Electric current — 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

1 (a) The current in a wire is I. Charge Q passes one point in the wire in time t. State (i) the relation between I, Q and t, … [1] (ii) which of the quantities I, Q and t are base quantities. … … [2] (b) The current in the wire is due to electrons, each with charge q, that move with speed v along the wire. There are n of these electrons per unit volume. For a wire having a cross-sectional area S, the current I is given by the equation I = nSqv k, where k is a constant. (i) State the units of I, n, S, q and v in terms of the base units. I … n … S … q … v … [3] (ii) By considering the homogeneity of the equation, determine the value of k. k = ………………………….. [2]

8 marks

Mark scheme: 1 (a) (i) Q = It (allow any subject for the equation) B1 [1] (ii) I B1 t B1 [2] (allow 1 mark only if all three quoted) (b) (i) base unit of I is A base unit of n is m–3 (not /m–3) base unit of S is m2 base unit of q is A s (not C) base unit of v is m s–1 (–1 for each error or omission) B3 [3] (ii) A = m–3 m2 A s (m s–1)k M1 e.g. for m: 0 = –3 + 2 + k k = 1 A1 [2] 2

This question in 9702/21 Oct/Nov 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 (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

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

9 marks

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

This question in 9702/22 May/June 2011

4 (a) Distinguish between potential difference (p.d.) and electromotive force (e.m.f.) in terms For of energy transformations. Examiner’s Use … … … … [2] (b) Two cells A and B are connected in series with a resistor R of resistance 5.5 Ω, as shown in Fig. 4.1. 4.4 V 2.3 Ω cell A R 5.5 Ω 2.1 V 1.8 Ω cell B Fig. 4.1 Cell A has e.m.f. 4.4 V and internal resistance 2.3 Ω. Cell B has e.m.f. 2.1 V and internal resistance 1.8 Ω. (i) State Kirchhoff’s second law. … … [1] (ii) Calculate the current in the circuit. current = … A [2] (iii) On Fig. 4.1, draw an arrow to show the direction of the current in the circuit. Label this arrow I. [1] (iv) Calculate For Examiner’s 1. the p.d. across resistor R, Use p.d. = … V [1] 2. the terminal p.d. across cell A, p.d. = … V [1] 3. the terminal p.d. across cell B. p.d. = … V [2]

10 marks

Mark scheme: 4 (a) p.d. = energy transformed from electrical to other forms unit charge B1 e.m.f. = energy transformed from other forms to electrical unit charge B1 [2] (b) (i) sum of e.m.f.s (in a closed circuit) = sum of potential differences B1 [1] (ii) 4.4 – 2.1 = I × (1.8 + 5.5 + 2.3) M1 I = 0.24 A A1 [2] (iii) arrow (labelled) I shown anticlockwise A1 [1] (iv) 1. V = I × R = 0.24 × 5.5 = 1.3(2) V A1 [1] 2. VA = 4.4 – (I × 2.3) = 3.8(5) V A1 [1] 3. either VB = 2.1 + (I × 1.8) or VB = 3.8 – 1.3 C1 = 2.5(3) V A1 [2] GCE AS/A LEVEL – October/November 2011 9702 23

This question in 9702/23 Oct/Nov 2011

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

10 marks

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

This question in 9702/22 May/June 2012

6 (a) Describe the structure of an atom of the nuclide 23592U. For Examiner’s … Use … … … [2] (b) The deflection of α-particles by a thin metal foil is investigated with the arrangement shown in Fig. 6.1. All the apparatus is enclosed in a vacuum. vacuum detector of _-particles D W _ source path of deflected X _-particles Y Fig. 6.1 The detector of α-particles, D, is moved around the path labelled WXY. (i) Explain why the apparatus is enclosed in a vacuum. … … [1] (ii) State and explain the readings detected by D when it is moved along WXY. … … … … … … [3] Question 6 continues on page 16. (c) A beam of α-particles produces a current of 1.5 pA. Calculate the number of α-particles For per second passing a point in the beam. Examiner’s Use number = … s–1 [3]

