Kirchhoff’s laws — Paper 2 · A Level Physics 9702

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

9 marks

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

This question in 9702/22 May/June 2011

5 A potentiometer circuit that is used as a means of comparing potential differences is shown For in Fig. 5.1. Examiner’s Use E1 r1 R1 H G I1 metal wire B J F I2 I3 A C r2 D E2 Fig. 5.1 A cell of e.m.f. E1 and internal resistance r1 is connected in series with a resistor of resistance R1 and a uniform metal wire of total resistance R2. A second cell of e.m.f. E2 and internal resistance r2 is connected in series with a sensitive ammeter and is then connected across the wire at BJ. The connection at J is halfway along the wire. The current directions are shown on Fig. 5.1. (a) Use Kirchhoff’s laws to obtain the relation (i) between the currents I1, I2 and I3, … [1] (ii) between E1, R1, R2, r1, I1 and I2 in loop HBJFGH, … [1] (iii) between E1, E2, r1, r2, R1, R2, I1 and I3 in the loop HBCDJFGH. … [2] (b) The connection at J is moved along the wire. Explain why the reading on the ammeter changes. … … … … [2]

6 marks

Mark scheme: 5 (a) (i) І1 + І3 = І2 A1 [1] (ii) E1 = І2R2 + І1R2 + І1R1+ І1r1 A1 [1] 2 2 (iii) E1 – E2 B1 = –І3r2 + І1 (R1 + r1 + R2 / 2) B1 [2] (b) p.d. across BJ of wire changes / resistance of BJ changes B1 there is a difference in p.d across wire and p.d. across cell E2 B1 [2]

This question in 9702/22 Oct/Nov 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

5 (a) (i) State Kirchhoff’s second law. For Examiner’s … Use … [1] (ii) Kirchhoff’s second law is linked to the conservation of a certain quantity. State this quantity. … [1] (b) The circuit shown in Fig. 5.1 is used to compare potential differences. cell A 2.0 V 0.50 Ω C D I R 0.90 m X J Y E r uniform resistance wire length 1.00 m cell B Fig. 5.1 The uniform resistance wire XY has length 1.00 m and resistance 4.0 Ω. Cell A has e.m.f. 2.0 V and internal resistance 0.50 Ω. The current through cell A is I. Cell B has e.m.f. E and internal resistance r. The current through cell B is made zero when the movable connection J is adjusted so that the length of XJ is 0.90 m. The variable resistor R has resistance 2.5 Ω. (i) Apply Kirchhoff’s second law to the circuit CXYDC to determine the current I. I = … A [2] (ii) Calculate the potential difference across the length of wire XJ. For Examiner’s Use potential difference = … V [2] (iii) Use your answer in (ii) to state the value of E. E = … V [1] (iv) State why the value of the internal resistance of cell B is not required for the determination of E. … … [1]

8 marks

Mark scheme: 5 (a) (i) sum of e.m.f.’s = sum of p.d.’s around a loop/circuit B1 [1] (ii) energy B1 [1] (b) (i) 2.0 = I × (4.0 + 2.5 + 0.5) C1 I = 0.286 A (allow 2 s.f.) A1 [2] (If total resistance is not 7 Ω, 0/2 marks) (ii) R = [0.90 / 1.0] × 4 (= 3.6) C1 V = I R = 0.286 × 3.6 = 1.03 V A1 [2] (If factor of 0.9 not used, then 0/2 marks) (iii) E = 1.03 V A1 [1] (iv) either no current through cell B or p.d. across r is zero B1 [1]

This question in 9702/21 May/June 2012

5 (a) (i) State Kirchhoff’s first law. For Examiner’s … Use … [1] (ii) Kirchhoff’s first law is linked to the conservation of a certain quantity. State this quantity. … [1] (b) A variable resistor of resistance R is used to control the current in a circuit, as shown in Fig. 5.1. 20 V 0.50 Ω + – G R 12 V 0.10 Ω Fig. 5.1 The generator G has e.m.f. 20 V and internal resistance 0.50 Ω. The battery has e.m.f. 12 V and internal resistance 0.10 Ω. The current in the circuit is 2.0 A. (i) Apply Kirchhoff’s second law to the circuit to determine the resistance R. R = … Ω [2] (ii) Calculate the total power generated by G. power = … W [2] (iii) Calculate the power loss in the total resistance of the circuit. For Examiner’s Use power = … W [2] (iv) The circuit is used to supply energy to the battery from the generator. Determine the efficiency of the circuit. efficiency = … [2]

