ExaminerPrep › Question bank › IGCSE Physics (Edexcel) › 2 Electricity
International GCSE Physics · Pearson Edexcel 4PH1 · Topic 2
363 free, high-quality past-paper-style questions on Electricity for Edexcel IGCSE Physics, each with a full mark scheme. Covers: (b) Mains Electricity; (c) Energy and Voltage in Circuits; (d) Electric Charge. For when you have finished the past papers, or want to practise one topic at a time.Notes 2(b) →Notes 2(c) →Notes 2(d) →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
A wireless doorbell chime is connected to a battery. A charge of 36 C flows through the chime circuit in 1.5 minutes.
What is the average current in the circuit?
| Marking point | Mark | Notes |
|---|---|---|
| Use charge = current × time, so current = charge ÷ time. The time must be in seconds: 1.5 minutes = 90 s. Current = 36 ÷ 90 = 0.40 A. 24 A comes from dividing by 1.5 without converting minutes to seconds. 54 A comes from multiplying charge by time in minutes. 3240 A comes from multiplying charge by time in seconds. | ✓ 1 | Answer A |
Answer: A
Common error: Always convert time into seconds before using Q = I × t. Minutes or hours give an answer that is wrong by a factor of 60 or 3600.
A student is testing a length of conductive thread that will be sewn into a jacket to connect small LED lights. She wants to find the resistance of the thread using a battery, an ammeter and a voltmeter.
State how the ammeter and the voltmeter should be connected in the circuit.
State the equation linking voltage, current and resistance.
The voltmeter reads 4.5 V and the ammeter reads 0.18 A.
Calculate the resistance of the thread.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The ammeter is connected in series with the thread | ✓ 1 | |
| The voltmeter is connected in parallel with the thread, across its two ends | ✓ 1 | |
| Part (b) | ||
| voltage = current × resistance, or V = I × R | ✓ 1 | |
| Part (c) | ||
| R = V ÷ I = 4.5 ÷ 0.18 | ✓ 1 | |
| R = 25 Ω | ✓ 1 | |
Common error: Do not put the voltmeter in series. A voltmeter measures the voltage across a component, so its leads must connect to both ends of that component.
A heart pacemaker draws a steady current of 10 μA from its battery.
How much charge flows from the battery in 1 hour?
| Marking point | Mark | Notes |
|---|---|---|
| Use Q = I × t with SI units. The current is 10 μA = 10 × 10−6 A and the time is 1 hour = 3600 s. Q = 10 × 10−6 × 3600 = 0.036 C. The option 6.0 × 10−4 C uses 60 s (minutes, not seconds). The option 36 C treats μA as mA. The option 2.8 × 10−9 C divides current by time instead of multiplying. | ✓ 1 | Answer C |
Answer: C
Common error: Convert both quantities to SI units before substituting: μ means ×10⁻⁶, and time must be in seconds.
An espresso machine in a café has a power rating of 1.5 kW and is connected to the 230 V mains supply.
Which fuse is the most suitable for the plug of this machine?
| Marking point | Mark | Notes |
|---|---|---|
| The normal current is I = P / V = 1500 / 230 = 6.5 A. A fuse should be rated slightly above the normal working current, so 10 A is correct. A 3 A or 5 A fuse is below 6.5 A and would melt during normal use. A 13 A fuse would let the current rise to nearly twice the normal value before melting, so it gives poorer protection. | ✓ 1 | Answer C |
Answer: C
Exam tip: Always calculate the normal current first. Then choose the smallest fuse rating that is still above it.
A family goes camping. They take a battery-powered torch and also use a mains hook-up point at the campsite, which provides alternating current (ac).
Describe the difference between alternating current (ac) and direct current (dc).
State which type of current the battery-powered torch uses.
The mains supply has a frequency of 50 Hz.
Calculate the time taken for one complete cycle of the alternating current.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Direct current flows in one direction only . | ✓ 1 | |
| Alternating current repeatedly reverses its direction . | ✓ 1 | |
| Part (b) | ||
| Direct current / dc | ✓ 1 | |
| Part (c) | ||
| T = 1 / f = 1 / 50 | ✓ 1 | |
| T = 0.02 s | ✓ 1 | |
Exam tip: For 'describe the difference' questions, say something about BOTH types of current. Describing only one of them usually earns only one mark.
In a hospital, a blood-warming unit warms donated blood before a transfusion. It is connected to the 230 V mains supply and draws a current of 1.8 A.
State the equation linking power, current and voltage.
Fuses of 3 A, 5 A and 13 A are available.
Suggest which fuse should be fitted. Give a reason for your answer.
Calculate the power of the blood-warming unit.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Power = current × voltage, P = I × V | ✓ 1 | |
| Part (b) | ||
| 3 A | ✓ 1 | |
| It is just above the normal current of 1.8 A, so it would melt quickly if a fault caused a larger current . | ✓ 1 | |
| Part (c) | ||
| P = 1.8 × 230 | ✓ 1 | |
| P = 414 W | ✓ 1 | |
Exam tip: When asked to choose a fuse, compare its rating with the normal current. Do not compare it with the power.
A treadmill in a gym is connected to the mains supply. The building's wiring is protected by circuit breakers, and the treadmill's plug also contains a fuse.
State the approximate voltage of the UK mains supply.
A fault in the treadmill motor causes a very large current.
Describe how the fuse protects the treadmill and its wiring.
Give two advantages of a circuit breaker compared with a fuse.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 230 V | ✓ 1 | |
| Part (b) | ||
| The current is larger than the fuse rating, so the fuse wire heats up and melts . | ✓ 1 | |
| This breaks the circuit and disconnects the live supply, so the wiring cannot overheat or catch fire . | ✓ 1 | |
| Part (c) | ||
| Any two different valid advantages, 1 mark each. Acceptable answers: it can be reset and does not need replacing; it switches off faster; it can be more sensitive to small changes in current. | — | |
| First valid advantage . | ✓ 1 | |
| Second, different valid advantage . | ✓ 1 | |
Common error: Writing only that a fuse 'blows' does not earn the description mark. Explain that the fuse wire heats up and melts, which breaks the circuit.
A heart pacemaker contains a battery with a voltage of 2.8 V. During one year, 300 C of charge passes through the pacemaker circuit.
How much energy does the battery transfer to the circuit in this year?
| Marking point | Mark | Notes |
|---|---|---|
| Energy transferred = charge × voltage, so E = 300 × 2.8 = 840 J. A voltage of 2.8 V means 2.8 J is transferred for every coulomb of charge, and 300 C flows. 107 J comes from dividing charge by voltage. 0.0093 J comes from dividing voltage by charge. 2.8 J treats the voltage as the total energy rather than the energy per coulomb. | ✓ 1 | Answer A |
Answer: A
Exam tip: Read voltage as 'joules per coulomb'. Multiplying by the number of coulombs then gives the total energy.
A hearing aid is powered by a small button cell with a voltage of 1.4 V. When the hearing aid is switched on, the current in it is 1.2 mA.
Describe what is meant by a voltage of 1.4 V.
Calculate the charge that flows through the hearing aid in 1.0 hour.
Calculate the energy transferred by the cell in 1.0 hour.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| each coulomb of charge passing through the cell | ✓ 1 | |
| is given / transfers 1.4 J of energy | ✓ 1 | |
| Part (b) | ||
| Q = I × t = 1.2×10−3 × 3600 | ✓ 1 | |
| = 4.3 C (4.32 C) | ✓ 1 | |
| Part (c) | ||
| E = Q × V = 4.32 × 1.4 | ✓ 1 | |
| = 6.0 J | ✓ 1 | |
Common error: 1.2 mA is 1.2×10⁻³ A, not 1.2 A. Convert milliamps to amps before you calculate.
At a music festival, three identical strings of decorative lights are connected in parallel to a 12 V power supply.
State the voltage across each string of lights.
The current in each string is 0.50 A.
Calculate the current supplied by the power supply.
Calculate the resistance of one string of lights.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 12 V | ✓ 1 | |
| Part (b) | ||
| 3 × 0.50 = 1.5 A | ✓ 1 | |
| Part (c) | ||
| R = V ÷ I = 12 ÷ 0.50 | ✓ 1 | |
| = 24 Ω | ✓ 1 | |
Exam tip: In parallel, every branch has the full supply voltage, and the branch currents add up to give the supply current.
A camera drone is used to film a cycling race. The drone's motors have a total power of 120 W, and the drone flies for 15 minutes.
State the equation linking energy transferred, power and time.
Calculate the energy transferred to the motors during the flight.
The drone's battery has a voltage of 15 V.
Calculate the charge that flows from the battery during the flight.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy transferred = power × time (E = P × t) | ✓ 1 | |
| Part (b) | ||
| t = 15 × 60 = 900 s | ✓ 1 | |
| E = 120 × 900 = 1.1×105 J (108 000 J) | ✓ 1 | |
| Part (c) | ||
| Q = E ÷ V = 108 000 ÷ 15 | ✓ 1 | |
| = 7200 C | ✓ 1 | |
Exam tip: Rearrange E = Q × V to Q = E ÷ V. Use your unrounded answer from (b) so that rounding errors do not build up.
An electrician is fitting a fuse into the plug of a pressure washer used to clean sports-ground seating.
Which wire in the plug must the fuse be connected to?
| Marking point | Mark | Notes |
|---|---|---|
| The fuse goes in the live wire, because the live wire is at a high voltage. When the fuse melts it cuts the appliance off from that high voltage, so no part of the appliance stays live. A fuse in the neutral wire would break the circuit but leave the appliance connected to the live supply. The earth wire normally carries no current, so a fuse there would never melt in normal use and would stop the earth giving any protection. | ✓ 1 | Answer C |
Answer: C
Common error: Breaking the neutral wire stops the current, but the appliance can still be live. Safety devices must always disconnect the live wire.
A coffee machine in a hospital staff room has a power of 1.2 kW. It takes 45 s to make one cup of coffee.
How much energy does the coffee machine transfer in this time?
| Marking point | Mark | Notes |
|---|---|---|
| Use energy transferred = power × time, with power in watts and time in seconds. E = 1200 W × 45 s = 54 000 J. 54 J comes from leaving the power in kW. 900 J comes from wrongly changing 45 s into minutes. 3 240 000 J comes from treating 45 s as 45 minutes. | ✓ 1 | Answer C |
Answer: C
Exam tip: Change kW to W (× 1000) and minutes to seconds (× 60) before you substitute into E = P × t.
A gardener is fitting a new three-pin plug to the cable of an electric lawnmower.
State the colour of the insulation on each of these wires:
(i) live
(ii) neutral
(iii) earth
State the function of the neutral wire.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (i) brown | ✓ 1 | |
| (ii) blue | ✓ 1 | |
| (iii) green and yellow stripes | ✓ 1 | |
| Part (b) | ||
| Any one of: it completes the circuit / it carries the current back to the supply / it provides the return path at (about) 0 V | ✓ 1 | |
Common error: The earth wire is green and yellow stripes, not just green. In normal use the neutral wire carries the same current as the live wire, but it is at about 0 V.
A tropical fish tank uses an electric heater that is connected to the 230 V mains supply. The power of the heater is 150 W.
State the colour of the insulation on the live wire in a three-pin plug.
Calculate the current in the heater when it is working normally.
Fuses with ratings of 1 A, 3 A and 13 A are available.
State which fuse should be fitted to the heater's plug. Explain your choice.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Brown | ✓ 1 | |
| Part (b) | ||
| Uses I = P ÷ V, so I = 150 ÷ 230 | ✓ 1 | |
| I = 0.65 A | ✓ 1 | |
| Part (c) | ||
| 1 A fuse | ✓ 1 | |
| Its rating is just above the normal current of 0.65 A, so the fuse does not blow in normal use but melts quickly if the current rises in a fault. The 3 A and 13 A fuses would allow a dangerously large current before melting | ✓ 1 | |
Exam tip: Choose the smallest fuse rating that is still above the normal working current. Justify your choice by stating both the calculated current and the chosen rating.
An electric bike battery is recharged using a charger. The charger supplies a current of 2.0 A at a voltage of 42 V for 3.5 hours.
State the equation linking energy transferred, current, voltage and time.
Calculate the energy transferred to the battery in this time.
Give your answer in joules.
The charger is plugged into the mains, which supplies alternating current (a.c.). The charger supplies direct current (d.c.) to the battery.
State the difference between alternating current and direct current.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy transferred = current × voltage × time, or E = I × V × t | ✓ 1 | |
| Part (b) | ||
| Converts the time: 3.5 h = 3.5 × 3600 = 12 600 s | ✓ 1 | |
| Substitutes: E = 2.0 × 42 × 12 600 | ✓ 1 | |
| E = 1.06 × 106 J, accept 1 058 400 J | ✓ 1 | |
| Part (c) | ||
| Alternating current repeatedly changes direction, while direct current flows in one direction only | ✓ 1 | |
Common error: If you leave the time in hours, your energy will be 3600 times too small. Convert to seconds before you substitute.
A small underwater survey drone is powered by a 14.8 V rechargeable battery. During one dive, the motor draws a steady current of 1.8 A for 25 minutes.
State what is meant by an electric current.
Calculate the charge that flows through the motor during the dive.
Calculate the energy transferred by this charge.
Use the equation energy transferred = charge × voltage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| An electric current is the rate of flow of charge | ✓ 1 | |
| Part (b) | ||
| Uses Q = I × t with t = 25 × 60 = 1500 s, so Q = 1.8 × 1500 | ✓ 1 | |
| Q = 2700 C | ✓ 1 | |
| Part (c) | ||
| E = 2700 × 14.8 | ✓ 1 | |
| E = 40 000 J, accept 39 960 J or 4.0 × 104 J | ✓ 1 | |
Exam tip: In a multi-step calculation, carry forward your answer from the earlier part. You can still earn the later marks even if the earlier value was wrong.
In a library, a 2.4 A current in the main cable reaches a junction and splits into three parallel lighting circuits. The currents in two of the circuits are 0.9 A and 0.7 A. What is the current in the third circuit?
| Marking point | Mark | Notes |
|---|---|---|
| The currents leaving a junction add up to the current entering it: 2.4 − 0.9 − 0.7 = 0.8 A, so B is correct. A (1.6 A) is the sum of the two known currents; C (1.5 A) subtracts only the 0.9 A; D (4.0 A) adds all the currents instead of subtracting. | ✓ 1 | Answer B |
Answer: B
Common error: At a junction in a parallel circuit, the current entering equals the total current leaving; do not add the entering current to the branch currents.
At a wind farm, a technician connects a 12 V supply across three parallel heating elements of resistance 40 Ω, 60 Ω and 120 Ω.
State the potential difference across the 60 Ω element.
Calculate the current in the 60 Ω element.
The currents in the 40 Ω and 120 Ω elements are 0.30 A and 0.10 A. Calculate the current drawn from the supply, using your answer to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 12 V | ✓ 1 | |
| Part (b) | ||
| I = V/R = 12 ÷ 60 | ✓ 1 | |
| 0.20 A | ✓ 1 | |
| Part (c) | ||
| current from supply = sum of the currents in the branches | ✓ 1 | |
| 0.30 + 0.20 + 0.10 (ECF from (b)) | ✓ 1 | |
| 0.60 A | ✓ 1 | |
Common error: In a parallel circuit each branch has the same pd as the supply; the branch currents add to give the supply current.
A kettle in a swimming pool changing room is labelled 230 V, 2300 W. The kettle has a metal case.
Calculate the current in the kettle during normal use.
The plug can be fitted with a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason.
Explain how the earth wire and the fuse protect a person if the live wire touches the metal case.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| power = current × pd, I = P/V | ✓ 1 | |
| I = 2300 ÷ 230 = 10 A | ✓ 1 | |
| Part (b) | ||
| 13 A | ✓ 1 | |
| its rating is just above the normal current of 10 A, so it does not melt in normal use but does if a larger current flows | ✓ 1 | |
| Part (c) | ||
| current flows through the (low resistance) earth wire | ✓ 1 | |
| the large current makes the fuse melt (blow) | ✓ 1 | |
| the appliance is disconnected from the live supply so the case is not live | ✓ 1 | |
Exam tip: Choose the fuse whose rating is just above the normal current, not the largest one available.
A library has an emergency lamp powered by a 12 V battery. The current in the lamp is 0.25 A.
State what is meant by an electric current.
Calculate the charge that passes through the lamp in 2.0 hours.
Calculate the power of the lamp.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the rate of flow of charge | ✓ 1 | |
| Part (b) | ||
| time = 2.0 × 3600 = 7200 s | ✓ 1 | |
| Q = It = 0.25 × 7200 | ✓ 1 | |
| 1800 C | ✓ 1 | |
| Part (c) | ||
| P = IV | ✓ 1 | |
| 0.25 × 12 = 3.0 W | ✓ 1 | |
Common error: Change hours to seconds before using Q = It; leaving the time in hours gives a charge in the wrong units.
A cable carrying 10.0 A splits at a junction in a data centre into three parallel branches. Two of the branches carry 4.0 A and 2.5 A. What is the current in the third branch?
| Marking point | Mark | Notes |
|---|---|---|
| Current entering a junction equals current leaving: 10.0 − 4.0 − 2.5 = 3.5 A, so B is correct. A subtracts the two known currents from each other (4.0 − 2.5). C adds the two known currents and stops. D adds all three numbers together. | ✓ 1 | Answer B |
Answer: B
Common error: Charge is conserved at a junction, so the currents leaving must add up to the current entering.
An archaeologist rubs a plastic brush handle with a dry cloth. The handle becomes negatively charged. Which statement describes what has happened?
| Marking point | Mark | Notes |
|---|---|---|
| A body becomes negatively charged by gaining electrons. Electrons are transferred by friction from the cloth to the handle, so C is correct. B has the electrons moving the wrong way, which would leave the brush positive. A and D are wrong because protons are fixed in the nuclei and do not move when objects are rubbed. | ✓ 1 | Answer C |
Answer: C
Common error: Only electrons transfer when objects are charged by friction. Negative charge means a gain of electrons.
An archaeology team uses a 12 V battery to power a floodlight and a radio. The two are connected in parallel. The floodlight takes 1.5 A and the radio takes 0.50 A.
State the current in the battery.
Calculate the charge that flows through the battery in 10 minutes.
A second identical floodlight is connected in parallel with the first. State the new current in the battery and give a reason for your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2.0 A | ✓ 1 | |
| Part (b) | ||
| Q = I t = 2.0 × 600 (ECF from (a)) | ✓ 1 | |
| 1200 C | ✓ 1 | |
| Part (c) | ||
| 3.5 A | ✓ 1 | |
| Each parallel branch still takes its own current / the currents in the branches add at the junction | ✓ 1 | |
Common error: In a parallel circuit each branch keeps the same p.d., so adding a branch adds its current to the total.
An archaeology team uses a 230 V mains-powered floodlight rated at 920 W at a dig site.
Calculate the current in the floodlight when it works normally.
The plug can be fitted with a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason.
The lamp has a metal case, so its plug has an earth wire connected to the case. Explain how the earth wire makes the lamp safe if the live wire touches the case.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P ÷ V = 920 ÷ 230 | ✓ 1 | |
| 4.0 A | ✓ 1 | |
| Part (b) | ||
| 5 A | ✓ 1 | |
| It is just above the normal current of 4.0 A, so it does not blow normally but melts if the current becomes too large | ✓ 1 | |
| Part (c) | ||
| A large current flows through the earth wire to earth | ✓ 1 | |
| The fuse melts and disconnects the lamp so the case cannot stay live | ✓ 1 | |
Exam tip: Choose the fuse with the smallest rating that is still above the normal operating current.
A bedside monitor in a hospital contains an LED indicator in series with a resistor. The circuit is connected to a low-voltage a.c. supply. Which statement describes the LED?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: an LED conducts in one direction only, so it carries current only in the half-cycles that make it forward-biased. A is wrong because a.c. reverses direction regularly. C is wrong because the LED still conducts in the forward half-cycles, whatever the average. D is wrong because current in the reverse direction is blocked, not added. | ✓ 1 | Answer B |
Answer: B
Common error: An LED allows current in one direction only, so on a.c. it conducts for roughly half of each cycle.
A portable infusion pump in a hospital runs from a battery that supplies a current of 0.25 A for 3.0 hours. How much charge passes through the pump in this time?
| Marking point | Mark | Notes |
|---|---|---|
| Q = It with t = 3.0 × 3600 = 10 800 s, so Q = 0.25 × 10 800 = 2700 C. A comes from using t = 3 without converting hours to seconds. B converts hours to minutes (180) instead of seconds. D divides t by I instead of multiplying. | ✓ 1 | Answer C |
Answer: C
Common error: Time must be in seconds when using Q = It, so change hours by multiplying by 3600.
A charger supplies a constant current of 1.6 A to a cordless drill battery for 90 minutes until it is full.
State what is meant by electric current.
Calculate the charge that flows into the battery.
The full battery has a p.d. of 18 V. Assume this does not change. Calculate the energy stored in the battery.
The drill motor takes a current of 6.0 A from the full battery. Assume all the charge from (b) can be delivered. Calculate how many minutes the drill can run.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 rate of flow of charge | ✓ 1 | |
| Part (b) | ||
| M1 Q = I × t = 1.6 × (90 × 60) | ✓ 1 | |
| A1 8640 C | ✓ 1 | |
| Part (c) | ||
| M1 E = Q × V = 8640 × 18 (ECF from (b)) | ✓ 1 | |
| A1 155 520 J | ✓ 1 | |
| Part (d) | ||
| M1 t = Q ÷ I = 8640 ÷ 6.0 = 1440 s | ✓ 1 | |
| A1 24 minutes | ✓ 1 | |
Exam tip: Link the parts together: charge delivered when discharging equals the charge put in when charging (if no losses).
In a Formula 1 pit lane, a mechanic rubs a plastic fuel-hose cover with a dry cloth and it becomes negatively charged.
Explain how the cover becomes negatively charged.
Explain why a spark near petrol vapour is dangerous.
The refuelling rig is connected to earth by a metal wire. Explain how this reduces the risk of sparks.
A charge leaks to earth with a current of 0.45 μA for 6.0 s. Calculate the charge that flows. Give your answer in μC.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 electrons transfer from the cloth to the plastic | ✓ 1 | |
| B1 the plastic gains an excess of electrons (so has a negative charge) | ✓ 1 | |
| Part (b) | ||
| B1 the spark can ignite the vapour (causing a fire or explosion) | ✓ 1 | |
| Part (c) | ||
| B1 charge flows through the wire to earth | ✓ 1 | |
| B1 so charge cannot build up to a high p.d. (large enough to cause a spark) | ✓ 1 | |
| Part (d) | ||
| M1 Q = I × t = 0.45 × 6.0 | ✓ 1 | |
| A1 2.7 μC | ✓ 1 | |
Common error: Only electrons move when objects are charged by friction; protons stay fixed in the nuclei.