9 marks

Mark scheme: 6 (a) 92 protons in the nucleus and 92 electrons around nucleus B1 143 neutrons (in the nucleus) B1 [2] (b) (i) α-particle travels short distance in air B1 [1] (ii) very small proportion in backwards direction / large angles B1 majority pass through with no /small deflections B1 either most of mass is in very small volume (nucleus) and is charged or most of atom is empty space B1 [3] (c) I = Q / t C1 n / t = (1.5 × 10–12) /( 2 × 1.6 × 10–19) C1 n / t = 4.7 × 106 s–1 A1 [3]

This question in 9702/21 Oct/Nov 2012

5 Fig. 5.1 shows a 12 V power supply with negligible internal resistance connected to a uniform For metal wire AB. The wire has length 1.00 m and resistance 10 Ω. Two resistors of resistance Examiner’s 4.0 Ω and 2.0 Ω are connected in series across the wire. Use 12 V I1 I2 C metal wire A B 40 cm I3 4.0 1 2.0 1 D Fig. 5.1 Currents І1, І2 and І3 in the circuit are as shown in Fig. 5.1. (a) (i) Use Kirchhoff’s first law to state a relationship between І1, І2 and І3. … [1] (ii) Calculate І1. І1 = … A [3] (iii) Calculate the ratio x, where power in metal wire x = . power in series resistors x = … [3] (b) Calculate the potential difference (p.d.) between the points C and D, as shown in Fig. 5.1. The distance AC is 40 cm and D is the point between the two series resistors. p.d. = … V [3]

10 marks

Mark scheme: 5 (a) (i) І1 = І2 + І3 B1 [1] (ii) І = V / R or І2 = 12 / 10 (= 1.2 A) C1 R = [1/6 + 1 / 10]–1 [total R = 3.75 Ω] or І3 = 12 / 6 (= 2.0 A) C1 І1 = 12 / 3.75 = 3.2 A or І1 = 1.2 + 2.0 = 3.2 A A1 [3] (iii) power = VІ or І2R or V2 / R C1 power in wire I 22 R w V I 2 V 2 / R w x = = or or C1 power in series resistors I 32 R s V I 3 V 2 / R s x = 12 × 1.2 / 12 × 2.0 = 0.6(0) allow 3 / 5 or 3:5 A1 [3] (b) p.d. BC: 12 – 12 × 0.4 = 7.2 (V) / p.d. AC = 4.8 (V) C1 p.d. BD: 12 – 12 × 4 / 6 = 4.0 (V) / p.d. AD = 8.0 (V) C1 p.d. = 3.2 V A1 [3]

This question in 9702/22 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 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

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

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

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

11 marks

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

This question in 9702/22 Oct/Nov 2014

2 A microphone detects a musical note of frequency f. The microphone is connected to a cathode- ray oscilloscope (c.r.o.). The signal from the microphone is observed on the c.r.o. as illustrated in Fig. 2.1. 1.0 cm 1.0 cm Fig. 2.1 The time-base setting of the c.r.o. is 0.50 ms cm–1. The Y-plate setting is 2.5 mV cm–1. (a) Use Fig. 2.1 to determine (i) the amplitude of the signal, amplitude = … mV [2] (ii) the frequency f, f = … Hz [3] (iii) the actual uncertainty in f caused by reading the scale on the c.r.o. actual uncertainty = … Hz [2] (b) State f with its actual uncertainty. f = … ± … Hz [1]

8 marks

Mark scheme: 2 (a) (i) amplitude scale reading 2.2 (cm) C1 amplitude = 2.2 × 2.5 = 5.5 mV A1 [2] (ii) time period scale reading = 3.8 (cm) C1 time period = 3.8 × 0.5 × 10–3 = 0.0019 (s) C1 frequency f = 1 / 0.0019 = 530 (526) Hz A1 [3] (iii) uncertainty in reading = ± 0.2 in 3.8 (cm) or 5.3% or 0.2 in 7.6 (cm) or 2.6% [allow other variations of the distance on the x-axis] M1 actual uncertainty = 5.3% of 526 = 27.7 or 28 Hz or 2.6% of 526 = 13 or 14 A1 [2] (b) frequency = 530 ± 30 Hz or 530 ± 10 Hz A1 [1]

