Resistance and resistivity — Paper 4 · A Level Physics 9702

9 A metal wire strain gauge is firmly fixed across a crack in a wall, as shown in Fig. 9.1, so that the growth of the crack may be monitored. strain crack gauge Fig. 9.1 (a) Explain why, as the crack becomes wider, the resistance of the strain gauge increases. … … … … [3] (b) The strain gauge has an initial resistance of 143.0 Ω and, after being fixed in position across the crack for several weeks, the resistance is found to be 146.2 Ω. The change in the area of cross-section of the strain gauge wire is negligible. Calculate the percentage increase in the width of the crack. Explain your working. increase = … % [3]

6 marks

Mark scheme: 9 (a) resistance of wire = ρL / A … B1 as crack widens, L increases … M1 and A decreases … M1 so resistance increases … A0 [3] (b) ∆L / L = ∆R / R … B1 = (146.2 – 143.0) / 143.0 × 100 … C1 ∆L / L = 2.24% … A1 [3] [Total: 6]

This question in 9702/41 Oct/Nov 2009

3 (a) The resistance of a thermistor at 0 °C is 3840 Ω. At 100 °C the resistance is 190 Ω. For When the thermistor is placed in water at a particular constant temperature, its resistance Examiner’s is 2300 Ω. Use (i) Assuming that the resistance of the thermistor varies linearly with temperature, calculate the temperature of the water. temperature = … °C [2] (ii) The temperature of the water, as measured on the thermodynamic scale of temperature, is 286 K. By reference to what is meant by the thermodynamic scale of temperature, comment on your answer in (i). … … … … [3] (b) A polystyrene cup contains a mass of 95 g of water at 28 °C. A cube of ice of mass 12 g is put into the water. Initially, the ice is at 0 °C. The water, of specific heat capacity 4.2 × 103 J kg–1 K–1, is stirred until all the ice melts. Assuming that the cup has negligible mass and that there is no heat exchange with the atmosphere, calculate the final temperature of the water. The specific latent heat of fusion of ice is 3.3 × 105 J kg–1. temperature = … °C [4]

9 marks

Mark scheme: 3 (a) (i) 1 deg C corresponds to (3840 – 190) / 100 Ω C1 for resistance 2300 Ω, temperature is 100 × (2300 – 3840) / (190 – 3840) temperature is 42 °C A1 [2] (ii) either 286 K ≡ 13 °C or 42 °C ≡ 315 K B1 thermodynamic scale does not depend on the property of a substance M1 so change in resistance (of thermistor) with temperature is non-linear A1 [3] (b) heat gained by ice in melting = 0.012 × 3.3 × 105 J C1 = 3960 J heat lost by water = 0.095 × 4.2 × 103 × (28 – θ) C1 3960 + (0.012 × 4.2 × 103 × θ) = 0.095 × 4.2 × 103 × (28 – θ) C1 θ = 16°C A1 [4] (answer 18°C – melted ice omitted – allow max 2 marks) (use of (θ – T) then allow max 1 mark) 2

This question in 9702/42 May/June 2010

9 (a) Describe the structure of a metal wire strain gauge. You may draw a diagram if you wish. … … … … [3] (b) A strain gauge S is connected into the circuit of Fig. 9.1. +4.5 V RF strain gauge S +9 V R – R + –9 V V1 VOUT 1.0 kΩ V2 RF Fig. 9.1 The operational amplifier (op-amp) is ideal. The output potential VOUT of the circuit is given by the expression RF VOUT = × (V2 – V1). R RF (i) State the name given to the ratio R . … [1] (ii) The strain gauge S has resistance 125 Ω when not under strain. For Examiner’s Calculate the magnitude of V1 such that, when the strain gauge S is not strained, the output VOUT is zero. Use V1 = … V [3] (iii) In a particular test, the resistance of S increases to 128 Ω. V1 is unchanged. RF The ratio is 12. R Calculate the magnitude of VOUT . VOUT = … V [2]