A beekeeper heats a hive with two identical heating pads, each of resistance 6.0 Ω, connected in parallel to a 12 V battery.
Calculate the current in each pad.
State the total current drawn from the battery.
Give one advantage of connecting the pads in parallel rather than in series.
Calculate the total resistance of the circuit.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = V ÷ R = 12 ÷ 6.0 | ✓ 1 | |
| A1 2.0 A | ✓ 1 | |
| Part (b) | ||
| B1 4.0 A (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| B1 if one pad fails the other keeps working (or each pad has the full 12 V p.d.) | ✓ 1 | |
| Part (d) | ||
| M1 R = V ÷ I = 12 ÷ 4.0 | ✓ 1 | |
| A1 3.0 Ω | ✓ 1 | |
Common error: The total resistance of two parallel resistors is smaller than either one, not larger.
A sports scientist monitors a cooling pad using an NTC thermistor connected in series with a cell and an ammeter. The pad is then warmed. Which row describes what happens?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: C. The resistance of an NTC thermistor falls as its temperature rises, and with the same cell p.d. the current (I = V/R) increases. A is wrong because it pairs a rising resistance with a rising current. B treats the thermistor like a metal wire, whose resistance rises with temperature. D has a falling resistance but a falling current, which contradicts I = V/R. | ✓ 1 | Answer C |
Answer: C
Common error: Do not assume every component behaves like a metal wire. For an NTC thermistor, higher temperature means lower resistance.
An electric shower is rated at 9.2 kW when connected to the 230 V mains supply. What current does it draw?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: C. I = P/V = 9200 W / 230 V = 40 A. A comes from inverting the equation (V/P). B forgets that 9.2 kW is 9200 W (9.2/230). D multiplies power and voltage instead of dividing. | ✓ 1 | Answer C |
Answer: C
Common error: Convert kilowatts to watts before using P = IV, and check that your answer is sensible for a household appliance.
A model lighthouse contains a lamp in series with a resistor, connected to a 9.0 V supply. The lamp works normally when the p.d. across it is 6.0 V and the current in it is 0.50 A.
State the p.d. across the resistor when the lamp works normally.
Calculate the resistance of the resistor.
Calculate the charge that passes through the lamp in 2.0 minutes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 3.0 V | ✓ 1 | |
| Part (b) | ||
| R = V / I = 3.0 / 0.50 | ✓ 1 | |
| 6.0 Ω | ✓ 1 | |
| Part (c) | ||
| Q = I t = 0.50 × 120 | ✓ 1 | |
| 60 C | ✓ 1 | |
Exam tip: In a series circuit the p.d. values across the components add up to the supply p.d. and the current is the same everywhere. Convert minutes to seconds before using Q = It.
An astronomy club tests a light sensor for a telescope dome. A lamp is placed at different distances d from an LDR and the resistance R of the LDR is found each time.
d = 20 cm, R = 300 Ω
d = 40 cm, R = 600 Ω
d = 60 cm, R = 870 Ω
d = 80 cm, R = 1210 Ω
d = 100 cm, R = 1500 Ω
Describe how the resistance of the LDR changes as the distance increases.
The LDR at 40 cm is connected across a 3.0 V cell. Calculate the current in the LDR.
Suggest why the experiment is done in a darkened room.
Explain why the resistance of the LDR is greater when the lamp is further away.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The resistance increases as distance increases | ✓ 1 | |
| Part (b) | ||
| I = V / R = 3.0 / 600 | ✓ 1 | |
| 0.0050 A (5.0 mA) | ✓ 1 | |
| Part (c) | ||
| So that light from other sources does not affect the resistance of the LDR | ✓ 1 | |
| Part (d) | ||
| The light intensity at the LDR is lower because the light spreads over a larger area | ✓ 1 | |
| The resistance of an LDR is greater in dim light | ✓ 1 | |
Exam tip: A description of a trend needs a direction (increases or decreases) for the quantity asked about; explanations then need the link between cause and effect.
A fashion designer sews 30 identical small lamps into a stage costume. The lamps are connected in series to a battery pack. One lamp's filament breaks. What happens to the other lamps?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: in a series circuit there is only one path, so a break anywhere stops the current in every lamp. B is wrong because no current flows at all, so no lamp is brighter. C is wrong because current is the same everywhere in the loop, so lamps before the break also stop. D is wrong because a series circuit has no other path (a parallel circuit would). | ✓ 1 | Answer A |
Answer: A
Common error: Do not confuse a break in a series circuit with a break in one branch of a parallel circuit; only the parallel arrangement lets the other branches keep working.
A nylon scarf becomes negatively charged when it is rubbed with a woollen glove. Which statement describes what happened?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: only electrons are free to move between the surfaces, and gaining electrons makes the scarf negative. B and D are wrong because protons are fixed in the nuclei and do not transfer. C has electrons moving the wrong way, which would leave the scarf positive. | ✓ 1 | Answer A |
Answer: A
Common error: Charging by friction always involves electrons moving; a positively charged object has lost electrons, not gained protons.
A fashion boutique decorates its window with a string of 20 identical lamps connected in series to the 230 V mains supply.
Calculate the p.d. across each lamp.
Explain why all the lamps go out if one lamp fails.
Domestic room lighting is wired in parallel. State two advantages of parallel wiring for room lights.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Identical lamps share the supply p.d. equally | ✓ 1 | |
| 230 ÷ 20 = 11.5 V | ✓ 1 | |
| Part (b) | ||
| Series circuit has only one path for the current | ✓ 1 | |
| A broken lamp breaks the circuit, so no current flows in any lamp | ✓ 1 | |
| Part (c) | ||
| Each lamp gets the full mains p.d. | ✓ 1 | |
| If one lamp fails the others keep working (or lamps can be switched independently) | ✓ 1 | |
Exam tip: In a series circuit the supply p.d. is shared; in a parallel circuit each branch has the full supply p.d.
A cordless drill battery needs 4500 C of charge to be fully recharged. The charger supplies a steady current of 2.5 A. How long does the recharging take?
| Marking point | Mark | Notes |
|---|---|---|
| t = Q ÷ I = 4500 ÷ 2.5 = 1800 s, and 1800 s ÷ 60 = 30 min, so 30 min is correct. 1800 min is the time in seconds wrongly labelled as minutes. 187.5 min comes from multiplying Q by I (11 250 s) and then converting. 108 000 min comes from multiplying 1800 by 60 instead of dividing. | ✓ 1 | Answer A |
Answer: A
Common error: Charge, current and time must be in coulombs, amperes and seconds. Divide seconds by 60 to get minutes.
A heated blanket in a mountain refuge is rated at 150 W and is used on the 230 V mains supply.
Calculate the current in the blanket. Give your answer to 2 significant figures.
The plug can be fitted with a 0.5 A, a 3 A or a 13 A fuse. State which fuse should be fitted and explain your choice.
The blanket is switched on for 8.0 hours. Electricity costs €0.30 per kWh. Calculate the cost. Give your answer in euros to 2 decimal places.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Use of P = I V (or I = P ÷ V) | ✓ 1 | |
| I = 150 ÷ 230 = 0.65 A | ✓ 1 | |
| Part (b) | ||
| 3 A fuse | ✓ 1 | |
| The normal current (0.65 A) is below 3 A so the fuse does not melt in normal use, but 3 A is much lower than 13 A so a fault current is cut off sooner (0.5 A would melt in normal use) | ✓ 1 | |
| Part (c) | ||
| Energy = 0.150 kW × 8.0 h = 1.2 kWh | ✓ 1 | |
| Cost = 1.2 × 0.30 = €0.36 | ✓ 1 | |
Common error: Use power in kW when time is in hours to get energy in kWh. Using 150 W with hours does not give kWh.
A cordless drill has a rechargeable battery. A charger supplies a steady current of 1.8 A to the battery for 50 minutes. The potential difference across the battery is 18 V.
State the equation linking charge, current and time.
Calculate the charge that flows into the battery.
Calculate the energy transferred to the battery. Give your answer in kJ to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| charge = current × time (Q = I t) | ✓ 1 | |
| Part (b) | ||
| Time = 50 × 60 = 3000 s | ✓ 1 | |
| Q = 1.8 × 3000 = 5400 C | ✓ 1 | |
| Part (c) | ||
| E = Q V = 5400 × 18 (ECF from (b)) | ✓ 1 | |
| E = 97 200 J = 97.2 kJ | ✓ 1 | |
Common error: Do not forget to convert minutes to seconds before calculating charge.
A positively charged rod attracts a small suspended ball. Which statement about the ball must be correct?
| Marking point | Mark | Notes |
|---|---|---|
| A positive rod attracts a ball that is negative, and also attracts an uncharged ball because the ball's electrons are drawn towards the near side (induction), so the ball is negative or uncharged: C. A is wrong because like charges repel. B is wrong because it ignores induction on a neutral ball. D is wrong because it ignores a negatively charged ball. | ✓ 1 | Answer C |
Answer: C
Common error: Attraction does not prove the second object is charged; only repulsion does.
A cordless drill battery is recharged with a constant current of 1.6 A for 2.5 hours. The charger operates at a potential difference of 20 V.
Calculate the charge that flows through the charger in this time.
Calculate the energy transferred by the charger. Use the value of charge from (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| convert 2.5 h to 9000 s | ✓ 1 | |
| Q = I × t | ✓ 1 | |
| 14400 C | ✓ 1 | |
| Part (b) | ||
| E = Q × V | ✓ 1 | |
| 288000 J (accept 2.88 × 105 J) | ✓ 1 | |
| ECF from (a) | — | |
Common error: Change hours into seconds before using Q = It.
A student rubs a polythene rod with a dry cloth. The rod becomes negatively charged.
Explain, in terms of electrons, why the rod becomes negatively charged.
Two identical rods are charged in the same way and hung close together on threads. The rods move apart. State what this shows about the charges on the rods.
A charged rod is brought near, but not touching, a small uncharged metal sphere hanging on an insulating thread. The sphere moves towards the rod. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| electrons are transferred from the cloth to the rod | ✓ 1 | |
| the rod gains electrons so has an excess of negative charge | ✓ 1 | |
| Part (b) | ||
| the charges are of the same type / both negative (like charges repel) | ✓ 1 | |
| Part (c) | ||
| electrons in the sphere are repelled to the far side | ✓ 1 | |
| the near side becomes positively charged | ✓ 1 | |
| attraction to the near (positive) side is greater than repulsion from the far side because it is closer | ✓ 1 | |
Exam tip: In static electricity questions, say which charges move (electrons) and where they end up.
A power tool has a plastic casing and is marked as double insulated. It has no earth wire. Why is this safe?
| Marking point | Mark | Notes |
|---|---|---|
| Double insulation means there is no accessible metal that could become live, so no earth wire is needed, option D. Plastic is an insulator, not a conductor (A). Insulation, not a circuit breaker, is the reason (B). Earthing depends on whether there is exposed metal, not on power (C). | ✓ 1 | Answer D |
Answer: D
Common error: Double insulation does not mean the appliance has no fuse; the fuse still protects against overload.
An archaeologist cleans a dry pottery fragment with a nylon brush. The brush becomes negatively charged. What has happened?
| Marking point | Mark | Notes |
|---|---|---|
| Only electrons move in rubbing; the brush gained electrons from the pottery, so it is negative, option B. Protons are fixed in nuclei (A, D). If the brush lost electrons it would be positive (C). | ✓ 1 | Answer B |
Answer: B
Common error: Only electrons transfer when objects are rubbed together; protons stay in the nuclei.
An archaeologist rubs a dry nylon brush against a dry pottery fragment to remove dust. The brush becomes charged and then picks up small dust particles.
Explain how the brush becomes negatively charged.
Explain why the charged brush attracts uncharged dust particles.
The brush is touched by an earthed metal rod. State what happens to the brush and why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 friction transfers electrons from the pottery to the brush | ✓ 1 | |
| B1 the brush gains electrons, so it has an excess of negative charge | ✓ 1 | |
| Part (b) | ||
| B1 the brush induces an opposite charge on the near side of each dust particle | ✓ 1 | |
| B1 unlike charges attract, so the dust is pulled towards the brush | ✓ 1 | |
| Part (c) | ||
| B1 it loses its charge (is discharged) | ✓ 1 | |
| B1 electrons flow from the brush to earth through the metal rod | ✓ 1 | |
Common error: Dust is attracted because charge is induced on it, not because it already carries the opposite charge.
A guitar amplifier is connected to the mains supply, which is an alternating current (a.c.) supply. Which statement about a.c. is correct?
| Marking point | Mark | Notes |
|---|---|---|
| The current in an a.c. supply repeatedly reverses direction, so this is the correct statement. A current that flows one way but changes in size is a varying d.c. supply, not a.c. A current that flows one way at a constant size is the steady d.c. from a cell or battery. In a.c. the p.d. reverses as well as the current, so it cannot stay constant while the current reverses. | ✓ 1 | Answer B |
Answer: B
Common error: Do not describe any changing current as a.c.: it is the reversing of direction that makes a current alternating.
A solar-powered pump on a farm carries a current of 0.40 A for 2.5 minutes. How much charge passes through the pump in this time?
| Marking point | Mark | Notes |
|---|---|---|
| Time = 2.5 × 60 = 150 s, so Q = I × t = 0.40 × 150 = 60 C. The value 1.0 C comes from multiplying by 2.5 without converting minutes to seconds. The value 0.0027 C comes from dividing the current by the time. The value 375 C comes from dividing the time by the current. | ✓ 1 | Answer D |
Answer: D
Common error: Convert minutes to seconds before using Q = I t, and check that your answer is the right order of magnitude.
A car seat has a heater made from two heating elements connected in series to the 12 V car battery. The elements have resistances of 1.0 Ω and 3.0 Ω.
Calculate the total resistance of the two elements.
Calculate the current in the 3.0 Ω element.
Calculate the power dissipated in the 3.0 Ω element.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| total resistance = 1.0 + 3.0 = 4.0 Ω | ✓ 1 | |
| Part (b) | ||
| I = V ÷ R | ✓ 1 | |
| = 12 ÷ 4.0 = 3.0 A (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| P = I2R or P = I V with V = 9.0 V | ✓ 1 | |
| P = 3.02 × 3.0 = 27 W (ECF from (b)) | ✓ 1 | |
Common error: Do not put the full 12 V across one element in a series circuit: the same current flows through both, but the p.d. is shared.
A keyboard player uses an amplifier that is plugged into the UK mains supply. The keyboard can also run from a battery.
Describe the difference between alternating current (a.c.) and direct current (d.c.).
State the frequency and the potential difference of the UK mains supply.
Calculate the number of complete cycles of the mains supply in 1 minute.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| d.c. flows in one direction only | ✓ 1 | |
| a.c. repeatedly reverses direction | ✓ 1 | |
| Part (b) | ||
| frequency = 50 Hz | ✓ 1 | |
| potential difference = 230 V | ✓ 1 | |
| Part (c) | ||
| number of cycles = frequency × time | ✓ 1 | |
| = 50 × 60 = 3000 | ✓ 1 | |
Common error: Give the mains frequency in hertz and the p.d. in volts, and convert minutes to seconds when you use a frequency.
A road tanker unloads fuel through a hose into an underground storage tank. The tanker is joined to the ground by a metal cable. What is the purpose of the cable?
| Marking point | Mark | Notes |
|---|---|---|
| The flowing fuel rubs against the hose and the tanker gains charge. The cable is a conductor, so charge leaks safely to earth and no large potential difference (and so no spark) develops near the fuel vapour. Slowing the fuel is a separate precaution and the cable does not do it. Charging the tanker positively would increase the danger, not remove it. The cable cannot stop electrons moving between fuel and hose; it removes the charge that results. | ✓ 1 | Answer C |
Answer: C
Common error: Earthing does not prevent charge being produced by friction; it removes the charge before it can build up.
A light-beam alarm contains a light-dependent resistor (LDR) in a series circuit with a battery. An intruder blocks the beam, so less light reaches the LDR. What happens?
| Marking point | Mark | Notes |
|---|---|---|
| Less light means the resistance of an LDR increases. With the same battery pd, a larger resistance gives a smaller current. The other options pair the wrong change in resistance with the light level, or pair a rise in resistance with a rise in current, which contradicts V = IR. | ✓ 1 | Answer D |
Answer: D
Common error: An LDR has high resistance in the dark and low resistance in bright light, not the other way round.
A fish-processing plant uses a freezer rated at 920 W. The freezer is connected to the 230 V mains supply by a three-core cable and a plug containing a fuse.
State what happens to a fuse when the current in it becomes too large.
Calculate the current in the freezer when it works normally.
Fuses rated 3 A, 5 A and 13 A are available. Explain which one should be fitted.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The wire (filament) in the fuse heats up and melts | ✓ 1 | |
| This breaks the circuit so the current stops | ✓ 1 | |
| Part (b) | ||
| Use of P = IV (or I = P ÷ V) | ✓ 1 | |
| I = 920 ÷ 230 = 4.0 A | ✓ 1 | |
| Part (c) | ||
| 5 A | ✓ 1 | |
| It is just above the normal current of 4.0 A so the freezer works normally, but it melts if a fault causes a larger current (a 3 A fuse would melt in normal use) ECF from (b) | ✓ 1 | |
Exam tip: Choose the fuse rating just above the normal working current; too high a rating does not protect the appliance.
A fashion designer finds that a polyester dress clings to her skirt after the two are rubbed together. The dress becomes negatively charged.
Explain how the dress becomes negatively charged.
The charged dress attracts small pieces of lint that have no overall charge. Explain why.
Two identical negatively charged dresses are held close together. State what happens.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Friction between the materials transfers electrons | ✓ 1 | |
| Electrons move onto the dress (from the other material), so it gains a surplus of electrons | ✓ 1 | |
| Part (b) | ||
| The charged dress repels electrons in the lint to the far side | ✓ 1 | |
| The near side of the lint becomes positive (the charge is induced) | ✓ 1 | |
| The attraction to the near side is greater than the repulsion from the far side | ✓ 1 | |
| Part (c) | ||
| They repel each other | ✓ 1 | |
Common error: Only electrons move when objects are charged by rubbing; positive charges stay fixed in the atoms.
A polythene rod is rubbed with a dry cloth and the rod becomes negatively charged. What happens?
| Marking point | Mark | Notes |
|---|---|---|
| A negative charge means an excess of electrons. Only electrons are free to move, so they transfer from the cloth to the rod. A and C are wrong because protons are fixed in the nuclei and do not transfer. B would leave the rod positively charged. | ✓ 1 | Answer D |
Answer: D
Common error: Only electrons transfer when objects are charged by friction.
An electric lawnmower is rated at 1.5 kW and is used on a 230 V mains supply. Plugs for the mower may be fitted with a 3 A, 5 A, 13 A or 30 A fuse. Which fuse is the most suitable?
| Marking point | Mark | Notes |
|---|---|---|
| The working current is I = P / V = 1500 / 230 = 6.5 A. A suitable fuse has a rating just above this current, so 13 A is correct. A 3 A or 5 A fuse is below 6.5 A, so it would melt during normal use. A 30 A fuse is far above the working current, so a fault current could flow for too long and overheat the cable before the fuse melted. | ✓ 1 | Answer C |
Answer: C
Exam tip: Work out the normal current first, then pick the smallest standard fuse rating that is larger than it.
A dry cloth is rubbed on an acetate rod and the rod becomes positively charged. Which statement describes what happened?
| Marking point | Mark | Notes |
|---|---|---|
| Only electrons can move between solid insulators. The rod ends up with fewer electrons than protons, so it is positive; the electrons it lost went to the cloth. Protons are held in nuclei and do not move, so the proton statement is wrong. Gaining electrons would make the rod negative. Charge is never created, only separated. | ✓ 1 | Answer A |
Answer: A
Common error: Positive charging is caused by losing electrons, not by gaining protons.
An electric lawnmower has a power rating of 1200 W and is used on a 230 V mains supply. Its plug contains a fuse.
Calculate the current in the lawnmower when it is working normally.
The plug can be fitted with a 3 A, 5 A or 13 A fuse. State which fuse should be used and explain your choice.
State which wire in the plug the fuse is connected to.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P / V | ✓ 1 | |
| A1 I = 1200 / 230 = 5.22 A, so 5.2 A | ✓ 1 | |
| Part (b) | ||
| B1 13 A | ✓ 1 | |
| B1 3 A and 5 A fuses are below the normal current of 5.2 A so would melt in normal use; 13 A is the lowest rating above this current | ✓ 1 | |
| Part (c) | ||
| B1 live wire | ✓ 1 | |
Exam tip: A fuse rating must be just above the normal working current, not simply the highest available.
A warning light on a footbridge is powered by a 12 V battery. The light draws a current of 0.50 A. The battery can supply 1.0 A for 36 hours.
Calculate the power of the light.
Calculate how long the battery can power the light.
Calculate the charge that passes through the light in 1.0 minute.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 P = I V | ✓ 1 | |
| A1 P = 0.50 × 12 = 6.0 W | ✓ 1 | |
| Part (b) | ||
| M1 time = 36 / 0.50 | ✓ 1 | |
| A1 72 hours | ✓ 1 | |
| Part (c) | ||
| M1 Q = I t = 0.50 × 60 | ✓ 1 | |
| A1 30 C | ✓ 1 | |
Exam tip: Check the units of time asked for; hours for the battery life but seconds when you use Q = I t.
A beekeeper's wax melter has a switch. Why must the switch be connected in the live wire rather than the neutral wire?
| Marking point | Mark | Notes |
|---|---|---|
| The live wire is at a high potential (about 230 V) compared with earth. A switch in the live wire disconnects the appliance from that potential when it is off, so touching a fault inside is much less dangerous. A switch in the neutral wire would leave the appliance connected to the live supply. The switch does not change the size of the current, does not replace the earth wire for a metal case, and does not affect where the fuse must be: the fuse also belongs in the live wire. | ✓ 1 | Answer A |
Answer: A
Common error: Switches and fuses go in the live wire; a switch in the neutral wire leaves the appliance live when it is 'off'.