This question in 9702/23 Oct/Nov 2014

5 (a) On Fig. 5.1, sketch the temperature characteristic of a thermistor. resistance 0 0 100 temperature / °C Fig. 5.1 [2] (b) A potential divider circuit is shown in Fig. 5.2. X Y 12 V Z Fig. 5.2 The battery of electromotive force (e.m.f.) 12 V and negligible internal resistance is connected in series with resistors X and Y and thermistor Z. The resistance of Y is 15 kΩ and the resistance of Z at a particular temperature is 3.0 kΩ. The potential difference (p.d.) across Y is 8.0 V. (i) Explain why the power transformed in the battery equals the total power transformed in X, Y and Z. … [1] (ii) Calculate the current in the circuit. current = … A [2] (iii) Calculate the resistance of X. resistance = … Ω [3] (iv) The temperature of Z is increased. State and explain the effect on the potential difference across Z. … … … … [2]

10 marks

Mark scheme: 5 (a) curved line showing decreasing gradient with temperature rise M1 smooth line not touching temperature axis, not horizontal or vertical anywhere A1 [2] (b) (i) (no energy lost in battery because) no/negligible internal resistance B1 [1] (ii) I = V / R = 8 / 15 × 103 or 1.6 / 3.0 × 103 or 2.4 / 4.5 × 103 or 12 / 22.5 × 103 C1 = 0.53 × 10–3 A A1 [2] (iii) p.d. across X = 12 – 8.0 – 3.0 × 103 × 0.53 × 10–3 (= 2.4 V) C1 RX = 2.4 / (0.53 × 10–3) C1 or Rtot = 12 / 0.53 × 10–3 (= 22.5 × 103 Ω) (C1) RX = (22.5 – 15.0 – 3.0) × 103 (C1) 4.5(2) × 103 Ω A1 [3] (iv) resistance decreases hence current (in circuit) is greater M1 p.d. across X and Y is greater hence p.d across Z decreases A1 or explanation in terms of potential divider: RZ decreases so RZ / (RX + RY + RZ) is less (M1) therefore p.d. across Z decreases (A1) [2]

This question in 9702/22 May/June 2015

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

10 marks

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

This question in 9702/23 May/June 2015

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

12 marks

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

This question in 9702/21 Oct/Nov 2015

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

9 marks

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

This question in 9702/22 Oct/Nov 2015

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

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

10 marks

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

This question in 9702/22 May/June 2016

6 (a) Define 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

2 A ball of mass 0.030 kg moves along a curved track, as shown in Fig. 2.1. ball mass 0.030 kg speed 1.3 m s–1 A wall 0.31 m B Fig. 2.1 The speed of the ball is 1.3 m s–1 when it is at point A at a height of 0.31 m. The ball moves down the track and collides with a vertical wall at point B. The ball then rebounds back up the track. It may be assumed that frictional forces are negligible. (a) Calculate the change in gravitational potential energy of the ball in moving from point A to point B. change in gravitational potential energy = … J [2] (b) Show that the ball hits the wall at B with a speed of 2.8 m s–1. [3] (c) The change in momentum of the ball due to the collision with the wall is 0.096 kg m s–1. The ball is in contact with the wall for a time of 20 ms. Determine, for the ball colliding with the wall, (i) the speed immediately after the collision, speed = … m s–1 [2] (ii) the magnitude of the average force on the ball. force = … N [2] (d) State and explain whether the collision is elastic or inelastic. … … [1] (e) In practice, frictional effects are significant so that the actual increase in kinetic energy of the ball in moving from A to B is 76 mJ. The length of the track between A and B is 0.60 m. Use your answer in (a) to determine the average frictional force acting on the ball as it moves from A to B. frictional force = … N [2] [Total: 12]

12 marks

Mark scheme: 2 (a) ∆E = mg∆h C1 = 0.030 × 9.81 × (–)0.31 = (–)0.091 J A1 [2] (b) E = ½mv 2 C1 (initial) E = ½ × 0.030 × 1.32 (= 0.0254) C1 0.5 × 0.030 × v 2 = (0.5 × 0.030 × 1.32) + (0.030 × 9.81 × 0.31) so v = 2.8 m s–1 or 0.5 × 0.030 × v 2 = (0.0254) + (0.091) so v = 2.8 m s–1 A1 [3] (c) (i) 0.096 = 0.030 (v + 2.8) C1 v = 0.40 m s–1 A1 [2] (ii) F = ∆p / (∆)t or F = ma = 0.096 / 20 × 10–3 or 0.030 (0.40 + 2.8) / 20 × 10–3 C1 = 4.8 N A1 [2] (d) kinetic energy (of ball and wall) decreases/changes/not conserved, so inelastic or (relative) speed of approach (of ball and wall) not equal to/greater than (relative) speed of separation, so inelastic. B1 [1] (e) force = work done / distance moved = (0.091 – 0.076) / 0.60 C1 = 0.025 N A1 [2]

