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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Resistance, resistivity & I-V characteristics · Core · Paper 1 style · 1 mark
(a)The current in a filament lamp is increased from zero to its normal working value. Which statement describes and explains the change in its resistance?[1]
Show answer
(a) B The resistance increases because the filament gets hotter and the lattice ions vibrate more.
As current rises the filament temperature rises; increased lattice vibration impedes the electrons, so R = V/I increases (the I–V graph curves towards the V-axis). A describes an NTC thermistor, not a metal filament.
Q2Resistance, resistivity & I-V characteristics · Core · Paper 2 style · 4 marks
A length of nichrome wire used in a hair-dryer heating element is 1.50 m long and has a diameter of 0.32 mm. The resistivity of nichrome is 1.10 × 10⁻⁶ Ω m.
(a)Calculate the cross-sectional area of the wire.[1]
(b)Calculate the resistance of the wire.[2]
(c)The wire is connected to a 6.0 V supply of negligible internal resistance. Calculate the current.[1]
Show answer
(a) 8.04 × 10⁻⁸ m²
A1: A = π(0.16 × 10⁻³)² = 8.04 × 10⁻⁸ m².
(b) 20.5 Ω
C1: R = ρL/A = 1.10 × 10⁻⁶ × 1.50 / 8.04 × 10⁻⁸. A1: R = 20.5 Ω.
(c) 0.292 A
A1: I = 6.0/20.5 = 0.292 A.
Q3Resistance, resistivity & I-V characteristics · Core · Paper 2 style · 5 marks
A student investigates a fixed resistor that obeys Ohm's law.
(a)Define the ohm.[1]
(b)State Ohm's law.[2]
(c)Calculate the resistance of the resistor when the p.d. across it is 3.0 V and the current is 250 mA. Describe the shape of its I–V graph.[2]
Show answer
(a) One ohm is the resistance of a component in which a current of one ampere flows when the potential difference across it is one volt (1 Ω = 1 V A⁻¹).
B1: volt per ampere / p.d. of 1 V produces a current of 1 A.
(b) The current in a metallic conductor is directly proportional to the potential difference across it, provided its temperature (physical conditions) remains constant.
B1: current ∝ p.d. B1: at constant temperature / physical conditions.
(c) 12.0 Ω; a straight line through the origin (for both positive and negative p.d.)
A1: R = 3.0/0.250 = 12.0 Ω. B1: straight line through origin with constant gradient, extending into the third quadrant.
Q4Resistance, resistivity & I-V characteristics · Core · Paper 2 style · 5 marks
(a)State how the resistance of an NTC thermistor changes as its temperature increases.[1]
(b)State how the resistance of a light-dependent resistor (LDR) changes as the light intensity falling on it increases.[1]
(c)A thermistor is used to monitor the temperature of milk in a pasteurisation tank. Its resistance is 4.7 kΩ at 25 °C. Suggest whether its resistance at 72 °C is greater or less than 4.7 kΩ, and explain whether its I–V characteristic would be a straight line if it were allowed to heat up as the current increases.[3]
Show answer
(a) It decreases (non-linearly).
B1: decreases.
(b) It decreases.
B1: decreases.
(c) Less than 4.7 kΩ. Not a straight line: as current increases the thermistor heats up, its resistance falls, so the current rises more than proportionally (graph curves towards the I-axis).
B1: less than 4.7 kΩ (higher temperature). B1: not straight / not ohmic. B1: self-heating lowers R, so I/V increases with V (gradient of I–V graph increases).
Q5Resistance, resistivity & I-V characteristics · Core · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) D 4R
Doubling L at constant volume halves A, so R = ρL/A becomes ρ(2L)/(A/2) = 4R. C ignores the change in cross-sectional area.
Q6Resistance, resistivity & I-V characteristics · Standard · Paper 2 style · 6 marks
A silicon diode is connected in forward bias. Readings of p.d. V and current I are:V / V: 0.20, 0.50, 0.60, 0.65, 0.70I / mA: 0, 2, 10, 25, 60
(a)Calculate the resistance of the diode when the p.d. is 0.60 V and when it is 0.70 V.[2]
(b)Describe the I–V characteristic of the diode for both directions of p.d.[3]
(c)Suggest why a diode is connected in series with a phone-charging circuit fed from a car battery.[1]
Show answer
(a) 60.0 Ω at 0.60 V; 11.7 Ω at 0.70 V
C1: R = V/I (not the gradient). A1: 0.60/0.010 = 60.0 Ω; 0.70/0.060 = 11.7 Ω.