10 marks

Mark scheme: 5 (a) (i) sum of currents into a junction = sum of currents out of junction B1 [1] (ii) charge B1 [1] (b) (i) ΣE = ΣIR 20 – 12 = 2.0(0.6 + R) (not used 3 resistors 0/2) C1 R = 3.4 Ω A1 [2] (ii) P = EI C1 = 20 × 2 = 40 W A1 [2] (iii) P = I2R C1 P = (2)2 × (0.1 + 0.5 + 3.4) = 16 W A1 [2] (iv) efficiency = useful power / output power C1 24 / 40 = 0.6 or 12 × 2 / 20 × 2 or 60% A1 [2] GCE AS/A LEVEL – May/June 2012 9702 23

This question in 9702/23 May/June 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

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

11 marks

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

This question in 9702/23 Oct/Nov 2012

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

11 marks

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

This question in 9702/22 Oct/Nov 2014

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

10 marks

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

This question in 9702/23 Oct/Nov 2014

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

12 marks

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

This question in 9702/22 Oct/Nov 2016

5 Three cells of electromotive forces (e.m.f.) E1, E2 and E3 are connected into a circuit, as shown in Fig. 5.1. X Y I3 R4 I1 E3 R1 E2 R3 R2 E1 I2 W Z Fig. 5.1 The circuit contains resistors of resistances R1, R2, R3 and R4. The currents in the different parts of the circuit are I1, I2 and I3. The cells have negligible internal resistance. Use Kirchhoff’s laws to state an equation relating (a) I1, I2 and I3, … [1] (b) E1, E3, R1, R3, R4, I1 and I3 in loop WXYZW, … … [1] (c) E1, E2, R1, R2, I1 and I2 in loop YZWY. … … [1] [Total: 3]

3 marks

Mark scheme: 5(a) B1 5(b) E1 + E3 = I1R1 + I3R3 + I3R4 [any subject] B1 5(c) E1 – E2 = I1R1 – I2R2 [any subject] B1

This question in 9702/21 Oct/Nov 2017

5 (a) State Kirchhoff’s second law. … … [2] (b) Two batteries, each of electromotive force (e.m.f.) 6.0 V and negligible internal resistance, are connected in series with three resistors, as shown in Fig. 5.1. R 4.0 Ω X 6.0 V V 6.0 V Y 1.5 Ω I Fig. 5.1 Resistor X has resistance 4.0 Ω and resistor Y has resistance 1.5 Ω. (i) The resistance R of the variable resistor is changed until the voltmeter in the circuit reads zero. Calculate 1. the current I in the circuit, I = … A [1] 2. the resistance R. R = … Ω [2] (ii) Resistors X and Y are wires made from the same material. The diameter of the wire of X is twice the diameter of the wire of Y. Determine the ratio average drift speed of free electrons in X . average drift speed of free electrons in Y ratio = … [2] (iii) The resistance R of the variable resistor is now increased. State and explain the effect of the increase in R on the power transformed by each of the batteries. … … … … [3] [Total: 10]

10 marks

Mark scheme: 5(a) sum of e.m.f.(s) = sum of p.d.(s) M1 around a loop / around a closed circuit A1 5(b)(i) 1 6.0 – 4.0I = 0 I = 1.5 A A1 2 6.0 + 6.0 = I (4.0 + R + 1.5) 12 = 1.5 (4.0 + R + 1.5) C1 R = 2.5 Ω A1 or 6.0 = I (R + 1.5) 6.0 = 1.5 (R + 1.5) (C1) R = 2.5 Ω (A1) or combines 6 = 4I and 6 = I(R + 1.5) to give 4 = R + 1.5 (C1) R = 2.5 Ω (A1) Question Answer Marks 5(b)(ii) I = Anvq ratio = 12 / 22 C1 = 0.25 A1 5(b)(iii) total (circuit) resistance increases B1 current / I decreases or P ∝ I or P ∝ 1 / (total resistance) M1 power (transformed) decreases A1