A hospital infusion pump draws a steady current of 0.25 A for 8.0 minutes. What charge flows through the pump?
| Marking point | Mark | Notes |
|---|---|---|
| Charge = current × time, with time in seconds: 8.0 minutes = 480 s, so Q = 0.25 × 480 = 120 C. Using 8.0 as the time in minutes gives 2.0 C, dividing 8.0 by 0.25 gives 32 C, and 480 C is just the time in seconds with the current left out. | ✓ 1 | Answer D |
Answer: D
Common error: Convert minutes to seconds before using Q = It.
An electric blanket is rated at 115 W when connected to the 230 V mains supply. A student uses it for 6.0 hours each night. Energy is charged at 30 p per kWh.
Calculate the current in the blanket when it is working at its rated power.
Calculate the energy transferred by the blanket in 6.0 hours, in kWh.
Calculate the cost of using the blanket for one night, giving your answer in pence. Use the unrounded value from (b), or 0.69 kWh.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 current = power ÷ potential difference, I = P/V = 115 ÷ 230 | ✓ 1 | |
| A1 0.50 A | ✓ 1 | |
| Part (b) | ||
| M1 energy = power × time = 0.115 kW × 6.0 h | ✓ 1 | |
| A1 0.69 kWh | ✓ 1 | |
| Part (c) | ||
| M1 cost = energy × price = 0.69 × 30 | ✓ 1 | |
| A1 20.7 p | ✓ 1 | |
Exam tip: Work in kW and hours for kWh, or in W and seconds for joules, but do not mix the two.
A pacemaker monitor in a hospital ward is connected to a patient by a lead in which the current is 0.040 A.
State what is meant by electric current.
Calculate the charge that flows through the lead in 5.0 minutes.
A different lead carries 0.030 A. Calculate the time taken for a charge of 15 C to flow through it.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 the rate of flow of charge | ✓ 1 | |
| Part (b) | ||
| M1 Q = It with time in seconds, 0.040 × 300 | ✓ 1 | |
| A1 12 C | ✓ 1 | |
| Part (c) | ||
| M1 t = Q/I = 15 ÷ 0.030 | ✓ 1 | |
| A1 500 s | ✓ 1 | |
Exam tip: Rearrange Q = It before you substitute: t = Q/I.
A tailor rubs an acetate rod with a wool cloth. The rod becomes positively charged. Which statement describes what has happened?
| Marking point | Mark | Notes |
|---|---|---|
| Only electrons are free to move when materials are rubbed together. The rod ends up positive because it loses electrons, and the cloth gains those same electrons so it becomes negative. Protons are fixed in the nuclei and do not transfer, which rules out both proton options. The option with electrons moving to the rod would make the rod negative, which contradicts the information given. | ✓ 1 | Answer C |
Answer: C
Common error: Positive charging does not mean positive charges moved in. The object has lost electrons.
A fashion designer charges an acetate rod by rubbing it with a silk cloth.
State what is transferred between the rod and the cloth when the rod is rubbed.
The rod becomes positively charged. Explain why.
The charged rod is hung from a thread. A second rod charged in the same way is brought close to it. State what happens to the hanging rod.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Electrons | ✓ 1 | |
| Part (b) | ||
| Electrons move from the rod to the cloth | ✓ 1 | |
| The rod has lost negative charge, so it has more positive than negative charge | ✓ 1 | |
| Part (c) | ||
| It is repelled (moves away) | ✓ 1 | |
Exam tip: Say what moves (electrons) and in which direction when explaining charging by friction.
A student rubs a polythene rod with a woollen cloth in a dry room. The rod becomes negatively charged.
Explain why the woollen cloth becomes positively charged.
The rod carries a charge of 45 nC. This charge leaks away steadily into the air in 15 s. Calculate the average current in nA.
(1 nC = 10−9 C)
A metal rod held in the hand cannot be charged by rubbing it with a cloth. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The electrons gained by the rod came from the cloth, which now has a shortage of electrons | ✓ 1 | |
| Part (b) | ||
| M1 current = charge ÷ time, 45 ÷ 15 | ✓ 1 | |
| A1 3.0 nA | ✓ 1 | |
| Part (c) | ||
| Metal is a conductor, so charge is free to move | ✓ 1 | |
| Any charge gained flows through the person's body to earth | ✓ 1 | |
Common error: Keep prefixes in mind: 45 nC divided by 15 s gives 3.0 nA, not 3.0 A.
A fish-processing plant uses an electric steamer that has a metal case and is connected to the mains by a three-core cable.
State the colour of the insulation on the live wire.
The live wire inside the steamer becomes loose and touches the metal case. Explain how the earth wire and the fuse together protect a person who touches the case.
State why the switch must be fitted in the live wire.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Brown | ✓ 1 | |
| Part (b) | ||
| A large current flows through the earth wire | ✓ 1 | |
| The fuse melts (blows) because of the large current | ✓ 1 | |
| The appliance is disconnected from the live supply so the case is no longer live | ✓ 1 | |
| Part (c) | ||
| So that the appliance is not live (at high voltage) when the switch is off | ✓ 1 | |
Exam tip: Link the three ideas in order: large current, fuse melts, case disconnected from the live supply.
The plastic shell of a game controller is rubbed with a dry cloth and becomes negatively charged.
What happens during the rubbing?
| Marking point | Mark | Notes |
|---|---|---|
| A negative charge means the shell has gained electrons, and these can only come from the cloth, so electrons move from the cloth to the shell. Electrons moving from the shell to the cloth would leave the shell positive. Protons are fixed in the nuclei of atoms in a solid and do not move during rubbing, so neither proton statement is correct. | ✓ 1 | Answer A |
Answer: A
Common error: Only electrons transfer when objects are charged by friction; gaining electrons gives a negative charge and losing them gives a positive charge.
A fixed resistor of resistance 20 Ω has a current of 0.30 A in it. Its temperature stays constant. The potential difference across it is then doubled.
What is the new current?
| Marking point | Mark | Notes |
|---|---|---|
| For a resistor at constant temperature, current is proportional to potential difference (I = V ÷ R). Doubling V with R unchanged doubles the current to 0.60 A. A current of 0.15 A would need the pd to be halved. A current of 0.30 A would mean doubling the pd had no effect. A current of 1.2 A suggests the current was quadrupled, confusing it with power, which quadruples when the pd doubles. | ✓ 1 | Answer C |
Answer: C
Exam tip: For an ohmic resistor at constant temperature, current and potential difference change by the same factor.
A student walks across a nylon carpet and then touches a metal door handle. A spark passes between her finger and the handle, transferring a charge of 0.60 µC in 0.20 ms. The potential difference between her finger and the handle just before the spark was 12 kV.
Explain how walking across the carpet can charge the student.
Calculate the average current in the spark. Give your answer in A, in standard form.
Calculate the resistance of the air gap during the spark.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| friction between her shoes and the carpet | ✓ 1 | |
| electrons are transferred from one surface to the other, leaving the student with a net charge | ✓ 1 | |
| Part (b) | ||
| M1 I = Q ÷ t = 0.60 × 10−6 ÷ 0.20 × 10−3 | ✓ 1 | |
| A1 3.0 × 10−3 A | ✓ 1 | |
| Part (c) | ||
| M1 R = V ÷ I = 12 000 ÷ 3.0 × 10−3 | ✓ 1 | |
| A1 4.0 × 106 Ω (4000000 Ω) (ECF from (b)) | ✓ 1 | |
Common error: Convert µC to C and ms to s before substituting; leaving the prefixes in gives answers wrong by powers of ten.
In a pharmacy laboratory a plastic funnel is used to pour a dry powder into a bottle. The powder grains and the funnel become charged by friction.
Explain why some powder sticks to the funnel.
Two powder grains with the same type of charge are close together. State what happens to them.
The funnel is replaced by a metal funnel held in the hand of a person who is connected to earth. State why the metal funnel does not stay charged.
State one safety risk of a build-up of static charge near a flammable powder.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| friction transfers electrons between the powder and the funnel, so they gain equal and opposite charges | ✓ 1 | |
| opposite charges attract | ✓ 1 | |
| Part (b) | ||
| they repel each other | ✓ 1 | |
| Part (c) | ||
| electrons flow through the metal and the person to earth, removing the charge | ✓ 1 | |
| Part (d) | ||
| a spark could occur, igniting the powder or vapour | ✓ 1 | |
Exam tip: Charge on a conductor can flow away to earth, but charge on an insulator stays where it is.
A steady current flows in a copper wire connected to a battery. What moves along the wire?
| Marking point | Mark | Notes |
|---|---|---|
| Copper contains free electrons that are not bound to single atoms. The battery's p.d. makes these negatively charged electrons drift towards the positive terminal while the copper ions stay fixed in the lattice. Positive ions do not move through a solid metal. Conventional current is defined in the opposite direction to electron flow, so the second statement is wrong. Atoms stay in place, so the last statement is wrong. | ✓ 1 | Answer C |
Answer: C
Common error: Conventional current is drawn from positive to negative, but the particles that actually move in a metal are electrons going the other way.
The metal casing of an electric shower is connected to the earth wire. What is the purpose of this connection?
| Marking point | Mark | Notes |
|---|---|---|
| If the live wire comes loose and touches the casing, the earth connection gives a low-resistance path, so a large current flows and melts the fuse, which disconnects the live supply and makes the casing safe to touch. The heating element is powered through the live and neutral wires, so the earth wire does not supply current to it. The earth wire does not change the element's resistance. The neutral wire, not the earth wire, carries the return current in normal use. | ✓ 1 | Answer B |
Answer: B
Exam tip: Link the three steps: fault, large current through the earth wire, fuse melts and the appliance is isolated.
An archaeologist uses a copper cable to connect a ground-survey instrument to a battery. A steady current of 0.50 A passes through the cable for 2.0 minutes.
State the name of the particles in the copper that carry the current.
State the direction in which these particles move compared with the conventional current.
Calculate the charge that passes any point in the cable in the 2.0 minutes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (Free or delocalised) electrons | ✓ 1 | |
| Part (b) | ||
| In the opposite direction | ✓ 1 | |
| Part (c) | ||
| M1 charge = current × time with time = 2.0 × 60 = 120 s | ✓ 1 | |
| A1 0.50 × 120 = 60 C | ✓ 1 | |
Common error: Time must be in seconds when you use charge = current × time. Two minutes is 120 s, not 2.0 s.
A cyclist's rear lamp is connected to a 6.0 V dynamo by copper wire. A steady current of 0.25 A flows for 4.0 minutes.
Calculate the charge that flows through the lamp.
Calculate the energy transferred to the lamp.
State what is meant by a current of 0.25 A.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 charge = current × time with time = 4.0 × 60 = 240 s | ✓ 1 | |
| A1 0.25 × 240 = 60 C | ✓ 1 | |
| Part (b) | ||
| M1 energy = charge × potential difference | ✓ 1 | |
| A1 60 × 6.0 = 360 J (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| A charge of 0.25 C passes a point every second, or the rate of flow of charge is 0.25 C per second | ✓ 1 | |
Exam tip: Energy transferred, charge and p.d. are linked by energy = charge × potential difference. Find the charge first.
A current of 0.50 A flows through a lamp for 3.0 minutes. How much charge passes through the lamp in this time?
| Marking point | Mark | Notes |
|---|---|---|
| Charge = current × time = 0.50 A × 180 s = 90 C, with the time converted to seconds. 1.5 C comes from using 3.0 minutes without converting to seconds. 360 C comes from dividing time by current. 0.0028 C comes from dividing current by time. | ✓ 1 | Answer C |
Answer: C
Common error: Always convert minutes or hours into seconds before using Q = It.
A mountain hut uses an electric blanket rated 120 W at 230 V. The blanket is switched on for 8.0 hours overnight.
Calculate the current in the blanket when it works at its rated power. Give your answer to 2 significant figures.
Calculate the energy transferred by the blanket in 8.0 hours. Give your answer in joules to 2 significant figures.
Electricity costs 25 p per kilowatt-hour. Calculate the cost, in pence, of running the blanket for 8.0 hours.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P/V = 120 / 230 | ✓ 1 | |
| A1 0.52 A | ✓ 1 | |
| Part (b) | ||
| M1 t = 8.0 × 3600 = 28 800 s | ✓ 1 | |
| M1 E = Pt = 120 × 28 800 | ✓ 1 | |
| A1 3.5 × 106 J (3 456 000 J before rounding) | ✓ 1 | |
| Part (c) | ||
| M1 energy = 0.120 kW × 8.0 h = 0.96 kWh | ✓ 1 | |
| M1 cost = 0.96 × 25 | ✓ 1 | |
| A1 24 p | ✓ 1 | |
Exam tip: Kilowatt-hours need power in kW and time in hours; do not mix them with watts and seconds.
A kettle is rated 2.3 kW at 230 V and is connected to the mains through a three-pin plug. The plug can be fitted with a 3 A, 5 A or 13 A fuse.
State the colours of the insulation on the earth wire.
The live wire inside the kettle comes loose and touches the metal casing. Explain how the earth wire and the fuse make the kettle safe.
Calculate the current in the kettle at its rated power.
State which fuse should be fitted.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| green and yellow | ✓ 1 | |
| Part (b) | ||
| a large current flows through the casing and earth wire | ✓ 1 | |
| this large current makes the fuse melt | ✓ 1 | |
| which disconnects the live supply so the casing is no longer live | ✓ 1 | |
| Part (c) | ||
| M1 I = P/V = 2300 / 230 | ✓ 1 | |
| A1 10 A | ✓ 1 | |
| Part (d) | ||
| 13 A | ✓ 1 | |
Exam tip: Choose the fuse value just above the working current; a fuse that is too large will not protect the wiring.
A glass rod is rubbed with a silk cloth. The rod becomes positively charged. What happens during the rubbing?
| Marking point | Mark | Notes |
|---|---|---|
| Only electrons can move between solid objects. The glass loses electrons to the silk, leaving the glass with a positive charge and the silk negative. Protons are fixed in nuclei. Electrons moving to the glass would make it negative. Charge is not created; it is transferred. | ✓ 1 | Answer C |
Answer: C
Common error: A positively charged object has lost electrons; no positive particles have been added.
In a café kitchen, a toaster (800 W), a kettle (2200 W) and a microwave oven (1000 W) are all plugged into one extension lead. The mains supply is 230 V and the lead is rated at 13 A.
Calculate the current in the kettle. Give your answer to 3 significant figures.
Calculate the total current in the lead when all three appliances are on. Give your answer to 3 significant figures.
Explain why it is unsafe to use the lead in this way.
Suggest one change that would make the kitchen safer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P/V = 2200 / 230 | ✓ 1 | |
| A1 9.57 A | ✓ 1 | |
| Part (b) | ||
| M1 total power = 800 + 2200 + 1000 = 4000 W | ✓ 1 | |
| A1 I = 4000 / 230 = 17.4 A | ✓ 1 | |
| Part (c) | ||
| the current (17.4 A) is greater than the 13 A rating | ✓ 1 | |
| the lead (and wires) would overheat, which could cause a fire | ✓ 1 | |
| Part (d) | ||
| plug the appliances into separate sockets / do not use the three appliances at once / use a lead with a higher rating | ✓ 1 | |
Exam tip: Add the powers first and divide by the voltage once; the currents in appliances on one lead also add.
A shop sells a floodlight rated 575 W at 230 V. It is supplied with a three-pin plug that can be fitted with a 3 A, 5 A or 13 A fuse.
State the colour of the insulation on the live wire.
Explain why the fuse is fitted in the live wire rather than the neutral wire.
Calculate the current in the floodlight at its rated power.
State which fuse should be fitted.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| brown | ✓ 1 | |
| Part (b) | ||
| if the fuse melts in the live wire, the appliance is disconnected from the live supply | ✓ 1 | |
| in the neutral wire the appliance would stay connected to the live wire when the fuse melts, so the floodlight casing could still be dangerous | ✓ 1 | |
| Part (c) | ||
| M1 I = P/V = 575 / 230 | ✓ 1 | |
| A1 2.5 A | ✓ 1 | |
| Part (d) | ||
| 3 A | ✓ 1 | |
Exam tip: The fuse rating must be just above the normal current, so the fuse melts when a fault raises the current.
A string of 23 identical decorative lamps is connected in series to a 230 V mains supply. What is the potential difference across each lamp?
| Marking point | Mark | Notes |
|---|---|---|
| The supply p.d. is shared equally between identical lamps in series, so each has 230 ÷ 23 = 10 V. 230 V would be the p.d. across each lamp only if they were in parallel. 23 V comes from dividing by 10 instead of by the number of lamps. 0.10 V comes from dividing the number of lamps by the supply p.d. | ✓ 1 | Answer C |
Answer: C
Common error: In a series circuit the supply p.d. is divided between the components; in parallel branches each component gets the full supply p.d.
An electric bicycle battery is charged with a constant current of 2.0 A for 30 minutes. How much charge flows through the battery?
| Marking point | Mark | Notes |
|---|---|---|
| Charge = current × time, with time in seconds: 2.0 A × (30 × 60) s = 3600 C. 60 C comes from using 30 minutes without converting to seconds. 1.1 C comes from dividing the current by the time in seconds. 0.067 C comes from dividing the current by 30. | ✓ 1 | Answer B |
Answer: B
Common error: Convert minutes and hours to seconds before using Q = I t.
A string of 20 identical decorative lamps is connected in series to the 230 V mains supply. The current in the string is 0.30 A.
State the potential difference across each lamp.
Calculate the charge that passes through one lamp in 2.0 minutes.
One lamp in the string fails and its filament breaks. Explain why all the other lamps go out.
Give two reasons why lamps in a house are connected in parallel rather than in series.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 11.5 V | ✓ 1 | |
| Part (b) | ||
| M1 use of charge = current × time with time = 120 s | ✓ 1 | |
| A1 36 C | ✓ 1 | |
| Part (c) | ||
| a break in the filament means the circuit is no longer complete | ✓ 1 | |
| in a series circuit the same single path carries the current, so no current in any lamp | ✓ 1 | |
| Part (d) | ||
| each lamp has the full mains p.d. (230 V) so is at normal brightness | ✓ 1 | |
| each lamp can be switched on or off independently / if one fails the others stay on | ✓ 1 | |
Exam tip: Write the unit of time you are using in the working: 2.0 minutes is 120 s.
In a sports hall a gymnast walks across a synthetic floor mat and becomes negatively charged. She then touches a metal door handle and a spark jumps across the gap.
Explain how the gymnast became negatively charged.
The spark transfers 0.80 μC of charge in 0.20 μs. Calculate the average current in the spark.
State the direction in which electrons move in the spark.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| friction between her shoes and the mat transfers electrons | ✓ 1 | |
| she gains electrons so has an excess of negative charge | ✓ 1 | |
| Part (b) | ||
| M1 use of current = charge ÷ time | ✓ 1 | |
| A1 4.0 A | ✓ 1 | |
| Part (c) | ||
| from the gymnast to the door handle | ✓ 1 | |
Common error: Electrons are the charges that move when objects are charged by friction; protons stay fixed in the nuclei.
An electric bicycle battery is charged from a charger. For the first 40 minutes the charging current is a constant 3.0 A. For the next 20 minutes it is a constant 1.0 A. Then the charger switches off.
Calculate the charge that flows during the first 40 minutes.
Calculate the charge that flows during the next 20 minutes.
Calculate the total charge that flows during the 60 minutes of charging.
Calculate the average current over the whole 60 minutes. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 charge = current × time with time = 2400 s | ✓ 1 | |
| A1 7200 C | ✓ 1 | |
| Part (b) | ||
| M1 charge = 1.0 × 1200 | ✓ 1 | |
| A1 1200 C | ✓ 1 | |
| Part (c) | ||
| 8400 C , ECF from (a) and (b) | ✓ 1 | |
| Part (d) | ||
| M1 average current = total charge ÷ total time = 8400 ÷ 3600 | ✓ 1 | |
| A1 2.33 A , ECF | ✓ 1 | |
Exam tip: The average current is not the mean of 3.0 A and 1.0 A unless the two stages last the same time.
A marine biologist uses a sensor connected by a copper cable. Which statement describes what happens in the copper wire when there is a current?
| Marking point | Mark | Notes |
|---|---|---|
| Copper is a metal with free (delocalised) electrons. A p.d. makes these electrons drift along the wire, and that flow of charge is the current. The copper ions stay fixed in the lattice and only vibrate. Protons are inside the nuclei and do not move through a wire. The atoms of the wire stay in place. | ✓ 1 | Answer C |
Answer: C
Common error: Only electrons move in a metal wire; the positive ions stay fixed in the lattice.
A cordless drill on an architect's building site is recharged with a current of 1.5 A for 40 minutes.
State what is meant by a current of 1.5 A.
Calculate the charge that flows during the 40 minutes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1.5 coulombs of charge flow past a point each second | ✓ 1 | |
| Part (b) | ||
| 40 × 60 = 2400 s | ✓ 1 | |
| Q = I × t = 1.5 × 2400 = 3600 C | ✓ 1 | |
Common error: Time must be in seconds in Q = I × t.
A marine biologist runs a copper cable from a boat to an underwater camera.
State what moves through the copper wire to make a current.
State the direction in which these particles move compared with the direction of conventional current.
Explain why the cable becomes slightly warm when there is a current.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| free (delocalised) electrons | ✓ 1 | |
| Part (b) | ||
| opposite direction | ✓ 1 | |
| Part (c) | ||
| moving electrons collide with the ions (atoms) in the lattice | ✓ 1 | |
| transferring energy to the ions, so they vibrate more and the wire heats up | ✓ 1 | |
Exam tip: Conventional current goes from positive to negative, the opposite way to the electrons.
In a logistics warehouse, plastic film becomes charged as it is pulled from a roll, and it then attracts dust.
Explain how the film becomes negatively charged when it rubs against the roller.
Explain why the charged film attracts neutral dust.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| electrons are transferred from the roller to the film (by friction) | ✓ 1 | |
| the film gains extra negative charge | ✓ 1 | |
| Part (b) | ||
| the charge on the film induces an opposite charge on the near side of the dust, so there is attraction | ✓ 1 | |
Common error: Only electrons move when an insulator is charged by rubbing; positive charges do not transfer.
What carries the current in a copper wire connected to a battery?
| Marking point | Mark | Notes |
|---|---|---|
| Copper contains free (delocalised) electrons. The battery makes them drift from the negative terminal towards the positive terminal through the fixed lattice of positive ions. Protons and positive ions stay fixed in the metal and do not carry the current. Electrons bound to atoms that only vibrate cannot carry charge along the wire. | ✓ 1 | Answer D |
Answer: D
Common error: Conventional current is drawn from positive to negative, which is opposite to the direction in which the electrons actually move.
