References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
or see the official Cambridge past-papers page; for other papers, ask your
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.
Q1Electromagnetic effects · Core · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) B 100
Vs/Vp = Ns/Np, so Ns = 2000 × 11.5 ÷ 230 = 100. D comes from inverting the ratio.
Q2Electromagnetic effects · Core · Paper 4 style · 5 marks
The north pole of a bar magnet is pushed quickly into a coil of wire connected to a centre-zero galvanometer. The pointer deflects to the right.
(a)State what is observed on the galvanometer when the magnet is held still inside the coil.[1]
(b)The magnet is pulled back out of the coil. Describe what the galvanometer shows.[1]
(c)State three ways of increasing the size of the induced e.m.f.[3]
Show answer
(a) No deflection (zero reading).
B1 no reading – no change in magnetic field linking the coil.
(b) A deflection in the opposite direction to when the magnet was pushed in.
B1 deflection in opposite direction.
(c) Move the magnet faster; use a stronger magnet; use a coil with more turns.
1 mark each: move faster; stronger magnet; more turns on the coil.
Q3Electromagnetic effects · Standard · Paper 4 style · 7 marks
A power station supplies 66 MW to a distant city through cables with a total resistance of 0.50 Ω.
(a)Calculate the current in the cables when the power is transmitted at 132 kV.[2]
(b)Calculate the power wasted as heat in the cables at 132 kV.[2]
(c)Calculate the power that would be wasted if the same power were transmitted at 11 kV.[2]
(d)Use your answers to explain why electrical energy is transmitted at high voltage.[1]
Show answer
(a) 500 A
I = P/V C1; = 66 × 10⁶ ÷ 132 × 10³ = 500 A A1.
(b) 125000 W (0.125 MW)
P = I²R C1; = 500² × 0.50 = 125 000 W A1.
(c) 18 MW
I = 66 × 10⁶ ÷ 11 × 10³ = 6000 A C1; P = 6000² × 0.50 = 1.8 × 10⁷ W A1.
(d) For the same power, a higher voltage means a smaller current, so much less energy is wasted as heat in the cables (loss ∝ I²).
B1 smaller current → smaller I²R losses.
Q4Electromagnetic effects · Standard · Paper 4 style · 6 marks
A simple d.c. motor has a rectangular coil that can rotate between the poles of a magnet. It is connected to a battery through a split-ring commutator and brushes.
(a)Explain why the coil rotates when there is a current in it.[2]
(b)State the purpose of the split-ring commutator.[2]
(c)State two ways of making the motor turn faster.[2]
Show answer
(a) The current in each side of the coil is in a magnetic field, so each side experiences a force; the currents in the two sides are in opposite directions, so the forces are opposite and produce a turning effect.
B1 current-carrying wire in a magnetic field experiences a force B1 forces on opposite sides are in opposite directions – turning effect (moment).
(b) It reverses the direction of the current in the coil every half turn, so the coil keeps turning in the same direction.
B1 reverses current in the coil B1 every half turn (when coil is vertical) so rotation continues in the same direction.
(c) Any two of: increase the current; use a stronger magnet; increase the number of turns on the coil.
1 mark each for any two of: larger current / higher voltage; stronger magnet; more turns.
Q5Electromagnetic effects · Standard · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) B west
Using Fleming's left-hand rule (first finger = field downward, second finger = current northward), the thumb points west. The force is always at right angles to both current and field, so C and D are impossible.
Q6Electromagnetic effects · Challenge · Paper 4 style · 7 marks
A simple a.c. generator has a coil rotating at a steady rate in a uniform magnetic field. The ends of the coil are connected to slip rings.
(a)Describe how the e.m.f. induced in the coil varies during one complete revolution.[3]
(b)Explain why the e.m.f. is largest when the plane of the coil is parallel to the magnetic field.[1]
(c)The coil is rotated twice as fast. State the effect on the peak e.m.f. and on the frequency of the output.[2]
(d)State the purpose of the slip rings.[1]
Show answer
(a) The e.m.f. varies smoothly (sinusoidally): it is zero when the plane of the coil is at right angles to the field, rises to a maximum when the plane of the coil is parallel to the field, falls to zero, then reaches a maximum in the opposite direction and returns to zero.