This question in 9702/22 Oct/Nov 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 Ω S R3 6.0 Ω 0–24 Ω 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 9702/23 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

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

10 marks

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

This question in 9702/22 May/June 2017

5 (a) Define the coulomb. … [1] (b) Two vertical metal plates in a vacuum have a separation of 4.0 cm. A potential difference of 2.0 × 102 V is applied between the plates. Fig. 5.1 shows a side view of this arrangement. 4.0 cm smoke particle weight 3.9 × 10–15 N charge –8.0 × 10–19 C metal plate metal plate +2.0 × 102 V s Fig. 5.1 A smoke particle is in the uniform electric field between the plates. The particle has weight 3.9 × 10–15 N and charge –8.0 × 10–19 C. (i) Show that the electric force acting on the particle is 4.0 × 10–15 N. [2] (ii) On Fig. 5.1, draw labelled arrows to show the directions of the two forces acting on the smoke particle. [1] (iii) The resultant force acting on the particle is F. Determine 1. the magnitude of F, magnitude = … N 2. the angle of F to the horizontal. angle = … ° [3] (c) The electric field in (b) is switched on at time t = 0 when the particle is at a horizontal displacement s = 2.0 cm from the left-hand plate. At time t = 0 the horizontal velocity of the particle is zero. The particle is then moved by the electric field until it hits a plate at time t = T. On Fig. 5.2, sketch the variation with time t of the horizontal displacement s of the particle from the left-hand plate. 4.0 s / cm 2.0 0 0 T t Fig. 5.2 [2] [Total: 9]

9 marks

Mark scheme: 5(a) (coulomb is) ampere second B1 5(b)(i) E = V / d or E = F / Q C1 F = VQ / d F = (2.0 × 102 × 8.0 × 10–19) / 4.0 × 10–2 = 4.0 × 10–15 N A1 5(b)(ii) arrow pointing to the left labelled ‘electric force’ and arrow pointing downwards labelled ‘weight’ B1 5(b)(iii) 1. resultant force = √ [(3.9 × 10–15)2 + (4.0 × 10–15)2] C1 = 5.6 × 10–15 N A1 2. angle = tan–1 (3.9 × 10–15 / 4.0 × 10–15) = 44° A1 5(c) downward sloping line from (0, 2.0) M1 magnitude of gradient of line increases with time and line ends at (T, 0) A1

This question in 9702/22 Oct/Nov 2017

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

12 marks

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

This question in 9702/22 Oct/Nov 2017

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

10 marks

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

This question in 9702/23 Oct/Nov 2017

6 A sample of a radioactive isotope emits a beam of β– radiation. (a) State the change, if any, to the number of neutrons in a nucleus of the sample that emits a β– particle. … [1] (b) The number of β– particles passing a fixed point in the beam in a time of 2.0 minutes is 9.8 × 1010. Calculate the current, in pA, produced by the beam of β– particles. current = … pA [3] (c) Suggest why the β– particles are emitted with a range of kinetic energies. … … … … [2] [Total: 6]

6 marks

Mark scheme: 6(a) –1 / decreases by 1 A1 6(b) I = Q / t or Ne / t C1 = (9.8×1010 × 1.6×10–19) / (2.0 × 60) = 1.3 × 10–10 (A) C1 = 130 pA A1 6(c) antineutrino(s) (emitted) / other particle(s) (emitted) C1 energy / momentum shared with antineutrino(s) A1

This question in 9702/22 Feb/March 2018

7 (a) The current I in a metal wire is given by the expression I = Anve. State what is meant by the symbols A and n. A: … n: … [2] (b) The diameter of a wire XY varies linearly with distance along the wire as shown in Fig. 7.1. X Y current I current I d d 2 drift speed vx Fig. 7.1 There is a current I in the wire. At end X of the wire, the diameter is d and the average drift d speed of the free electrons is vx. At end Y of the wire, the diameter is . 2 On Fig. 7.2, sketch a graph to show the variation of the average drift speed with position along the wire between X and Y. 5vx 4vx 3vx average drift speed 2vx vx 0 X Y position along wire Fig. 7.2 [2]