This question in 9702/22 Feb/March 2018

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

10 marks

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

This question in 9702/21 May/June 2018

6 (a) (i) State Kirchhoff’s first law. … … [1] (ii) Kirchhoff’s first law is linked to the conservation of a certain quantity. State this quantity. … [1] (b) A battery of electromotive force (e.m.f.) 8.0 V and internal resistance 2.0 Ω is connected to a resistor X and a wire Y, as shown in Fig. 6.1. 8.0 V 2.0 Ω 2.5 A 15 Ω X RY wire Y Fig. 6.1 The resistance of X is 15 Ω. The resistance of Y is RY. The current in the battery is 2.5 A. (i) Calculate 1. the thermal energy dissipated in the battery in a time of 5.0 minutes, energy = … J [2] 2. the terminal potential difference of the battery. terminal potential difference = … V [1] (ii) Determine the resistance RY. RY = … Ω [3] (iii) A new wire Z has the same length but less resistance than wire Y. 1. State two possible differences between wire Z and wire Y that would separately cause wire Z to have less resistance than wire Y. first difference: … … second difference: … … [2] 2. Wire Y is replaced in the circuit by wire Z. By considering the current in the battery, state and explain the effect of changing the wires on the total power produced by the battery. … … … [2] [Total: 12]

12 marks

Mark scheme: 6(a)(i) 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(a)(ii) charge B1 6(b)(i)1. E = I2Rt or E = VIt or E = (V2/ R)t C1 E = 2.52 × 2.0 × 5.0 × 60 or 5.0 × 2.5 × 5.0 × 60 or (5.02 / 2.0) × 5.0 × 60 = 3800 J A1 6(b)(i)2. p.d. = 8.0 – (2.0 × 2.5) = 3.0 V A1 6(b)(ii) IX = 3.0 / 15 = 0.20 (A) C1 IY = 2.5 – 0.20 = 2.3 (A) C1 RY = 3.0 / 2.3 = 1.3 Ω A1 or RT = 3.0 / 2.5 = 1.2 (Ω) or (8.0 / 2.5) – 2.0 = 1.2 (Ω) (C1) 1 / 1.2 = 1 / 15 + 1 / RY (C1) RY = 1.3 Ω (A1) Question Answer Marks 6(b)(iii)1. Z has larger radius/diameter/(cross-sectional) area B1 Z has (material of) smaller resistivity/greater conductivity B1 6(b)(iii)2. current/I (in battery) increases M1 (P = EI so) power/P (produced by battery) increases A1

This question in 9702/22 May/June 2018

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) State Kirchhoff’s second law. … … … [2] (b) An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force (e.m.f.) 9.0 V and negligible internal resistance, as shown in Fig. 6.1. 9.0 V 2.4 Ω wire X V 1.2Ω wire Y Fig. 6.1 Wire X has a resistance of 2.4 Ω and wire Y has a resistance of 1.2 Ω. A voltmeter is connected in parallel with the wires. A variable resistor is used to adjust the power dissipated in wires X and Y. The variable resistor is adjusted so that the voltmeter reads 6.0 V. (i) Calculate the resistance of the variable resistor. resistance = … Ω [3] (ii) Calculate the power dissipated in wire X. power = … W [2] (iii) The cross-sectional area of wire X is three times the cross-sectional area of wire Y. Assume that the resistivity and the number density of free electrons for the metal of both wires are the same. Determine the ratio length of wire X 1. , length of wire Y ratio = … [2] average drift velocity of free electrons in wire X 2. . average drift velocity of free electrons in wire Y ratio = … [2] [Total: 11]