A temperature sensor in a coffee roaster carries a current of 0.40 A for 2.5 minutes. How much charge passes through the sensor?
| Marking point | Mark | Notes |
|---|---|---|
| Charge = current × time. The time must be in seconds: 2.5 min = 150 s, so Q = 0.40 × 150 = 60 C. 1.0 C comes from using the time in minutes. 375 C comes from dividing time by current. 0.0027 C comes from dividing current by time. | ✓ 1 | Answer B |
Answer: B
Exam tip: Convert minutes to seconds before using Q = I t.
A student investigates what happens to charge in a copper wire.
State which particles carry the current in a copper wire.
State the direction in which these particles move relative to the conventional current.
The current in the wire is 2.4 A. Calculate the charge passing a point in 20 s.
Each electron has a charge of 1.6 × 10−19 C. Calculate the number of electrons passing the point in the 20 s.
A lamp lights almost at once when a switch is closed, although each electron drifts very slowly. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 free (delocalised) electrons | ✓ 1 | |
| Part (b) | ||
| B1 opposite to the conventional current | ✓ 1 | |
| Part (c) | ||
| M1 Q = I t | ✓ 1 | |
| A1 48 C | ✓ 1 | |
| Part (d) | ||
| M1 N = Q ÷ e = 48 ÷ 1.6 × 10−19 | ✓ 1 | |
| A1 3.0 × 1020 (ECF from (c)) | ✓ 1 | |
| Part (e) | ||
| electrons are already present throughout the wire and filament | ✓ 1 | |
| the p.d. sets up an electric field along the wire almost instantly so all the electrons begin to move together | ✓ 1 | |
Common error: Electrons do not travel from the switch to the lamp in a flash; the whole electron 'flow' starts together.
A three-pin mains plug has three wires connected to its pins.
State the colour of the live wire, the neutral wire and the earth wire.
State the function of the fuse and the wire in which it is fitted.
Explain why the earth wire is connected to the metal case of an appliance.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 live: brown | ✓ 1 | |
| B1 neutral: blue | ✓ 1 | |
| B1 earth: green and yellow | ✓ 1 | |
| Part (b) | ||
| B1 melts (breaks the circuit) if the current is too large | ✓ 1 | |
| B1 fitted in the live wire | ✓ 1 | |
| Part (c) | ||
| if the live wire touches the case a large current flows to earth | ✓ 1 | |
| this blows the fuse so the case is not left live (no electric shock) | ✓ 1 | |
Exam tip: Fuses and switches always go in the live wire so that the appliance is isolated when they break the circuit.
An egg incubator has three identical heating elements, each of resistance 6.0 Ω. Two of the elements are connected in parallel with each other, and this pair is connected in series with the third element.
What is the total resistance of the three elements?
| Marking point | Mark | Notes |
|---|---|---|
| Two identical 6.0 Ω resistors in parallel have a combined resistance of half of one, 3.0 Ω, which is less than either resistor alone. This 3.0 Ω is in series with 6.0 Ω, so the total is 3.0 + 6.0 = 9.0 Ω. 18 Ω treats all three as being in series. 2.0 Ω treats all three as being in parallel. 4.0 Ω is the result for the reverse arrangement, where two in series (12 Ω) are in parallel with one (6.0 Ω). | ✓ 1 | Answer C |
Answer: C
Exam tip: Work out the parallel part first and replace it with one equivalent resistor. Then add the series resistances. A parallel combination is always smaller than the smallest resistor in it.
A gardener builds a light sensor to control a grow-light in a greenhouse. A light-dependent resistor (LDR) is connected in series with a 2.0 kΩ fixed resistor and a 9.0 V battery.
Describe how the resistance of an LDR changes as the light intensity increases.
In dim light the resistance of the LDR is 7.0 kΩ.
Calculate the current in the circuit.
Calculate the voltage across the fixed resistor in dim light.
The light becomes brighter.
Explain what happens to the voltage across the fixed resistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The resistance decreases as the light intensity increases | ✓ 1 | |
| Part (b) | ||
| Total resistance = 2.0 kΩ + 7.0 kΩ = 9.0 kΩ, which is 9000 Ω | ✓ 1 | |
| I = V ÷ R = 9.0 ÷ 9000 | ✓ 1 | |
| I = 1.0 × 10−3 A, or 1.0 mA | ✓ 1 | |
| Part (c) | ||
| V = I × R = 1.0 × 10−3 × 2000 | ✓ 1 | |
| V = 2.0 V | ✓ 1 | |
| Part (d) | ||
| The LDR resistance falls, so the total resistance falls and the current increases | ✓ 1 | |
| The voltage across the fixed resistor increases, because V = IR and the fixed resistor now takes a larger share of the 9.0 V | ✓ 1 | |
Common error: Convert kΩ to Ω before using V = IR. Leaving the resistance as 9.0 gives a current that is 1000 times too large.
The motor of a delivery drone transfers 54 J of energy when 4.5 C of charge passes through it.
What is the voltage across the motor?
| Marking point | Mark | Notes |
|---|---|---|
| Energy transferred = charge × voltage, so voltage = energy ÷ charge = 54 ÷ 4.5 = 12 V. Voltage is the energy transferred per coulomb of charge. The option 0.083 V divides charge by energy. The option 58.5 V adds the two values. The option 243 V multiplies them. | ✓ 1 | Answer A |
Answer: A
Exam tip: Think of voltage as 'joules per coulomb'. The unit itself tells you to divide energy by charge.
A hospital defibrillator delivers a short pulse of electrical energy to a patient's heart. In one pulse it transfers 200 J of energy at an average voltage of 2000 V.
State the equation linking energy transferred, charge and voltage.
Calculate the charge that flows during one pulse.
The pulse lasts 5.0 ms.
Calculate the average current during the pulse.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy transferred = charge × voltage (E = Q × V) | ✓ 1 | |
| Part (b) | ||
| Q = E ÷ V = 200 ÷ 2000 | ✓ 1 | |
| Q = 0.10 C | ✓ 1 | |
| Part (c) | ||
| t = 5.0 ms = 0.0050 s | ✓ 1 | |
| I = Q ÷ t = 0.10 ÷ 0.0050 | ✓ 1 | |
| I = 20 A | ✓ 1 | |
Common error: Milliseconds must be divided by 1000 to give seconds. Using 5.0 instead of 0.0050 gives a current 1000 times too small.
A city tram takes electricity from an overhead cable at a voltage of 750 V.
State what a voltage of 750 V means in terms of energy and charge.
On one journey between two stops, the tram motor transfers 4.5 MJ of energy.
Calculate the charge that flows through the motor.
The average current in the motor during this journey is 200 A.
Calculate the time taken for the journey.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| For each coulomb of charge | ✓ 1 | |
| 750 J of energy is transferred | ✓ 1 | |
| Part (b) | ||
| Q = E ÷ V = 4.5 × 106 ÷ 750 | ✓ 1 | |
| Q = 6000 C | ✓ 1 | |
| Part (c) | ||
| t = Q ÷ I = 6000 ÷ 200 | ✓ 1 | |
| t = 30 s | ✓ 1 | |
Exam tip: MJ means ×10⁶ J. Write the conversion clearly before substituting so the method mark is easy to see.
A dental surgery uses an examination lamp with a metal case. The case is connected to the earth wire, and the plug contains a correctly rated fuse.
A fault makes the live wire touch the metal case. What happens?
| Marking point | Mark | Notes |
|---|---|---|
| The earth wire gives the case a low-resistance path to earth. When the live wire touches the case, a large current flows from live, through the case, to earth. This current is much bigger than the fuse rating, so the fuse wire overheats and melts. This breaks the connection to the live supply, and the case is no longer at a dangerous voltage. A fuse does not lower the voltage; it simply breaks the circuit. The earth path does not keep the case live until the lamp is switched off. | ✓ 1 | Answer B |
Answer: B
Common error: The earth wire does not 'soak up' electricity. It works because it lets a large current flow, which makes the fuse melt.
A home charger for an electric car is connected to the 230 V mains supply and draws a current of 32 A.
How much energy is transferred in 2.5 hours of charging?
| Marking point | Mark | Notes |
|---|---|---|
| Use E = I × V × t with the time in seconds: t = 2.5 × 3600 = 9000 s. E = 32 × 230 × 9000 = 6.6 ×107 J. The distractors come from common mistakes:
| ✓ 1 | Answer D |
Answer: D
Common error: In E = IVt, time must be in seconds. Convert hours by multiplying by 3600.
A 3D printer in a school design lab is connected to the 230 V mains supply. During one print, its average current is 0.80 A and it transfers 1.0 MJ of energy.
State the equation linking energy transferred, current, voltage and time.
Calculate the time taken for the print, in seconds.
Convert your answer to the previous part into hours.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Energy transferred = current × voltage × time, E = I × V × t | ✓ 1 | |
| Part (b) | ||
| Rearrangement: t = E / (I × V) | ✓ 1 | |
| Substitution: t = 1.0 ×106 / (0.80 × 230) | ✓ 1 | |
| t = 5400 s (accept 5430 s, 5435 s or 5.4 ×103 s) | ✓ 1 | |
| Part (c) | ||
| t ÷ 3600 = 1.5 h (accept 1.51 h; allow follow-through from the previous part) | ✓ 1 | |
Common error: 1.0 MJ means 1.0 ×10⁶ J. Using 1.0 ×10³ J gives an answer one thousand times too small.
A temporary floodlight at a cricket ground is powered through a long extension cable. The total resistance of the cable's conductors is 0.15 Ω, and the current in the cable is 12 A.
Explain why a current in the cable causes it to heat up.
Calculate the power wasted as heat in the cable.
The cable is stored on a drum. Suggest why it is dangerous to use the cable while it is still tightly wound on the drum.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Electrons collide with the ions / atoms in the metal . | ✓ 1 | |
| This transfers energy to the ions, so they vibrate more and the thermal energy store of the wire increases . | ✓ 1 | |
| Part (b) | ||
| P = I2 × R | ✓ 1 | |
| P = 122 × 0.15 | ✓ 1 | |
| P = 22 W (accept 21.6 W) | ✓ 1 | |
| Part (c) | ||
| The wound layers trap heat, and less surface is exposed to transfer energy to the surroundings . | ✓ 1 | |
| So the temperature rises a lot, which could melt the insulation or cause a fire . | ✓ 1 | |
Exam tip: When you are given current and resistance, use P = I²R directly. Remember to square the current before multiplying.
A greenhouse sensor circuit has a 9.0 V battery connected in series with three resistors of 1.0 kΩ, 2.0 kΩ and 6.0 kΩ.
What is the voltage across the 6.0 kΩ resistor?
| Marking point | Mark | Notes |
|---|---|---|
| The total resistance is 9.0 kΩ, so the current is I = 9.0 ÷ 9000 = 1.0×10−3 A. The voltage across the 6.0 kΩ resistor is V = IR = 1.0×10−3 × 6000 = 6.0 V. In series, the supply voltage is shared in proportion to resistance, so the 6.0 kΩ resistor takes 6/9 of 9.0 V. 3.0 V assumes the voltage is shared equally. 9.0 V assumes each component gets the full supply voltage, which is only true in parallel. 1.5 V wrongly gives the largest resistor the smallest share. | ✓ 1 | Answer A |
Answer: A
Common error: In a series circuit the largest resistor takes the largest share of the voltage, not the smallest.
A small satellite is powered by a solar panel made of solar cells. Each cell produces a voltage of 0.55 V. The satellite's electronics need a supply of at least 28 V.
Calculate the minimum number of cells that must be connected in series.
Explain why the cells are connected in series, not in parallel, to provide this voltage.
The panel delivers a current of 4.0 A at 28 V. In each orbit, the satellite is in sunlight for 55 minutes.
Calculate the energy delivered by the panel during this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 28 ÷ 0.55 = 50.9 | ✓ 1 | |
| so 51 cells are needed (round up) | ✓ 1 | |
| Part (b) | ||
| the voltages of cells in series add together | ✓ 1 | |
| cells in parallel give the same voltage as one cell, 0.55 V | ✓ 1 | |
| Part (c) | ||
| P = I × V = 4.0 × 28 = 112 W | ✓ 1 | |
| t = 55 × 60 = 3300 s | ✓ 1 | |
| E = 112 × 3300 = 3.7×105 J | ✓ 1 | |
Common error: Rounding 50.9 down to 50 gives only 27.5 V, which is too low. When a minimum is needed, round up.
A car's rear-window demister is a thin metal heating wire. The wire has a resistance of 1.2 Ω, and the current in it is 10 A.
Calculate the power transferred by the heating wire.
Calculate the voltage across the heating wire.
Explain, in terms of electrons, why the wire becomes hot when there is a current in it.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P = I2 × R = 102 × 1.2 | ✓ 1 | |
| = 120 W | ✓ 1 | |
| Part (b) | ||
| V = I × R = 10 × 1.2 | ✓ 1 | |
| = 12 V | ✓ 1 | |
| Part (c) | ||
| electrons collide with the ions / atoms of the metal | ✓ 1 | |
| the electrons transfer energy to the ions | ✓ 1 | |
| the ions vibrate more, so the temperature of the wire increases | ✓ 1 | |
Common error: In P = I² × R, square the current before multiplying by the resistance. 10² = 100, not 20.
A student compares two mobile phone chargers.
Charger A: output 5.0 V, 2.0 A
Charger B: output 9.0 V, 2.0 A
The phone battery needs 4.0×104 J of energy to charge fully.
Calculate the output power of each charger.
Calculate the time taken for charger B to charge the battery fully.
Give your answer in minutes.
Suggest why the actual charging time is longer than your answer to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A: P = I × V = 2.0 × 5.0 = 10 W | ✓ 1 | |
| B: P = I × V = 2.0 × 9.0 = 18 W | ✓ 1 | |
| Part (b) | ||
| t = E ÷ P | ✓ 1 | |
| = 4.0×104 ÷ 18 = 2200 s (2222 s) | ✓ 1 | |
| = 37 minutes | ✓ 1 | |
| Part (c) | ||
| Any one of: | — | |
| • some energy is wasted by heating in the charger, cable or battery, so energy is stored in the battery at a rate of less than 18 W | — | |
| • the phone draws less than 2.0 A, for example as the battery nears full charge, so the power delivered is less than 18 W | — | |
| ✓ 1 | ||
Common error: Do not stop at seconds when the question asks for minutes. Divide by 60 for the final mark.
Which statement about the mains electricity supply to a house is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Mains electricity is alternating current (a.c.). The direction of the current reverses repeatedly, many times each second. The voltage also changes all the time, so it is not fixed at 230 V; 230 V is only a quoted value for the supply. In a.c. the electrons move back and forth by tiny distances, so they do not travel from the power station. The current flows round a complete circuit through both the live and the neutral wires. | ✓ 1 | Answer C |
Answer: C
Common error: Direct current (d.c.), for example from a cell, flows in one direction only. Alternating current (a.c.) keeps changing direction.
A home charger for an electric car is connected to the 230 V mains supply. The charger has a power of 7.4 kW.
Calculate the current in the charger cable.
The charger is connected to its own circuit using thick copper cables. Explain why thick cables are used.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Uses power = current × voltage, rearranged to I = P / V . | ✓ 1 | |
| I = 7400 / 230 . | ✓ 1 | |
| I = 32 A . | ✓ 1 | |
| Part (b) | ||
| The current is large, and a current in a resistance causes heating . Thick cables have a lower resistance, so they heat up less and are less likely to overheat or cause a fire . | ✓ 2 | |
Exam tip: Convert 7.4 kW to 7400 W before you divide by the voltage, or your answer will be 1000 times too small.
An intensive-care ward in a hospital uses circuit breakers instead of fuses to protect its electrical equipment.
Describe how a fuse protects an appliance when a fault causes too large a current.
Suggest two reasons why circuit breakers are better than fuses on the hospital ward.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The current becomes larger than the rating of the fuse . The thin wire in the fuse heats up and melts . This breaks the circuit in the live wire, so the current stops . | ✓ 3 | |
| Part (b) | ||
| First valid reason . | ✓ 1 | |
| Second, different valid reason . | ✓ 1 | |
| Acceptable reasons (each idea credited once only): | — | |
| • A circuit breaker can be reset after it trips, with no need to find and fit a replacement fuse, so life-support equipment is back in use sooner. | — | |
| • A circuit breaker switches off the current faster than a fuse wire melts, giving better protection to patients and sensitive equipment. | — | |
| • It is easy to see which circuit breaker has tripped, so the faulty circuit is identified quickly. | — | |
Common error: Saying a fuse 'blows' is not enough. Explain that the wire heats up and melts because the current is too large.
A tropical fish tank has an electric heater connected to the 230 V mains. The current in the heater is 0.65 A.
Explain why the temperature of the heater wire increases when there is a current in it.
Calculate the power of the heater.
The water needs 1.8 × 105 J of energy to warm to the correct temperature. Calculate the time the heater takes to transfer this energy.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The electrons that make up the current collide with the ions (atoms) in the metal wire . | ✓ 1 | |
| Energy is transferred to the ions, so they vibrate more and the thermal energy store and temperature of the wire increase . | ✓ 1 | |
| Part (b) | ||
| P = I × V = 0.65 × 230 . | ✓ 1 | |
| P = 149.5 W, accept 150 W . | ✓ 1 | |
| Part (c) | ||
| Uses E = I × V × t, or E = P × t with the power from (b), rearranged for t . | ✓ 1 | |
| t = 180 000 / (0.65 × 230), or 180 000 / 149.5 (allow 180 000 / 150) . | ✓ 1 | |
| t = 1204 s, accept 1200 s; allow an answer carried forward from an incorrect power in (b) . | ✓ 1 | |
Exam tip: In part (c) you can reuse your power from part (b) in E = P × t. Keeping the unrounded 149.5 W gives 1204 s, and rounding to 150 W gives 1200 s. Both answers are acceptable.
A football stadium is lit by floodlights connected in parallel to a 230 V mains supply. Each floodlight has a power of 2.0 kW.
Suggest one reason why the floodlights are connected in parallel.
Calculate the current in one floodlight.
During one evening match the floodlights are switched on for 90 minutes. Calculate the charge that flows through one floodlight in this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Any one from: each floodlight receives the full 230 V / if one floodlight fails the others stay on / each floodlight can be switched separately . | ✓ 1 | |
| Part (b) | ||
| I = P / V = 2000 / 230 . | ✓ 1 | |
| I = 8.7 A . | ✓ 1 | |
| Part (c) | ||
| Converts time: t = 90 × 60 = 5400 s . | ✓ 1 | |
| Q = I × t = 8.7 × 5400 . | ✓ 1 | |
| Q = 4.7 × 104 C (accept 46 900 to 47 000 C) . | ✓ 1 | |
Common error: Forgetting to convert 90 minutes into 5400 s gives an answer 60 times too small.
A 3D printer is plugged into the mains. An adapter inside the printer changes the mains a.c. into a low-voltage d.c. supply for the heated printing bed.
State two differences between the mains a.c. supply and a d.c. supply.
The current in the heated bed is 5.0 A. In 4.0 minutes, the bed transfers 28 800 J of energy. Calculate the voltage across the heated bed.
Suggest why the printing bed uses a low-voltage supply rather than mains voltage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| In a.c. the current keeps changing direction, but in d.c. it flows in one direction only . | ✓ 1 | |
| In a.c. the voltage keeps changing, but in d.c. it stays constant . | ✓ 1 | |
| Part (b) | ||
| Uses E = I × V × t, rearranged to V = E / (I × t) . | ✓ 1 | |
| V = 28 800 / (5.0 × 240) . | ✓ 1 | |
| V = 24 V . | ✓ 1 | |
| Part (c) | ||
| A low voltage is less likely to give the user a dangerous electric shock if they touch the exposed bed . | ✓ 1 | |
Exam tip: In part (b), the time must be in seconds: 4.0 minutes = 240 s.
A designer is choosing components for a vintage-style display lamp. She tests a filament lamp and records two pairs of readings:
at 1.0 V the current is 0.20 A;
at 6.0 V the current is 0.50 A.
Calculate the resistance of the filament lamp at each voltage.
Explain why the resistance of the filament increases as the voltage increases.
The designer also tests a diode. She connects it the opposite way round, in reverse.
Describe the current through the diode in reverse and give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| At 1.0 V: R = 1.0 ÷ 0.20 = 5.0 Ω | ✓ 1 | |
| At 6.0 V: R = 6.0 ÷ 0.50 = 12 Ω | ✓ 1 | |
| Part (b) | ||
| A larger current heats the filament, so its temperature rises | ✓ 1 | |
| The ions in the metal vibrate more, so electrons collide with them more often and the flow of charge is opposed more | ✓ 1 | |
| Part (c) | ||
| The current is zero, or almost zero | ✓ 1 | |
| This is because the diode has a very high resistance in the reverse direction, so it only allows current in one direction | ✓ 1 | |
Common error: Do not calculate resistance from the change in voltage divided by the change in current. Resistance at a point is V ÷ I using that point's own readings.
A plastic rod hangs from a thread and carries a negative charge. Rod P is brought near one end and the hanging rod moves away from it. Rod Q is brought near the same end and the hanging rod moves towards it. Which statement must be correct?
| Marking point | Mark | Notes |
|---|---|---|
| Repulsion only happens between like charges, so P must be negative. Attraction happens between unlike charges, but a charged rod also attracts an uncharged object, so Q is positive or uncharged: C is correct. A and B wrongly make P positive. D allows P to be uncharged, but an uncharged rod cannot repel. | ✓ 1 | Answer C |
Answer: C
Exam tip: Repulsion is the only reliable test for charge; attraction can also occur with an uncharged object.
A student plots a current–potential difference graph for a filament lamp. The line curves so that the current increases more slowly at higher pd. What is the explanation?
| Marking point | Mark | Notes |
|---|---|---|
| A larger current heats the filament; the metal ions vibrate more, collide more often with the flowing electrons and the resistance rises, so each extra volt gives less extra current: D is correct. A is false because the pd is set by the supply; B is false because heating does not remove free electrons; C is false because an ammeter has negligible resistance. | ✓ 1 | Answer D |
Answer: D
Exam tip: Link the shape of the graph to what changes in the lamp: temperature, then ion vibrations, then resistance.
A pupil investigates how the current in a 6.0 V filament lamp depends on the potential difference across it. The results are shown.
| Potential difference (V) | 0 | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 | 6.0 |
| Current (A) | 0 | 0.10 | 0.18 | 0.24 | 0.28 | 0.31 | 0.33 |
State how the ammeter and the voltmeter are connected in the circuit.
Explain why the circuit contains a variable resistor.
The resistance of a filament lamp increases as the current increases. Explain this, using the idea of temperature.