4 marks

Mark scheme: 7(a) A: (cross-sectional) area (of wire) B1 n: number of free electrons per unit volume or number density of free electrons B1 7(b) line drawn between (X, vx) and (Y, 4vx) M1 line has increasing gradient A1

This question in 9702/22 Oct/Nov 2018

6 (a) Define the coulomb. … … [1] (b) An electric current is a flow of charge carriers. In the following list, underline the possible charges for a charge carrier. 8.0 × 10–19 C 4.0 × 10–19 C 1.6 × 10–19 C 1.6 × 10–20 C [1] (c) The diameter of a wire ST varies linearly with distance along the wire as shown in Fig. 6.1. S T current I current I d 2d drift speed vs Fig. 6.1 There is a current I in the wire. At end S of the wire, the diameter is d and the average drift speed of the free electrons is vs. At end T of the wire, the diameter is 2d. On Fig. 6.2, sketch a graph to show the variation of the average drift speed with position along the wire between S and T. 1.00vs 0.75vs average drift 0.50vs speed 0.25vs 0 S T position along wire Fig. 6.2 [2] [Total: 4]

4 marks

Mark scheme: 6(a) (coulomb is an) ampere second B1 6(b) 8.0 × 10–19 C and 1.6 × 10–19 C both underlined (and no others underlined) B1 6(c) line drawn between (S, 1.00vs) and (T, 0.25vs) M1 line with decreasing magnitude of gradient A1

This question in 9702/23 Oct/Nov 2018

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

10 marks

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

This question in 9702/21 May/June 2019

7 A stationary nucleus of a radioactive isotope X decays by emitting an α-particle to produce a nucleus of neptunium-237 and 5.5 MeV of energy. The decay is represented by α + 5.5 MeV. X 23973Np + (a) Calculate the number of protons and the number of neutrons in a nucleus of X. number of protons = … number of neutrons = … [2] (b) Explain why the energy transferred to the α-particle as kinetic energy is less than the 5.5 MeV of energy released in the decay process. … … [1] (c) A sample of X is used to produce a beam of α-particles in a vacuum. The number of α-particles passing a fixed point in the beam in a time of 30 s is 6.9 × 1011. (i) Calculate the average current produced by the beam of α-particles. current = … A [2] (ii) Determine the total power, in W, that is produced by the decay of 6.9 × 1011 nuclei of X in a time of 30 s. power = … W [2] [Total: 7]

7 marks

Mark scheme: 7(a) number of protons = 95 A1 number of neutrons = 146 A1 7(b) Np/neptunium (nucleus) has kinetic energy or gamma/γ-radiation produced B1 7(c)(i) I = NQ / t C1 I = (6.9 × 1011 × 2 × 1.60 × 10–19) / 30 = 7.4 × 10–9 A A1 7(c)(ii) P = (6.9 × 1011 × 5.5 × 106 × 1.60 × 10–19) / 30 C1 = 0.020 W A1

This question in 9702/23 Oct/Nov 2019

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

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

6 The current I in a metal wire is given by the expression I = Anve where v is the average drift speed of the free electrons in the wire and e is the elementary charge. (a) State what is meant by the symbols A and n. A: … n: … [2] (b) Use the above expression to determine the SI base units of e. Show your working. base units … [2] (c) Two lamps P and Q are connected in series to a battery, as shown in Fig. 6.1. P Q Fig. 6.1 The radius of the filament wire of lamp P is twice the radius of the filament wire of lamp Q. The filament wires are made of metals with the same value of n. Calculate the ratio average drift speed of free electrons in filament wire of P . average drift speed of free electrons in filament wire of Q ratio = … [2] [Total: 6]

6 marks

Mark scheme: 6(a) A: cross-sectional area B1 n: number density of free electrons B1 6(b) units of I: A and units of A: m2 and units of v: m s–1 B1 units of e: A / (m2 m–3 m s–1) = A s A1 6(c) ratio = AQ / AP C1 = [πr2] / [π(2r2)] = 0.25 A1

This question in 9702/23 May/June 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

