 Ω and is connected to a power supply of 240 V. The heater is switched on for 2.60 Ms. Calculate (i) the power transformed in the heater, power = … W [2] (ii) the current in the heater, current = … A [1] (iii) the charge passing through the heater in this time, charge = … C [2] (iv) the number of electrons per second passing a given point in the heater. number = … s–1 [2]

8 marks

Mark scheme: 6 (a) charge = current × time B1 [1] (b) (i) P = V 2 / R C1 = (240)2 / 18 = 3200 W A1 [2] (ii) I = V / R = 240 / 18 = 13.3 A A1 [1] (iii) charge = It = 13.3 × 2.6 × 106 C1 = 3.47 × 107 C A1 [2] (iv) number of electrons = 3.47 × 107 / 1.6 × 10–19 (= 2.17 × 1026) C1 number of electrons per second = 2.17 × 1026 / 2.6 × 106 = 8.35 × 1019 A1 [2] GCE AS/A LEVEL – May/June 2013 9702 22

This question in 9702/22 May/June 2013

6 (a) Define potential difference (p.d.). For Examiner’s … [1] Use (b) A battery of electromotive force 20 V and zero internal resistance is connected in series with two resistors R1 and R2, as shown in Fig. 6.1. 20 9 R1 R2 0 – 400 1 600 1 Fig. 6.1 The resistance of R2 is 600 Ω. The resistance of R1 is varied from 0 to 400 Ω. Calculate (i) the maximum p.d. across R2, maximum p.d. = … V [1] (ii) the minimum p.d. across R2. minimum p.d. = … V [2] (c) A light-dependent resistor (LDR) is connected in parallel with R2, as shown in Fig. 6.2. For Examiner’s Use 20 9 R1 R2 LDR R2 0 – 400 1 600 1 Fig. 6.2 When the light intensity is varied, the resistance of the LDR changes from 5.0 kΩ to 1.2 kΩ. (i) For the maximum light intensity, calculate the total resistance of R2 and the LDR. total resistance = … Ω [2] (ii) The resistance of R1 is varied from 0 to 400 Ω in the circuits of Fig. 6.1 and Fig. 6.2. State and explain the difference, if any, between the minimum p.d. across R2 in each circuit. Numerical values are not required. … … … [2] Please turn over for Question 7.

8 marks

Mark scheme: 6 (a) p.d. = work done / energy transformed (from electrical to other forms) charge B1 [1] (b) (i) maximum 20 V A1 [1] (ii) minimum = (600 / 1000) × 20 C1 = 12 V A1 [2] GCE AS/A LEVEL – May/June 2013 9702 23 (c) (i) use of 1.2 kΩ M1 1/1200 + 1/600 = 1/R, R = 400 Ω A1 [2] (ii) total parallel resistance (R2 + LDR) is less than R2 M1 (minimum) p.d. is reduced A1 [2]

This question in 9702/23 May/June 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 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 battery connected in series with a resistor R of resistance 5.0 Ω is shown in Fig. 6.1. For Examiner’s Use r 9.0 V R 5.0 Ω Fig. 6.1 The electromotive force (e.m.f.) of the battery is 9.0 V and the internal resistance is r. The potential difference (p.d.) across the battery terminals is 6.9 V. (a) Use energy considerations to explain why the p.d. across the battery is not equal to the e.m.f. of the battery. … … … [2] (b) Calculate (i) the current in the circuit, current = … A [2] (ii) the internal resistance r. r = … Ω [2] (c) Calculate, for the battery in the circuit, For Examiner’s (i) the total power produced, Use power = … W [2] (ii) the efficiency. efficiency = … [2]

10 marks

Mark scheme: 6 (a) e.m.f. = total energy available (per unit charge) B1 some (of the available energy) is used/lost/wasted/given out in the internal resistance of the battery (hence p.d. available less than e.m.f.) B1 [2] (b) (i) V = IR C1 I = 6.9 / 5.0 = 1.4 (1.38) A A1 [2] (ii) r = lost volts / current C1 r = (9– 6.9) / 1.38 = 1.5(2) Ω A1 [2] (c) (i) P = EI (not P = VI if only this line given or 9 V not used in second line) C1 P = 9 × 1.38 = 12 (12.4) W A1 [2] (ii) efficiency = output power / total power C1 efficiency = VI / EI = 6.9 / 9 or (9.52) / (12.4) = 0.767 / 76.7% A1 [2] GCE AS/A LEVEL – October/November 2013 9702 23

This question in 9702/23 Oct/Nov 2013

2 (a) (i) Define power. … [1] (ii) Use your definition in (i) to show that power may also be expressed as the product of force and velocity. [2] (b) A lorry moves up a road that is inclined at 9.0° to the horizontal, as shown in Fig. 2.1. 8.5 m s–1 road 9.0° Fig. 2.1 The lorry has mass 2500 kg and is travelling at a constant speed of 8.5 m s−1. The force due to air resistance is negligible. (i) Calculate the useful power from the engine to move the lorry up the road. power = … kW [3] (ii) State two reasons why the rate of change of potential energy of the lorry is equal to the power calculated in (i). 1. … … 2. … … [2]

8 marks

Mark scheme: 2 (a) (i) work (done) / time (taken) B1 [1] (ii) work = force × displacement (in direction of force) B1 power = force × displacement / time (taken) = force × velocity B1 [2] (b) (i) weight = mg C1 P = Fv = 2500 × 9.81 × sin 9° × 8.5 (or use cos 81°) C1 = 33 (32.6) kW A1 [3] (ii) no gain or loss of KE B1 no work (done) against air resistance B1 [2]

This question in 9702/21 May/June 2014

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

7 (a) A cell with internal resistance supplies a current. Explain why the terminal potential difference (p.d.) is less than the electromotive force (e.m.f.) of the cell. … … … [1] (b) A battery of e.m.f. 12 V and internal resistance 0.50 Ω is connected to a variable resistor X and a resistor Y of constant resistance, as shown in Fig. 7.1. 12 V 0.50 1 X Y Fig. 7.1 The resistance R of X is increased from 2.0 Ω to 16 Ω. The variation with R of the current І in the circuit is shown in Fig. 7.2. 3.0 I / A 2.0 1.0 0 0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 R / 1 Fig. 7.2 Calculate, for І = 1.2 A, (i) the p.d. across X, p.d. = … V [2] (ii) the resistance of Y, resistance = … Ω [3] (iii) the power dissipated in the battery. power = … W [2] (c) Use Fig. 7.2 to explain the variation in the terminal p.d. of the battery as the resistance R of X is increased. … … [1]

9 marks

Mark scheme: 7 (a) due to the lost volts in internal resistance / cell or energy losses in the internal resistance / cell B1 [1] (b) (i) V = ІR C1 = 1.2 × 6 = 7.2 V A1 [2] (ii) p.d. across Y and internal resistance r = 4.8 (V) [12 – 7.2] C1 resistance of Y + r = 4.8 / 1.2 = 4 (Ω) C1 resistance of Y = 4 – 0.5 = 3.5 Ω A1 [3] or Rtotal = 12 / 1.2 = 10 (Ω) (C1) X + r = 6.5 (Ω) (C1) resistance of Y = 3.5 Ω (A1) (iii) P = І2r C1 = (1.2)2 × 0.5 = 0.72 W A1 [2] (c) terminal p.d. increases as R is increased current decreases so there are less lost volts B1 [1]

This question in 9702/21 Oct/Nov 2014

1 (a) Use the definition of power to show that the SI base units of power are kg m2 s–3. [2] (b) Use an expression for electrical power to determine the SI base units of potential difference. units … [2]

4 marks

Mark scheme: 1 (a) power = work / time or energy / time or (force × distance) / time B1 = kg m s–2 × m s–1 = kg m2 s–3 A1 [2] (b) power = VI [or V2 / R and V = IR or I 2R and V = IR] B1 (units of V:) kg m2 s–3 A–1 B1 [2]

This question in 9702/21 May/June 2015

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

3 (a) Define power. … … [1] (b) Fig. 3.1 shows a car travelling at a speed of 22 m s–1 on a horizontal road. speed 22 m s–1 1200 N resistive force horizontal road Fig. 3.1 The car has a mass of 1500 kg. A resistive force of 1200 N acts on the car. Calculate (i) the force F required from the car to produce an acceleration of 0.82 m s–2, F = … N [3] (ii) the power required to produce this acceleration. power = … W [2] (c) The resistive force on the car is proportional to v 2, where v is the speed of the car. Suggest why the car has a maximum speed. … … [1]

7 marks

Mark scheme: 3 (a) (power =) work done / time (taken) or rate of work done A1 [1] (b) (i) F – R = ma C1 F = 1500 × 0.82 + 1200 C1 = 2400 (2430) N A1 [3] (ii) P = Fv C1 = (2430 × 22) = 53 000 (53 500) W A1 [2] (c) (there is maximum power from car and) resistive force = force produced by car hence no acceleration or suggestion in terms of power produced by car and power wasted to overcome resistive force B1 [1]

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

6 A 12 V battery with internal resistance 0.50 Ω is connected to two identical filament lamps L1 and L2 as shown in Fig. 6.1. 12 V 0.50 1 S1 L1 S2 L2 Fig. 6.1 The lamps are connected to the battery via switches S1 and S2. The power rating of each lamp is 48 W for a potential difference of 12 V. (a) S1 is closed and S2 open. State and explain whether the power transformed in L1 is 48 W. … … … [2] (b) S2 is now also closed. (i) State and explain the effect on the current in L1. … … … [1] (ii) State and explain the effect on the resistance of L1. … … … [1]

4 marks

Mark scheme: 6 (a) internal resistance causes lost volts B1 p.d. across lamp is less than 12 V, power is less than 48 W B1 [2] (b) (i) greater lost volts or p.d. across cell/lamp reduced, less current in lamp B1 [1] (ii) p.d. across lamp/current in lamp decreases, hence resistance decreases B1 [1]

This question in 9702/22 Oct/Nov 2015

5 (a) (i) State what is meant by an electric current. … … [1] (ii) Define electric potential difference (p.d.). … … [1] (b) A power supply of electromotive force (e.m.f.) 8.7 V and negligible internal resistance is connected by two identical wires to three filament lamps, as shown in Fig. 5.1. connecting wires power supply 8.7 V 0.30 A Fig. 5.1 (not to scale) The power supply provides a current of 0.30 A to the circuit. The filament lamps are identical. The I–V characteristic for one of the lamps is shown in Fig. 5.2. 0.40 I / A 0.30 0.20 0.10 0 0 1.0 2.0 3.0 4.0 V / V Fig. 5.2 (i) Show that the resistance of each connecting wire is 2.0 Ω. [2] (ii) The resistivity of the metal of the connecting wires does not vary with temperature. On Fig. 5.2, sketch the I–V characteristic for one of the connecting wires. [2] (iii) Calculate the power loss in one of the connecting wires. power = … W [2] (iv) Some data for the connecting wires are given below. cross-sectional area = 0.40 mm2 resistivity = 1.7 × 10−8 Ω m number density of free electrons = 8.5 × 1028 m−3 Calculate 1. the length of one of the connecting wires, length = … m [2] 2. the drift speed of a free electron in the connecting wires. drift speed = … m s−1 [2] [Total: 12]

12 marks

Mark scheme: 5 (a) (i) movement / flow of charge carriers B1 work (done) or energy (transformed)(from electrical to other forms) (ii) B1 charge (b) (i) p.d. across one lamp = 2.5 V C1 resistance = [(8.7 – 7.5) / 0.3] / 2 = 2.0 (Ω) A1 (ii) straight line through the origin M1 with gradient of 0.5 A1 (iii) P = I 2R or P = VI and V= IR or P = V 2 / R and V= IR C1 = 0.302 × 2.0 = 0.60 × 0.30 = 0.602 / 2.0 = 0.18 (W) A1 (iv) 1 R = ρl / A C1 l = (2.0 × 0.40 × 10–6) / 1.7 × 10–8 = 47 (m) A1 2 I = Anvq v = 0.30 / (0.40 × 10–6 × 8.5 × 1028 × 1.6 × 10–19) C1 = 5.5 × 10–5 (m s–1) A1 1

This question in 9702/22 Feb/March 2016

6 (a) Define the coulomb. … [1] (b) A resistor X is connected to a cell as shown in Fig. 6.1. I ; $ l Fig. 6.1 The resistor is a wire of cross-sectional area A and length l. The current in the wire is I. Show that the average drift speed v of the charge carriers in X is given by the equation I v = nAe where e is the charge on a charge carrier and n is the number of charge carriers per unit volume in X. [3] (c) A 12 V battery with negligible internal resistance is connected to two resistors Y and Z, as shown in Fig. 6.2. 12 V Y Z Fig. 6.2 The resistors are made from wires of the same material. The wire of Y has a diameter d and length l. The wire of Z has a diameter 2d and length 2l. (i) Determine the ratio average drift speed of the charge carriers in Y . average drift speed of the charge carriers in Z ratio = … [3] (ii) Show that resistance of Y = 2. resistance of Z [2] (iii) Determine the potential difference across Y. potential difference = … V [2] (iv) Determine the ratio power dissipated in Y . power dissipated in Z ratio = … [1] [Total: 12]

12 marks

Mark scheme: 6 (a) (coulomb is) ampere second B1 [1] (b) (total) charge or Q = nAle M1 I = Q / t and l / t = v M1 I = nAle / t = nAve therefore v = I / nAe A1 [3] (c) (i) ratio = (I / nAYe) / (I / nAZe) C1 = AZ / AY or 4A / A or πd2 / (πd2 / 4) C1 = 4 A1 [3] (ii) R = ρl / A or R = 4ρl / πd2 B1 RY = ρl / A and RZ = ρ(2l) / 4A so RY / RZ = 2 or RY = 4ρl /πd2 and RZ = 4ρ(2l) /π4d2 or 2ρl / πd2 so RY / RZ = 2 A1 [2] (iii) V = 12RY / (RY + RZ) or I = 12 / (RY + RZ) and V = IRY C1 V = 12 × 2/3 = 8(.0) V A1 [2] (iv) ratio = I2RY / I2RZ or (VY2 / RY) / (VZ 2 / RZ) or (VYI) / (VZI) = 2 A1 [1]

This question in 9702/21 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 electric potential difference (p.d.). … … [1] (b) A battery of electromotive force (e.m.f.) 14 V and negligible internal resistance is connected to a resistor network, as shown in Fig. 6.1. 14 V R2 R1 12 1 S R3 6.0 1 0–24 1 Fig. 6.1 R1 and R2 are fixed resistors of resistances 6.0 Ω and 12 Ω respectively. R3 is a variable resistor. Switch S is closed. (i) Calculate the current in the battery when the resistance of R3 is set 1. at zero, current = … A [2] 2. at 24 Ω. current = … A [2] (ii) Use your answers in (b)(i) to calculate the change in the total power produced by the battery when the resistance of R3 is changed from zero to 24 Ω. change in power = … W [2] (c) Switch S in Fig. 6.1 is now opened. Resistors R1 and R2 are made from metal wires. Some data for these resistors are shown in Fig. 6.2. R1 R2 cross-sectional area of wire A 1.8 A number of free electrons per unit volume in metal n 0.50 n Fig. 6.2 Determine the ratio average drift speed of free electrons in R1 . average drift speed of free electrons in R2 ratio = … [2] [Total: 9]

9 marks

Mark scheme: work done or energy (transform ed) (from electrical to other forms) 6 (a) B1 [1] charge (b) (i) 1. V = IR or E = IR C1 I = 14 / 6.0 = 2.3 (2.33) A A1 [2] 2. total resistance of parallel resistors = 8.0 Ω C1 current = 14 / (6.0 + 8.0) = 1.0 A A1 [2] (ii) P = EI (allow P = VI) or P = V2 / R or P = I2R C1 change in power = (14 × 2.33) – (14 × 1.0) or (142 / 6.0) – (142 / 14) or (2.332 × 6.0) – (1.02 × 14) = 19 W (18 W if 2.3 A used) A1 [2] (c) I = Anvq ratio = (0.50n / n) × (1.8 A / A) or ratio = 0.50 × 1.8 C1 = 0.90 A1 [2]

This question in 9702/21 Oct/Nov 2016

5 (a) State Kirchhoff’s second law. … … … … [2] (b) A battery is connected in parallel with two lamps A and B, as shown in Fig. 5.1. 6.8 V U A B Fig. 5.1 The battery has electromotive force (e.m.f.) 6.8 V and internal resistance r. The I–V characteristics of lamps A and B are shown in Fig. 5.2. 0.40 I / A ODPS % 0.30 0.20 ODPS $ 0.10 0 0 2.0 4.0 6.0 8.0 9 / V Fig. 5.2 The potential difference across the battery terminals is 6.0 V. (i) Use Fig. 5.2 to show that the current in the battery is 0.40 A. [2] (ii) Calculate the internal resistance r of the battery. r = … Ω [2] (iii) Determine the ratio resistance of lamp A . resistance of lamp B ratio = … [2] (iv) Determine 1. the total power produced by the battery, power = … W [2] 2. the efficiency of the battery in the circuit. efficiency = … [2] [Total: 12]

12 marks

Mark scheme: 5 (a) total/sum of electromotive forces or e.m.f.s = total/sum of potential differences or p.d.s M1 around a loop/(closed) circuit A1 [2] (b) (i) (current in battery =) current in A + current in B or IA + IB C1 (I =) 0.14 + 0.26 = 0.40 A A1 [2] (ii) E = V + Ir 6.8 = 6.0 + 0.40r or 6.8 = 0.40 (15 + r) C1 r = 2.0 Ω A1 [2] (iii) R = V / I C1 ratio (= RA / RB) = (6.0 / 0.14) / (6.0 / 0.26) = 42.9 / 23.1 or 0.26 / 0.14 = 1.9 (1.86) A1 [2] (iv) 1. P = EI or VI or P = I 2R or P = V 2 / R C1 = 6.8 × 0.40 = 0.402 × 17 = 6.82 / 17 = 2.7 W (2.72 W) A1 [2] 2. output power = VI = 6.0 × 0.40 (= 2.40 W) C1 efficiency = (6.0 × 0.40) / (6.8 × 0.40) = 2.40 / 2.72 = 0.88 or 88% (allow 0.89 or 89%) A1 [2]

This question in 9702/22 Oct/Nov 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) (i) Define power. … … [1] (ii) Show that the SI base units of power are kg m2 s–3. [1] (b) All bodies radiate energy. The power P radiated by a body is given by P = kAT 4 where T is the thermodynamic temperature of the body, A is the surface area of the body and k is a constant. (i) Determine the SI base units of k. base units … [2] (ii) On Fig. 1.1, sketch the variation with T 2 of P. The quantity A remains constant. P 0 0 T 2 Fig. 1.1 [1] [Total: 5]

5 marks

Mark scheme: 1(a)(i) work (done) / time (taken) or energy (transferred) / time (taken) B1 1(a)(ii) Correct substitution of base units of all quantities into any correct equation for power. Examples: (P = E / t or W / t gives) kg m2 s–2 / s = kg m2 s–3 (P = Fs / t or mgh / t gives) kg m s–2 m / s = kg m2 s–3 (P = ½mv2/ t gives) kg (m s–1)2 / s = kg m2 s–3 (P = Fv gives) kg m s–2 m s–1 = kg m2 s–3 (P = VI gives) kg m2 s–2 A–1 s–1 A = kg m2 s–3 A1 1(b)(i) units of A: m2 and units of T: K C1 units of k: kg m2 s–3 / m2 K4 = kg s–3 K–4 A1 1(b)(ii) curve from the origin with increasing gradient B1

This question in 9702/23 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) Define the volt. … [1] (b) A battery of electromotive force (e.m.f.) 4.5 V and negligible internal resistance is connected to two filament lamps P and Q and a resistor R, as shown in Fig. 6.1. 4.5 V R P Q Fig. 6.1 The current in lamp P is 0.15 A. The I–V characteristics of the filament lamps are shown in Fig. 6.2. 0.20 P I / A 0.15 Q 0.10 0.05 0 0 1.0 2.0 3.0 4.0 V / V Fig. 6.2 (i) Use Fig. 6.2 to determine the current in the battery. Explain your working. current = … A [2] (ii) Calculate the resistance of resistor R. resistance = … Ω [2] (iii) The filament wires of the two lamps are made from material with the same resistivity at their operating temperature in the circuit. The diameter of the wire of lamp P is twice the diameter of the wire of lamp Q. Determine the ratio length of filament wire of lamp P length of filament wire of lamp Q. ratio = … [3] (iv) The filament wire of lamp Q breaks and stops conducting. State and explain, qualitatively, the effect on the resistance of lamp P. … … … … [2] [Total: 10]

10 marks

Mark scheme: 6(a) joule / coulomb B1 6(b)(i) lamps have same p.d./lamps have p.d. of 2.7 V B1 current = 0.15 + 0.090 = 0.24 A A1 6(b)(ii) R = (4.5 – 2.7) / 0.24 or RP = 18 (Ω) and RQ = 30 (Ω) I / RT = 1 / 18 + 1 / 30 and so RT = 11.25 4.5 = 0.24 × (R + 11.25) C1 R = 7.5 Ω A1 Question Answer Marks 6(b)(iii) R = ρl / A C1 RP / RQ = [(2.7 / 0.15) / (2.7 / 0.09)] (= 0.60) C1 ratio = 0.60 × 22 = 2.4 A1 6(b)(iv) less p.d. across resistor/greater p.d. across P B1 greater current through P and so resistance (of P) increases B1

This question in 9702/21 May/June 2018

5 A solid cylinder is lifted out of oil by a wire attached to a motor. Fig. 5.1 shows two different positions X and Y of the cylinder during the lifting process. beam motor wire cylinder at position Y velocity surface of oil 0.020 m s–1 cylinder at position X oil Fig. 5.1 The motor is fixed to an overhead beam. The cylinder has cross-sectional area 0.018 m2, length 1.2 m and weight 560 N. The density of the oil is 940 kg m–3. Throughout the lifting process, the cylinder moves vertically upwards with a constant velocity of 0.020 m s–1. The viscous force of the oil acting on the cylinder is negligible. (a) Calculate the density of the cylinder. density = … kg m–3 [2] (b) For the cylinder at position X, show that the upthrust due to the oil is 200 N. [2] (c) Calculate, for the moving cylinder at position X, (i) the tension in the wire, tension = … N [1] (ii) the power output of the motor. power = … W [2] (d) The cylinder is raised with constant velocity from position X to position Y. (i) State and explain the variation, if any, of the power output of the motor as the cylinder is raised. Numerical values are not required. … … … … … [3] (ii) The rate of energy output of the motor is less than the rate of increase of gravitational potential energy of the cylinder. Without calculation, explain this difference. … … [1] [Total: 11]

11 marks

Mark scheme: 5(a) C1 = (560 / 9.81) / (1.2 × 0.018) = 2600 kg m–3 A1 5(b) (∆)p = 940 × 9.81 × 1.2 C1 (upthrust =) 940 × 9.81 × 1.2 × 0.018 = 200 N A1 5(c)(i) tension = 560 – 200 = 360 N A1 5(c)(ii) P = Fv C1 = 360 × 0.020 = 7.2 W A1 5(d)(i) upthrust decreases B1 tension (in wire) increases M1 power (output of motor) increases A1 5(d)(ii) there is work done (on the cylinder) by the upthrust or GPE of oil decreases (as it fills the space left by cylinder and so total energy is conserved) B1

This question in 9702/22 May/June 2018

1 (a) An analogue voltmeter is used to take measurements of a constant potential difference across a resistor. For these measurements, describe one example of (i) a systematic error, … … [1] (ii) a random error. … … [1] (b) The potential difference across a resistor is measured as 5.0 V ± 0.1 V. The resistor is labelled as having a resistance of 125 Ω ± 3%. (i) Calculate the power dissipated by the resistor. power = … W [2] (ii) Calculate the percentage uncertainty in the calculated power. percentage uncertainty = … % [2] (iii) Determine the value of the power, with its absolute uncertainty, to an appropriate number of significant figures. power = … ± … W [2] [Total: 8]

8 marks

Mark scheme: 1(a)(i) zero error or wrongly calibrated scale B1 1(a)(ii) reading scale from different angles or wrongly interpolating between scale readings/divisions B1 1(b)(i) P = V 2 / R or P = VI and V = IR C1 P = 5.02 / 125 or 5.0 × 0.04 or (0.04)2 × 125 = 0.20 W A1 1(b)(ii) %V = 2% or ∆V / V = 0.02 C1 %P = (2 × 2%) + 3% or %P = (2 × 0.02 + 0.03) × 100 = 7% A1 1(b)(iii) absolute uncertainty in P = (7 / 100) × 0.20 = 0.014 C1 power = 0.20 ± 0.01 W or (2.0 ± 0.1) × 10–1 W A1

This question in 9702/23 May/June 2018

3 (a) (i) Define power. … … [1] (ii) State what is meant by gravitational potential energy. … … [1] (b) An aircraft of mass 1200 kg climbs upwards with a constant velocity of 45 m s–1, as shown in Fig. 3.1. velocity thrust force 45 m s–1 2.0 × 103 N path of aircraft aircraft mass 1200 kg Fig. 3.1 (not to scale) The aircraft’s engine produces a thrust force of 2.0 × 103 N to move the aircraft through the air. The rate of increase in height of the aircraft is 3.3 m s–1. (i) Calculate the power produced by the thrust force. power = … W [2] (ii) Determine, for a time interval of 3.0 minutes, 1. the work done by the thrust force to move the aircraft, work done = … J [2] 2. the increase in gravitational potential energy of the aircraft, increase in gravitational potential energy = … J [2] 3. the work done against air resistance. work done = … J [1] (iii) Use your answer in (b)(ii) part 3 to calculate the force due to air resistance acting on the aircraft. force = … N [1] (iv) With reference to the motion of the aircraft, state and explain whether the aircraft is in equilibrium. … … … [2] [Total: 12]

12 marks

Mark scheme: 3(a)(i) work (done) / time (taken) B1 3(a)(ii) energy of a mass due to its position in a gravitational field B1 3(b)(i) P = Fv C1 = 2.0 × 103 × 45 = 9.0 × 104 W A1 3(b)(ii) 1. W = (2.0 × 103) × (45 × 3.0 × 60) or W = 9.0 × 104 × 3.0 × 60 C1 W = 1.6 × 107 J A1 2. (∆)EP = mg(∆)h C1 = 1200 × 9.81 × 3.3 × 3.0 × 60 = 7.0 × 106 J A1 3. W = 1.6 × 107 – 7.0 × 106 = 9.0 × 106 J A1 3(b)(iii) force = (9.0 × 106) / (45 × 3.0 × 60) = 1.1 × 103 N A1 3(b)(iv) constant velocity so no resultant force B1 no resultant force so in equilibrium B1

This question in 9702/21 Oct/Nov 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) Using energy transformations, describe the electromotive force (e.m.f.) of a battery and the potential difference (p.d.) across a resistor. e.m.f.: … … p.d.: … … [2] (b) A battery of e.m.f. 6.0 V and negligible internal resistance is connected to a network of resistors and a voltmeter, as shown in Fig. 6.1. Z V 32 Ω 6.0 V Y X 24 Ω Fig. 6.1 Resistor Y has a resistance of 24 Ω and resistor Z has a resistance of 32 Ω. (i) The resistance RX of the variable resistor X is adjusted until the voltmeter reads 4.8 V. Calculate: 1. the current in resistor Z current = … A [1] 2. the total power provided by the battery power = … W [2] 3. the number of conduction electrons that move through the battery in a time interval of 25 s number = … [2] 4. the total resistance of X and Y connected in parallel total resistance = … Ω [2] 5. the resistance RX. Ω [2] RX = … (ii) The resistance RX is now decreased. State and explain the change, if any, to the reading on the voltmeter. … … … [2] [Total: 13]

13 marks

Mark scheme: 6(a) e.m.f.: energy transferred from chemical to electrical (per unit charge) B1 p.d.: energy transferred from electrical to thermal (per unit charge) B1 Question Answer Marks 6(b)(i) 1 I = 4.8 / 32 = 0.15 A A1 2 P = EI or P = VI or P = I 2R or P = V 2 / R = 6.0 × 0.15 or 0.152 × 40 or 6.02 / 40 C1 = 0.90 W A1 3 number = It / e = [0.15 × 25] / 1.6 × 10–19 C1 = 2.3 × 1019 A1 or Q = 0.15 × 25 (= 3.75) number = 3.75 / 1.6 × 10–19 (C1) = 2.3 × 1019 (A1) 4 4.8 / 6.0 = 32 / (RXY + 32) or 1.2 / 6.0 = RXY / (RXY + 32) or 4.8 / 1.2 = 32 / RXY C1 RXY = 8.0 Ω A1 Alternative methods: RXY = (6.0 – 4.8) / 0.15 or (C1) = 8.0 Ω (A1) or 6.0 = 0.15 (32 + RXY) (C1) RXY = 40 – 32 = 8.0 Ω (A1) Question Answer Marks 6(b)(i) 5 1 / 8.0 = 1 / RX + 1 / 24 C1 Rx = 12 Ω A1 Alternative method: IZ = 4.8 / 32 = 0.15 and IY = 1.2 / 24 = 0.05 IX = 0.15 – 0.05 (= 0.10) (C1) RX = 1.2 / 0.10 = 12 Ω (A1) 6(b)(ii) total resistance decreases M1 (so voltmeter) reading increases A1

This question in 9702/22 Feb/March 2019

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

3 A cylindrical disc of mass 0.24 kg has a circular cross-sectional area A, as shown in Fig. 3.1. cross-sectional force X area A 8.9 N constant 30° speed 0.60 m s–1 disc, disc mass 0.24 kg ground Fig. 3.1 Fig. 3.2 The disc is on horizontal ground, as shown in Fig. 3.2. A force X of magnitude 8.9 N acts on the disc in a direction of 30° to the horizontal. The disc moves at a constant speed of 0.60 m s−1 along the ground. (a) Determine the rate of doing work on the disc by the force X. rate of doing work = … W [2] (b) The force X and the weight of the disc exert a combined pressure on the ground of 3500 Pa. Calculate the cross-sectional area A of the disc. A = … m2 [3] (c) Newton’s third law describes how forces exist in pairs. One such pair of forces is the weight of the disc and another force Y. State: (i) the direction of force Y … [1] (ii) the name of the body on which force Y acts. … [1] [Total: 7]

7 marks

Mark scheme: 3(a) P = Fv C1 P = 8.9 cos 30° × 0.60 = 4.6 W A1 3(b) p = F / A C1 F = 8.9 sin 30° + (0.24 × 9.81) ( = 6.80 N) C1 A = 6.80 / 3500 = 1.9 × 10–3 m2 A1 3(c)(i) upwards/up B1 3(c)(ii) the Earth/planet B1

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 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

6 (a) A battery of electromotive force (e.m.f.) 7.8 V and internal resistance r is connected to a filament lamp, as shown in Fig. 6.1. 7.8 V r Fig. 6.1 A total charge of 750 C moves through the battery in a time interval of 1500 s. During this time the filament lamp dissipates 5.7 kJ of energy. The e.m.f. of the battery remains constant. (i) Explain, in terms of energy and without a calculation, why the potential difference across the lamp must be less than the e.m.f. of the battery. … … [1] (ii) Calculate: 1. the current in the circuit current = … A [2] 2. the potential difference across the lamp potential difference = … V [2] 3. the internal resistance of the battery. internal resistance = … Ω [2] (b) A student is provided with three resistors of resistances 90 Ω, 45 Ω and 20 Ω. (i) Sketch a circuit diagram showing how two of these three resistors may be connected together to give a combined resistance of 30 Ω between the terminals shown. Label the values of the resistances on your diagram. [1] (ii) A potential divider circuit is produced by connecting the three resistors to a battery of e.m.f. 9.0 V and negligible internal resistance. The potential divider circuit provides an output potential difference VOUT of 3.6 V. The circuit diagram is shown in Fig. 6.2. 9.0 V Fig. 6.2 On Fig. 6.2, label the resistances of all three resistors and the potential difference V OUT. [2] [Total: 10]

10 marks

Mark scheme: 6(a)(i) energy is dissipated in the internal resistance/r B1 6(a)(ii) 1. I = Q / t C1 = 750 / 1500 = 0.50 A A1 2. V = W /Q or V = W / It C1 = 5700 / 750 or 5700 / (0.50 × 1500) = 7.6 V A1 or V = P / I and P = W / t (C1) V = 3.8 / 0.50 = 7.6 V (A1) 3. r = (7.8 – 7.6) / 0.50 C1 = 0.40 Ω A1 6(b)(i) 90 Ω and 45 Ω resistors shown connected in parallel B1 6(b)(ii) the resistors connected in parallel labelled as 90 Ω and 45 Ω with the other resistor labelled as 20 Ω M1 VOUT or 3.6 V labelled across the 20 Ω resistor A1

This question in 9702/22 May/June 2020

5 (a) Define the volt. … … [1] (b) Fig. 5.1 shows a network of three resistors. 300 Ω 55 Ω X Y 100 Ω Fig. 5.1 Calculate: (i) the combined resistance of the two resistors connected in parallel combined resistance = … Ω [1] (ii) the total resistance between terminals X and Y. total resistance = … Ω [1] (c) The network in (b) is connected to a power supply so that there is a potential difference between terminals X and Y. The power dissipated in the resistor of resistance 55 Ω is 0.20 W. (i) Calculate the current in the resistor of resistance: 1. 55 Ω current = … A 2. 300 Ω. current = … A [3] (ii) Calculate the potential difference between X and Y. potential difference = … V [1] [Total: 7]

7 marks

Mark scheme: 5(a) joule per coulomb B1 5(b)(i) 1 / R = 1 / R1 + 1 / R2 = 1 / 300 + 1 / 200 R = 75 Ω A1 5(b)(ii) R = 75 + 55 = 130 Ω A1 5(c)(i) 1. P = I2R or P = VI and V = IR C1 I = (0.20 / 55)0.5 = 0.060 A A1 2. I = 0.060 / 4 = 0.015 A A1 5(c)(ii) potential difference = 130 × 0.060 = 7.8 V A1 or potential difference = (300 × 0.015) + (55 × 0.060) = 7.8 V (other valid methods are also possible) (A1)

This question in 9702/23 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

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

2 A person uses a trolley to move suitcases at an airport. The total mass of the trolley and suitcases is 72 kg. (a) The person pushes the trolley and suitcases along a horizontal surface with a constant speed of 1.4 m s–1 and then releases the trolley. The released trolley moves in a straight line and comes to rest. Assume that a constant total resistive force of 18 N opposes the motion of the trolley and suitcases. (i) Calculate the power required to overcome the total resistive force on the trolley and suitcases when they move with a constant speed of 1.4 m s–1. power = … W [2] (ii) Calculate the time taken for the trolley to come to rest after it is released. time = … s [3] (b) At another place in the airport, the trolley and suitcases are on a slope, as shown in Fig. 2.1. trolley and suitcases 18 N F, 54 N X slope 9.5 m Y Fig. 2.1 (not to scale) The person releases the trolley from rest at point X. The trolley moves down the slope in a straight line towards point Y. The distance along the slope between points X and Y is 9.5 m. The component F of the weight of the trolley and suitcases that acts along the slope is 54 N. Assume that a constant total resistive force of 18 N opposes the motion of the trolley and suitcases. (i) Calculate the speed of the trolley at point Y. speed = … m s–1 [3] (ii) Calculate the work done by F for the movement of the trolley from X to Y. work done = … J [1] (iii) The trolley is released at point X at time t = 0. On Fig. 2.2, sketch a graph to show the variation with time t of the work done by F for the movement of the trolley from X to Y. Numerical values of the work done and t are not required. work done 0 0 t Fig. 2.2 [2] (c) The angle of the slope in (b) is constant. The frictional forces acting on the wheels of the moving trolley are also constant. Explain why, in practice, it is incorrect to assume that the total resistive force opposing the motion of the trolley and suitcases is constant as the trolley moves between X and Y. … … [1] [Total: 12]

12 marks

Mark scheme: 2(a)(i) P = Fv C1 = 18 × 1.4 = 25 W A1 2(a)(ii) a = F / m C1 a = 18 / 72 = 0.25 (m s–2) t = 1.4 / 0.25 C1 = 5.6 s A1 2(b)(i) a = (54 – 18) / 72 or 36 / 72 (= 0.50 m s–2) C1 v 2 = 2 × 0.50 × 9.5 C1 v = 3.1 m s–1 A1 2(b)(ii) W = 54 × 9.5 = 510 J A1 2(b)(iii) curved line from the origin M1 gradient of line increases A1 2(c) (force due to) air resistance increases/changes/not constant or air resistance increases with speed B1