Calculate the resistance of the lamp when the pd is 6.0 V. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ammeter in series with the lamp | ✓ 1 | |
| voltmeter in parallel with (across) the lamp | ✓ 1 | |
| Part (b) | ||
| to change the current in (or pd across) the lamp | ✓ 1 | |
| so that a range of readings can be taken | ✓ 1 | |
| Part (c) | ||
| a bigger current makes the filament hotter | ✓ 1 | |
| ions in the filament vibrate more, so electrons collide with them more often (resistance increases) | ✓ 1 | |
| Part (d) | ||
| resistance = pd ÷ current, R = V/I | ✓ 1 | |
| R = 6.0 ÷ 0.33 = 18.2 Ω | ✓ 1 | |
Exam tip: Calculate resistance from the pd and current at the same point on the table; do not use the gradient of the curve.
Two identical plastic rods, A and B, are rubbed with a dry cloth. Rod A is hung from a thread and is found to have gained a negative charge.
Explain how rubbing gave rod A a negative charge.
Rod B is brought near rod A and rod A moves away from it. State the type of charge on rod B.
A third rod C is brought near rod A and rod A moves towards C. A pupil says this proves that C is positively charged. Explain why the pupil's conclusion is not certain.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| electrons are transferred from the cloth to the rod | ✓ 1 | |
| the rod has gained negative charge (more electrons than protons) | ✓ 1 | |
| Part (b) | ||
| negative | ✓ 1 | |
| Part (c) | ||
| a charged object is also attracted to an uncharged object | ✓ 1 | |
| so C could be uncharged (charges in C are moved by the charged rod) | ✓ 1 | |
Exam tip: In static electricity questions, say that electrons move, not that positive charge moves.
A 12 V supply is connected in series with a 10 Ω resistor and a lamp, to light a swimming pool lane marker. The current in the circuit is 0.50 A.
Calculate the pd across the resistor.
Calculate the resistance of the lamp.
Calculate the power of the lamp.
The lamp is replaced by one of higher resistance. Explain what happens to the current.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V = IR = 0.50 × 10 | ✓ 1 | |
| 5.0 V | ✓ 1 | |
| Part (b) | ||
| pd across lamp = 12 − 5.0 = 7.0 V (ECF from (a)) | ✓ 1 | |
| R = V/I = 7.0 ÷ 0.50 | ✓ 1 | |
| 14 Ω | ✓ 1 | |
| Part (c) | ||
| P = IV = 7.0 × 0.50 | ✓ 1 | |
| 3.5 W | ✓ 1 | |
| Part (d) | ||
| the current decreases | ✓ 1 | |
| total resistance increases while the supply pd is unchanged | ✓ 1 | |
Exam tip: In a series circuit the supply pd is shared: subtract the known pd from the supply to find the other component's pd.
An e-bike battery is charged for one hour. The charging current is a steady 4.0 A for the first 30 minutes, then falls at a steady rate to zero over the next 30 minutes. How much charge flows into the battery?
| Marking point | Mark | Notes |
|---|---|---|
| Charge is the area under the current–time graph. First 30 minutes: 4.0 × 1800 = 7200 C. Next 30 minutes: ½ × 4.0 × 1800 = 3600 C. Total = 10800 C, so C is correct. A uses minutes instead of seconds (4.0 × 60). B ignores the falling part. D treats the current as 4.0 A for the whole hour (4.0 × 3600). | ✓ 1 | Answer C |
Answer: C
Common error: Convert time to seconds before using charge = current × time, and include every section of the graph.
A car-seat heater has two heating elements of resistance 4.0 Ω and 12 Ω connected in series to a 12 V supply. What is the power transferred in the 12 Ω element?
| Marking point | Mark | Notes |
|---|---|---|
| Total resistance = 16 Ω, so current = 12 ÷ 16 = 0.75 A. Power = I²R = 0.75² × 12 = 6.75 W, so B is correct. A is the power in the 4.0 Ω element. C uses the total resistance of 16 Ω in P = I²R. D uses the full 12 V across the 12 Ω element (V²/R), forgetting that the p.d. is shared in series. | ✓ 1 | Answer B |
Answer: B
Common error: In a series circuit the supply p.d. is shared between the elements, so find the current in the whole circuit first.
A rack of servers in a data centre is supplied through a cable carrying 10.0 A at 230 V. At a junction the current divides between four servers connected in parallel. Three of the servers each take 2.0 A.
Calculate the current in the fourth server.
Calculate the power of the fourth server.
The fourth server runs continuously for 24 hours. Electrical energy costs €0.25 per kilowatt-hour. Calculate the cost.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 10.0 − 3 × 2.0 | ✓ 1 | |
| 4.0 A | ✓ 1 | |
| Part (b) | ||
| P = V I = 230 × 4.0 (ECF from (a)) | ✓ 1 | |
| 920 W | ✓ 1 | |
| Part (c) | ||
| Power = 0.92 kW | ✓ 1 | |
| Energy = 0.92 × 24 = 22.08 kWh | ✓ 1 | |
| 22.08 × 0.25 = €5.52 | ✓ 1 | |
Common error: Energy in kilowatt-hours needs power in kilowatts and time in hours. Do not mix watts with hours.
A car-seat heater has two heating elements of resistance 10.0 Ω and 13.0 Ω connected in series to the car battery, which supplies 13.8 V.
Calculate the total resistance of the heater.
Calculate the current in the heater.
Calculate the power transferred in the 13.0 Ω element.
The heater is switched on for 25 minutes. Calculate the total energy transferred by the heater.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 23.0 Ω | ✓ 1 | |
| Part (b) | ||
| I = V ÷ R = 13.8 ÷ 23.0 | ✓ 1 | |
| 0.60 A | ✓ 1 | |
| Part (c) | ||
| P = I²R = 0.60² × 13.0 (ECF from (b)) | ✓ 1 | |
| 4.68 W | ✓ 1 | |
| Part (d) | ||
| Total power = 13.8 × 0.60 = 8.28 W | ✓ 1 | |
| Time = 1500 s | ✓ 1 | |
| 8.28 × 1500 = 12420 J | ✓ 1 | |
Exam tip: The same current passes through every component in series. Use the supply p.d. only with the total resistance.
An engineer in a data centre walks across a nylon carpet in rubber-soled shoes and then receives a small spark on touching a metal server cabinet.
Explain how the engineer becomes charged while walking.
Engineers in the data centre wear a wrist strap connected to earth. Explain how this stops sparks.
In one spark, 8.0 × 10−7 C of charge flows in 2.0 × 10−3 s. Calculate the average current in the spark.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Friction between the shoes and carpet transfers electrons from one to the other | ✓ 1 | |
| The rubber soles insulate the engineer, so the net charge stays on the body | ✓ 1 | |
| Part (b) | ||
| The strap provides a conducting path to earth | ✓ 1 | |
| Charge flows away as it builds up, so no large p.d. develops between the engineer and the cabinet | ✓ 1 | |
| Part (c) | ||
| I = Q ÷ t = 8.0 × 10−7 ÷ 2.0 × 10−3 | ✓ 1 | |
| 4.0 × 10−4 A (= 0.0004 A) | ✓ 1 | |
Common error: Charging by friction moves electrons only; do not say that charge is created or that protons move.
A car-seat heater has two identical heating elements connected in series to a 12 V battery. The total power of the heater is 24 W. One element is disconnected and the other is connected directly to the same battery. What is the power of the single element?
| Marking point | Mark | Notes |
|---|---|---|
| Series: R = V²/P = 144/24 = 6.0 Ω, so each element is 3.0 Ω. One element alone: P = 144/3.0 = 48 W. A (12 W) wrongly assumes removing an element halves the power. B (24 W) assumes the power does not change. D (96 W) assumes the power is four times larger, using both the doubled current and doubled p.d. | ✓ 1 | Answer C |
Answer: C
Exam tip: Use P = V²/R when the p.d. is fixed: halving the resistance doubles the power.
A honey extractor used by a beekeeper has a motor rated at 690 W. It is connected to the UK mains supply of 230 V through a plug containing a fuse.
State the colours of the insulation on the earth wire.
Calculate the current in the motor when it operates normally.
The plug can be fitted with a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason.
The motor runs for 20 minutes. Electricity costs 30 p per kWh. Calculate the cost of running the motor for this time. Give your answer in pence.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 green and yellow (stripes) | ✓ 1 | |
| Part (b) | ||
| M1 I = P ÷ V = 690 ÷ 230 | ✓ 1 | |
| A1 3.0 A | ✓ 1 | |
| Part (c) | ||
| B1 5 A (ECF from (b)) | ✓ 1 | |
| B1 the rating is just above the normal current of 3.0 A, so the fuse blows if the current rises above a safe value (a 3 A fuse could blow in normal use and a 13 A fuse would allow too much current) | ✓ 1 | |
| Part (d) | ||
| M1 energy = 0.69 kW × (20/60) h | ✓ 1 | |
| A1 0.23 kWh | ✓ 1 | |
| A1 0.23 × 30 = 6.9 p | ✓ 1 | |
Exam tip: Use kW with hours to get kWh. Convert 690 W to 0.69 kW before multiplying.
A hospital bed monitor has an LED indicator. The LED is connected in series with a resistor to a 6.0 V battery. When the LED glows, the p.d. across it is 2.0 V and the current is 20 mA.
State the type of particle that carries charge in the connecting wires.
The battery supplies direct current (d.c.). Mains electricity is alternating current (a.c.). State the difference between a.c. and d.c.
The LED does not glow when the battery is connected the opposite way round. Explain why.
Calculate the resistance of the resistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 electrons | ✓ 1 | |
| Part (b) | ||
| B1 d.c. flows in one direction only, a.c. repeatedly reverses direction | ✓ 1 | |
| Part (c) | ||
| B1 an LED only allows current to pass in one direction | ✓ 1 | |
| B1 reversed, it has very high resistance so (almost) no current passes | ✓ 1 | |
| Part (d) | ||
| M1 p.d. across resistor = 6.0 − 2.0 = 4.0 V | ✓ 1 | |
| M1 R = V ÷ I = 4.0 ÷ 0.020 | ✓ 1 | |
| A1 200 Ω | ✓ 1 | |
Common error: The 20 mA must be changed to 0.020 A before dividing, and the p.d. used is the one across the resistor only, not the 6.0 V of the battery.
A hospital steriliser cabinet has a power of 1.2 kW and is connected to the 230 V mains supply. The hospital pays £0.28 for each kWh of electrical energy.
Calculate the current in the steriliser.
The steriliser is used for 4.5 hours each day. Calculate the cost of this use per day. Give your answer to the nearest penny.
A second steriliser has a power of 0.80 kW. Over one week it costs £10.08 to run. Calculate how many hours it was used in the week.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P ÷ V = 1200 ÷ 230 | ✓ 1 | |
| A1 5.2 A | ✓ 1 | |
| Part (b) | ||
| M1 energy = 1.2 × 4.5 = 5.4 kWh and cost = 5.4 × 0.28 | ✓ 1 | |
| A1 £1.51 | ✓ 1 | |
| Part (c) | ||
| M1 energy = 10.08 ÷ 0.28 | ✓ 1 | |
| A1 36 kWh | ✓ 1 | |
| A1 36 ÷ 0.80 = 45 hours | ✓ 1 | |
Exam tip: Cost = energy in kWh × price per kWh, so reverse the steps when the cost is given.
A pit-lane test rig uses a filament lamp. The table shows current I for different p.d. V across the lamp.
V (V): 2.0, 4.0, 6.0, 8.0
I (A): 0.40, 0.70, 0.90, 1.00
Calculate the resistance of the lamp when the p.d. is 8.0 V.
Explain why the resistance of the filament increases as the p.d. increases.
Describe the shape of the graph of current against p.d. for the lamp.
Calculate the power of the lamp when the p.d. is 6.0 V.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V ÷ I = 8.0 ÷ 1.00 | ✓ 1 | |
| A1 8.0 Ω | ✓ 1 | |
| Part (b) | ||
| B1 the filament gets hotter as more power is transferred | ✓ 1 | |
| B1 metal ions vibrate more so electrons collide with them more often, which increases resistance | ✓ 1 | |
| Part (c) | ||
| B1 a curve through the origin | ✓ 1 | |
| B1 the gradient decreases as p.d. increases (curve bends towards the V-axis) | ✓ 1 | |
| Part (d) | ||
| M1 P = V × I = 6.0 × 0.90 | ✓ 1 | |
| A1 5.4 W | ✓ 1 | |
Common error: The resistance at a point is V ÷ I at that point, not the gradient of the graph.
A plastic cover for a telescope is rubbed with a dry cloth and becomes positively charged. Which statement explains this?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: C. Only electrons are free to move between insulating surfaces; the cover loses electrons and so is left with more positive than negative charge. A has the electrons moving the wrong way, which would charge the cover negatively. B is wrong because protons are fixed in nuclei and do not transfer. D is wrong because charge is conserved, not created. | ✓ 1 | Answer C |
Answer: C
Common error: A surface becomes positive by losing electrons, never by gaining protons.
In a school laboratory a lamp is connected in series with an 8.0 Ω resistor and a power supply. The ammeter reads 0.25 A. A voltmeter connected across the lamp reads 2.5 V.
Calculate the p.d. across the resistor.
State the p.d. of the power supply.
Calculate the resistance of the lamp.
The circuit is switched on for 2.0 minutes. Calculate the energy transferred by the lamp.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V = I R = 0.25 × 8.0 | ✓ 1 | |
| 2.0 V | ✓ 1 | |
| Part (b) | ||
| 4.5 V (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| R = V / I = 2.5 / 0.25 | ✓ 1 | |
| 10 Ω | ✓ 1 | |
| Part (d) | ||
| E = V I t = 2.5 × 0.25 × 120 | ✓ 1 | |
| 75 J | ✓ 1 | |
Common error: Use the p.d. across the lamp, not the supply p.d., when finding the lamp's resistance or energy transfer.
A sports scientist uses an NTC thermistor to measure skin temperature. The thermistor is connected across a 6.0 V cell with an ammeter in series (ignore the resistance of the cell and ammeter). At 20 °C the current is 2.0 mA. At 36 °C the current is 6.0 mA.
Calculate the resistance of the thermistor at 20 °C.
Calculate the resistance of the thermistor at 36 °C.
Explain why the current increases when the temperature of the thermistor rises.
Calculate the power dissipated in the thermistor at 36 °C.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| R = V / I = 6.0 / 0.0020 | ✓ 1 | |
| 3000 Ω | ✓ 1 | |
| Part (b) | ||
| R = 6.0 / 0.0060 | ✓ 1 | |
| 1000 Ω | ✓ 1 | |
| Part (c) | ||
| The resistance of the thermistor decreases as temperature rises | ✓ 1 | |
| The p.d. is constant, so the current (I = V/R) is larger | ✓ 1 | |
| Part (d) | ||
| P = V I = 6.0 × 0.0060 | ✓ 1 | |
| 0.036 W (36 mW) | ✓ 1 | |
Exam tip: Change milliamps to amps by dividing by 1000 before substituting into V = IR or P = VI.
An electric shower is rated at 6.9 kW when connected to the 230 V mains supply. Give the answers to parts (a) and (b) as indicated; use the unrounded value of the current in part (b).
Calculate the current in the shower.
Calculate the resistance of the heating element, giving your answer to 2 significant figures.
The shower is used for 12 minutes. The electricity costs 25 p per kWh. Calculate the cost of this use, in pence.
The shower circuit has a fuse. Fuses of 3 A, 13 A and 45 A are available. State which fuse is suitable and explain your choice.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P / V = 6900 / 230 | ✓ 1 | |
| 30 A | ✓ 1 | |
| Part (b) | ||
| R = V / I = 230 / 30 | ✓ 1 | |
| 7.7 Ω | ✓ 1 | |
| Part (c) | ||
| time = 0.20 h | ✓ 1 | |
| energy = 6.9 × 0.20 = 1.38 kWh | ✓ 1 | |
| cost = 1.38 × 25 = 34.5 p | ✓ 1 | |
| Part (d) | ||
| 45 A | ✓ 1 | |
| It is the smallest rating that is above the normal current of 30 A, so it will not melt in normal use but will protect the wiring if the current rises (13 A would melt) | ✓ 1 | |
Common error: A fuse must have a rating just above the normal current. A fuse that is too low blows in normal use, and one that is too high gives no protection.
A trainee in a railway workshop investigates an LDR connected to a 6.0 V cell and an ammeter in series. A lamp is placed at different distances d from the LDR.
d = 10 cm, current = 8.0 mA
d = 20 cm, current = 4.0 mA
d = 30 cm, current = 2.5 mA
d = 40 cm, current = 2.0 mA
Calculate the resistance of the LDR when the distance is 30 cm.
Calculate the power dissipated in the LDR when the distance is 20 cm.
Suggest two ways to make the resistance values more reliable.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| R = V / I = 6.0 / 0.0025 | ✓ 1 | |
| 2400 Ω | ✓ 1 | |
| Part (b) | ||
| P = V I = 6.0 × 0.0040 | ✓ 1 | |
| 0.024 W | ✓ 1 | |
| Part (c) | ||
| Repeat the readings at each distance (and calculate a mean) | ✓ 1 | |
| Keep the room light the same (or use a dark room) for every reading | ✓ 1 | |
Common error: Current readings in milliamps must be converted to amps before using V = IR; forgetting this gives a resistance 1000 times too small.
A robot's motor runs from a 12 V supply and draws a steady current of 3.0 A for 40 s. How much energy does the supply transfer to the motor?
| Marking point | Mark | Notes |
|---|---|---|
| D is correct: charge Q = It = 3.0 × 40 = 120 C, then energy E = QV = 120 × 12 = 1440 J. A (120) is only the charge, forgetting to multiply by the p.d. B (36) is the power in watts, not the energy. C (480) comes from multiplying p.d. by time and omitting the current. | ✓ 1 | Answer D |
Answer: D
Exam tip: Keep track of units: coulombs from I × t, watts from I × V, joules from Q × V or P × t.
A car seat heater has two heating elements of resistance 5.0 Ω and 7.0 Ω connected in series to the 12 V car battery.
Calculate the total resistance of the heater.
Calculate the current in the heater.
Calculate the power dissipated in the 7.0 Ω element.
A switch is closed to short-circuit the 7.0 Ω element so that the current passes only through the 5.0 Ω element. Calculate the total power of the heater now.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 5.0 + 7.0 = 12 Ω | ✓ 1 | |
| Part (b) | ||
| Use I = V ÷ R | ✓ 1 | |
| I = 12 ÷ 12 = 1.0 A | ✓ 1 | |
| Part (c) | ||
| Use P = I2R | ✓ 1 | |
| P = 1.02 × 7.0 = 7.0 W | ✓ 1 | |
| Part (d) | ||
| Current = 12 ÷ 5.0 = 2.4 A | ✓ 1 | |
| Use P = IV (or V2/R) | ✓ 1 | |
| P = 2.4 × 12 = 28.8 W | ✓ 1 | |
Exam tip: Use P = I²R when the same current flows through a component, and P = V²/R when the p.d. is known.
A fish hatchery uses a pond heater rated 2.3 kW that operates from the 230 V mains supply.
Calculate the current in the heater.
The plug can be fitted with a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason.
The heater runs for 5.0 hours. Calculate the energy transferred, in kWh.
Electricity costs £0.24 per kWh. Calculate the cost of running the heater for 5.0 hours. Give your answer in pounds (£), to the nearest penny.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Use I = P ÷ V | ✓ 1 | |
| I = 2300 ÷ 230 = 10 A | ✓ 1 | |
| Part (b) | ||
| 13 A fuse | ✓ 1 | |
| The rating is just above the normal current of 10 A, so it does not blow normally but will blow if a fault current flows | ✓ 1 | |
| Part (c) | ||
| Use energy = power × time | ✓ 1 | |
| 2.3 × 5.0 = 11.5 kWh | ✓ 1 | |
| Part (d) | ||
| Cost = energy × price per kWh | ✓ 1 | |
| 11.5 × 0.24 = £2.76 | ✓ 1 | |
Exam tip: For cost questions, the energy must be in kWh: power in kW multiplied by time in hours.
A robot arm's control board has two components, of resistance 8.0 Ω and 12 Ω, connected in parallel to a 6.0 V battery of negligible internal resistance.
Calculate the current in the 8.0 Ω component.
Calculate the total current drawn from the battery.
Explain why the combined resistance of the two components is less than 8.0 Ω.
Calculate the combined resistance of the two components.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Use I = V ÷ R | ✓ 1 | |
| I = 6.0 ÷ 8.0 = 0.75 A | ✓ 1 | |
| Part (b) | ||
| Current in 12 Ω component = 6.0 ÷ 12 = 0.50 A | ✓ 1 | |
| Total = 0.75 + 0.50 = 1.25 A | ✓ 1 | |
| Part (c) | ||
| The battery supplies current through both branches (more paths) | ✓ 1 | |
| The total current is larger than for either component alone, so V ÷ I is smaller | ✓ 1 | |
| Part (d) | ||
| Use R = V ÷ I | ✓ 1 | |
| R = 6.0 ÷ 1.25 = 4.8 Ω (ECF from (b)) | ✓ 1 | |
Exam tip: Adding a parallel branch always lowers the total resistance of the circuit.
A synthetic dress made of polyester clings to the wearer after it is rubbed against wool. Later, a spark passes between the dress and a metal door handle.
Explain why the dress becomes charged when it is rubbed against the wool.
Explain why a charged dress attracts small pieces of uncharged paper.
The spark transfers 0.30 μC of charge in 2.0 ms. Calculate the average current in the spark.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Electrons are transferred between the surfaces by friction | ✓ 1 | |
| The dress gains (or loses) electrons so it has an overall charge, and the wool has the opposite charge | ✓ 1 | |
| Part (b) | ||
| The charge on the dress repels like charges in the paper and attracts opposite charges, so the paper pieces become polarised (induced charges) | ✓ 1 | |
| The nearer opposite charge is attracted more strongly than the repelled charge | ✓ 1 | |
| Part (c) | ||
| Convert units: 0.30 μC = 3.0 × 10−7 C and 2.0 ms = 2.0 × 10−3 s | ✓ 1 | |
| I = Q ÷ t = 3.0 × 10−7 ÷ 2.0 × 10−3 = 1.5 × 10−4 A | ✓ 1 | |
Common error: Convert μC and ms to C and s before using I = Q ÷ t.