11 marks

Mark scheme: 6(a) sum of e.m.f.(s) equal to sum of p.d.(s) M1 around a loop/around a closed circuit A1 6(b)(i) current in variable resistor = (6.0 / 2.4) + (6.0 / 1.2) (= 7.5 A) C1 p.d. across variable resistor = 9.0 – 6.0 (= 3.0 V) C1 R = 3.0 / 7.5 = 0.40 Ω A1 or 1 1 1 2.4 1.2 T R = + RT = 0.80 (Ω) (C1) ( ) 3 9 0.80 R R = + or 3 6 0.8 R = (C1) R = 0.40 Ω (A1) 6(b)(ii) P = V2 / R or P = I2R or P = IV C1 P = 6.02 / 24 or 2.52 × 2.4 or 6.0 × 2.5 = 15 W A1 Question Answer Marks 6(b)(iii) 1. L R A ρ = C1 ratio = (2.4 / 1.2) × (3 / 1) = 6.0 A1 2. (I = nAvq) IX / IY = 2.5 / 5.0 or 1.2 / 2.4 or 0.5 C1 ratio = (2.5 / 5.0) × (1 / 3) or (1.2 / 2.4) × (1 / 3) = 0.17 A1

This question in 9702/21 Oct/Nov 2018

7 (a) State Kirchhoff’s first law. … … [1] (b) A potentiometer is connected to a battery of electromotive force (e.m.f.) 9.6 V and negligible internal resistance, as shown in Fig. 7.1. 9.6 V 800 Ω X Y slider 400Ω R Fig. 7.1 The maximum resistance of the potentiometer is 800 Ω. A resistor R of resistance 400 Ω is connected between the slider and end X of the potentiometer. (i) State the potential difference across resistor R when the slider is positioned 1. at end X of the potentiometer, potential difference = … V 2. at end Y of the potentiometer. potential difference = … V [2] (ii) Calculate the potential difference across resistor R when the slider is positioned half-way between X and Y. potential difference = … V [3] [Total: 6]

6 marks

Mark scheme: 7(a) sum of current(s) in(to) junction = sum of current(s) out of junction or (algebraic) sum of current(s) at a junction is zero B1 7(b)(i) 1. potential difference = 0 A1 2. potential difference = 9.6 V A1 7(b)(ii) for resistance in parallel: (1 / RT) = (1 / 400) + (1 / 400) RT = 200 (Ω) C1 V / 9.6 = 200 / 600 C1 V = 3.2 V A1

This question in 9702/23 Oct/Nov 2018

5 (a) State Kirchhoff’s second law. … … [2] (b) A battery of electromotive force (e.m.f.) 5.6 V and internal resistance r is connected to two external resistors, as shown in Fig. 5.1. 5.6 V r V 90 18 Fig. 5.1 The reading on the voltmeter is 4.8 V. (i) Calculate: 1. the combined resistance of the two resistors connected in parallel combined resistance = … Ω [2] 2. the current in the battery. current = … A [2] (ii) Show that the internal resistance r is 2.5 Ω. [2] (iii) Determine the ratio power dissipated by internal resistance r . total power produced by battery ratio = … [3] (c) The battery in (b) is now connected to a battery of e.m.f. 7.2 V and internal resistance 3.5 Ω. The new circuit is shown in Fig. 5.2. 5.6 V 2.5 3.5 7.2 V Fig. 5.2 Determine the current in the circuit. current = … A [2] [Total: 13]

13 marks

Mark scheme: 5(a) sum of e.m.f.(s) = sum of p.d.(s) M1 around a loop/around a closed circuit A1 5(b)(i) 1. 1 / R = 1 / R1 + 1 / R2 1 / R = 1 / 90 + 1 / 18 C1 R = 15 Ω A1 2. I = V / R C1 I = 4.8 / 15 or I = 4.8 / 90 + 4.8 / 18 I = 0.32 A A1 5(b)(ii) E = V + Ir or E = I(R + r) C1 5.6 = 4.8 + 0.32 r so r = 2.5 (Ω) or 5.6 = 0.32 × (15 + r) so r = 2.5 (Ω) A1 5(b)(iii) P = EI or P = VI or P = I2R or P = V2 / R C1 ratio = (0.322 × 2.5) / (5.6 × 0.32) or 0.256 / 1.792 C1 = 0.14 A1 Question Answer Marks 5(c) 7.2 – 5.6 – 2.5I – 3.5I = 0 C1 I = 0.27 A A1