B1 alternating / reverses direction twice per revolution B1 zero when coil plane perpendicular to field B1 maximum when coil plane parallel to field (sides cutting field lines fastest).
(b) The sides of the coil are then moving at right angles to the field lines, so they cut the field lines at the greatest rate.
B1 sides cut field lines at maximum rate.
(c) The peak e.m.f. doubles and the frequency doubles.
B1 peak e.m.f. doubles B1 frequency doubles.
(d) They keep the coil connected to the external circuit through the brushes while it rotates, without twisting the wires.
B1 continuous connection between rotating coil and external circuit.
Q7Electromagnetic effects · Challenge · Paper 4 style · 7 marks
A transformer connected to the 230 V a.c. mains supplies a 12 V, 36 W lamp in a model railway set. The lamp works normally.
(a)Calculate the current in the lamp.[1]
(b)Calculate the current in the primary coil. Assume the transformer is 100% efficient.[2]
(c)The primary coil has 1150 turns. Calculate the number of turns on the secondary coil.[2]
(d)Explain why the transformer would not work if connected to a 230 V d.c. supply.[2]
Show answer
(a) 3.0 A
I = P/V = 36 ÷ 12 = 3.0 A B1.
(b) 0.16 A
VpIp = VsIs C1; Ip = 36 ÷ 230 = 0.16 A A1.
(c) 60
Ns = 1150 × 12 ÷ 230 C1; = 60 A1.
(d) A direct current produces a constant magnetic field in the core, so the field linking the secondary coil does not change and no e.m.f. is induced.
B1 d.c. gives a steady / unchanging magnetic field B1 induction needs a changing field – no e.m.f. in secondary.
Q8Circuits & components · Core · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) B 2.0 Ω
1/R = 1/6 + 1/3 = 1/2, so R = 2.0 Ω (less than the smallest resistor). D is the series total; A is 1/R not inverted.
Q9Circuits & components · Core · Paper 4 style · 6 marks
A 9.0 V battery is connected in series with a 4.0 Ω resistor and an 8.0 Ω resistor.
(a)Calculate the combined resistance of the two resistors.[1]
(b)Calculate the current in the circuit.[2]
(c)Calculate the potential difference across each resistor.[2]
(d)State how the current in the 4.0 Ω resistor compares with the current in the 8.0 Ω resistor.[1]
Show answer
(a) 12 Ω
R = 4.0 + 8.0 = 12 Ω B1.
(b) 0.75 A
I = V/R C1; = 9.0 ÷ 12 = 0.75 A A1.
(c) 3.0 V across 4.0 Ω; 6.0 V across 8.0 Ω
V = IR: 0.75 × 4.0 = 3.0 V B1; 0.75 × 8.0 = 6.0 V B1 (sum = 9.0 V).
(d) It is the same (0.75 A).
B1 current is the same everywhere in a series circuit.
Q10Circuits & components · Standard · Paper 4 style · 8 marks
A 12 V battery is connected to a 2.0 Ω resistor in series with two lamps that are in parallel with each other. One lamp has resistance 12 Ω and the other 6.0 Ω. Treat the lamp resistances as constant.
(a)Calculate the combined resistance of the two lamps in parallel.[2]
(b)Calculate the current from the battery.[2]
(c)Calculate the current in each lamp.[2]
(d)The 12 Ω lamp breaks. State and explain what happens to the brightness of the 6.0 Ω lamp.[2]
Show answer
(a) 4.0 Ω
1/R = 1/12 + 1/6 C1; R = 4.0 Ω A1.
(b) 2.0 A
total R = 2.0 + 4.0 = 6.0 Ω C1; I = 12 ÷ 6.0 = 2.0 A A1.
(c) 12 Ω lamp: 0.67 A; 6.0 Ω lamp: 1.33 A
p.d. across lamps = 12 − 2.0 × 2.0 = 8.0 V C1; 8.0 ÷ 12 = 0.67 A and 8.0 ÷ 6.0 = 1.33 A A1.
(d) It gets brighter: the total resistance increases to 8.0 Ω, the current falls to 1.5 A, so less p.d. is across the 2.0 Ω resistor and the p.d. across the lamp rises from 8.0 V to 9.0 V.
B1 brighter B1 resistance of parallel part increases, so a larger share of the 12 V (9.0 V) is across the lamp.