In electrostatic paint spraying, paint droplets leave a charged nozzle and are directed at an earthed metal car panel. Which statement gives a correct explanation of why a fine, even coat of paint forms?
| Marking point | Mark | Notes |
|---|---|---|
| All the droplets are charged with the same sign, so they repel and spread out. Electrons move between Earth and the panel, leaving the panel with an opposite charge near the droplets, so they are attracted to it: option A is correct. B is wrong because the droplets have like charges, not unlike. C is wrong because the droplets are charged and the attraction is electrostatic. D is wrong because only electrons move in a solid conductor; protons cannot flow from Earth. | ✓ 1 | Answer A |
Answer: A
Common error: In solid conductors only electrons move. A positive charge arises from a loss of electrons, never from protons arriving.
A mountaineer uses a heated sleeping-bag liner powered from a 12 V battery pack. The liner has a power of 36 W. The capacity of the pack is 9.0 A h, which means it can supply 9.0 A for 1.0 hour.
Calculate the current in the liner.
Show that the pack can supply a total charge of about 3.2 × 104 C.
Calculate how long the pack can run the liner. Give your answer in hours.
The mountaineer wants the liner to run for 8.0 hours on one charge. Calculate the minimum capacity of pack needed, in A h.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P ÷ V | ✓ 1 | |
| I = 36 ÷ 12 = 3.0 A | ✓ 1 | |
| Part (b) | ||
| 1.0 hour = 3600 s | ✓ 1 | |
| Q = I t = 9.0 × 3600 = 32 400 C ≈ 3.2 × 104 C | ✓ 1 | |
| Part (c) | ||
| t = Q ÷ I = 32 400 ÷ 3.0 = 10 800 s | ✓ 1 | |
| 10 800 s ÷ 3600 = 3.0 h (ECF from (a)) | ✓ 1 | |
| Part (d) | ||
| Q = I t = 3.0 × 8.0 × 3600 = 86 400 C (ECF from (a)) | ✓ 1 | |
| Capacity = 86 400 ÷ 3600 = 24 A h | ✓ 1 | |
Exam tip: An ampere-hour is a unit of charge. Convert it to coulombs by multiplying by 3600 before using Q = I t.
A battery bank at a remote mobile-phone mast stores charge for use during power cuts. The battery bank is charged by a 24 V charger that supplies a steady current of 12 A. It must receive a total charge of 5.4 × 105 C. Use 1 kWh = 3.6 × 106 J.
Calculate the time needed to charge the battery bank. Give your answer in hours.
Calculate the energy transferred to the battery bank, in kWh.
The charger takes 4.5 kWh from the mains while charging the battery bank. Calculate the efficiency of the charger as a percentage.
Electricity costs £0.24 per kWh. Calculate the cost of the 4.5 kWh taken from the mains.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| t = Q ÷ I = 5.4 × 105 ÷ 12 = 45 000 s | ✓ 1 | |
| 45 000 ÷ 3600 = 12.5 h | ✓ 1 | |
| Part (b) | ||
| E = Q V = 5.4 × 105 × 24 = 1.296 × 107 J | ✓ 1 | |
| 1.296 × 107 ÷ 3.6 × 106 = 3.6 kWh | ✓ 1 | |
| Part (c) | ||
| efficiency = useful energy out ÷ total energy in (× 100) | ✓ 1 | |
| (3.6 ÷ 4.5) × 100 = 80% (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| cost = energy in kWh × price per kWh | ✓ 1 | |
| 4.5 × 0.24 = £1.08 | ✓ 1 | |
Exam tip: Efficiency compares useful energy out with total energy in, so use the mains energy for the cost, not the energy stored.
In an electrostatic paint-spraying system a car panel is coated with paint droplets. The droplets leave a charged nozzle and the metal panel is connected to Earth.
Explain how the paint droplets become positively charged as they leave the nozzle.
Explain why the droplets spread out to form a fine mist.
Explain why the droplets are attracted to the earthed panel.
During one spraying run the droplets deliver a total charge of 0.36 mC to the panel in 2.0 minutes. Calculate the average current. Give your answer in µA.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Electrons are transferred from the droplets to the nozzle | ✓ 1 | |
| so the droplets have fewer electrons than protons, giving them a positive charge | ✓ 1 | |
| Part (b) | ||
| All the droplets carry the same sign of charge | ✓ 1 | |
| like charges repel, so the droplets push each other apart | ✓ 1 | |
| Part (c) | ||
| Electrons flow from Earth to the panel, so the panel gains a negative charge near the droplets | ✓ 1 | |
| unlike charges attract, so the positive droplets are pulled towards the panel | ✓ 1 | |
| Part (d) | ||
| I = Q ÷ t with t = 120 s and Q = 0.36 × 10−3 C | ✓ 1 | |
| I = 3.0 × 10−6 A = 3.0 µA | ✓ 1 | |
Common error: Positive charge comes from removing electrons. Say that electrons move, and name where they move from and to.
A telecoms engineer is working in a site cabin. The engineer plugs a fan heater of power 1300 W, a kettle of power 1150 W and a drill of power 770 W into one extension lead. The lead has a 13 A fuse and the mains supply is 230 V.
Calculate the total power of the three appliances.
Calculate the total current when all three appliances are on. Give your answer to 3 significant figures.
State and explain what happens to the fuse in the lead.
The kettle has a metal case connected to the earth wire. Explain how the earth wire makes the kettle safe if the live wire touches the case.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1300 + 1150 + 770 = 3220 W | ✓ 1 | |
| Part (b) | ||
| I = P ÷ V | ✓ 1 | |
| I = 3220 ÷ 230 = 14.0 A (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| The fuse blows (melts) | ✓ 1 | |
| because the current (14.0 A) is greater than the 13 A rating, which breaks the circuit | ✓ 1 | |
| Part (d) | ||
| A large current flows through the earth wire to Earth (low-resistance path) | ✓ 1 | |
| this melts the fuse and disconnects the live supply, so the case does not stay live | ✓ 1 | |
Common error: A fuse protects the wiring, so compare the total current with the fuse rating. Do not compare the power of each appliance with it.
A radio mast is on a mountain summit. Charge builds up in storm clouds above it, and lightning can strike the mast.
Explain how charge builds up in a storm cloud when ice particles collide.
The mast is connected to Earth by a thick copper cable. Explain why copper and a thick cable are used.
A lightning strike transfers 15 C of charge in 2.0 ms. Calculate the average current. Give your answer in A.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Friction between the ice particles transfers electrons from one to the other | ✓ 1 | |
| the particle gaining electrons becomes negative and the one losing electrons becomes positive | ✓ 1 | |
| Part (b) | ||
| Copper is a good conductor (low resistance) so the charge flows to Earth through the cable | ✓ 1 | |
| a thick cable has even lower resistance and does not overheat or melt with a large current | ✓ 1 | |
| Part (c) | ||
| I = Q ÷ t with t = 0.0020 s | ✓ 1 | |
| I = 15 ÷ 0.0020 = 7500 A | ✓ 1 | |
Exam tip: Convert milliseconds to seconds before using I = Q ÷ t: 2.0 ms = 0.0020 s.
A cordless drill motor draws a constant current of 15 A for 40 s. How much charge flows through the motor?
| Marking point | Mark | Notes |
|---|---|---|
| Q = It = 15 × 40 = 600 C (D). A comes from I ÷ t. B comes from t ÷ I. C comes from converting 40 s into minutes and then multiplying, which is not needed because the time is already in seconds. | ✓ 1 | Answer D |
Answer: D
Common error: Charge = current × time, with time in seconds; do not divide when you should multiply.
An electric lawnmower is rated 1200 W and works from the 230 V mains supply. The plug contains a fuse. Fuses of 3 A, 5 A, 10 A and 13 A are available.
Calculate the current in the lawnmower when it works normally.
Select the most suitable fuse and give a reason for your choice.
A fault makes the live wire touch the metal casing of the lawnmower. Explain how the fuse protects the user.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P = I × V | ✓ 1 | |
| I = 1200 ÷ 230 | ✓ 1 | |
| 5.2 A | ✓ 1 | |
| Part (b) | ||
| 10 A | ✓ 1 | |
| 5 A would melt in normal use as the current is above 5 A / the rating is the lowest above the working current | ✓ 1 | |
| ECF from (a) | — | |
| Part (c) | ||
| a large current flows (through the live wire to earth) | ✓ 1 | |
| the fuse wire heats up and melts | ✓ 1 | |
| the circuit is broken, so the casing is no longer live | ✓ 1 | |
Exam tip: A fuse goes in the live wire so that it cuts the appliance off from the supply when it melts.
A different lawnmower motor takes a current of 6.5 A from the 230 V mains supply.
Calculate the power of the motor.
The lawnmower is used for 25 minutes. Calculate the energy transferred, in kWh. Give your answer to 3 significant figures.
Electricity costs 30 p per kWh. Calculate the cost of using the lawnmower for 25 minutes. Use 0.623 kWh.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P = I × V | ✓ 1 | |
| 1495 W | ✓ 1 | |
| Part (b) | ||
| convert power to 1.495 kW and time to 25/60 h | ✓ 1 | |
| E = P × t | ✓ 1 | |
| 0.623 kWh | ✓ 1 | |
| ECF from (a) | — | |
| Part (c) | ||
| cost = 0.623 × 30 | ✓ 1 | |
| 18.69 p (accept 18.7 p) | ✓ 1 | |
Common error: Keep kW with hours (or W with seconds); do not mix them.
An electric train takes a current of 80 A from a 750 V overhead supply for 4.0 minutes while it accelerates. The charge on one electron is 1.6 × 10−19 C.
Calculate the charge that passes through the train's motors in this time.
Calculate the energy transferred to the train. Use the value of charge from (a).
Calculate the number of electrons that pass through the motors in this time.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| time = 240 s | ✓ 1 | |
| Q = I × t = 19200 C | ✓ 1 | |
| Part (b) | ||
| E = Q × V | ✓ 1 | |
| 14400000 J (accept 1.44 × 107 J) | ✓ 1 | |
| ECF from (a) | — | |
| Part (c) | ||
| number = Q ÷ e = 19200 ÷ (1.6 × 10−19) | ✓ 1 | |
| 1.2 × 1023 | ✓ 1 | |
| ECF from (a) | — | |
Common error: Convert minutes into seconds before using Q = It.
A spacecraft hull collects a net negative charge of 3.2 μC from the solar wind. The charge on one electron is 1.6 × 10−19 C.
Explain why the hull becomes negatively charged.
Calculate the number of electrons gained by the hull.
The charge flows to a metal frame in 0.80 ms. Calculate the average current.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the hull gains electrons (from the solar wind) | ✓ 1 | |
| electrons carry negative charge, so there is an excess of negative charge | ✓ 1 | |
| Part (b) | ||
| number = 3.2 × 10−6 ÷ (1.6 × 10−19) | ✓ 1 | |
| 2.0 × 1013 | ✓ 1 | |
| Part (c) | ||
| I = Q ÷ t = 3.2 × 10−6 ÷ 0.80 × 10−3 | ✓ 1 | |
| 4.0 × 10−3 A (0.004 A) | ✓ 1 | |
Exam tip: Convert μC and ms to C and s before you substitute.
A sports-science team builds a pressure-sensor circuit for a gym mat. A 12 Ω resistor and a 6.0 Ω resistor are connected in parallel across a 9.0 V cell of negligible internal resistance.
Calculate the total resistance of the two resistors.
Calculate the current in the cell.
Calculate the power dissipated in the 12 Ω resistor. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1/R = 1/12 + 1/6.0 | ✓ 1 | |
| 1/R = 0.25 | ✓ 1 | |
| 4.0 Ω | ✓ 1 | |
| accept (12 × 6.0) ÷ (12 + 6.0) for the first two marks | — | |
| Part (b) | ||
| I = V ÷ R = 9.0 ÷ 4.0 | ✓ 1 | |
| 2.25 A | ✓ 1 | |
| ECF from (a) | — | |
| Part (c) | ||
| P = V2 ÷ R = 9.02 ÷ 12 (or I = 0.75 A and P = IV) | ✓ 1 | |
| 6.75 W | ✓ 1 | |
Common error: In parallel, each branch has the full cell potential difference; the total resistance is smaller than the smallest branch.
At a data centre, a server rack is supplied through three parallel cables. The current in the main supply is 3.0 A. The currents in two of the cables are 0.8 A and 1.5 A. What is the current in the third cable?
| Marking point | Mark | Notes |
|---|---|---|
| At a junction the current entering equals the current leaving, so 3.0 − 0.8 − 1.5 = 0.7 A, option A. 2.3 A forgets to subtract the second cable. 5.3 A adds all three currents. 1.5 A copies one of the given values. | ✓ 1 | Answer A |
Answer: A
Common error: Currents in parallel branches add up to the supply current; they do not each equal it.
A cycling workshop uses a 2.3 kW heater on the 230 V mains supply. Which fuse is most suitable for the plug?
| Marking point | Mark | Notes |
|---|---|---|
| I = P/V = 2300/230 = 10 A, so the fuse must be rated just above 10 A: 13 A, option C. 3 A and 5 A are below the normal current and would blow immediately. 30 A is far too large and would not protect the cable. | ✓ 1 | Answer C |
Answer: C
Exam tip: Choose the smallest standard fuse rated above the normal working current.
A bicycle's lighting circuit has a front lamp rated 6.0 V, 2.4 W and a rear lamp rated 6.0 V, 0.60 W. They are connected in parallel to a 6.0 V dynamo and both work at their rated values.
Calculate the current in each lamp.
State the current in the dynamo.
Calculate the energy transferred by both lamps together in 30 minutes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P ÷ V | ✓ 1 | |
| A1 0.40 A (front) and 0.10 A (rear) | ✓ 1 | |
| Part (b) | ||
| B1 0.50 A (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 total power 3.0 W and time 1800 s | ✓ 1 | |
| A1 E = 3.0 × 1800 = 5400 J | ✓ 1 | |
Exam tip: The total power of parallel lamps is the sum of their powers, since each has the same p.d.
A data centre has a 920 W dehumidifier connected to a 230 V mains supply through a plug containing a fuse.
Calculate the normal current in the dehumidifier.
Fuses of 3 A, 5 A and 13 A are available. State which fuse should be fitted and give a reason.
Explain how a fuse protects the appliance when a fault causes too large a current.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P ÷ V = 920 ÷ 230 | ✓ 1 | |
| A1 4.0 A | ✓ 1 | |
| Part (b) | ||
| B1 5 A | ✓ 1 | |
| B1 rating just above the normal current (4.0 A), so it will not blow in normal use but protects the circuit if the current rises | ✓ 1 | |
| Part (c) | ||
| B1 the large current makes the fuse wire heat up and melt | ✓ 1 | |
| B1 this breaks the circuit, stopping the current | ✓ 1 | |
Exam tip: The fuse sits in the live wire so that the appliance is disconnected from the live supply when it blows.
A metal-cased drying oven in a laboratory at an archaeological site is connected to the mains with a three-core cable. The earth wire has become disconnected inside the oven.
Explain why the oven is dangerous if the live wire then touches the metal casing.
Explain how a correctly connected earth wire would prevent the danger.
State why a double-insulated oven would not need an earth wire.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 the casing becomes live (at mains potential) | ✓ 1 | |
| B1 a person touching it gives a path to earth, so current passes through the body causing an electric shock | ✓ 1 | |
| Part (b) | ||
| B1 a large current flows from the casing through the low-resistance earth wire | ✓ 1 | |
| B1 the fuse melts | ✓ 1 | |
| B1 the live supply is disconnected, so the casing is no longer live | ✓ 1 | |
| Part (c) | ||
| B1 it has no exposed metal part that a user can touch | ✓ 1 | |
| B1 so even a live fault inside cannot make a touchable part live | ✓ 1 | |
Common error: The earth wire does not carry current in normal use; it only matters when a fault makes the casing live.
An archaeologist's head torch has a bulb marked 6.0 V, 0.30 A in normal use. It is run from a 9.0 V battery with a resistor connected in series with the bulb.
Calculate the resistance of the bulb in normal use.
Calculate the resistance of the series resistor needed for the bulb to work normally.
Calculate the power dissipated in the resistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V ÷ I = 6.0 ÷ 0.30 | ✓ 1 | |
| A1 20 Ω | ✓ 1 | |
| Part (b) | ||
| M1 p.d. across resistor = 9.0 − 6.0 = 3.0 V | ✓ 1 | |
| M1 R = 3.0 ÷ 0.30 | ✓ 1 | |
| A1 10 Ω | ✓ 1 | |
| Part (c) | ||
| M1 P = V I = 3.0 × 0.30 | ✓ 1 | |
| A1 0.90 W | ✓ 1 | |
Exam tip: In series the current is the same everywhere, and the p.d.s across the components add up to the supply p.d.
A data-centre cooling unit has a power of 3.2 kW and runs continuously for 24 hours. Electricity costs £0.20 per kWh.
Calculate the energy transferred in kWh in 24 hours.
Calculate the cost of running the unit for 24 hours.
A more efficient unit has a power of 2.4 kW. Calculate how much money is saved in 24 hours by using the efficient unit instead. Give your answer in £.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 E = P × t = 3.2 × 24 | ✓ 1 | |
| A1 76.8 kWh | ✓ 1 | |
| Part (b) | ||
| M1 76.8 × 0.20 | ✓ 1 | |
| A1 £15.36 | ✓ 1 | |
| Part (c) | ||
| M1 (3.2 − 2.4) × 24 × 0.20 | ✓ 1 | |
| A1 £3.84 | ✓ 1 | |
Exam tip: Use power in kW and time in hours to get energy in kWh; do not convert to joules for costs.
A student investigates a metal wire used in a cycle sensor. The table shows the p.d. across the wire and the current in it.
p.d. / V: 1.0, 2.0, 3.0, 4.0
Current / A: 0.20, 0.40, 0.60, 0.80
Calculate the resistance of the wire when the p.d. is 3.0 V.
Explain how the data show that the wire is an ohmic conductor.
A second wire of the same material and cross-sectional area is twice as long. Calculate the current in the second wire when the p.d. is 4.0 V. Use the resistance from (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V ÷ I = 3.0 ÷ 0.60 | ✓ 1 | |
| A1 5.0 Ω | ✓ 1 | |
| Part (b) | ||
| B1 the current is directly proportional to the p.d. (doubling p.d. doubles current) | ✓ 1 | |
| B1 V ÷ I is constant (5.0 Ω) for all readings | ✓ 1 | |
| Part (c) | ||
| M1 resistance = 2 × 5.0 = 10 Ω, I = 4.0 ÷ 10 | ✓ 1 | |
| A1 0.40 A | ✓ 1 | |
Common error: Doubling the length of a wire doubles its resistance, so the current at the same p.d. halves.
A heated glove for a crop-spraying pilot has two heating elements of resistance 4.0 Ω and 2.0 Ω connected in series to a 12 V supply. What is the power dissipated in the 4.0 Ω element?
| Marking point | Mark | Notes |
|---|---|---|
| Total resistance = 6.0 Ω, so the current is 12 ÷ 6.0 = 2.0 A. The p.d. across the 4.0 Ω element is 2.0 × 4.0 = 8.0 V, so P = 8.0 × 2.0 = 16 W. The value 36 W comes from using the full 12 V across this element alone. The value 24 W is the total power of both elements together. The value 8 W is the power in the 2.0 Ω element (4.0 V × 2.0 A). | ✓ 1 | Answer A |
Answer: A
Exam tip: In a series circuit the supply p.d. is shared, so find the current first and then the p.d. across the component you need.
A light-beam alarm protects a fuel store at a small airfield. A narrow beam of light shines on a light-dependent resistor (LDR). The LDR is connected in series with a fixed resistor of resistance 1000 Ω and a 6.0 V battery. When the beam is unbroken the LDR has a resistance of 500 Ω. When the beam is broken the LDR has a resistance of 9000 Ω. The alarm sounds when the current in the circuit falls below 1.0 mA.
Calculate the current in the circuit when the beam is unbroken.
Explain, using the properties of the LDR, why the current falls when the beam is broken.
Calculate the current when the beam is broken.
Calculate the resistance of the LDR at which the alarm just sounds.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| total resistance = 500 + 1000 = 1500 Ω | ✓ 1 | |
| I = 6.0 ÷ 1500 = 0.0040 A (4.0 mA) | ✓ 1 | |
| Part (b) | ||
| less light falls on the LDR so its resistance increases | ✓ 1 | |
| total resistance of the circuit increases | ✓ 1 | |
| battery p.d. is constant so current = V ÷ R decreases | ✓ 1 | |
| Part (c) | ||
| total resistance = 9000 + 1000 = 10 000 Ω | ✓ 1 | |
| I = 6.0 ÷ 10 000 = 6.0 × 10−4 A (0.60 mA) | ✓ 1 | |
| Part (d) | ||
| total resistance = V ÷ I = 6.0 ÷ 0.0010 = 6000 Ω | ✓ 1 | |
| LDR resistance = total − fixed resistor = 6000 − 1000 | ✓ 1 | |
| = 5000 Ω | ✓ 1 | |
Exam tip: In a series circuit with a fixed resistor, find the total resistance first, then subtract the known resistance to find the unknown one.
The indicator panel in an aircraft cockpit has light-emitting diodes (LEDs). Each LED is connected to a 5.0 V d.c. supply through a resistor, using copper wires.
State what carries the current in the copper wires.
Explain why an LED does not light when it is connected the wrong way round.
Suggest why a resistor is connected in series with each LED.
An LED carries a current of 15 mA for 8.0 minutes. Calculate the charge that passes through it.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (free) electrons | ✓ 1 | |
| Part (b) | ||
| an LED (diode) only allows current in one direction | ✓ 1 | |
| in the reverse direction its resistance is very high so (almost) no current flows | ✓ 1 | |
| Part (c) | ||
| to limit the current | ✓ 1 | |
| to stop the LED overheating / being damaged | ✓ 1 | |
| Part (d) | ||
| Q = I t | ✓ 1 | |
| = 0.015 × 480 = 7.2 C | ✓ 1 | |
Common error: Convert milliamperes to amperes and minutes to seconds before using Q = I t.
A farmer uses a 2.4 kW water-trough heater connected to the 230 V mains supply through a plug containing a fuse. The heater has a metal casing. Fuses of rating 3 A, 5 A, 13 A and 30 A are available.
Calculate the current in the heater. Give your answer to 3 significant figures.
State which fuse should be used and give a reason for your choice.
Explain how the earth wire and the fuse together protect a person if the live wire becomes loose and touches the metal casing.