This question in 9702/22 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) 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) 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

6 (a) State Kirchhoff’s first law. … … [1] (b) A battery of electromotive force (e.m.f.) 12.0 V and internal resistance r is connected to a filament lamp and a resistor, as shown in Fig. 6.1. 12.0 V r 3.6 A 2.1 A Fig. 6.1 The current in the battery is 3.6 A and the current in the resistor is 2.1 A. The I-V characteristic for the lamp is shown in Fig. 6.2. 2.0 1.5 I / A 1.0 0.5 0 0 2.0 4.0 6.0 V / V Fig. 6.2 (i) Determine the resistance of the lamp in Fig. 6.1. resistance = … Ω [3] (ii) Determine the internal resistance r of the battery. r = … Ω [2] (iii) The initial energy stored in the battery is 470 kJ. Assume that the e.m.f. and the current in the battery do not change with time. Calculate the time taken for the energy stored in the battery to become 240 kJ. time = … s [2] (iv) The filament wire of the lamp is connected in series with the adjacent copper connecting wire of the circuit, as illustrated in Fig. 6.3. filament wire copper wire Fig. 6.3 (not to scale) Some data for the filament wire and the adjacent copper connecting wire are given in Table 6.1. Table 6.1 filament wire copper wire cross-sectional area A 360 A number density of free electrons n 2.5 n Calculate the ratio average drift speed of free electrons in filament wire . average drift speed of free electrons in copper wire ratio = … [2] [Total: 10]

10 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) I = 3.6 – 2.1 = 1.5 C1 V = 4.4 C1 R = 4.4 / 1.5 = 2.9 Ω A1 6(b)(ii) 12.0 = 4.4 + 3.6r or 12.0 = 3.6 (1.2 + r ) C1 r = 2.1 Ω A1 6(b)(iii) t = (470 × 103 – 240 × 103 ) / (12 × 3.6) C1 = 5300 s A1 6(b)(iv) I = Anvq ratio = (360A / A) × (2.5n / n) or 360 × 2.5 C1 = 900 A1

This question in 9702/22 Feb/March 2021

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

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

11 marks

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

This question in 9702/21 Oct/Nov 2021

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

11 marks

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

This question in 9702/21 May/June 2022

5 (a) State Kirchhoff’s second law. … … … [2] (b) Three identical cells, each of electromotive force (e.m.f.) 1.5 V and internal resistance 590 mΩ, are connected in parallel across a conductor, as shown in Fig. 5.1. 1.5 V 590 mΩ 1.5 V 590 mΩ 1.5 V 590 mΩ conductor A B Fig. 5.1 The conductor is composed of two cylindrical sections A and B. The total resistance of the circuit is 2.2 Ω. (i) Show that the resistance of the conductor is 2.0 Ω. [2] (ii) Calculate the current in the conductor. current = … A [2] (c) The two cylindrical sections A and B of the conductor in Fig. 5.1 are made from the same material and have the same length.