(b) Forward: almost no current below about 0.5 V (threshold), then current rises steeply as resistance falls. Reverse: current is (almost) zero, very high resistance.
B1: negligible current below a threshold of about 0.5–0.6 V. B1: current rises rapidly / resistance decreases above threshold. B1: reverse bias: very high resistance, (almost) zero current.
(c) It allows current in one direction only, protecting the circuit if the battery is connected the wrong way round.
B1: conducts in one direction only / protects against reverse polarity.
Q7Resistance, resistivity & I-V characteristics · Standard · Paper 2 style · 5 marks
Two wires A and B are made of the same alloy. Wire B is twice as long as wire A and has three times the diameter.
(a)Show that the resistance of B is 2/9 of the resistance of A.[2]
(b)The resistance of A is 18 Ω. A and B are connected in series to a 12 V supply of negligible internal resistance. Calculate the current and the p.d. across B.[3]
Show answer
(a) R_B/R_A = (2L/9A)/(L/A) = 2/9
M1: tripling diameter multiplies area by 9. A1: R_B/R_A = 2/9 (doubling length doubles R).
(b) current 0.545 A; p.d. across B 2.18 V
C1: R_B = 4.0 Ω, total 22 Ω. A1: I = 12/22 = 0.545 A. A1: V_B = 0.545 × 4.0 = 2.18 V.
Q8Resistance, resistivity & I-V characteristics · Standard · Paper 2 style · 5 marks
An electricity company in Pakistan compares copper and aluminium cables for an overhead line. Resistivity: copper 1.72 × 10⁻⁸ Ω m, aluminium 2.82 × 10⁻⁸ Ω m. Density: copper 8960 kg m⁻³, aluminium 2700 kg m⁻³.
(a)Show that, for the same length and resistance, the aluminium cable must have a cross-sectional area 1.64 times that of the copper cable.[2]
(b)Hence calculate the ratio (mass of aluminium cable) / (mass of copper cable).[2]
(c)Suggest why aluminium is preferred to copper for overhead power lines between pylons.[1]
Show answer
(a) 1.64 times
M1: R = ρL/A with R and L equal, so A ∝ ρ. A1: A_Al/A_Cu = 2.82/1.72 = 1.64.
(b) 0.494
C1: mass = density × A × L, so ratio = 1.64 × 2700/8960. A1: 0.494.
(c) It is about half the mass for the same resistance, so less load on the pylons (and it is cheaper).
B1: lighter for same resistance (or cheaper).
Q9Resistance, resistivity & I-V characteristics · Standard · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) B 8.0 Ω
Resistance is defined as V/I = 2.0/0.25 = 8.0 Ω at that point. C (12.5 Ω) is 1/gradient, which is not the resistance for a non-ohmic component.
Q10Resistance, resistivity & I-V characteristics · Challenge · Paper 5 style · 8 marks
A student measures the resistance R of different lengths L of a uniform metal wire. The diameter of the wire is 0.274 mm.L / m: 0.200, 0.400, 0.600, 0.800, 1.000R / Ω: 1.68, 3.30, 4.99, 6.66, 8.29
(a)Explain why a graph of R against L should be a straight line through the origin, and state an expression for its gradient.[2]
(b)Determine the gradient of the graph.[2]
(c)Calculate the resistivity of the metal.[2]
(d)The diameter is measured as 0.274 ± 0.002 mm and the uncertainty in the gradient is ±2.0%. Calculate the percentage uncertainty in the resistivity.[2]
Show answer
(a) R = (ρ/A)L, with ρ/A constant, so R ∝ L; gradient = ρ/A.
B1: R = ρL/A, compare with y = mx (no intercept). B1: gradient = ρ/A.
(b) 8.26 Ω m⁻¹
C1: use points far apart on the line, e.g. (0.200, 1.68) and (1.000, 8.29). A1: gradient = 6.61/0.800 = 8.26 Ω m⁻¹.
(c) 4.87 × 10⁻⁷ Ω m
C1: A = π(0.137 × 10⁻³)² = 5.90 × 10⁻⁸ m². A1: ρ = gradient × A = 4.87 × 10⁻⁷ Ω m.
(d) ±3.5%
C1: %U in A = 2 × (0.002/0.274) × 100 = 1.5%. A1: total = 1.5 + 2.0 = 3.5%.
Q11Resistance, resistivity & I-V characteristics · Challenge · Paper 2 style · 6 marks
A copper water pipe used as an earthing path is 2.5 m long, with an outer diameter of 15 mm and an inner diameter of 13 mm. The resistivity of copper is 1.72 × 10⁻⁸ Ω m.
(a)Calculate the cross-sectional area of the copper in the pipe.[2]
(b)Calculate the resistance of the pipe between its ends.[2]
(c)During a fault, a current of 50 A flows along the pipe. Calculate the p.d. between its ends and comment on whether a person touching both ends is in danger.[2]
Show answer
(a) 4.40 × 10⁻⁵ m²
C1: A = π(r_out² − r_in²) = π(7.5² − 6.5²) × 10⁻⁶. A1: 4.40 × 10⁻⁵ m².
A1: V = IR = 0.0489 V. B1: p.d. far too small to drive a harmful current through a person.
Q12Resistance, resistivity & I-V characteristics · Challenge · Paper 2 style · 6 marks
A 12 V, 2.4 W tungsten filament lamp is used in a motorcycle tail-light.
(a)Calculate the resistance of the filament at its normal operating temperature.[1]
(b)At room temperature the resistance of the filament is 4.5 Ω. Calculate the current at the instant the lamp is switched on, and explain why lamps most often fail at switch-on.[3]
(c)The filament has a diameter of 25 µm, and the resistivity of tungsten at the operating temperature is 5.6 × 10⁻⁷ Ω m. Calculate the length of the filament.[2]
Show answer
(a) 60.0 Ω
A1: R = V²/P = 144/2.4 = 60.0 Ω.
(b) 2.67 A; the switch-on current is about 13 times the working current (0.20 A), causing large, rapid heating stresses in the thin filament
A1: I = 12/4.5 = 2.67 A. B1: working current = 2.4/12 = 0.20 A, much smaller. B1: large initial current → large power / rapid heating → filament breaks.
(c) 0.0526 m
C1: L = RA/ρ = 60 × π(12.5 × 10⁻⁶)² / 5.6 × 10⁻⁷. A1: L = 0.0526 m.
Q13Resistance, resistivity & I-V characteristics · Challenge · Paper 2 style · 6 marks
A 0.60 m length of constantan wire and a 2.0 m length of copper wire, both of diameter 0.50 mm, are joined in series and connected across a 1.5 V cell of negligible internal resistance. Resistivity: constantan 4.9 × 10⁻⁷ Ω m, copper 1.72 × 10⁻⁸ Ω m.
(a)Calculate the resistance of each wire.[3]
(b)Calculate the current in the circuit and the p.d. across the constantan wire.[2]
(c)Explain why the copper wire is much less hot than the constantan wire even though the current in them is the same.[1]
Show answer
(a) constantan 1.50 Ω; copper 0.175 Ω
C1: A = π(0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m². A1: constantan: 4.9 × 10⁻⁷ × 0.60 / A = 1.50 Ω. A1: copper: 1.72 × 10⁻⁸ × 2.0 / A = 0.175 Ω.
(b) 0.897 A; 1.34 V
C1: I = 1.5/(1.50 + 0.175). A1: I = 0.897 A and V = I × 1.50 = 1.34 V.
(c) Power = I²R and the copper wire has a much smaller resistance, so it dissipates much less power.
B1: same I, P = I²R, R_copper ≪ R_constantan.
Q14Resistance, resistivity & I-V characteristics · Challenge · Paper 2 style · 5 marks
A strain gauge consists of a fine metal wire glued to a bridge beam. Its unstretched resistance is 120.0 Ω. When the beam bends, the wire stretches.
(a)Show that, if the volume of the wire stays constant, its resistance is proportional to L².[2]
(b)The strain gauge wire is stretched so that its length increases by 0.10%. Calculate the new resistance and the percentage increase in resistance.[3]
Show answer
(a) R = ρL/A and A = V/L, so R = ρL²/V ∝ L²
M1: A = volume/L. A1: R = ρL²/volume, with ρ and volume constant, so R ∝ L².
(b) The wire is already full of free electrons; when the switch is closed an electric field is set up along the whole circuit almost instantly, so electrons everywhere (including in the lamp filament) start drifting at the same time.
B1: free electrons are present throughout the circuit. B1: the electric field/energy is transmitted very quickly, so all electrons begin to drift together.
Q18Current, charge & drift speed (I = Anvq) · Standard · Paper 2 style · 5 marks
Two copper wires, X of diameter 0.60 mm and Y of diameter 0.30 mm, are joined end to end and connected in series to a power supply.
(a)State, with a reason, how the current in X compares with the current in Y.[1]
(b)Calculate the ratio (drift speed in Y)/(drift speed in X).[2]
(c)Explain, in terms of the electrons, why the thinner wire becomes hotter.[2]
Show answer
(a) Equal — they are in series, and charge is conserved.
B1: same current, series / charge conserved.
(b) 4.0
C1: v = I/(nAq) with I, n, q equal, so v ∝ 1/A ∝ 1/d². A1: (0.60/0.30)² = 4.0.
(c) Electrons drift faster in the thinner wire and collide more often/energetically with the lattice ions, transferring more energy per second (larger resistance, P = I²R).
B1: higher drift speed / more collisions with lattice ions per unit length. B1: more energy transferred to the lattice per second (higher resistance → more power).
Q19Current, charge & drift speed (I = Anvq) · Standard · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) A v/4
v = I/(nAq); doubling diameter makes A four times larger, so v becomes v/4. B wrongly treats A as proportional to d.
Q20Current, charge & drift speed (I = Anvq) · Standard · Paper 2 style · 5 marks
An aluminium wire has a diameter of 2.0 mm. Aluminium has density 2700 kg m⁻³ and molar mass 0.027 kg mol⁻¹, and each atom contributes 3 free electrons. The Avogadro constant is 6.02 × 10²³ mol⁻¹.
(a)Show that the number density of free electrons in aluminium is about 1.8 × 10²⁹ m⁻³.[2]
(b)Calculate the drift speed of the electrons when the wire carries a current of 5.0 A.[3]
C1: A = π(1.0 × 10⁻³)² = 3.14 × 10⁻⁶ m². C1: v = I/(nAq). A1: v = 5.51 × 10⁻⁵ m s⁻¹.
Q21Current, charge & drift speed (I = Anvq) · Standard · Paper 2 style · 5 marks
In a research accelerator, a beam of protons striking a target is equivalent to a current of 2.5 µA.
(a)Calculate the number of protons arriving at the target each second.[2]
(b)In a second experiment the beam is changed to α-particles and the beam current is 0.40 µA. Calculate the number of α-particles arriving each second.[2]
(c)Calculate the total charge delivered by the α-particle beam in one hour.[1]
C1: charge on α = 2e = 3.20 × 10⁻¹⁹ C. A1: N = 0.40 × 10⁻⁶ / 3.20 × 10⁻¹⁹ = 1.25 × 10¹² s⁻¹.
(c) 0.00144 C
A1: Q = It = 0.40 × 10⁻⁶ × 3600 = 0.00144 C.
Q22Current, charge & drift speed (I = Anvq) · Challenge · Paper 2 style · 6 marks
A strip of doped silicon is 1.0 mm wide and 0.50 mm thick. It carries a current of 1.0 mA. The number density of charge carriers is 2.0 × 10²³ m⁻³ and each carrier has charge 1.60 × 10⁻¹⁹ C. The number density of free electrons in copper is 8.5 × 10²⁸ m⁻³.
(a)Calculate the drift speed of the charge carriers in the silicon strip.[2]
(b)A copper strip of the same dimensions carries the same current. Calculate the ratio (drift speed in silicon)/(drift speed in copper).[2]
(c)Suggest why semiconductors rather than metals are used in Hall probes, using your answer to (b).[2]
Show answer
(a) 0.0625 m s⁻¹
C1: A = 1.0 × 10⁻³ × 0.50 × 10⁻³ = 5.0 × 10⁻⁷ m². A1: v = I/(nAq) = 0.0625 m s⁻¹.
(b) 4.25 × 10⁵
C1: v ∝ 1/n for the same I, A and q. A1: ratio = 8.5 × 10²⁸ / 2.0 × 10²³ = 4.25 × 10⁵.
(c) Carriers in a semiconductor drift much faster for a given current, so the magnetic force on them and hence the Hall voltage are much larger and easier to measure.
B1: larger drift speed (smaller n). B1: larger magnetic force on carriers → larger, measurable Hall voltage.
Q23Current, charge & drift speed (I = Anvq) · Challenge · Paper 2 style · 6 marks
When a small electric motor for a toy car is switched on, the current in it increases uniformly from 0 to 3.0 A in 5.0 s. The current then stays at 3.0 A for a further 10 s.
(a)Calculate the total charge that passes through the motor in the 15 s.[3]
(b)Calculate the average current over the 15 s.[1]
(c)Calculate the number of electrons that pass through the motor.[1]
(d)Explain why the drift speed of electrons in the supply wire increases during the first 5.0 s.[1]
Show answer
(a) 37.5 C
C1: charge = area under I–t graph. C1: ½ × 3.0 × 5.0 + 3.0 × 10 = 7.5 + 30. A1: 37.5 C.
(b) 2.50 A
A1: 37.5/15 = 2.50 A.
(c) 2.34 × 10²⁰
A1: 37.5/1.60 × 10⁻¹⁹ = 2.34 × 10²⁰.
(d) I = nAvq with n, A and q constant, so v is proportional to I, which is increasing.
B1: v ∝ I since n, A, q fixed.
Q24Potential difference & power · Core · Paper 2 style · 4 marks
An electric kettle is rated at 2.2 kW for use on the 230 V mains supply.
(a)Define the potential difference between two points in a circuit.[1]
(b)Calculate the current in the kettle and its resistance.[2]
(c)Calculate the energy transferred by the kettle in 3.0 minutes.[1]
Show answer
(a) The energy transferred from electrical to other forms per unit charge passing between the two points.
B1: energy transferred (electrical → other forms) per unit charge.
(b) 9.57 A; 24.0 Ω
A1: I = P/V = 2200/230 = 9.57 A. A1: R = V/I (or V²/P) = 24.0 Ω.
(c) 396 kJ
A1: E = Pt = 2200 × 180 = 396 kJ.
Q25Potential difference & power · Core · Paper 1 style · 1 mark
(a)What are the SI base units of the volt?[1]
Show answer
(a) B kg m² s⁻³ A⁻¹
V = W/Q = (kg m² s⁻²)/(A s) = kg m² s⁻³ A⁻¹. A forgets that coulomb = A s, so the time power is wrong.
Q26Potential difference & power · Core · Paper 2 style · 4 marks
A mobile-phone charger supplies a steady current of 2.1 A at a p.d. of 5.0 V for 1.5 hours.
(a)Calculate the charge that flows into the phone in this time.[2]
(b)Calculate the energy transferred from the charger.[2]
Show answer
(a) 11300 C
C1: Q = It = 2.1 × 1.5 × 3600 = 11 340 C. A1: 11 300 C (3 s.f.).
(b) 56.7 kJ
C1: W = QV = 11 340 × 5.0 (or W = VIt). A1: 56.7 kJ.
Q27Potential difference & power · Standard · Paper 1 style · 1 mark
(a)[1]
Show answer
(a) A P/8
P = V²/R: (V/2)²/(2R) = V²/(8R) = P/8. B ignores the change in resistance.
Q28Potential difference & power · Standard · Paper 2 style · 6 marks
A 1.0 kW room heater for use on a 230 V supply has a heating element made of nichrome wire of diameter 0.45 mm. The resistivity of nichrome at the working temperature is 1.10 × 10⁻⁶ Ω m.
(a)Calculate the resistance of the element when working normally.[1]
(b)Calculate the length of wire needed.[3]
(c)State and explain what would happen to the power if a shorter length of the same wire were used.[2]
Show answer
(a) 52.9 Ω
A1: R = V²/P = 230²/1000 = 52.9 Ω.
(b) 7.65 m
C1: A = π(0.225 × 10⁻³)² = 1.59 × 10⁻⁷ m². C1: L = RA/ρ. A1: L = 52.9 × 1.59 × 10⁻⁷ / 1.10 × 10⁻⁶ = 7.65 m.
(c) Power increases: shorter wire has smaller resistance, and P = V²/R with V fixed.
B1: power increases. B1: R ∝ L decreases, P = V²/R at constant 230 V.
Q29Potential difference & power · Challenge · Paper 2 style · 6 marks
In an electron gun, electrons start from rest at a heated cathode and are accelerated through a potential difference of 2500 V. The mass of an electron is 9.11 × 10⁻³¹ kg.
(a)Calculate the kinetic energy gained by each electron, in joules.[1]
(b)Calculate the speed of the electrons as they reach the anode.[2]
(c)The beam current is 0.80 mA. Calculate the power delivered to the beam and state where this energy ends up when the electrons hit the screen.[3]
Show answer
(a) 4.00 × 10⁻¹⁶ J
A1: W = QV = 1.60 × 10⁻¹⁹ × 2500 = 4.00 × 10⁻¹⁶ J.