9 marks

Mark scheme: 9 (a) thin / fine metal wire B1 lay-out shown as a grid B1 encased in plastic B1 [3] (b) (i) gain (of amplifier) B1 [1] (ii) for VOUT = 0, then V + = V – or V1 = V2 C1 V1 = (1000/1125) × 4.5 C1 V1 = 4.0 V A1 [3] (iii) V2 = (1000 / 1128) × 4.5 = 3.99 V C1 VOUT = 12 × (3.99 – 4.00) = (–) 0.12 V A1 [2]

This question in 9702/42 May/June 2011

9 (a) The resistance of a light-dependent resistor (LDR) is approximately 500 Ω in daylight. Suggest an approximate value for the resistance of the LDR in darkness. resistance = … Ω [1] (b) An electronic light-meter is used to warn when light intensity becomes low. A light-dependent resistor is connected into the circuit of Fig. 9.1. +4.5 V 1.7 kΩ +9 V – P +2.5 V + R R –9 V red green Fig. 9.1 The operational amplifier (op-amp) is ideal. The resistors R are to ensure that the light-emitting diodes (LEDs) do not over-heat. (i) On Fig. 9.1, mark the polarity of the point P for the red LED to be emitting light. [1] (ii) The LDR is in daylight and has a resistance of 500 Ω. State and explain which diode, red or green, will be emitting light. … … … … [3] (iii) The intensity of the light decreases and the LDR is in darkness. State and explain the effect on the LEDs of this change in intensity. … … … [2]

7 marks

Mark scheme: 9 (a) any value greater than, or equal to, 5 kΩ B1 [1] (b) (i) ‘positive’ shown in correct position B1 [1] (ii) V + = (500/2200) × 4.5 ≈ 1 V B1 V – > V + so output is negative M1 green LED on, (red LED off) A1 [3] (allow full ecf of incorrect value of V +) (iii) either V + increases or V + > V – M1 green LED off, red LED on A1 [2]

This question in 9702/43 Oct/Nov 2011

3 (a) Define specific heat capacity. … … … [2] (b) A student carries out an experiment to determine the specific heat capacity of a liquid using the apparatus illustrated in Fig. 3.1. liquid out, tube temperature 25.5 °C heating liquid in, coil temperature 19.5 °C Fig. 3.1 Liquid enters the tube at a constant temperature of 19.5 °C and leaves the tube at a temperature of 25.5 °C. The mass of liquid flowing through the tube per unit time is m. Electrical power P is dissipated in the heating coil. The student changes m and adjusts P until the final temperature of the liquid leaving the tube is 25.5 °C. The data shown in Fig. 3.2 are obtained. m / g s–1 P/ W 1.11 33.3 1.58 44.9 Fig. 3.2 (i) Suggest why the student obtains data for two values of m, rather than for one value. … … [1] (ii) Calculate the specific heat capacity of the liquid. Show your working. specific heat capacity = … J kg–1 K–1 [3] (c) When the heating coil in (b) dissipates 33.3 W of power, the potential difference V across the coil is given by the expression V = 27.0 sin (395t). The potential difference is measured in volts and the time t is measured in seconds. Determine the resistance of the coil. resistance = … Ω [3] [Total: 9]

9 marks

Mark scheme: 3 (a) the (thermal) energy per unit mass to raise the temperature M1 of a substance by one degree A1 [2] (If ratio not clear for M1 mark, allow 1 / 2 marks for an otherwise correct answer) (b) (i) to allow for / determine / cancel heat transfer to / from tube / surroundings B1 [1] (do not allow ‘to stop / prevent’ heat loss) (ii) either P = mc∆θ ± h or 44.9 = 1.58 × 10–3 × c × (25.5 – 19.5) ± h or 33.3 = 1.11 × 10–3 × c × (25.5 – 19.5) ± h B1 (44.9 – 33.3) = (1.58 – 1.11) × 10–3 × c × (25.5 – 19.5) C1 c = 4100 (4110) J kg–1 K–1 A1 [3] (allow 1 / 3 for use of only 33.3 W, 1.11 g s–1 leading to 5000 J kg –1 K –1) (allow 1 / 3 for use of only 44.9 W, 1.58 g s–1 leading to 4740 J kg –1 K –1) (c) V0 = 27 or Vrms = 19.1 C1 33.3 = 272 / 2R or 33.3 = 19.12 / R C1 R = 11 Ω A1 [3]

This question in 9702/42 Feb/March 2016

11 Use band theory to explain why, unlike a copper wire, the resistance of an intrinsic semiconductor decreases with an increase of temperature. … … … … … … … … … … … … [5] [Total: 5]

5 marks

Mark scheme: 11 any five from: • electrons need energy to enter conduction band (from valence band) • (positively-charged) holes are left in valence band • moving charge carriers / holes / electrons are current • (increase of temperature leads to) more (positive and negative) charge carriers / more holes / more electrons so more current • more charge carriers / holes / electrons gives rise to less resistance • (increase of temperature causes) greater (amplitude of) vibrations of atoms / ions / lattice • effect of more charge carriers/holes/electrons is greater than effect of greater vibrations (and so resistance decreases) B5

This question in 9702/42 Feb/March 2017

11 Some electron energy bands in a solid are shown in Fig. 11.1. conduction band gap between conduction band and valence band (forbidden band) valence band Fig. 11.1 The width of the forbidden band and the number density of charge carriers occupying each band depends on the nature of the solid. Use band theory to explain why (a) the resistance of a metal at room temperature increases gradually with temperature, … … … … … [3] (b) the resistance, at constant temperature, of a light-dependent resistor (LDR) decreases with increasing light intensity. … … … … … … [4] [Total: 7]

7 marks

Mark scheme: 11(a) no forbidden band / valence and conduction bands overlap B1 no change in number of charge carriers (as temperature rises) B1 increased lattice vibrations so resistance increases B1 11(b) photons captured / absorbed by electrons in valence band B1 electrons promoted to conduction band B1 leaving holes in the valence band B1 more holes and / or electrons so resistance decreases B1

This question in 9702/42 Feb/March 2018

7 (a) Use band theory to explain why the resistance of an intrinsic semiconductor decreases as its temperature rises. … … … … … … … … [5] (b) The variation with temperature t of the resistance R of a thermistor is shown in Fig. 7.1. 3.5 3.0 R / kΩ 2.5 2.0 1.5 1.0 0 5 10 15 20 25 30 t / °C Fig. 7.1 The thermistor is connected into the circuit shown in Fig. 7.2. 12.0 kΩ 9.00 V A R B Fig. 7.2 The battery has electromotive force (e.m.f.) 9.00 V and negligible internal resistance. When the temperature of the thermistor is 25 °C, the potential difference between the terminals A and B is 1.00 V. The temperature of the thermistor changes from 25 °C to 10 °C. Determine, to two significant figures, the change in potential difference between A and B. change = … V [3] (c) The temperature of the thermistor in (b) changes from 25 °C to 10 °C at a constant rate. State two reasons why the potential difference between A and B does not change at a constant rate. 1. … … 2. … … [2] [Total: 10]

10 marks

Mark scheme: 7(a) Any five from: • (as temperature rises) energy of electrons increases • electrons (have enough energy to) cross forbidden band • electrons enter conduction band • leaving holes in valence band • both holes and electrons act as charge carriers • more charge carriers results in lower resistance • increased lattice vibrations outweighed by increase in (number of) charge carriers B5 7(b) (at 10 °C resistance is) 2.55 kΩ C1 new potential difference = 9.00 × 2.55 / (2.55 + 12.0) = 1.58 V C1 change in p.d. = 0.58 V A1 7(c) change of resistance with temperature is not linear B1 change in potential with resistance is not linear or potential divider equation is non-linear B1

This question in 9702/42 May/June 2019

10 (a) The upper electron energy bands in an intrinsic semiconductor material are illustrated in Fig. 10.1. conduction band forbidden band valence band Fig. 10.1 Use band theory to explain why the resistance of an intrinsic semiconductor material decreases as its temperature increases. … … … … … … … … [4] (b) A comparator circuit incorporating an ideal operational amplifier (op-amp) is shown in Fig. 10.2. +3.0 V +5 V 1.50 kΩ RT –5 V VOUT 1.20 kΩ 1.76 kΩ Fig. 10.2 The variation with temperature θ of the resistance RT of the thermistor is shown in Fig. 10.3. 3.5 3.0 RT / kΩ 2.5 2.0 1.5 1.0 0 5 10 15 20 25 θ/ °C Fig. 10.3 (i) Determine the temperature at which the light-emitting diode (LED) in Fig. 10.2 switches on or off. temperature = … °C [4] (ii) State and explain whether the thermistor is above or below the temperature calculated in (i) for the LED to emit light. … … … … [3] [Total: 11]