The heater is on for 8.0 hours. Electricity costs £0.28 per kWh. Calculate the cost, to the nearest penny. Use unrounded values in your working.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P ÷ V | ✓ 1 | |
| = 2400 ÷ 230 = 10.4 A | ✓ 1 | |
| Part (b) | ||
| 13 A | ✓ 1 | |
| its rating is above the normal current (10.4 A) so it will not melt in normal use, but it is as close as possible so it melts if the current becomes too large | ✓ 1 | |
| Part (c) | ||
| a large current flows from the casing to earth through the earth wire | ✓ 1 | |
| the large current makes the fuse melt | ✓ 1 | |
| this breaks the circuit so the casing cannot stay live and a shock is prevented | ✓ 1 | |
| Part (d) | ||
| energy = 2.4 × 8.0 = 19.2 kWh | ✓ 1 | |
| cost = 19.2 × 0.28 | ✓ 1 | |
| = £5.38 | ✓ 1 | |
Exam tip: When asked to choose a fuse, pick the rating just above the normal working current, not simply the largest one available.
A person touches a door handle and a spark passes. The charge transferred is 4.5 µC and the spark lasts 1.5 µs. What is the average current in the spark?
| Marking point | Mark | Notes |
|---|---|---|
| Current = charge ÷ time = 4.5 × 10−6 C ÷ 1.5 × 10−6 s = 3.0 A. The prefixes cancel, so the answer is a simple ratio. 30 A comes from slipping a power of ten. 0.33 A comes from dividing time by charge. 6.8 × 10−12 A comes from multiplying charge by time. | ✓ 1 | Answer A |
Answer: A
Exam tip: Convert µ to ×10⁻⁶ before substituting into Q = It and check that the powers of ten cancel sensibly.
A lamp is marked 6.0 V, 2.0 W. It is connected in series with a resistor to a 12 V supply so that the lamp works at its marked values. What is the resistance of the resistor?
| Marking point | Mark | Notes |
|---|---|---|
| Current in the lamp = P ÷ V = 2.0 ÷ 6.0 = 0.33 A, and this is also the current in the resistor. The resistor must take 12 − 6.0 = 6.0 V, so R = 6.0 ÷ 0.33 = 18 Ω. 36 Ω uses the whole 12 V across the resistor. 3.0 Ω comes from dividing the voltage by the power. 0.50 Ω comes from multiplying power and voltage to get a current of 12 A and then dividing. | ✓ 1 | Answer B |
Answer: B
Exam tip: In a series circuit the current is the same everywhere and the pds add up to the supply pd, so find the current first.
A student walks across a nylon carpet and then touches a metal door handle. A spark jumps between the student's finger and the handle. The student's body is positively charged before the spark.
Explain, in terms of potential difference, why a spark jumps.
The spark transfers 6.0 × 10−7 C in 1.5 × 10−6 s. Calculate the average current in the spark.
State the direction in which electrons move during the spark and explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The charged body has a large potential difference relative to the handle | ✓ 1 | |
| The air between them breaks down (ionises) and conducts, so charge flows | ✓ 1 | |
| Part (b) | ||
| Use of I = Q ÷ t | ✓ 1 | |
| I = 6.0 × 10−7 ÷ 1.5 × 10−6 = 0.40 A | ✓ 1 | |
| Part (c) | ||
| Electrons move from the handle to the student | ✓ 1 | |
| A positive body has a shortage of electrons, so electrons are attracted to it | ✓ 1 | |
Exam tip: Current is the rate of flow of charge, so a very small charge in a very short time can still give a large current.
A model lighthouse has a lamp and a resistor connected in series to a 9.0 V battery. The lamp works normally with a potential difference of 6.0 V across it when the current is 0.50 A.
State the current in the resistor.
Calculate the power of the lamp.
Calculate the resistance of the resistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0.50 A (same as the lamp because they are in series) | ✓ 1 | |
| Part (b) | ||
| Use of P = IV | ✓ 1 | |
| P = 0.50 × 6.0 = 3.0 W | ✓ 1 | |
| Part (c) | ||
| pd across resistor = 9.0 − 6.0 = 3.0 V | ✓ 1 | |
| Use of R = V ÷ I | ✓ 1 | |
| R = 3.0 ÷ 0.50 = 6.0 Ω | ✓ 1 | |
Common error: Do not use the battery voltage with the resistor alone; the resistor only takes the share of the pd that the lamp does not.
A thermistor is connected in series with a fixed resistor of 1000 Ω and a 12 V battery. At 20 °C the thermistor has a resistance of 4000 Ω. At 60 °C its resistance is 800 Ω. What happens to the potential difference across the fixed resistor when the thermistor is heated from 20 °C to 60 °C?
| Marking point | Mark | Notes |
|---|---|---|
| Heating lowers the resistance of the thermistor, so the total resistance falls and the current rises (2.4 mA to 6.7 mA). The pd across the fixed resistor is I × 1000 Ω, which increases from 2.4 V to 6.7 V. B gives the pd across the thermistor, not across the fixed resistor. C wrongly assumes a fixed resistance means a fixed pd. D is wrong because the thermistor still has resistance at 60 °C, so the fixed resistor cannot have the whole 12 V. | ✓ 1 | Answer A |
Answer: A
Exam tip: In a series circuit the pd is shared in proportion to the resistances; when one resistance falls, the other component takes a bigger share.
An electric blanket is rated 90 W. It is used for 10 hours. How much energy does it transfer?
| Marking point | Mark | Notes |
|---|---|---|
| Energy = power × time = 0.090 kW × 10 h = 0.90 kWh. A is wrong because it uses 0.090 kW × 1 h, ignoring the time. C is wrong because it multiplies by 100 instead of 10 (or mis-converts the power). D is wrong because 900 is the value in watt-hours, not kilowatt-hours. | ✓ 1 | Answer B |
Answer: B
Exam tip: Convert watts to kilowatts before multiplying by hours to get kWh.
An electric blanket is connected to the 230 V mains supply and has a power of 120 W. It is switched on for 8.0 hours.
Calculate the current in the blanket. Give your answer to 2 significant figures.
Calculate the energy transferred in 8.0 hours, in joules. Give your answer to 2 significant figures.
Electricity costs €0.25 per kilowatt-hour. Calculate the cost of using the blanket for 8.0 hours.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P ÷ V = 120 ÷ 230 | ✓ 1 | |
| = 0.52 A | ✓ 1 | |
| Part (b) | ||
| E = P × t with t = 8.0 × 3600 s | ✓ 1 | |
| = 3.5×106 J | ✓ 1 | |
| Part (c) | ||
| energy = 0.120 kW × 8.0 h = 0.96 kWh | ✓ 1 | |
| cost = 0.96 × 0.25 = €0.24 | ✓ 1 | |
Exam tip: Use joules with seconds, and kWh with hours; do not mix the two sets of units.
A thermistor is placed in a poultry incubator. The table shows the resistance of the thermistor at different temperatures.
| Temperature / °C | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|
| Resistance / Ω | 3200 | 2000 | 1250 | 800 | 500 |
Describe how the resistance of the thermistor changes as the temperature rises.
The thermistor is connected to a 6.0 V cell of negligible internal resistance. Calculate the current in the thermistor at 40 °C. Give your answer in mA.
Calculate the increase in current when the thermistor is heated from 40 °C to 60 °C.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the resistance decreases as the temperature increases | ✓ 1 | |
| the decrease gets smaller for each 10 °C step (not a straight line / not directly proportional) | ✓ 1 | |
| Part (b) | ||
| I = V ÷ R = 6.0 ÷ 1250 | ✓ 1 | |
| = 4.8 mA | ✓ 1 | |
| Part (c) | ||
| current at 60 °C = 6.0 ÷ 500 = 12 mA | ✓ 1 | |
| increase = 12 − 4.8 = 7.2 mA (ECF from (b)) | ✓ 1 | |
Exam tip: When the pd is fixed, halving the resistance doubles the current; use this to check your answers.
A warning light on a tractor dashboard uses an LED. The LED needs a pd of 2.0 V and a current of 20 mA to work correctly. It is connected in series with a resistor to a 5.0 V supply.
The LED lights when connected one way round but not when reversed. Explain why.
Calculate the resistance of the series resistor.
The warning light stays on for 5.0 minutes. Calculate the charge that flows through the LED.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| an LED allows current in one direction only | ✓ 1 | |
| in the reverse direction its resistance is very high so no current flows (the LED does not light) | ✓ 1 | |
| Part (b) | ||
| pd across the resistor = 5.0 − 2.0 = 3.0 V | ✓ 1 | |
| R = V ÷ I with I = 0.020 A | ✓ 1 | |
| = 150 Ω | ✓ 1 | |
| Part (c) | ||
| Q = I × t with t = 300 s | ✓ 1 | |
| = 6.0 C | ✓ 1 | |
Common error: Change milliamps to amps (20 mA = 0.020 A) before using R = V ÷ I.
An amplifier used in a school music room is connected to the 230 V mains supply and has a power of 460 W. The fuse in the plug can be chosen from 1 A, 3 A, 5 A and 13 A.
Calculate the current in the amplifier.
State which fuse should be used and give a reason.
Explain why the fuse is connected in the live wire and not in the neutral wire.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P ÷ V = 460 ÷ 230 | ✓ 1 | |
| = 2.0 A | ✓ 1 | |
| Part (b) | ||
| 3 A fuse | ✓ 1 | |
| its rating is just above the normal current, so it does not blow in normal use but melts if a fault causes too much current (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| if the fuse blew in the neutral wire the appliance would still be connected to the live wire | ✓ 1 | |
| touching the inside of the appliance could then give an electric shock | ✓ 1 | |
Common error: Choose the fuse with the next rating above the normal current, not the largest one available.
The camera heater on a space probe draws a steady current of 0.40 A for 2.0 minutes. What charge passes through the heater in this time?
| Marking point | Mark | Notes |
|---|---|---|
| Charge Q = I t. The time must be in seconds: 2.0 min = 120 s, so Q = 0.40 × 120 = 48 C. Using 2.0 as the time without converting gives 0.80 C. Dividing time by current gives 300 C, which is not charge. Dividing current by time gives 0.0033 C, which is also not charge. | ✓ 1 | Answer B |
Answer: B
Common error: Always convert minutes to seconds before using Q = I t; the coulomb is defined using seconds.
A diode is tested in the forward direction. The table shows the current for different p.d.s across it.
p.d. / V: 0.0, 0.4, 0.6, 0.7, 0.8
current / mA: 0, 0, 2, 12, 40
State the largest p.d. in the table at which the diode does not conduct.
Calculate the resistance of the diode when the p.d. across it is 0.80 V.
The diode is now connected in series with a resistor and a 3.0 V cell that has negligible internal resistance. The current in the circuit is 40 mA. Calculate the resistance of the resistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 0.4 V | ✓ 1 | |
| Part (b) | ||
| M1 R = V / I with I = 40 mA = 0.040 A | ✓ 1 | |
| A1 R = 0.80 / 0.040 = 20 Ω | ✓ 1 | |
| Part (c) | ||
| B1 p.d. across the diode is 0.80 V (from the table) | ✓ 1 | |
| M1 p.d. across the resistor = 3.0 − 0.80 = 2.2 V | ✓ 1 | |
| A1 R = 2.2 / 0.040 = 55 Ω | ✓ 1 | |
Common error: Convert milliamps to amps before using R = V / I; leaving the current as 40 gives an answer 1000 times too small.
A student investigates a length of thin metal wire using a variable power supply, an ammeter and a voltmeter. The results are shown.
p.d. / V: 1.0, 2.0, 3.0, 4.0
current / A: 0.25, 0.44, 0.60, 0.74
Calculate the resistance of the wire when the p.d. is 3.0 V.
State how the resistance of the wire changes as the p.d. across it increases.
Explain, in terms of particles, why the resistance changes in this way.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V / I | ✓ 1 | |
| A1 R = 3.0 / 0.60 = 5.0 Ω | ✓ 1 | |
| Part (b) | ||
| B1 it increases | ✓ 1 | |
| Part (c) | ||
| B1 a larger current heats the wire | ✓ 1 | |
| B1 the metal ions vibrate with greater amplitude | ✓ 1 | |
| B1 more frequent collisions between electrons and ions, so the flow of charge is opposed more | ✓ 1 | |
Exam tip: In a resistance explanation, link the temperature rise to ion vibrations and then to collisions with the moving electrons.
A coffee roaster has a heating element rated at 2.3 kW on a 230 V supply. One batch of beans takes 12 minutes to roast.
Calculate the current in the heating element.
Calculate the energy transferred in one batch, in kilowatt-hours.
Electricity costs €0.30 per kWh. The roaster is used for 5 batches in a day. Calculate the cost of the energy for the day.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P / V | ✓ 1 | |
| A1 I = 2300 / 230 = 10 A | ✓ 1 | |
| Part (b) | ||
| M1 E = P × t = 2.3 kW × 0.20 h | ✓ 1 | |
| A1 0.46 kWh | ✓ 1 | |
| Part (c) | ||
| M1 cost = 0.46 × 5 × 0.30 | ✓ 1 | |
| A1 €0.69 | ✓ 1 | |
Common error: Time must be in hours for kilowatt-hours: 12 minutes is 0.20 h, not 12 h.
A crew spray-paints a steel bridge girder using an electrostatic sprayer. The paint droplets leave the nozzle with a positive charge. The girder is connected to earth.
Explain why the charged droplets spread out into a fine mist.
Explain why paint also reaches the back of the girder, which faces away from the nozzle.
Describe the movement of charge that keeps the girder uncharged as the droplets land.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 the droplets carry like charges, which repel each other | ✓ 1 | |
| Part (b) | ||
| B1 the earthed girder has opposite (negative) charge induced on its surface, so attracts the positive droplets | ✓ 1 | |
| B1 droplets are pulled towards the girder, even round to the far side | ✓ 1 | |
| Part (c) | ||
| B1 electrons flow from earth to the girder | ✓ 1 | |
Exam tip: In static questions name the charges involved and state clearly whether they attract or repel.
A hospital technician rubs a plastic rod with a dry cloth. The rod becomes charged. Which statement describes what has happened?
| Marking point | Mark | Notes |
|---|---|---|
| Rubbing transfers electrons from the cloth to the rod. The rod gains a negative charge and the cloth, having lost the same number of electrons, gains an equal positive charge, so total charge is conserved. Protons are fixed in nuclei and do not move, so the first option is wrong. The cloth cannot stay neutral if it loses electrons, so the third option is wrong. Electrons are not created, so the fourth option is wrong. | ✓ 1 | Answer B |
Answer: B
Common error: Charge is never created: whatever one object gains, the other loses.
In a light-beam alarm, a fixed resistor and a light-dependent resistor (LDR) are connected in series to a cell. An intruder blocks the light falling on the LDR. What happens to the resistance of the LDR and to the potential difference across it?
| Marking point | Mark | Notes |
|---|---|---|
| The resistance of an LDR rises as the light falls. In a series circuit the supply potential difference is shared in proportion to resistance, so the component with the larger share of the total resistance gets a larger share of the potential difference. The LDR therefore has a higher resistance and a greater potential difference across it. The other options either state that resistance falls in the dark, or state that the share of potential difference falls when the resistance rises, and both contradict these facts. | ✓ 1 | Answer C |
Answer: C
Exam tip: In a series circuit the largest resistance takes the largest share of the supply potential difference.
A Formula 1 team tests a sensor on a bench. A 12 V supply of negligible internal resistance is connected to a 6.0 Ω resistor in series with two 12 Ω resistors that are connected in parallel with each other.
Show that the combined resistance of the two 12 Ω resistors is 6.0 Ω.
Calculate the current from the supply.
Calculate the power transferred in the 6.0 Ω resistor.
A third 12 Ω resistor is added in parallel with the other two. State and explain what happens to the current from the supply.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 1/R = 1/12 + 1/12 | ✓ 1 | |
| A1 R = 6.0 Ω | ✓ 1 | |
| Part (b) | ||
| M1 total resistance = 6.0 + 6.0 = 12 Ω and I = V/R | ✓ 1 | |
| A1 1.0 A | ✓ 1 | |
| Part (c) | ||
| M1 P = I2R = 1.02 × 6.0 (or IV with V = 6.0 V) | ✓ 1 | |
| A1 6.0 W | ✓ 1 | |
| Part (d) | ||
| B1 the current increases | ✓ 1 | |
| B1 the total resistance of the circuit decreases, because the parallel combination now has a smaller resistance, with the supply potential difference unchanged | ✓ 1 | |
Exam tip: Adding a resistor in parallel always lowers the total resistance of the circuit.
A beekeeper uses a honey-warming cabinet with a metal case. It is rated 1.84 kW at 230 V and is connected to the mains supply by a three-pin plug fitted with a fuse.
Explain how the earth wire and the fuse protect a user if the live wire inside the cabinet comes loose and touches the metal case.
Calculate the normal current in the cabinet.
Fuses of 3 A, 5 A and 13 A are available. State which fuse should be fitted and give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 the case would become live, so it could give a shock | ✓ 1 | |
| B1 a large current flows through the earth wire to earth | ✓ 1 | |
| B1 the fuse melts (or overheats and breaks), disconnecting the cabinet from the live supply | ✓ 1 | |
| Part (b) | ||
| M1 I = P/V = 1840 ÷ 230 | ✓ 1 | |
| A1 8.0 A | ✓ 1 | |
| Part (c) | ||
| B1 13 A | ✓ 1 | |
| B1 it is the lowest rating above the normal current of 8.0 A, so it does not melt in normal use but will melt if the current becomes too large (ECF from (b)) | ✓ 1 | |
Common error: The fuse should be just above the normal current, not the smallest or largest available.
A hospital steriliser has a heating element of resistance 46 Ω that is connected to the 230 V mains supply.
Calculate the current in the element.
Calculate the power of the element.
Calculate the energy transferred by the element in 20 minutes, in MJ.
The element is replaced by one of lower resistance. State and explain the effect on the power.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = V/R = 230 ÷ 46 | ✓ 1 | |
| A1 5.0 A | ✓ 1 | |
| Part (b) | ||
| M1 P = IV = 5.0 × 230 | ✓ 1 | |
| A1 1150 W | ✓ 1 | |
| Part (c) | ||
| M1 E = Pt = 1150 × 1200 | ✓ 1 | |
| A1 1.38 MJ | ✓ 1 | |
| Part (d) | ||
| B1 the power increases | ✓ 1 | |
| B1 the same potential difference gives a larger current, and P = IV | ✓ 1 | |
Exam tip: Mains power questions often chain V = IR and P = IV; write down each step with its unit.
A robot's indicator LED is connected in series with a resistor to a low-voltage a.c. supply of frequency 50 Hz. Which statement is correct?
| Marking point | Mark | Notes |
|---|---|---|
| An LED allows current in one direction only. With a.c. the supply reverses direction repeatedly, so the LED carries current only during the half-cycles in which the supply pushes current the right way, and none during the other halves. It is not conducting in both directions. Electrons in an a.c. circuit vibrate back and forth rather than travelling round one way. A zero average does not stop the LED lighting during the half-cycles when it conducts. | ✓ 1 | Answer B |
Answer: B
Common error: An LED is a diode: reversing the supply does not make it conduct the other way.
A battery-powered pump in a fish hatchery draws a steady current of 0.40 A for 2.5 minutes. How much charge passes through the pump?
| Marking point | Mark | Notes |
|---|---|---|
| Charge = current × time, and the time must be in seconds: 2.5 minutes = 150 s, so Q = 0.40 × 150 = 60 C. Using 2.5 without converting gives 1.0 C. Dividing current by time gives 0.0027 C. Dividing time by current gives 375, which is not a charge. | ✓ 1 | Answer A |
Answer: A
Exam tip: Convert minutes and hours to seconds before using Q = I t.
A robot's workshop uses a mains supply of 230 V and frequency 50 Hz. The robot also contains a battery-powered control unit.
State one difference between the movement of charge in an a.c. circuit and in a d.c. circuit.
State whether the battery supplies a.c. or d.c.
An LED indicator lamp is connected to an a.c. supply. Explain why it lights for only part of each cycle.
Calculate the time for one complete cycle of the mains supply.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| In a.c. the charge flow repeatedly reverses direction, but in d.c. it flows in one direction only | ✓ 1 | |
| Part (b) | ||
| d.c. | ✓ 1 | |
| Part (c) | ||
| An LED only allows current in one direction | ✓ 1 | |
| The supply reverses, so the LED conducts for only the half of the cycle when current is in that direction | ✓ 1 | |
| Part (d) | ||
| M1 time = 1 ÷ frequency, 1 ÷ 50 | ✓ 1 | |
| A1 0.020 s | ✓ 1 | |
Exam tip: The period is the reciprocal of the frequency.
An indicator LED on a fishing-boat winch panel is connected in series with a 350 Ω resistor to a 9.0 V d.c. supply. The current in the circuit is 20 mA.
Calculate the potential difference across the resistor.
Calculate the potential difference across the LED.
Calculate the charge that passes through the LED in 5.0 minutes.
State what the charge carriers are in the copper connecting wires.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 V = I R, 0.020 × 350 | ✓ 1 | |
| A1 7.0 V | ✓ 1 | |
| Part (b) | ||
| M1 supply pd minus pd across resistor, 9.0 − 7.0 | ✓ 1 | |
| A1 2.0 V (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 Q = I t with t = 300 s, 0.020 × 300 | ✓ 1 | |
| A1 6.0 C | ✓ 1 | |
| Part (d) | ||
| Electrons | ✓ 1 | |
Common error: In series the pds add up to the supply pd, so subtract the resistor's pd from the supply to find the LED's.
A robot charger is plugged into the 230 V mains supply. It draws a current of 2.5 A for 6.0 hours. The fuses available are 3 A, 5 A and 13 A.
State which fuse should be fitted in the plug and give a reason for your choice.
Calculate the power of the charger.
Calculate the energy transferred in kilowatt-hours.
Electricity costs 28 pence per kWh. Calculate the cost of charging, using the unrounded value of the energy. Give your answer in pence to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 3 A | ✓ 1 | |
| It is the lowest rating above the normal current of 2.5 A, so the fuse blows if the current becomes too high | ✓ 1 | |
| Part (b) | ||
| M1 P = V I, 230 × 2.5 | ✓ 1 | |
| A1 575 W | ✓ 1 | |
| Part (c) | ||
| M1 energy = power in kW × time, 0.575 × 6.0 | ✓ 1 | |
| A1 3.45 kWh (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 3.45 × 28 | ✓ 1 | |
| A1 96.6 p (ECF from (c)) | ✓ 1 | |
Common error: Convert watts to kilowatts before multiplying by hours to get kWh.