5 (a) State Kirchhoff’s second law. … … … [2] (b) A battery has electromotive force (e.m.f.) 4.0 V and internal resistance 0.35 Ω. The battery is connected to a uniform resistance wire XY and a fixed resistor of resistance R, as shown in Fig. 5.1. 4.0 V 0.35 Ω R X Y uniform resistance wire Fig. 5.1 Wire XY has resistance 0.90 Ω. The potential difference across wire XY is 1.8 V. Calculate: (i) the current in wire XY current = … A [1] (ii) the number of free electrons that pass a point in the battery in a time of 45 s number = … [2] (iii) resistance R. R = … Ω [2] (c) A cell of e.m.f. 1.2 V is connected to the circuit in (b), as shown in Fig. 5.2. 4.0 V 0.35 Ω R P X Y 1.2 V Fig. 5.2 The connection P is moved along the wire XY. The galvanometer reading is zero when distance XP is 0.30 m. (i) Calculate the total length L of wire XY. L = … m [2] (ii) The fixed resistor is replaced by a different fixed resistor of resistance greater than R. State and explain the change, if any, that must be made to the position of P on wire XY so that the galvanometer reading is zero. … … … … [2] [Total: 11]

11 marks

Mark scheme: 5(a) sum of e.m.f.(s) = sum of p.d.(s) or (algebraic) sum of e.m.f.(s) and p.d.(s) is zero M1 around a loop/around a closed circuit A1 5(b)(i) I = 1.8 / 0.90 = 2.0 A A1 5(b)(ii) Q = It C1 number = (2.0 × 45) / 1.60 × 10–19 = 5.6 × 1020 A1 5(b)(iii) 4.0 = 1.8 + [2.0 × (0.35 + R)] or 4.0 = 2.0 × (0.90 + 0.35 + R) C1 R = 0.75 Ω A1 5(c)(i) 1.2 / 1.8 = 0.30 / L C1 L = 0.45 m A1 5(c)(ii) p.d. across XY decreases/p.d. across XP decreases B1 (so) P is moved towards Y/away from X/to the right B1

This question in 9702/21 May/June 2021

6 (a) One of the results of the α-particle scattering experiment is that a very small minority of the α-particles are scattered through angles greater than 90°. State what may be inferred about the structure of the atom from this result. … … … … [2] (b) An α-particle is made up of other particles. One of these particles is a proton. State and explain whether a proton is a fundamental particle. … … [1] (c) A radioactive source produces a beam of α-particles in a vacuum. The average current produced by the beam is 6.9 × 10–9 A. Calculate the average number of α-particles passing a fixed point in the beam in a time of 1.0 minute. number = … [3] (d) The α-particles in the vacuum in (c) enter a uniform electric field. The α-particles enter the field with their velocity in the same direction as the field. State and explain whether the magnitude of the acceleration of an α-particle due to the field decreases, increases or stays constant as the α-particle moves through the field. … … … [2] (e) A nucleus X is an isotope of a nucleus Y. The mass of nucleus X is greater than that of Y. Both of the nuclei are in the same uniform electric field. State and explain whether the magnitude of the electric force acting on nucleus X is greater than, less than or the same as that acting on nucleus Y. … … … [2] [Total: 10]

10 marks

Mark scheme: 6(a) the nucleus is charged B1 the majority of the mass (of atom) is in the nucleus B1 6(b) made up of quarks (so) not a fundamental particle B1 6(c) (Q =) 6.9 × 10–9 × 60 C1 number = (6.9 × 10–9 × 60) / (2 × 1.60 × 10–19) C1 = 1.3 × 1012 A1 6(d) (magnitude of electric) force is constant B1 (so magnitude of) acceleration is constant B1 6(e) (nuclei have) same charge/same number of protons B1 (so) same (magnitude of) force B1