11 marks

Mark scheme: 10(a) (as temperature rises) electrons in valence band gain energy B1 electrons jump to conduction band B1 holes are left in the valence band B1 increased number (density) of charge carriers causes lower resistance B1 10(b)(i) V– = V+ C1 1.50 / 1.20 = RT / 1.76 C1 RT = 2.2 (kΩ) C1 temperature = 14 °C A1 10(b)(ii) (For LED to conduct,) VOUT must be negative B1 V– > V+ B1 RT must be lower so temperature must be above (b)(i) value B1

This question in 9702/42 Oct/Nov 2019

7 (a) On Fig. 7.1, sketch the temperature characteristic of a negative temperature coefficient (n.t.c.) thermistor. Label the axes with quantity and unit. 0 0 Fig. 7.1 [2] (b) An n.t.c. thermistor and a resistor are connected as shown in Fig. 7.2. V Fig. 7.2 The temperature of the thermistor is increased. State and explain the change, if any, to the reading on the voltmeter. … … … [2] (c) The variation with the fractional change in length Δx /x of the fractional change in resistance ΔR /R for a strain gauge is shown in Fig. 7.3. 10 ∆R/R 8 10–2 6 4 2 0 0 0.5 1.0 1.5 2.0 2.5 3.0 ∆x/x 10–2 Fig. 7.3 The unstrained resistance of the gauge is 120 Ω. Calculate the new resistance of the gauge when it is extended to a strain of 0.020. resistance = … Ω [3] [Total: 7]

7 marks

Mark scheme: 7(a) axes labelled with resistance and temperature M0 concave curve not touching temperature axis A1 line with negative gradient throughout A1 7(b) resistance of thermistor decreases B1 total circuit resistance decreases so voltmeter reading increases or current increases so voltmeter reading increases or greater proportion of resistance in fixed resistor so voltmeter reading increases or p.d. across thermistor decreases so voltmeter reading increases B1 7(c) (0.020 strain means) ΔR / R = 0.090 C1 ΔR = 0.090 × 120 = 10.8 Ω C1 resistance = 120 + 10.8 = 130 Ω A1

This question in 9702/42 Feb/March 2020

10 (a) White light passes through a cloud of cool low-pressure gas, as illustrated in Fig. 10.1. cool gas white emergent light light Fig. 10.1 For light that has passed through the gas, its continuous spectrum is seen to contain a number of darker lines. Use the concept of discrete electron energy levels to explain the existence of these darker lines. … … … … … … [4] (b) The uppermost electron energy bands in a solid are illustrated in Fig. 10.2. conduction band (CB) forbidden band (FB) valence band (VB) Fig. 10.2 Use band theory to explain the dependence on light intensity of the resistance of a light-dependent resistor (LDR). … … … … … … … [5] [Total: 9]

9 marks

Mark scheme: 10(a) • photon gives energy to electron (in an inner shell) or electron (in an inner shell) absorbs a photon • electron moves (from lower) to higher energy level • energy (of photon) is equal to difference in energy levels • electron de-excites giving off photon (of same energy) • photons emitted in all directions Any four points, 1 mark each B4 10(b) (in light) photons gives energy to electrons in VB or (in light) electrons in VB absorb photons B1 electron crosses FB/jumps to CB B1 (positive) holes left/created in VB B1 low intensity: few electrons in CB/most electrons in VB or high intensity: more photons so more electrons in CB or electron-hole pairs are charge carriers B1 more charge carriers results in lower resistance B1

This question in 9702/41 May/June 2020

11 (a) The uppermost energy bands in a solid are known as the valence band (VB), the forbidden band (FB) and the conduction band (CB). A copper wire is at room temperature. Use band theory to explain why the resistance of the copper wire increases as its temperature increases. … … … … … … … [4] (b) The structure of a copper crystal is to be examined using electron diffraction. Electrons, having been accelerated from rest through a potential difference V, are incident on the crystal. The de Broglie wavelength λ of the electrons is 2.6 × 10–11 m. Calculate the accelerating potential difference V. V = … V [4] [Total: 8]

8 marks

Mark scheme: 11(a) conduction band and valence band overlap B1 number (density) of charge carriers does not vary B1 increase in temperature gives rise to increased lattice vibrations B1 (lattice) vibrations hinder movement of charge carriers so resistance increases B1 11(b) mv = h / λ C1 v = (6.63 × 10–34) / [(2.6 × 10–11) × (9.11 × 10–31)] ( = 2.80 × 107 m s–1) C1 qV = ½mv2 C1 V = [9.11 × 10–31 × (2.80 × 107)2] / [2 × 1.60 × 10–19] = 2.2 × 103 V A1

This question in 9702/42 May/June 2020

10 (a) White light passes through a cloud of cool low-pressure gas, as illustrated in Fig. 10.1. cool gas white emergent light light Fig. 10.1 For light that has passed through the gas, its continuous spectrum is seen to contain a number of darker lines. Use the concept of discrete electron energy levels to explain the existence of these darker lines. … … … … … … [4] (b) The uppermost electron energy bands in a solid are illustrated in Fig. 10.2. conduction band (CB) forbidden band (FB) valence band (VB) Fig. 10.2 Use band theory to explain the dependence on light intensity of the resistance of a light-dependent resistor (LDR). … … … … … … … [5] [Total: 9]

9 marks

Mark scheme: 10(a) • photon gives energy to electron (in an inner shell) or electron (in an inner shell) absorbs a photon • electron moves (from lower) to higher energy level • energy (of photon) is equal to difference in energy levels • electron de-excites giving off photon (of same energy) • photons emitted in all directions Any four points, 1 mark each B4 10(b) (in light) photons gives energy to electrons in VB or (in light) electrons in VB absorb photons B1 electron crosses FB/jumps to CB B1 (positive) holes left/created in VB B1 low intensity: few electrons in CB/most electrons in VB or high intensity: more photons so more electrons in CB or electron-hole pairs are charge carriers B1 more charge carriers results in lower resistance B1

This question in 9702/43 May/June 2020

2 (a) (i) State the magnitude and unit of absolute zero on the thermodynamic temperature scale. … [1] (ii) Explain why temperature measured using a laboratory liquid-in-glass thermometer does not give a measurement of thermodynamic temperature. … … [1] (b) Fig. 2.1 shows a simplified diagram of a type of thermometer called a platinum resistance thermometer. plastic strip platinum wire X Y large glass tube Fig. 2.1 The glass tube is immersed in the environment for which the temperature is to be determined. The resistance between the terminals X and Y is measured. Fig. 2.2 shows the variation of the resistivity ρ of platinum with thermodynamic temperature T. ρ 0 T Fig. 2.2 (i) Explain how Fig. 2.2 shows that platinum is a suitable metal for use in a resistance thermometer. … … … [2] (ii) Suggest a reason why a platinum resistance thermometer is not suitable for measuring a rapidly changing temperature. … … … [1] (iii) Suggest a type of thermometer that is suitable for measuring a rapidly changing temperature. … [1] (c) A negative temperature coefficient thermistor may be used as a type of resistance thermometer. State one way in which the variation with temperature of the resistance of a thermistor differs from that of a platinum wire. … … [1] [Total: 7]

7 marks

Mark scheme: 2(a)(i) 0 K B1 2(a)(ii) (measurement) depends on properties of the liquid B1 2(b)(i)  resistivity varies with temperature  variation with temperature is linear  unique value of resistivity for each (different value of) temperature Any two points, 1 mark each B2 2(b)(ii) thermometer has high heat capacity/specific heat capacity or energy transfer needed for thermometer to reach correct temperature or thermometer takes time to reach the correct temperature B1 2(b)(iii) thermocouple B1 2(c) (variation is) inverse or (variation is) non-linear B1

This question in 9702/41 May/June 2024

2 (a) (i) State the magnitude and unit of absolute zero on the thermodynamic temperature scale. … [1] (ii) Explain why temperature measured using a laboratory liquid-in-glass thermometer does not give a measurement of thermodynamic temperature. … … [1] (b) Fig. 2.1 shows a simplified diagram of a type of thermometer called a platinum resistance thermometer. plastic strip platinum wire X Y large glass tube Fig. 2.1 The glass tube is immersed in the environment for which the temperature is to be determined. The resistance between the terminals X and Y is measured. Fig. 2.2 shows the variation of the resistivity ρ of platinum with thermodynamic temperature T. ρ 0 T Fig. 2.2 (i) Explain how Fig. 2.2 shows that platinum is a suitable metal for use in a resistance thermometer. … … … [2] (ii) Suggest a reason why a platinum resistance thermometer is not suitable for measuring a rapidly changing temperature. … … … [1] (iii) Suggest a type of thermometer that is suitable for measuring a rapidly changing temperature. … [1] (c) A negative temperature coefficient thermistor may be used as a type of resistance thermometer. State one way in which the variation with temperature of the resistance of a thermistor differs from that of a platinum wire. … … [1] [Total: 7]

7 marks

Mark scheme: 2(a)(i) 0 K B1 2(a)(ii) (measurement) depends on properties of the liquid B1 2(b)(i)  resistivity varies with temperature  variation with temperature is linear  unique value of resistivity for each (different value of) temperature Any two points, 1 mark each B2 2(b)(ii) thermometer has high heat capacity/specific heat capacity or energy transfer needed for thermometer to reach correct temperature or thermometer takes time to reach the correct temperature B1 2(b)(iii) thermocouple B1 2(c) (variation is) inverse or (variation is) non-linear B1

This question in 9702/43 May/June 2024

6 Fig. 6.1 shows a circuit that rectifies an alternating input voltage VIN and produces an output voltage VOUT across a resistor R. W Y rectification VIN C R VOUT circuit X Z Fig. 6.1 The four terminals of the rectification circuit are labelled W, X, Y and Z. A capacitor C is connected in parallel with resistor R. (a) (i) State what is meant by rectification. … … [1] (ii) State the purpose of capacitor C. … … [1] (b) Fig. 6.2 shows the variations with time t of the potential differences (p.d.s) VIN and VOUT. 12 8 VOUT p.d. / V 4 0 0 10 20 30 40 t / ms –4 –8 VIN –12 Fig. 6.2 (i) The variation of VIN with t can be represented by VIN = A cos Bt where A and B are constants. Determine the values of A and B. Give a unit with your answer for A. A = … unit … B = … rad s–1 [2] (ii) Determine the type of rectification produced by the circuit in Fig. 6.1. … [1] (iii) On Fig. 6.3, draw the circuit diagram for the components inside the rectification circuit. W Y X Z Fig. 6.3 [2] (iv) Determine a value for the time constant for the discharge of the capacitor C through the resistor R in Fig. 6.1. time constant = … s [3] (c) The capacitor C has a capacitance of 570 μF. Use your answer in (b)(iv) to determine the resistance of resistor R. resistance = … Ω [2] [Total: 12]

12 marks

Mark scheme: 6(a)(i) conversion from a.c. to d.c. B1 6(a)(ii) smoothing B1 6(b)(i) A = 12 V A1 B = 2 / (20 × 10–3) A1 = 310 rad s–1 6(b)(ii) full-wave (rectification) B1 6(b)(iii) four diodes shown, with correct circuit symbols B1 four diodes correctly connected to form a bridge rectifier B1 6(b)(iv) V = V0 exp (–t / ) C1 or V = V0 exp (–t / RC) and  = RC 8.0 = 12 exp (– 7.3 × 10–3 / ) C1  = 0.018 s A1 6(c) time constant = RC C1 R = (0.018 / 570 × 10–6) A1 = 32 

This question in 9702/43 May/June 2025