Q11Circuits & components · Standard · Paper 4 style · 6 marks
A garden light in Islamabad is controlled by a potential divider. A 2.0 kΩ fixed resistor and a light-dependent resistor (LDR) are connected in series across a 6.0 V supply. The output is the p.d. across the LDR.
(a)Calculate the output voltage across the LDR at night, when its resistance is 10 kΩ.[2]
(b)Calculate the output voltage across the LDR in daylight, when its resistance is 500 Ω.[2]
(c)The lamp switches on when the output voltage is above 3.0 V. Explain how the circuit switches the lamp on automatically at dusk.[2]
Show answer
(a) 5.0 V
V = 6.0 × 10 ÷ (10 + 2.0) C1; = 5.0 V A1.
(b) 1.2 V
V = 6.0 × 0.50 ÷ (0.50 + 2.0) C1; = 1.2 V A1.
(c) As light fades the resistance of the LDR increases, so it takes a larger share of the 6.0 V; when the p.d. across it rises above 3.0 V the lamp switches on.
B1 LDR resistance increases as light intensity decreases B1 p.d. across LDR increases (above 3.0 V).
Q12Circuits & components · Standard · Paper 4 style · 5 marks
A diode is connected in series with a small lamp and a 6.0 V battery, so that the diode conducts.
(a)The lamp lights. The battery is then reversed. State and explain what happens to the lamp.[2]
(b)The battery is replaced by an alternating (a.c.) supply. Describe the current in the lamp.[2]
(c)State the name given to this use of a diode.[1]
Show answer
(a) The lamp goes out, because the diode now has a very high resistance in this direction and lets (almost) no current pass.
B1 lamp goes out B1 diode conducts in one direction only / very high resistance in reverse.
(b) Current flows in only one direction, during alternate half-cycles of the supply; it is zero for the other half-cycles (half-wave rectification).
B1 current flows in one direction only B1 only for half of each cycle / pulses (rectified).
(c) Rectification
B1 rectification / rectifier.
Q13Circuits & components · Challenge · Paper 4 style · 7 marks
A poultry farmer near Faisalabad uses an automatic fan. A thermistor and a 4.0 kΩ fixed resistor are connected in series across a 9.0 V supply. A relay coil is connected across the fixed resistor; the relay contacts switch a separate mains circuit containing the fan. Assume the relay coil takes negligible current.
(a)Calculate the p.d. across the fixed resistor when the thermistor resistance is 14 kΩ (at 20 °C) and when it is 2.0 kΩ (at 35 °C).[3]
(b)The relay switches on when the p.d. across its coil is 5.0 V. Calculate the resistance of the thermistor at which the fan switches on.[2]
(c)Explain why a relay is used rather than connecting the fan directly into the thermistor circuit.[2]
Show answer
(a) 20 °C: 2.0 V; 35 °C: 6.0 V
C1 V = 9.0 × 4.0 ÷ (4.0 + R_thermistor); A1 at 20 °C: 9.0 × 4.0 ÷ 18 = 2.0 V; A1 at 35 °C: 9.0 × 4.0 ÷ 6.0 = 6.0 V.
(b) 3.2 kΩ
p.d. across thermistor = 9.0 − 5.0 = 4.0 V; current = 5.0 ÷ 4.0 kΩ = 1.25 mA C1; R = 4.0 V ÷ 1.25 mA = 3.2 kΩ A1.
(c) The small current in the low-voltage sensing circuit can switch the large current needed by the mains fan; the relay keeps the high-voltage circuit separate (isolated) from the sensing circuit.
B1 small current/voltage controls a large current/voltage circuit B1 mains circuit isolated from low-voltage circuit / thermistor could not carry fan current.
Q14Circuits & components · Challenge · Paper 4 style · 7 marks
A 6.0 V supply is connected to a 10 Ω resistor in series with a parallel combination of a 20 Ω resistor and a 30 Ω resistor.
(a)Calculate the current from the supply.[3]
(b)Calculate the p.d. across the 30 Ω resistor.[1]
(c)The 30 Ω resistor is removed. Calculate the new p.d. across the 20 Ω resistor and state whether the current from the supply increases or decreases.[3]
Show answer
(a) 0.27 A
parallel part: 1/R = 1/20 + 1/30, R = 12 Ω C1; total = 22 Ω C1; I = 6.0 ÷ 22 = 0.27 A A1.