This question in 9702/21 May/June 2021

5 (a) Define the electromotive force (e.m.f.) of a source. … … … [2] (b) The circuit shown in Fig. 5.1 contains a battery of e.m.f. E that has internal resistance r, a variable resistor, a voltmeter and an ammeter. E r X Y A V I Fig. 5.1 Readings from the two meters are taken for different settings of the variable resistor. The variation with current I of the potential difference (p.d.) V across the terminals XY of the battery is shown in Fig. 5.2. 8 V / V 6 4 2 0 0 0.2 0.4 0.6 0.8 1.0 1.2 I / A Fig. 5.2 Explain why V is not constant. … … … … [3] (c) For the battery in (b), use Fig. 5.2 to determine: (i) the e.m.f. E E = … V [1] (ii) the maximum current that the battery can supply maximum current = … A [1] (iii) the internal resistance r. r = … Ω [2] (d) On Fig. 5.2, sketch a line to show a possible variation with I of V for a battery with a lower e.m.f. and a lower internal resistance than the battery in (b). Your line should extend over at least the same range of currents as the original line. [2] [Total: 11]

11 marks

Mark scheme: 5(a) energy per unit charge B1 energy transferred by source driving charge around the complete circuit or energy transferred from other forms to electrical energy B1 5(b) there is a p.d. across the internal resistance/r B1 change in current/I results in a change in p.d. across the internal resistance B1 V = E – p.d. across internal resistance or change in p.d. across r causes a change in V (as e.m.f. is constant) B1 5(c)(i) E = 7.4 V A1 5(c)(ii) maximum current = 0.92 A A1 5(c)(iii) r = E / IMAX or (–)gradient C1 e.g. r = 7.4 / 0.92 = 8.0 Ω A1 5(d) straight line with negative gradient that is smaller in magnitude than the original line B1 line which would have intercept on V-axis below the original line B1 Question Answer Marks

This question in 9702/23 May/June 2021

3 (a) Define power. … … [1] (b) A car of mass 1700 kg moves in a straight line along a slope that is at an angle θ to the horizontal, as shown in Fig. 3.1. B 25 m car, slope A θ mass 1700 kg horizontal Fig. 3.1 (not to scale) The car moves at constant velocity for a distance of 25 m from point A to point B. Air resistance and friction provide a total resistive force of 440 N that opposes the motion of the car. For the movement of the car from A to B: (i) state the change in the kinetic energy change in kinetic energy = … J [1] (ii) calculate the work done against the total resistive force. work done = … J [1] (c) The movement of the car in (b) from A to B causes its gravitational potential energy to increase by 4.8 × 104 J. Calculate: (i) the increase in vertical height h of the car for its movement from A to B h = … m [2] (ii) angle θ. θ = … ° [1] (d) The engine of the car in (b) produces an output power of 1.7 × 104 W to move the car along the slope. Calculate the time taken for the car to move from A to B. time = … s [2] [Total: 8]

8 marks

Mark scheme: 3(a) work (done) / time (taken) B1 3(b)(i) zero / 0 J A1 3(b)(ii) work done = 440 × 25 = 1.1 × 104 J A1 3(c)(i) (Δ)E(P) = mg(Δ)h C1 h = 4.8 × 104 / (1700 × 9.81) = 2.9 m A1 3(c)(ii) θ = sin–1 (2.9 / 25) = 6.7° A1 3(d) work done = 4.8 × 104 + 1.1 × 104 (= 5.9 × 104 J) C1 time = 5.9 × 104 / 1.7 × 104 = 3.5 s A1

This question in 9702/22 Oct/Nov 2021

6 A cell of electromotive force (e.m.f.) 0.48 V is connected to a metal wire X, as shown in Fig. 6.1. 0.48 V internal resistance 0.80 A wire X, resistance 0.40 Ω Fig. 6.1 The cell has internal resistance. The current in the cell is 0.80 A. Wire X has length 3.0 m, cross-sectional area 1.3 × 10–7 m2 and resistance 0.40 Ω. (a) Calculate the charge passing through the cell in a time of 7.5 minutes. charge = … C [2] (b) Calculate the percentage efficiency with which the cell supplies power to wire X. efficiency = … % [3] (c) There are 3.2 × 1022 free (conduction) electrons contained in the volume of wire X. For wire X, calculate: (i) the number density n of the free electrons n = … m–3 [1] (ii) the average drift speed of the free electrons. average drift speed = … m s–1 [2] (d) A wire Y has the same cross-sectional area as wire X and is made of the same metal. Wire Y is longer than wire X. Wire X in the circuit is now replaced by wire Y. Assume that wire Y has the same temperature as wire X. State and explain whether the average drift speed of the free electrons in wire Y is greater than, the same as, or less than that in wire X. … … … … … … [3] [Total: 11]

11 marks

Mark scheme: 6(a) C1 = 0.80 × 7.5 × 60 = 360 C A1 6(b) P = EI or P = VI or P = I2R or P = V2 / R C1 0.802 × 0.40 (= 0.256 W) or 0.48 × 0.80 (= 0.384 W) C1 efficiency = (0.256 / 0.384) × 100 = 67% A1 6(c)(i) n = 3.2 × 1022 / (1.3 × 10–7 × 3.0) = 8.2 × 1028 m–3 A1 6(c)(ii) I = Anvq v = 0.80 / (1.3 × 10–7 × 8.2 × 1028 × 1.60 × 10–19) C1 = 4.7 × 10–4 m s–1 A1 6(d) (wire Y has) larger resistance / resistance increases M1 (wire Y has) smaller current / current decreases M1 (average drift) speed is less (in wire Y) A1

This question in 9702/22 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

3 (a) (i) Define power. … … [1] (ii) A force F takes time t to move an object through a displacement x at constant velocity v in the direction of the force. The work done by the force is W. Use the definition of power to show that the power P transferred by the force is given by P = Fv. [2] (b) A block is pulled up a slope by a wire attached to a motor, as shown in Fig. 3.1. wire motor block slope 11° horizontal Fig. 3.1 (not to scale) The useful power output of the motor is 56 W. The block has a weight of 430 N and travels with constant velocity along the slope at an angle of 11° to the horizontal. Assume that there are no resistive forces opposing the motion of the block. (i) Calculate the tension T in the wire. T = … N [2] (ii) Calculate the speed of the block. speed = … m s–1 [1] (iii) The rate of increase of gravitational potential energy of the block is equal to the useful power output of the motor. One of the reasons for this is that there is no work done against resistive forces. By considering the motion of the block, state another reason for this. … … [1] (iv) The motor has an efficiency of 80%. Calculate the time taken for an input energy of 1.2 kJ to be supplied to the motor. time taken = … s [2] [Total: 9]

9 marks

Mark scheme: 3(a)(i) work (done) / time (taken) B1 3(a)(ii) W = Fx or E = Fx B1 (P =) Fx / t = Fv or P = Fvt / t = Fv B1 3(b)(i) T = 430 sin 11° or 430 cos 79° C1 = 82 N A1 3(b)(ii) speed = 56 / 82 = 0.68 m s–1 A1 3(b)(iii) no change in kinetic energy (of block) B1 3(b)(iv) (input power) = 56 / 0.80 ( = 70 W) C1 time taken = 1.2  103 / 70 = 17 s A1 or (useful energy) = 1200  80 / 100 ( = 960 J) (C1) time taken = 960 / 56 = 17 s (A1)

This question in 9702/21 May/June 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