This question in 9702/22 May/June 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 (a) A resistance wire of uniform cross-sectional area 3.3 × 10–7 m2 and length 2.0 m is made of metal of resistivity 5.0 × 10–7 Ω m. Show that the resistance of the wire is 3.0 Ω. [2] (b) The ends of the resistance wire in (a) are connected to the terminals X and Y in the circuit shown in Fig. 6.1. 1.50 V r X Y uniform metal wire, resistance 3.0 Ω Fig. 6.1 The cell has an electromotive force (e.m.f.) of 1.50 V and internal resistance r. The potential difference between X and Y is 1.20 V. Calculate: (i) the current in the circuit current = … A [1] (ii) the internal resistance r. r = … Ω [2] (c) A galvanometer and a cell of e.m.f. E with negligible internal resistance are connected to the circuit in (b), as shown in Fig. 6.2. 1.50 V r P X Y E Fig. 6.2 The resistance wire between X and Y has a length of 2.0 m. The galvanometer has a reading of zero when the connection P is adjusted so that the length XP is 1.4 m. Determine the e.m.f. E of the cell. E = … V [2] (d) The circuit in Fig. 6.2 is modified by replacing the original resistance wire with a second resistance wire. The second wire has the same length as the original wire and is made of the same metal. The second wire has a smaller cross-sectional area than the original wire. Connection P is adjusted on the second wire so that the galvanometer has a reading of zero. State and explain whether length XP for the second wire is shorter than, longer than or the same as length XP for the original wire when the galvanometer reading is zero. … … … … … [3]

10 marks

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

This question in 9702/23 Oct/Nov 2021

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

12 marks

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

This question in 9702/22 Feb/March 2022

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

11 marks

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

This question in 9702/23 May/June 2022

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 metal wire in a circuit has a length of 1.8 m and a cross‑sectional area of 1.5 × 10–6 m2. The total number of free electrons (charge carriers) in the wire is 2.3 × 1023. There is a current in the wire so that a charge of 172 C moves past a fixed point in the wire in a time of 2.5 minutes. (a) Show that the number density of the free electrons in the wire is 8.5 × 1028 m–3. [1] (b) Calculate the average drift speed of the free electrons. average drift speed = … m s–1 [3] [Total: 4]

4 marks

Mark scheme: 6(a) A1 6(b) I = Q / t or I = 172 / 2.5  60 or I = 1.1(5) C1 I = nAvq 172 / (2.5  60) = 1.5  10–6  8.5  1028  v  1.6  10–19 C1 v = 5.6  10–5 m s–1 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

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

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

12 marks

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

This question in 9702/22 Oct/Nov 2023

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

11 marks

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

This question in 9702/23 Oct/Nov 2023

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

11 marks

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

This question in 9702/22 May/June 2024

5 (a) Define electric potential difference (p.d.). … … [1] (b) A power supply, three resistors and a component X are connected in the circuit shown in Fig. 5.1. 230 V 7.0 A + – 0.86 Ω I1 2.4 Ω X I2 170 Ω Fig. 5.1 The power supply has an electromotive force (e.m.f.) of 230 V and negligible internal resistance. The current in the power supply is 7.0 A. (i) Identify component X. … [1] (ii) Show that the p.d. across the resistor of resistance 0.86 Ω is 6.0 V. [1] (iii) Determine the current I1. I1 = … A [2] (iv) Calculate the p.d. across component X. p.d. = … V [2] (v) Calculate the power dissipated in component X. power = … W [2] (vi) The purpose of the circuit is to provide power to component X. Determine the percentage efficiency of the circuit. efficiency = … % [2] (vii) The resistor of resistance 170 Ω is removed, leaving an open circuit in the lower branch of the circuit. There is no change to the resistance of component X. State whether the current in the power supply increases, decreases or remains the same. … [1] [Total: 12]

12 marks

Mark scheme: 5(a) energy transferred per unit charge (from electrical to other forms) B1 5(b)(i) heater A1 5(b)(ii) (V =) 7.0  0.86 = 6.0 (V) A1 5(b)(iii) I = (230 – 6.0) / 170 C1 ( = 1.3 A) I1 = 7.0 – 1.3 A1 = 5.7 A 5(b)(iv) V = 230 – 6.0 – (5.7  2.4) C1 = 210 V A1 or R = ((230 – 6.0) / 5.7) – 2.4 (C1) ( = 36.9 ) V = 5.7  36.9 (A1) = 210 V 5(b)(v) P = IV or P = I2R or P = V2 / R C1 P = 5.7  210 A1 or P = 5.72  (210 / 5.7) or P = 2102 / (210 / 5.7) P = 1200 W 5(b)(vi) % efficiency = useful power out / (total) power in ( 100) C1 = 1200 / (230  7.0) (100) A1 = 0.75 ( 100) = 75% or 74% (using 3 s.f. value from (v) gives 74%) 5(b)(vii) (current) decreases A1

This question in 9702/22 Oct/Nov 2024