6 marks

Mark scheme: 5(a) sum of e.m.f.(s) = sum of p.d.(s) M1 or (algebraic) sum of e.m.f.(s) and p.d.(s) is zero around a loop / around a closed circuit A1 5(b)(i) 1 / r(T) = 1 / 0.59 + 1 / 0.59 + 1 / 0.59 B1 (r(T) =) 0.197 () A1 (R =) 2.2 – 0.197 = 2.0  or I = 1.5 / 2.2 (= 0.68 A) and i = 0.68 / 3 (where I is the circuit current and i is the current from each cell) (B1) (E = IR + ir =) 1.5 = 0.68R + (0.68 / 3)  0.59 and R = 2.0  (A1) 5(b)(ii) current = 1.5 / 2.2 C1 = 0.68 A A1 or p.d. across cell = p.d. across conductor (C1) 1.5 – 0.59I = 3I  2.0 so I = 0.228 A (where I is current in cell) current = 3  0.228 = 0.68 A (A1) or current in conductor = 3  current in cell (C1) V / 2.0 = 3  (1.5 – V) / 0.59 (where V is p.d. across conductor) V = 1.37 V current = 1.37 / 2.0 = 0.68 A (A1) 5(c)(i) R = L / A C1 R = 4L / d 2 C1 (and L are the same so) RA / RB = 7.62 / 4.32 = 3.1 A1 5(c)(ii) I = Anvq and I, n, q are same / equal / constant B1 v A A B dB 2 A1 = = v B A A dA 2 ratio = 7.62 / 4.32 = 3.1 5(d) combined internal resistance (of the cells) will be greater B1 or total / circuit resistance (of circuit) greater (because a parallel resistance removed) more ‘lost volts’ (inside each cell) or internal resistances take a greater share of total p.d. or conductor gets a smaller share M1 of the total p.d. or current in conductor/total current decreases (so) potential difference (across conductor) decreases A1

This question in 9702/23 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) State Kirchhoff’s first law. … … [1] (b) A cell with internal resistance r is connected to two resistors of resistances R1 and R2 as shown in Fig. 6.1. r I R1 R2 Fig. 6.1 The potential differences (p.d.s) across R1 and R2 are V1 and V2 respectively. The terminal p.d. across the cell is V. The current in the circuit is I. Use Kirchhoff’s laws to show that the total resistance RT of the external circuit is given by RT = R1 + R2 . [2] (c) The electromotive force (e.m.f.) of the cell in Fig. 6.1 is 1.50 V. The values of R1 and R2 are 10 Ω and 15 Ω respectively. The terminal p.d. of the cell is 1.35 V. Calculate the internal resistance r of the cell. r = … Ω [3] (d) A resistor of resistance R3 is added to the circuit in Fig. 6.1, so that the circuit is as shown in Fig. 6.2. r R1 R2 R3 Fig. 6.2 State and explain the effect, if any, of this change on: (i) the current in the cell … … … [2] (ii) the terminal p.d. of the cell. … … … [2] [Total: 10]

10 marks

Mark scheme: 6(a) sum of current(s) entering a junction = sum of current(s) leaving (the same junction) or (algebraic) sum of current (s) at a junction is zero B1 6(b) (by Kirchhoff’s second law) V = V1 + V2 B1 so IRT = IR1 + IR2 (and cancelling I gives) RT = R1 + R2 or V / I = V1/ I + V2 / I (and substituting R gives) RT = R1 + R2 B1 6(c) current in circuit = 1.35 / (10 + 15) (= 0.054 A) C1 r = (E – V) / I C1 = (1.5 – 1.35) / 0.054 = 2.8  A1 or by potential divider principle  0.15 1.35 25 r (C2) r = 2.8  (A1) or I = 1.35 / (10 + 15) (= 0.054 A) (C1) total resistance = 1.50 / 0.054 (= 27.8 ) r = 27.8 – 25 (C1) r = 2.8  (A1) Question Answer Marks 6(d)(i) the (total) resistance (of the circuit) has decreased (and e.m.f. is unchanged) M1 (the current (in the cell) will) increase A1 6(d)(ii) (as the current is greater and so there is a) larger p.d. across the internal resistance M1 (terminal p.d. will) decrease A1

This question in 9702/21 May/June 2024

5 (a) State Kirchhoff’s second law. … … [1] (b) A battery of electromotive force (e.m.f.) 9.0 V and negligible internal resistance is connected in series with a variable resistor X and a thermistor Y as shown in Fig. 5.1. X 9.0 V Y Fig. 5.1 Fig. 5.2 shows the relationship between temperature and resistance for the thermistor. 200 resistance / Ω 150 100 50 0 0 100 200 300 400 temperature / °C Fig. 5.2 (i) The current in the circuit is 1.1 × 10–2 A. The potential difference across Y is 4.0 V. Calculate the resistance of X. resistance = … Ω [2] (ii) The temperature of Y is changed to 190 °C. The resistance of X remains unchanged. Determine the new potential difference across Y. potential difference = … V [3] (iii) The resistance of X is increased. The temperature of Y remains at 190 °C. By reference to the current in the circuit, state and explain the effect of this change, if any, on the potential difference across Y. … … … … … … [3] [Total: 9]

9 marks

Mark scheme: 5(a) sum of electromotive force(s) = sum of potential difference(s) around a (closed) loop. B1 5(b)(i) R = V / I C1 = (9.0 – 4.0) / 1.1  10–2 = 450  A1 5(b)(ii) resistance (of thermistor) = 25 () (from graph) C1 V = E  RY / (RY + RX) = 9  25 / (25 + 450) or I = E / RTotal = 9 / (25 + 450) = 1.89  10–2 A V = IR = 1.89  10–2  25 or V(X) = 9  450 / (25 + 450) = 8.53 V V = 9 – 8.53 C1 V = 0.47 V A1 5(b)(iii) (resistance of X increases so) the total resistance increases B1 current decreases B1 potential difference (across thermistor / Y) decreases B1

This question in 9702/22 May/June 2024

5 (a) (i) State Kirchhoff’s second law. … … … [1] (ii) State the conservation law that gives rise to Kirchhoff’s second law. … [1] (b) A circuit contains a cell of internal resistance r and two resistors of resistances R1 and R2, as shown in Fig. 5.1. r R1 I R2 V Fig. 5.1 The potential difference (p.d.) across the two resistors is V. The current in the cell is I. (i) Use Kirchhoff’s laws to show that the total resistance RT of the external circuit is given by 1 1 1 = + . RT R1 R2 [2] (ii) The electromotive force (e.m.f.) of the cell is 1.50 V. When the values of R1 and R2 are 10 Ω and 15 Ω respectively, the p.d. measured by the voltmeter is 1.38 V. Calculate the internal resistance r of the cell. r = … Ω [3] (c) A third resistor is added in parallel with R1 and R2 in the circuit in Fig. 5.1. State and explain the effect, if any, of this change on: (i) the current in the cell … … … [2] (ii) the p.d. measured by the voltmeter. … … … [2] [Total: 11]

11 marks

Mark scheme: 5(a)(i) sum of electromotive force(s) = sum of potential difference(s) around a (closed) loop or the (algebraic) sum of the p.d.(s) and e.m.f.(s) is zero around a (closed) loop B1 5(a)(ii) (law of conservation of) energy B1 5(b)(i) (by Kirchhoff’s first law) I = I1 + I2 B1 V / RT = V / R1 + V / R2 therefore 1 / RT = 1 / R1 + 1 / R2 B1 5(b)(ii) resistance of parallel combination = (15  10) / (15 + 10) (= 6.0 ) C1 r = (E – V) / I C1 I= 1.38 / 6.0 = 0.23 A r = (1.50 – 1.38) / 0.23 = 0.52  A1 or (by potential divider principle) r / RT = Ir / V (C1) r / 6.0 = 0.12 / 1.38 r = 0.52  (A1) or (by potential divider equation) V = E  RT / (RT + r) (C1) 1.38 = 1.5  6.0 / (6.0 + r) r = 0.52  (A1) Question Answer Marks 5(c)(i) as the (total) resistance has decreased (and e.m.f. is unchanged) M1 current will (in the cell) increase A1 5(c)(ii) (as greater current means a) bigger drop in p.d. across the internal resistance M1 p.d. (on voltmeter) will decrease A1

This question in 9702/23 May/June 2024

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

10 marks

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

This question in 9702/22 May/June 2025

5 Fig. 5.1 shows a circuit containing a battery, two fixed resistors X and Y, and a light-dependent resistor (LDR) Z. 5.0 V 4.7 Ω I1 100 Ω X Z I2 Y Fig. 5.1 The battery has electromotive force (e.m.f.) 5.0 V and internal resistance 4.7 Ω. The current in X is I1 and the current in Y is I2. The resistance of X is 100 Ω. The resistance of Z varies with the intensity of light incident on it as shown in Fig. 5.2. 500 400 resistance / Ω 300 200 100 0 0 50 100 150 200 250 intensity / W m–2 Fig. 5.2 (a) State Kirchhoff’s first law. … … [1] (b) The intensity of light incident on Z is 130 W m–2. The current in the battery is 38 mA. (i) Show that the terminal potential difference of the battery is 4.8 V. [2] (ii) Calculate the current I2 in Y. I2 = … A [3] (iii) Calculate the power dissipated in Y. power = … W [2] (iv) The intensity of the light incident on Z decreases. State and explain the effect on the terminal potential difference of the battery. … … … … … [3] [Total: 11]

11 marks

Mark scheme: 5(a) sum of current(s) into junction = sum of current(s) out junction or (algebraic) sum of current(s) at a junction is zero B1 5(b)(i) (V =) E – Ir C1 (V =) 5.0 – (38  10–3  4.7) = 4.8 (V) A1 5(b)(ii) R(Z) = 120  C1 I2 = 38 × 10–3 – (4.8 / (100 + 120)) C1 I2 = 0.016 A A1 or R(Z) = 120  (C1) R(EXT) = 4.8 / 38 × 10–3 (C1) (= 126 ) 1 / R(Y) = 1 / 126 – 1 / (120 + 100) (R(Y) = 297 ) I2 = 4.8 / 297 I2 = 0.016 A (A1) 5(b)(iii) P = IV or P = I2R or P = V2 / R C1 P = 0.016  4.8 A1 or P = 0.0162  (4.8 / 0.016) or P = 4.82 / (4.8 / 0.016) P = 0.077 W 5(b)(iv) resistance of the LDR / Z increases (as light intensity decreases) B1 total resistance (of the circuit) increases M1 or current in the battery decreases (potential difference across the internal resistor decreases so) the terminal potential difference increases A1

This question in 9702/22 Oct/Nov 2025

5 (a) State Kirchhoff’s first law. … … [1] (b) Fig. 5.1 shows a circuit containing a thermistor T that has a negative temperature coefficient. E r R T Fig. 5.1 (i) The thermistor has resistance R0 at a temperature of 0 °C. On Fig. 5.2, sketch a possible variation of the resistance of the thermistor with temperature between 0 °C and 100 °C. resistance R0 0 0 100 temperature / °C Fig. 5.2 [2] (ii) With reference to the current in the cell, explain why the current in resistor R decreases with increasing temperature of the thermistor. … … … … … [3] (c) The electromotive force (e.m.f.) E of the cell in Fig. 5.1 is 1.50 V. The internal resistance r of the cell is 0.12 Ω. Resistor R has a resistance of 6.00 Ω. At a particular temperature of the thermistor, the current in R is 0.200 A. For this temperature of the thermistor, determine: (i) the current in the cell current = … A [2] (ii) the resistance of the thermistor. resistance = … Ω [2] [Total: 10]

10 marks

Mark scheme: 5(a) sum of current(s) into junction = sum of current(s) out junction B1 or (algebraic) sum of current(s) at a junction is zero 5(b)(i) line with negative gradient C1 line from 0 °C to 100 °C, starting at (0, R0) with negative gradient throughout and never reaching R = 0 A1 5(b)(ii) (resistance of T decreases so) total resistance (of circuit) decreases B1 current in cell increases (so p.d. across internal resistance increases) B1 terminal p.d. decreases (and resistance of R is constant) so current in R decreases B1 5(c)(i) p.d. across r = 1.50 – (6.00  0.200) C1 (= 0.30 V) current in cell = 0.30 / 0.12 A1 = 2.5 A 5(c)(ii) current in thermistor = 2.5 – 0.200 C1 (= 2.3 A) resistance of T = (6.00  0.200) / 2.3 A1 = 0.52 

This question in 9702/23 Oct/Nov 2025