4 (a) A mass m moves a vertical distance Δh in a uniform gravitational field and gains gravitational potential energy ΔEP. The acceleration of free fall is g. Use the concept of work done to show that ΔEP = mgΔh. [2] (b) A 0.60 kg mass is attached to a string which is wrapped around the wheel of a generator, as shown in Fig. 4.1. generator wheel string resistor mass, 0.60 kg Fig. 4.1 The mass is held stationary above the floor. When released, the mass initially accelerates and then falls at a steady speed and spins the wheel. The generator causes a current in a resistor. Air resistance is negligible. State the main energy change when the mass is falling at a steady speed. … energy to … energy. [1] (c) When falling at a steady speed, the mass in (b) falls through a vertical distance of 1.4 m in a time of 4.0 s. This causes a current of 90 mA in the resistor. The resistance of the resistor is 47 Ω. Calculate: (i) the rate of work done by the falling mass rate of work done = … W [2] (ii) the power dissipated in the resistor power = … W [2] (iii) the efficiency of the generator. efficiency = … [2] [Total: 9]

9 marks

Mark scheme: 4(a) M1 (force = mg and distance = h) (so) work (done) = mgh and work = E(P) (so E(P) = mgh) A1 4(b) gravitational potential (energy) to heat/thermal (energy) B1 4(c)(i) P = mg()h / ()t or Fv C1 P = (0.60  9.81  1.4) / 4.0 or 0.60  9.81  (1.4 / 4.0) = 2.1 W A1 4(c)(ii) P = I2R or IV or V2 / R C1 = 0.092  47 or 0.09  4.23 or 4.232 / 47 = 0.38 W A1 4(c)(iii) efficiency = Pout / Pin ( 100) or Eout / Ein ( 100) C1 = 0.38 / 2.1 ( 100) or 0.38  4.0 / 2.1  4.0 ( 100) = 0.18 or 18% A1

This question in 9702/23 May/June 2022

3 (a) (i) Define power. … … [1] (ii) Mechanical power P can be calculated using the formula P = Fv. Use the concept of work and the definition of power to show how this formula is derived. [2] (b) The engine of a lorry provides 130 kW of power to the lorry’s wheels when it is travelling at a constant speed of 25 m s–1 along a straight horizontal road. Show that the resistive force opposing the forward motion of the lorry is 5200 N. [1] (c) The lorry in (b) travels up a straight section of road that is inclined at an angle θ to the horizontal, as shown in Fig. 3.1. lorry, mass m road θ horizontal Fig. 3.1 (not to scale) The lorry has mass m and the acceleration of free fall is g. (i) Determine an expression, in terms of m, g and θ, for the component of the weight of the lorry that acts parallel to the surface of the road. [1] (ii) The total resistive force remains unchanged at 5200 N and the engine now provides greater power to maintain the speed of 25 m s–1. The total mass m of the lorry is 36 000 kg. The angle θ is 1.4°. Determine the power, in kW, now provided by the engine. power = … kW [3] [Total: 8]

8 marks

Mark scheme: 3(a)(i) work done per unit time B1 3(a)(ii) W = Fs B1 P = Fs / t and (so) P = Fv B1 3(b) (F =) 130  103 / 25 = 5200 (N) A1 3(c)(i) (component of weight =) mg sin A1 3(c)(ii) F (along slope due to weight) = 36 000  9.81  sin 1.4° C1 ( = 8600 N) (total) F = 5200 + 36 000  9.81  sin 1.4° C1 ( = 13 800 N) P = 13 800  25 A1 = 350  103 (W) = 350 kW

This question in 9702/21 Oct/Nov 2022

6 (a) Define electric potential difference. … … [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. State and explain which resistor dissipates more power. … … … … … [3] (c) A battery of electromotive force (e.m.f.) 9.0 V and internal resistance r is connected to two resistors P and Q, as shown in Fig. 6.2. 9.0 V r P Q 4.5 A Fig. 6.2 A total charge of 650 C moves through resistor P in a time interval of 540 s. During this time resistor P dissipates 4800 J of energy. The current in resistor Q is 4.5 A. Assume that the e.m.f. of the battery remains constant. Calculate: (i) the current in resistor P current = … A [2] (ii) the potential difference across resistor P potential difference = … V [2] (iii) the internal resistance r of the battery. r = … Ω [2] [Total: 10]

10 marks

Mark scheme: 6(a) energy transferred per (unit) charge (from electrical to other forms) B1 6(b) same/equal current (in X and Y) B1 P = I2R (and RX > RY) M1 or P = VI and VX > VY (so) X (dissipates more power) A1 6(c)(i) I = Q / t C1 = 650 / 540 A1 = 1.2 A 6(c)(ii) V = W / Q or W / It C1 = 4800 / 650 or 4800 / (1.2  540) A1 = 7.4 V or V = P / I and P = W / t (C1) = 8.9 / 1.2 (A1) = 7.4 V 6(c)(iii) I = 4.5 + 1.2 (= 5.7 A) C1 9.0 = 7.4 + 5.7r or 9.0 = 5.7 (1.3 + r) A1 r = 0.28 

This question in 9702/22 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

1 (a) (i) Define power. … … [1] (ii) Use the definition of power to show that the SI base units of power are kg m2 s–3. [1] (b) The intensity I of a sound wave moving through a gas is given by I = f 2A2vk where f is the frequency of the wave, A is the amplitude of the wave, v is the speed of the wave and k is a constant that depends on the gas. Determine the SI base units of k. SI base units … [3] [Total: 5]

5 marks

Mark scheme: 1(a)(i) work done per unit time B1 1(a)(ii) (P = W / t gives) units: kg m2 s–2 / s = kg m2 s–3 B1 1(b) (I = P / A so) units of I: kg m2 s–3 / m2 or kg s–3 C1 units of f: s–1 and units of A: m and units of v: m s–1 C1 units of k: kg s–3 / [(s–1)2 m2 m s–1] = kg m–3 A1

This question in 9702/21 May/June 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

5 A train travels at a constant high speed along a straight horizontal track towards an observer standing adjacent to the track, as shown in Fig. 5.1. train observer track Fig. 5.1 The train sounds its horn continuously as it approaches the observer, from time t = 0 until it is well past the observer at time t = t2. The train passes the observer at time t = t1. The horn emits a sound wave of constant frequency fS. (a) On Fig. 5.2, sketch the variation of the frequency of sound heard by the observer with time t, from time t = 0 to t = t2. frequency fS 0 0 t1 t2 t Fig. 5.2 [1] (b) At a particular time, the sound waves at the observer have an intensity of 4.7 × 10–3 W m–2. The waves at the observer are incident at right angles on a circular detector of radius 2.8 cm. Calculate the power P of the waves incident on the detector. P = … W [3] [Total: 4]

4 marks

Mark scheme: 5(a) sketch: approximately horizontal line above horizontal dashed line from t = 0 to t = t1 A1 and approximately horizontal line below horizontal dashed line from t = t1 to t = t2 5(b) I = P / A C1 A =   0.0282 or   2.82 C1 ( = 2.46  10–3 or 24.6) P = 4.7  10–3  2.46  10–3 A1 = 1.2  10–5 W

This question in 9702/21 Oct/Nov 2023

1 (a) Table 1.1 lists some SI quantities. Complete the table by indicating with a tick (✓) which rows are SI base quantities. Table 1.1 quantity base quantity current energy force mass [1] (b) Use the definition of power to determine its SI base units. SI base units … [2] (c) A light meter is used to measure the intensity of light in a classroom. Daylight is incident normally on the sensor of the meter. The sensor has an area of 2.2 cm2. The reading on the meter is 950 W m–2. Calculate the power of the daylight incident on the sensor. power = … W [3] [Total: 6]

6 marks

Mark scheme: Question Answer Marks 1(a) current and mass only ticked A1 1(b) (power =) work (done) / time C1 units of power = J s–1 A1 = kg m2 s–2 / s = kg m2 s–3 1(c) power = intensity  area C1 = 950  2.2  10–4 C1 = 0.21 W A1

This question in 9702/22 Feb/March 2024

7 (a) Define electric potential difference. … … [1] (b) A cell of electromotive force (e.m.f.) 1.8 V and internal resistance r is connected in parallel with a resistor of resistance 6.0 Ω and a filament lamp, as shown in Fig. 7.1. 1.8 V r A 6.0 Ω S Fig. 7.1 The switch S is open. The ammeter reading is 0.25 A. Determine the internal resistance r of the cell. r = … Ω [3] (c) At time t1 switch S in Fig. 7.1 is closed. Fig. 7.2 shows the variation with time t of the ammeter reading I. I 0 0 t1 t Fig. 7.2 (i) State whether the e.m.f. of the cell after t1 is greater than, less than or the same as it was before t1. … [1] (ii) By considering the effect of the lamp on the total resistance of the circuit, explain the variation of the ammeter reading shown in Fig. 7.2. … … … … … … [3] [Total: 8]

8 marks

Mark scheme: 7(a) energy (transferred) per (unit) charge B1 7(b) V = 0.25  6 C1 =1.5 Ir = E – IR C1 Ir = 1.8 – 1.5 = 0.3 r = 0.3 / 0.25 A1 = 1.2  or (C1) (Total) R = 1.8 / 0.25 = 7.2 E / I = (R + r) 1.8/0.25 = 6 + r (C1) r = 7.2 – 6 (A1) = 1.2  7(c)(i) The same B1 7(c)(ii) Any 3 from: B3 • before t1 / when current constant, the (total) resistance is constant • at t1 / when current increases, the (total) resistance decreases (due to decrease of external resistance) • (after t1) temperature (of lamp) increases (so the resistance of the lamp increases) • (after t1) resistance of lamp increases (so total resistance increases so the current in the ammeter decreases)

This question in 9702/22 Feb/March 2024

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

3 (a) State what is meant by the work done by a force. … … [1] (b) A block of mass m is raised vertically at constant speed. The vertical height gained by the block is Δh, as shown in Fig. 3.1. block, mass m Δh Fig. 3.1 Derive an expression, in terms of m and Δh, for the change in gravitational potential energy ΔEP of the block. State the meaning of any other symbols you use. [2] (c) An electric motor has an input power of 900 W. The motor takes 1.0 minute to lift a load of weight 240 N at constant speed through a vertical height of 150 m. Resistive forces are negligible. (i) Show that the work done by the motor on the load in 1.0 minute is 36 kJ. [1] (ii) Determine the useful output power of the motor. power = … W [2] (iii) Use your answer in (c)(ii) to determine the efficiency of the motor. efficiency = … [2] (iv) Some of the power wasted in the motor is dissipated by the resistance of its coil. This dissipated power is 280 W. The coil of the motor is made from wire of total length 23 m. The wire has a cross- sectional area of 2.6 × 10–8 m2 and is made from metal of resistivity 1.7 × 10–8 Ω m. Calculate the current in the coil. current = … A [3] [Total: 11]

11 marks

Mark scheme: 3(a) product of force and displacement in direction of force B1 3(b) weight/force = mg, and g identified as acceleration of free fall B1 EP identified as work (done by lifting force/weight), and so = mg  h B1 3(c)(i) 240  150 = 36 kJ A1 3(c)(ii) P = W / t C1 = 36 000 / 60 A1 = 600 W 3(c)(iii) efficiency = useful output power / total input power C1 = 600 / 900 A1 = 0.67 3(c)(iv) P = I2R and R = L / A C1 280 = I2  (1.7  10–8  23) / (2.6  10–8) C1 I = 4.3 A A1

This question in 9702/21 Oct/Nov 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 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

6 (a) Define electric potential difference across a component. … … [1] (b) A circuit contains four resistors and a battery of electromotive force (e.m.f.) 8.0 V with negligible internal resistance. When the variable resistor has resistance R, the currents in the circuit are 0.030 A, I1 and I2, as shown in Fig. 6.1. 8.0 V 0.030 A I2 210 Ω R I1 430 Ω 240 Ω Fig. 6.1 (i) Determine the charge passing through the battery in a time of 4.0 minutes. charge = … C [2] (ii) Calculate I1. I1 = … A [2] (iii) Calculate I2. I2 = … A [1] (iv) Determine R. R = … Ω [2] (c) The variable resistor in (b) is fitted with a scale so that its resistance can be accurately determined. The resistor of resistance 240 Ω is now replaced by a new resistor X of unknown resistance. A galvanometer is connected as shown in Fig. 6.2. 8.0 V 210 Ω 430 Ω X Fig. 6.2 With reference to ratios of resistances, explain how this circuit can be used to determine the resistance of X. … … … … … [2] [Total: 10]

10 marks

Mark scheme: 6(a) energy transferred (to the component) per (unit) charge B1 6(b)(i) Q = It C1 = 0.030  4.0  60 = 7.2 C A1 6(b)(ii) I = V / R C1 I1 = 8.0 / (430+240) = 0.012 A A1 6(b)(iii) I2 = 0.030 – I1 A1 = 0.030 – 0.012 = 0.018 A 6(b)(iv) R = V / I2 C1 = (8.0 – (0.018  210)) / 0.018 = 230  A1 OR (C1) resistance of top branch = 8.0 / 0.018 R = 8.0 / 0.018 – 210 = 230  (A1) OR (C1) total circuit resistance = 8.0 / 0.030 = 267 1 / 267 = 1 / (210 + R) + 1 / (430 + 240) 1 / 267 – 1 / 670 = 1 / (210 +R) 210 + R = 443 R = 230  (A1) 6(c) (When) the galvanometer reads 0 (A) M1 The ratio of the resistances in the top branch will equal the ratio of the resistances in the bottom branch (so the resistance A1 of X can be determined) OR (A1) The ratio of the left pair of resistances will equal the ratio of the right pair of resistances

This question in 9702/22 May/June 2025

3 A car of mass 1500 kg is travelling along a straight horizontal road at constant velocity v. The car is subject to a total resistive force F, as shown in Fig. 3.1. v car, mass 1500 kg F horizontal road Fig. 3.1 (a) Show that the power P developed by the engine in overcoming the total resistive force is given by the equation P = Fv . [2] (b) The car now moves up a slope at a constant speed of 30 m s−1. The slope is at an angle to the horizontal of 6.0°, as shown in Fig. 3.2. 30 m s–1 car road 6.0° Fig. 3.2 The total resistive force acting on the car is 1600 N. (i) Show that the increase in gravitational potential energy of the car in a time of 1.0 s is 46 000 J. [2] (ii) Use the information in (b)(i) to determine the power developed by the engine to move the car up the slope. power = … W [2] (c) The car picks up a passenger and then continues up the slope at the same speed as in (b). State and explain the effect, if any, that the passenger has on: (i) the air resistance acting on the car … … [1] (ii) the power developed by the engine. … … [1] [Total: 8]

8 marks

Mark scheme: 3(a) W = Fd B1 P = Fd / t = Fv or P = Fvt / t = Fv B1 3(b)(i) ()E(P) = mg()h C1 increase of gravitational potential energy of car in 1.0 s A1 = 1500  9.81  30  sin 6.0 = 46 000 J 3(b)(ii) (Power to overcome total resistive forces) = 1600  30 C1 = 48 000 W power = 48 000 + 46 000 A1 = 9.4  104 W 3(c)(i) Air resistance is the same, as the speed is the same B1 3(c)(ii) Mass / weight has increased so (power will) increase B1

This question in 9702/23 May/June 2025