(b) 3.3 V
V = 0.273 × 12 = 3.3 V B1 (or 6.0 − 0.273 × 10).
(c) 4.0 V; the current decreases (to 0.20 A)
total R = 10 + 20 = 30 Ω C1; I = 0.20 A, decreases B1; V = 0.20 × 20 = 4.0 V A1.
Q15Current, p.d., e.m.f. & resistance · Core · Paper 4 style · 4 marks
A mobile phone is charged with a current of 1.5 A for 2.0 hours.
(a)State what is meant by electric current.[1]
(b)Calculate the charge that flows into the phone.[2]
(c)State the unit of charge in terms of base units.[1]
Show answer
(a) The rate of flow of charge.
B1 charge passing a point per unit time (I = Q/t).
(b) 10800 C
Q = It C1; = 1.5 × 2.0 × 3600 = 10 800 C A1.
(c) ampere second (A s)
B1 A s (1 C = 1 A s).
Q16Current, p.d., e.m.f. & resistance · Standard · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) B 2.0 Ω
R ∝ L/A. Doubling the diameter makes the cross-sectional area 4 times larger, so R = 4.0 × 2 ÷ 4 = 2.0 Ω. C comes from thinking the area only doubles.
Q17Current, p.d., e.m.f. & resistance · Standard · Paper 4 style · 6 marks
A torch battery of e.m.f. 1.5 V is connected in series with a lamp and a small resistor. Assume the battery itself has no resistance.
(a)Define the electromotive force (e.m.f.) of a source.[2]
(b)Calculate the energy transferred by the battery when 400 C of charge flows round the circuit.[2]
(c)The p.d. across the lamp is 1.2 V. Calculate the p.d. across the resistor and the energy transferred in the resistor when 400 C flows.[2]
Show answer
(a) The electrical work done by the source in moving a unit charge around a complete circuit.
B1 work done / energy transferred by the source B1 per unit charge, around the complete circuit.
(b) 600 J
W = QE C1; = 400 × 1.5 = 600 J A1.
(c) 0.30 V; 120 J
p.d. = 1.5 − 1.2 = 0.30 V B1; energy = QV = 400 × 0.30 = 120 J B1.
Q18Current, p.d., e.m.f. & resistance · Challenge · Paper 4 style · 6 marks
The current in a 6 V filament lamp is measured at different values of potential difference.p.d. / V: 0, 1.0, 2.0, 4.0, 6.0current / A: 0, 0.20, 0.32, 0.46, 0.55
(a)Calculate the resistance of the lamp at 2.0 V and at 6.0 V.[2]
(b)Describe how the resistance of the filament changes as the p.d. increases, and explain why.[3]
(c)Describe how the I–V graph for a fixed resistor at constant temperature would differ from the graph for this lamp.[1]
(b) The resistance increases. A larger current heats the filament; at higher temperature the metal ions vibrate more, so the free electrons collide with them more often.
B1 resistance increases B1 filament temperature increases B1 more frequent collisions of electrons with vibrating ions.
(c) It would be a straight line through the origin (current proportional to p.d.), not a curve that gets less steep.
B1 straight line through origin.
Q19Electrical energy & power · Core · Paper 4 style · 5 marks
A 1.5 kW air conditioner in a house in Hyderabad runs for 8 hours a day during a 30-day month.
(a)State what is meant by one kilowatt-hour (kW h).[1]
(b)Calculate the energy used by the air conditioner in 30 days, in kW h.[2]
(c)The electricity company charges Rs 62 per kW h. Calculate the cost of running the air conditioner for the 30 days.[1]
(d)Calculate the energy used in 30 days in joules.[1]
Show answer
(a) The energy transferred by a 1 kW appliance in 1 hour (3.6 × 10⁶ J).
B1 energy used by 1 kW in 1 h.
(b) 360 kW h
E = Pt = 1.5 × 8 × 30 C1; = 360 kW h A1.
(c) Rs 22,320
360 × 62 = Rs 22 320 B1.
(d) 1.3 × 10⁹ J
360 × 3.6 × 10⁶ = 1.3 × 10⁹ J B1.
Q20Electrical energy & power · Standard · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) D 720 J
E = IVt = 0.50 × 12 × 120 s = 720 J. B uses t in minutes (2.0) instead of seconds.
Q21Electrical energy & power · Challenge · Paper 4 style · 6 marks
A small electric motor in a school lift model is connected to a 12 V supply and takes a current of 2.5 A. In 8.0 s it raises a 6.0 kg load through a height of 2.4 m.
(a)Calculate the electrical energy supplied to the motor.[2]
(b)Calculate the gain in gravitational potential energy of the load. Take g = 9.8 N/kg.[1]
(c)Calculate the efficiency of the motor system.[2]
(d) As thermal energy in the motor coils (resistance) or by friction; also as sound.
B1 heating of wires / friction / sound.
Q22Magnetism · Core · Paper 4 style · 4 marks
A student tests three bars, P, Q and R, by bringing each end close to the north pole and then the south pole of a bar magnet.P: both ends are attracted by both poles.Q: one end is repelled by the north pole and attracted by the south pole.R: no force is felt with either pole.
(a)Identify which bar is a permanent magnet, which is an unmagnetised magnetic material, and which is non-magnetic. Give a reason for each.[3]
(b)Explain why attraction alone cannot prove that a bar is a magnet.[1]
Show answer
(a) Bar Q is a magnet (only a magnet can repel a magnet). Bar P is unmagnetised magnetic material (attracted by both poles). Bar R is non-magnetic (no force).
B1 Q magnet – repulsion B1 P magnetic but unmagnetised – attracted by both poles B1 R non-magnetic – no effect.
(b) A magnet also attracts unmagnetised magnetic materials (by inducing magnetism in them), so only repulsion is a sure test.
B1 magnetic materials are attracted too (induced magnetism); repulsion is the only test.
Q23Magnetism · Standard · Paper 2 style · 1 mark
(a)An electromagnet in a scrapyard crane must lift steel and then drop it as soon as the current is switched off. Which material should be used for its core, and why?[1]
Show answer
(a) B soft iron, because it is easily magnetised and demagnetised
Soft iron is a temporary magnetic material: it magnetises strongly and loses its magnetism when the current stops. Steel (A) keeps its magnetism, so the load would not drop. Copper and aluminium are non-magnetic.
Q24Magnetism · Challenge · Paper 4 style · 6 marks
Two identical bar magnets lie in line on a bench with their north poles facing each other, 10 cm apart.
(a)Describe the magnetic field pattern in the region between the two north poles.[2]
(b)State what is meant by the direction of a magnetic field at a point.[1]
(c)A small plotting compass is placed exactly halfway between the two north poles. Ignoring the Earth's field, state and explain how the compass behaves.[2]
(d)State how a field-line diagram shows where the magnetic field is strongest.[1]
Show answer
(a) Field lines leave each north pole and curve away sideways, away from the other pole; the lines from the two magnets do not cross, and there is a point midway between the poles where there are no lines.
B1 lines from each N pole curve away from each other (repulsion pattern) B1 neutral point midway with no field lines / lines never cross.
(b) The direction of the force on a north pole placed at that point.
B1 direction of force on a (free) N pole.
(c) It does not point in any particular direction, because the two magnets' fields cancel at this neutral point, so there is no resultant force on its needle.
B1 no fixed direction / needle not deflected B1 fields cancel – neutral point (zero field).
(d) The field lines are closest together where the field is strongest.
B1 lines closer together = stronger field.
Q25Electric charge & static · Core · Paper 2 style · 1 mark
(a)[1]
Show answer
(a) A Electrons have moved from the cloth to the rod.
Only electrons move when insulators are charged by friction; the rod gains electrons, so it becomes negative. Protons are fixed in nuclei (B), and charge is never created (D).
Q26Electric charge & static · Standard · Paper 4 style · 6 marks
(a)State what is meant by an electric field.[1]
(b)State how the direction of an electric field is defined.[1]
(c)Describe the electric field pattern around an isolated, positively charged metal sphere.[2]
(d)Two parallel metal plates are connected to a high-voltage supply; the top plate is positive and the bottom plate is negative. Describe the field between the plates, away from the edges.[2]
Show answer
(a) A region in which an electric charge experiences a force.
B1 region where a charge experiences a force.
(b) The direction of the force on a positive charge placed in the field.
B1 direction of force on a (small) positive charge.
(c) Field lines are radial (spread out from the centre like spokes), pointing outwards away from the sphere; they get further apart with distance.
B1 radial lines starting at the surface B1 arrows pointing outwards.
(d) Uniform field: straight, parallel, equally spaced lines from the top (positive) plate to the bottom (negative) plate.
B1 parallel, equally spaced straight lines (uniform) B1 directed from positive (top) to negative (bottom) plate.
Q27Electric charge & static · Challenge · Paper 4 style · 6 marks
An aircraft lands at Lahore airport after a flight through dry air and must be refuelled.
(a)Explain how the aircraft becomes charged while flying through dry air.[2]
(b)Explain why a conducting cable connects the aircraft to the ground before refuelling begins.[2]
(c)A charge of 4.5 µC flows to earth in 3.0 ms. Calculate the average current.[2]
Show answer
(a) Friction between the aircraft surface and the air (or dust/ice particles) transfers electrons, so the aircraft gains or loses electrons and becomes charged.
B1 friction / rubbing with air or particles B1 electrons transferred to or from the aircraft.
(b) Charge (electrons) flows through the conducting cable to or from earth, so the aircraft is discharged; this prevents a spark that could ignite the fuel vapour.
I = Q/t C1; = 4.5 × 10⁻⁶ ÷ 3.0 × 10⁻³ = 1.5 × 10⁻³ A A1.
Q28Electrical safety · Core · Paper 4 style · 6 marks
A hairdryer is rated 1.8 kW, 230 V.
(a)Calculate the normal current in the hairdryer.[2]
(b)Fuses rated at 3 A, 5 A and 13 A are available. State which fuse should be used and explain why.[2]
(c)The hairdryer has a plastic case and is double insulated. Explain why it does not need an earth wire.[2]
Show answer
(a) 7.8 A
I = P/V C1; = 1800 ÷ 230 = 7.8 A A1.
(b) 13 A: it is the lowest rating above the normal current of 7.8 A, so it does not blow in normal use but melts if the current becomes dangerously large.
B1 13 A B1 just above normal current – lower ratings would blow in normal use.
(c) The outer case is an insulator and no live part can touch it, so the case can never become live and give a shock.
B1 case is an insulator / not a conductor B1 cannot become live, so no earth path is needed.
Q29Digital electronics (logic gates, 2022 only) · Standard · Paper 4 style · 4 marks
A security lamp outside a shop in Rawalpindi must switch on only when it is dark AND movement is detected. A light sensor gives output 1 when it is dark; a motion sensor gives output 1 when movement is detected.
(a)Name the single logic gate that should be used.[1]
(b)Write the truth table for this gate, with inputs A (light sensor) and B (motion sensor) and output Q (lamp).[2]
(c)The owner now wants the lamp to come on when it is dark OR when a test switch C is pressed (C gives 1 when pressed), but the motion sensor is no longer used. Name the gate needed.[1]
Show answer
(a) AND gate
B1 AND.
(b) A B Q: 0 0 0; 0 1 0; 1 0 0; 1 1 1
B1 all four input combinations listed B1 outputs correct (only 1 1 gives 1).
(c) OR gate
B1 OR.
Q30Digital electronics (logic gates, 2022 only) · Challenge · Paper 4 style · 6 marks
Input A is connected to a NOT gate. The output of the NOT gate and input B are the two inputs of an OR gate, whose output is Q.
(a)Complete the truth table for this circuit for all four combinations of A and B.[3]
(b)State the only combination of inputs that makes Q = 0.[1]
(c)An extra NOT gate is added after the OR gate. State the new output for A = 1, B = 0, and describe in words when the new output is 1.[2]
Show answer
(a) A B Q: 0 0 1; 0 1 1; 1 0 0; 1 1 1
B1 NOT A column correct (1, 1, 0, 0) B1 first three outputs correct B1 all four correct.
(b) A = 1 and B = 0
B1 A = 1, B = 0.
(c) Output 1; the new output is 1 only when A = 1 and B = 0.
B1 output = 1 B1 output is 1 only when A is 1 and B is 0 (inverse of previous table).
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).