A fashion student sews an LED strip into a jacket. The strip has a power of 6.0 W and works from a 12 V battery.
Calculate the current in the strip.
Calculate the charge that flows through the strip in 2.0 hours.
Calculate the energy transferred by the strip in 2.0 hours. Give your answer in kJ.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P ÷ V, 6.0 ÷ 12 | ✓ 1 | |
| A1 0.50 A | ✓ 1 | |
| Part (b) | ||
| M1 Q = I t with t = 7200 s, 0.50 × 7200 | ✓ 1 | |
| A1 3600 C (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 E = P t, 6.0 × 7200 | ✓ 1 | |
| A1 43.2 kJ | ✓ 1 | |
Exam tip: Check your answer a second way: energy = charge × pd also gives 3600 × 12 J.
Three identical lamps are connected in series to a 12 V battery. A fourth identical lamp is then added in series with the others.
What happens to the current from the battery?
| Marking point | Mark | Notes |
|---|---|---|
| Adding a lamp in series increases the total resistance while the battery pd stays 12 V, so I = V ÷ R decreases. The current does not increase, because extra resistance in series opposes the current. It does not stay the same, because the same current in every lamp does not mean the current is unchanged when the circuit changes. The battery pd is fixed by the battery and does not fall when a lamp is added. | ✓ 1 | Answer D |
Answer: D
Common error: Adding components in series always raises the total resistance; adding branches in parallel lowers it.
A gaming console has a power of 150 W. It is used for 4.0 hours.
How much energy does it transfer?
| Marking point | Mark | Notes |
|---|---|---|
| Energy in kWh = power in kW × time in hours = 0.150 × 4.0 = 0.60 kWh. A value of 600 kWh comes from using 150 W without converting to kW. A value of 37.5 kWh comes from dividing the power by the time. A value of 0.027 kWh comes from dividing the time by the power. | ✓ 1 | Answer B |
Answer: B
Exam tip: Convert watts to kilowatts before multiplying by hours; the answer is then in kilowatt-hours.
A model lighthouse has a lamp and a resistor connected in series to a 6.0 V supply. The lamp works normally when the potential difference across it is 2.5 V and the current is 0.50 A.
Calculate the resistance of the lamp when it works normally.
State the current in the resistor.
Calculate the resistance of the resistor.
Calculate the total power supplied by the 6.0 V supply.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V ÷ I = 2.5 ÷ 0.50 | ✓ 1 | |
| A1 5.0 Ω | ✓ 1 | |
| Part (b) | ||
| A1 0.50 A, the same as in the lamp because they are in series | ✓ 1 | |
| Part (c) | ||
| M1 pd across resistor = 6.0 − 2.5 = 3.5 V | ✓ 1 | |
| M1 R = V ÷ I = 3.5 ÷ 0.50 | ✓ 1 | |
| A1 7.0 Ω | ✓ 1 | |
| Part (d) | ||
| M1 P = I × V = 0.50 × 6.0 | ✓ 1 | |
| A1 3.0 W | ✓ 1 | |
Exam tip: In a series circuit the supply pd is shared, so the resistor's pd is the supply pd minus the lamp's pd.
A submersible pump at a water-supply station is connected to the 230 V mains supply. Its power is 1.61 kW.
Calculate the current in the pump.
The plug for the pump can be fitted with a 3 A, 5 A or 13 A fuse. State which fuse should be used and explain your choice.
Electricity costs 20 pence per kWh. Calculate the cost of running the pump for 2.0 hours. Give your answer to the nearest penny.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P ÷ V = 1610 ÷ 230 | ✓ 1 | |
| A1 7.0 A | ✓ 1 | |
| Part (b) | ||
| 13 A fuse | ✓ 1 | |
| the 3 A and 5 A fuses are below the normal current of 7.0 A so they would melt in normal use, while 13 A is the lowest rating above it (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 energy = 1.61 × 2.0 = 3.22 kWh | ✓ 1 | |
| M1 cost = 3.22 × 20 = 64.4 pence | ✓ 1 | |
| A1 £0.644, so £0.64 (64 pence) to the nearest penny | ✓ 1 | |
Exam tip: A fuse should have the lowest standard rating that is still above the normal operating current.
A wireless game controller has a 3.7 V battery. When the vibration motor is on, the battery supplies a current of 0.40 A for 15 minutes.
Calculate the charge that flows through the motor in this time.
Calculate the energy transferred by the battery in this time. Use the unrounded value from (a).
Name the instrument used to measure the current and state how it is connected in the circuit.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 Q = I × t = 0.40 × (15 × 60) | ✓ 1 | |
| A1 360 C | ✓ 1 | |
| Part (b) | ||
| M1 E = Q × V = 360 × 3.7 | ✓ 1 | |
| A1 1332 J (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| ammeter | ✓ 1 | |
| in series with the motor and battery | ✓ 1 | |
Common error: Time must be in seconds for Q = I × t, so 15 minutes becomes 900 s.
A metal-cased pump at a water-supply station is connected to the mains by a three-core cable and a plug containing a fuse.
State the colour of the insulation on the live wire, the neutral wire and the earth wire.
Explain how the earth wire and the fuse together protect a person if the live wire touches the metal casing.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| live: brown | ✓ 1 | |
| neutral: blue | ✓ 1 | |
| earth: green and yellow stripes | ✓ 1 | |
| Part (b) | ||
| the casing is connected to earth by the earth wire | ✓ 1 | |
| a large current flows through the low-resistance earth wire | ✓ 1 | |
| the fuse melts and breaks the circuit, so the casing is no longer live | ✓ 1 | |
Common error: The fuse must be fitted in the live wire, so that the appliance is disconnected from the mains when it melts.
A car-seat heater has two heating elements, of resistance 6.0 Ω and 4.0 Ω, connected in series to a 12 V battery. What is the potential difference across the 4.0 Ω element?
| Marking point | Mark | Notes |
|---|---|---|
| The total resistance is 6.0 + 4.0 = 10 Ω, so the current is 12 ÷ 10 = 1.2 A. The p.d. across the 4.0 Ω element is 1.2 × 4.0 = 4.8 V. The value 7.2 V is the p.d. across the 6.0 Ω element. The value 6.0 V assumes the supply splits equally even though the resistances differ. The value 3.0 V comes from dividing 12 V by 4.0 Ω, which mixes up the quantities. | ✓ 1 | Answer A |
Answer: A
Common error: In a series circuit the supply p.d. is shared in proportion to the resistances. Do not split it equally unless the resistors are equal.
An electric fence energiser transfers 5.0 J of energy to the fence wire in each pulse. It produces 48 pulses every minute and is left switched on all the time. Electricity costs €0.25 per kWh. Ignore any energy wasted inside the energiser.
Calculate the mean power of the energiser output.
Calculate the energy transferred in 30 days. Give your answer in kWh.
Calculate the cost of running the energiser for 30 days. Give your answer in euros to 2 decimal places.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy per second = energy per pulse × pulses per second = 5.0 × 48 ÷ 60 | ✓ 1 | |
| A1 4.0 W | ✓ 1 | |
| Part (b) | ||
| M1 power = 0.0040 kW and time = 30 × 24 = 720 h | ✓ 1 | |
| A1 0.0040 × 720 = 2.88 kWh (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 cost = energy in kWh × price per kWh | ✓ 1 | |
| A1 2.88 × 0.25 = €0.72 (ECF from (b)) | ✓ 1 | |
Common error: Convert watts to kilowatts and days to hours before multiplying. Mixing units is the most frequent cause of an answer that is wrong by a factor of 1000.
A car-seat heater has two heating elements connected in series to a 12 V supply. One element has a resistance of 2.5 Ω. The current in the heater is 3.0 A.
Calculate the total resistance of the heater.
Calculate the resistance of the other element.
Explain why the current is the same in both elements.
Calculate the power transferred in the 2.5 Ω element.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 resistance = potential difference ÷ current | ✓ 1 | |
| A1 12 ÷ 3.0 = 4.0 Ω | ✓ 1 | |
| Part (b) | ||
| M1 total resistance is the sum of the resistances, so resistance = 4.0 − 2.5 | ✓ 1 | |
| A1 1.5 Ω (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| They are in series so there is only one path, and the same charge passes through each every second | ✓ 1 | |
| Part (d) | ||
| M1 power = I2R, or find p.d. = 3.0 × 2.5 = 7.5 V and then use power = current × p.d. | ✓ 1 | |
| A1 3.02 × 2.5 = 22.5 W | ✓ 1 | |
Exam tip: In a series circuit add the resistances and keep the current the same. Find the unknown resistance by subtraction.
The heating element of an electric shower has a resistance of 5.75 Ω. It is connected to the 230 V mains supply.
Calculate the current in the heating element.
Calculate the power of the shower.
The shower circuit can be protected by a fuse rated at 30 A, 45 A or 60 A. State which fuse is the most suitable and give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 current = potential difference ÷ resistance | ✓ 1 | |
| A1 230 ÷ 5.75 = 40 A | ✓ 1 | |
| Part (b) | ||
| M1 power = current × potential difference | ✓ 1 | |
| A1 40 × 230 = 9200 W (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| 45 A fuse | ✓ 1 | |
| It is above the normal current of 40 A so it will not melt in normal use, but it is the lowest rating that is safe so it melts if a fault causes a larger current (ECF from (a)) | ✓ 1 | |
Common error: A fuse must have a rating just above the normal current. A 60 A fuse would allow a dangerously large current before melting.
Which statement about the current in a copper wire connected to a cell is correct?
| Marking point | Mark | Notes |
|---|---|---|
| In a metal the charge carriers are free electrons, which are negative, so they drift in the opposite direction to the conventional current. The positive ions are fixed in the lattice and only vibrate. Electrons are not used up: the same current enters and leaves a component. The electrons drift slowly; it is the energy transfer along the circuit that is fast. | ✓ 1 | Answer A |
Answer: A
Common error: Conventional current is drawn from + to −, but the moving charges in a metal are electrons going the other way.
An electric blanket is rated 90 W when connected to the 230 V mains supply. What is the current in the blanket when it works at its rated power?
| Marking point | Mark | Notes |
|---|---|---|
| Current = power ÷ potential difference = 90 W ÷ 230 V = 0.39 A. 2.6 A comes from dividing 230 by 90. 2.1 × 104 A comes from multiplying power by potential difference. 1.7 × 10−3 A comes from dividing power by the potential difference squared. | ✓ 1 | Answer D |
Answer: D
Exam tip: Rearrange P = VI to I = P/V and check that the size of the answer is sensible for a household appliance.
A student investigates a length of thin metal wire. She connects the wire to a variable power supply, measures the potential difference V across it and the current I in it, and records the readings below.
| V / V | 0 | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 |
|---|---|---|---|---|---|---|
| I / A | 0 | 0.20 | 0.38 | 0.54 | 0.68 | 0.79 |
Calculate the resistance of the wire when the potential difference is 4.0 V. Give your answer to 2 significant figures.
Describe how the resistance of the wire changes as the potential difference increases. Use values from the table to support your answer.
Explain why the resistance changes in this way.
The wire carries the current shown in the table for a potential difference of 3.0 V. Calculate the charge that passes through the wire in 2.0 minutes. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V/I = 4.0 / 0.68 | ✓ 1 | |
| A1 5.9 Ω | ✓ 1 | |
| Part (b) | ||
| resistance increases (as the potential difference increases) | ✓ 1 | |
| e.g. 5.0 Ω at 1.0 V rising to 6.3 Ω at 5.0 V (accept any two correct values that show the increase) | ✓ 1 | |
| Part (c) | ||
| larger current heats the wire / the temperature of the wire rises | ✓ 1 | |
| metal ions vibrate more so electrons collide with them more often | ✓ 1 | |
| Part (d) | ||
| M1 time = 2.0 × 60 = 120 s | ✓ 1 | |
| M1 Q = It = 0.54 × 120 | ✓ 1 | |
| A1 64.8 C | ✓ 1 | |
Exam tip: When a table gives readings, say which values you used from it; a bare statement that resistance rises will not earn the data mark.
A food scientist heats 5.0 kg of water in a water bath using an electric heater rated 1.5 kW at 230 V. The water starts at 20 °C and is heated to 56 °C. The specific heat capacity of water is 4200 J/kg °C.
Calculate the energy transferred to the water.
Assume all the electrical energy from the heater is transferred to the water. Calculate the time the heater takes to warm the water.
Suggest why the heater takes longer than this in practice.
Calculate the current in the heater. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 E = mcΔT = 5.0 × 4200 × 36 | ✓ 1 | |
| A1 756 000 J | ✓ 1 | |
| Part (b) | ||
| M1 t = E/P = 756 000 / 1500 | ✓ 1 | |
| A1 504 s (accept 8.4 min) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| some energy is transferred to the surroundings / to the container / by evaporation | ✓ 1 | |
| Part (d) | ||
| M1 I = P/V = 1500 / 230 | ✓ 1 | |
| A1 6.52 A | ✓ 1 | |
Exam tip: Write the temperature change as a difference (56 − 20), not as the final temperature.
In a food-packaging plant, a student rubs a dry cloth on a strip of polythene. The polythene becomes negatively charged. The strip is then held just above some small uncharged pieces of paper and the pieces jump up to the strip.
Explain, in terms of electrons, how the polythene becomes negatively charged.
Explain why the uncharged pieces of paper are attracted to the charged strip.
A second negatively charged polythene strip is held near the first. State what happens.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| electrons are transferred from the cloth to the polythene (by friction) | ✓ 1 | |
| polythene has gained electrons (so a surplus of negative charge) / cloth is left positive | ✓ 1 | |
| Part (b) | ||
| the strip repels electrons in the paper, so the side nearer the strip becomes positive (charge induced) | ✓ 1 | |
| the attraction to the nearer positive charge is greater than the repulsion from the farther negative charge | ✓ 1 | |
| Part (c) | ||
| the strips repel each other | ✓ 1 | |
Common error: Say that electrons move; positive charges in a solid do not move from one object to another.
A mountaineer's head torch has three identical lamps connected in parallel to a 4.5 V battery. Each lamp carries a current of 0.20 A.
Calculate the current in the battery when all three lamps are lit.
One lamp fails. State the current in the battery now, and explain why the other two lamps stay lit at the same brightness.
Calculate the power of one lamp.
Calculate the energy transferred by all three lamps in 1.0 hour.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0.60 A | ✓ 1 | |
| Part (b) | ||
| 0.40 A | ✓ 1 | |
| each remaining lamp is still connected directly across the 4.5 V battery so its current is unchanged | ✓ 1 | |
| Part (c) | ||
| M1 P = VI = 4.5 × 0.20 | ✓ 1 | |
| A1 0.90 W | ✓ 1 | |
| Part (d) | ||
| M1 total power = 3 × 0.90 = 2.7 W (or I = 0.60 A) | ✓ 1 | |
| M1 E = Pt = 2.7 × 3600 | ✓ 1 | |
| A1 9720 J | ✓ 1 | |
Exam tip: In a parallel circuit the branch currents add to give the battery current.
A shop replaces 25 filament lamps of power 60 W with LED lamps of power 8.0 W that give the same light. The lamps are lit for 12 hours each day. Electricity costs 25 p per kilowatt-hour.
Calculate the total power saved when all 25 lamps are replaced.
Calculate the energy saved in 30 days. Give your answer in kilowatt-hours.
Calculate the money saved in 30 days. Give your answer in pounds (£).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 saving per lamp = 60 − 8.0 = 52 W | ✓ 1 | |
| A1 1300 W (1.3 kW) | ✓ 1 | |
| Part (b) | ||
| M1 1.3 × (12 × 30) | ✓ 1 | |
| A1 468 kWh (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 468 × 25 p = 11 700 p | ✓ 1 | |
| A1 £117 | ✓ 1 | |
Common error: Convert pence to pounds by dividing by 100 at the end, and keep the energy in kWh when using a price per kWh.
A student plots a graph of current against potential difference for a filament lamp. Which description of the graph, and its explanation, is correct?
| Marking point | Mark | Notes |
|---|---|---|
| As the potential difference increases, the current heats the filament. Its resistance rises, so each extra volt gives less extra current and the curve flattens. A straight line would mean constant resistance, which a hot filament does not have. A steeper curve would mean falling resistance, which is wrong for a metal. The current is not constant, as it increases with potential difference. | ✓ 1 | Answer D |
Answer: D
Common error: A filament lamp's resistance increases with temperature, so it does not obey Ohm's law.
An electric fence energiser sends out a pulse in which 0.0040 C of charge is moved through a potential difference of 5000 V. How much energy is transferred in one pulse?
| Marking point | Mark | Notes |
|---|---|---|
| Energy transferred = charge × potential difference = 0.0040 C × 5000 V = 20 J. 8.0 × 10−7 J comes from dividing charge by potential difference. 1.25 × 106 J comes from dividing potential difference by charge. 4.0 × 10−3 J ignores the potential difference. | ✓ 1 | Answer A |
Answer: A
Exam tip: Energy = charge × potential difference: a joule is a coulomb moved through a volt.
A student investigates a 12 V filament lamp. She increases the potential difference V across the lamp and records the current I in it.
| V / V | 0 | 2.0 | 4.0 | 6.0 | 8.0 | 10.0 | 12.0 |
|---|---|---|---|---|---|---|---|
| I / A | 0 | 0.10 | 0.17 | 0.22 | 0.26 | 0.29 | 0.31 |
State where the student should connect the voltmeter and the ammeter in relation to the lamp.
Calculate the resistance of the lamp when the potential difference is 12.0 V. Give your answer to 3 significant figures.
Describe how the resistance of the lamp changes as the potential difference increases, and explain why.
Calculate the power of the lamp when the potential difference is 6.0 V.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| voltmeter in parallel with the lamp | ✓ 1 | |
| ammeter in series with the lamp | ✓ 1 | |
| Part (b) | ||
| M1 R = V/I = 12.0 / 0.31 | ✓ 1 | |
| A1 38.7 Ω | ✓ 1 | |
| Part (c) | ||
| resistance increases | ✓ 1 | |
| the filament gets hotter | ✓ 1 | |
| ions vibrate more, so more collisions with the electrons | ✓ 1 | |
| Part (d) | ||
| M1 P = VI = 6.0 × 0.22 | ✓ 1 | |
| A1 1.32 W | ✓ 1 | |
Exam tip: For each reading take V and I from the same column of the table before calculating.
A rural clinic stores vaccines in a refrigerator rated 48 W that runs from a 12 V battery charged by solar panels. Assume the refrigerator works at its rated power all day.
Calculate the current in the refrigerator.
Calculate the charge that passes through the refrigerator in 24 hours. Give your answer to 2 significant figures.
Calculate the energy used by the refrigerator in 24 hours, in kilowatt-hours. Give your answer to 3 significant figures.
The solar panels deliver their rated power for 6.0 hours each day. Calculate the minimum rated power of the panels needed to supply the refrigerator for 24 hours. Use the exact energy 1.152 kWh and ignore energy losses.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P/V = 48 / 12 | ✓ 1 | |
| A1 4.0 A | ✓ 1 | |
| Part (b) | ||
| M1 Q = It = 4.0 × (24 × 3600) | ✓ 1 | |
| A1 3.5 × 105 C (345 600 C before rounding) | ✓ 1 | |
| Part (c) | ||
| M1 0.048 kW × 24 h | ✓ 1 | |
| A1 1.15 kWh (1.152 kWh before rounding) | ✓ 1 | |
| Part (d) | ||
| M1 P = E/t = 1.152 / 6.0 | ✓ 1 | |
| A1 0.192 kW (192 W) | ✓ 1 | |
Exam tip: Give power in the unit the question uses (W or kW) and check that the energy unit is consistent.
A community wind turbine supplies a test circuit at 12 V. Two heating resistors, of resistance 20 Ω and 30 Ω, are connected in parallel across the supply.
Calculate the current in the 20 Ω resistor.
Calculate the current in the 30 Ω resistor.
Calculate the total current taken from the supply.
Calculate the total resistance of the circuit. Use the exact current from part (c).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = V/R = 12 / 20 | ✓ 1 | |
| A1 0.60 A | ✓ 1 | |
| Part (b) | ||
| 0.40 A | ✓ 1 | |
| Part (c) | ||
| 1.0 A (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| M1 R = V/I = 12 / 1.0 | ✓ 1 | |
| A1 12 Ω (ECF from (c)) | ✓ 1 | |
Common error: The total resistance of resistors in parallel is smaller than the smallest resistor.
During a spark from a door handle, 6.0 μC of charge passes in 2.0 μs. What is the average current in the spark?
| Marking point | Mark | Notes |
|---|---|---|
| Current = charge ÷ time = 6.0 × 10−6 C ÷ 2.0 × 10−6 s = 3.0 A. Multiplying the two values gives 12 × 10−12, not 12 A, so 12 A is wrong. 0.33 A comes from dividing time by charge. 3.0 × 10−12 A comes from forgetting that the prefixes cancel. | ✓ 1 | Answer A |
Answer: A
Exam tip: Write micro-prefixes as powers of ten before dividing; the ×10⁻⁶ factors cancel here.
In a model lighthouse a 9.0 V supply drives a current of 0.50 A through a lamp in series with a 12 Ω resistor. What is the power of the lamp?
| Marking point | Mark | Notes |
|---|---|---|
| The resistor has p.d. = 0.50 × 12 = 6.0 V, so the lamp has 9.0 − 6.0 = 3.0 V. Lamp power = 3.0 × 0.50 = 1.5 W. 3.0 W is the power of the resistor (6.0 V × 0.50 A). 4.5 W is the power of the whole circuit (9.0 V × 0.50 A). 6.75 W comes from wrongly applying the full supply p.d. across the resistor alone (9.0² ÷ 12). | ✓ 1 | Answer D |
Answer: D
Common error: Use the p.d. across the component you are asked about, not the supply p.d., when components are in series.
A model lighthouse has a lamp connected in series with a resistor to a 9.0 V supply. The lamp is rated 6.0 V, 3.0 W and operates at its rated values.
Calculate the current in the lamp.
Calculate the potential difference across the resistor.
Calculate the resistance of the resistor.
Calculate the power dissipated in the resistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 current = power ÷ potential difference | ✓ 1 | |
| A1 0.50 A | ✓ 1 | |
| Part (b) | ||
| 3.0 V | ✓ 1 | |
| Part (c) | ||
| M1 resistance = potential difference ÷ current | ✓ 1 | |
| A1 6.0 Ω , ECF from (a) and (b) | ✓ 1 | |
| Part (d) | ||
| M1 power = potential difference × current | ✓ 1 | |
| A1 1.5 W , ECF | ