Resistance and resistivity — Paper 2 · A Level Physics 9702

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

11 marks

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

This question in 9702/23 May/June 2017

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

10 marks

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

This question in 9702/21 Oct/Nov 2017

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

12 marks

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

This question in 9702/22 Oct/Nov 2017

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

8 marks

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

This question in 9702/23 May/June 2018

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

9 marks

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

This question in 9702/22 Oct/Nov 2018

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

6 marks

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

This question in 9702/23 May/June 2019

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

10 marks

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

This question in 9702/23 May/June 2019

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

11 marks

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

This question in 9702/21 Oct/Nov 2019

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

11 marks

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

This question in 9702/22 Oct/Nov 2019

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

11 marks

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

This question in 9702/23 Oct/Nov 2019

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

12 marks

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

This question in 9702/22 Feb/March 2020

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

12 marks

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

This question in 9702/21 May/June 2020

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

10 marks

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

This question in 9702/23 Oct/Nov 2020

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

8 marks

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

This question in 9702/22 May/June 2021

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

11 marks

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

This question in 9702/21 Oct/Nov 2021

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

11 marks

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

This question in 9702/21 Oct/Nov 2022

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

12 marks

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

This question in 9702/22 Feb/March 2023

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

5 marks

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

This question in 9702/22 May/June 2023

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

10 marks

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

This question in 9702/23 May/June 2023

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

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

10 marks

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

This question in 9702/23 Oct/Nov 2024

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

13 marks

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

This question in 9702/22 Feb/March 2025

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

8 marks

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

This question in 9702/23 May/June 2025

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

9 marks

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

This question in 9702/23 Oct/Nov 2025

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

10 marks

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

This question in 9702/24 Oct/Nov 2025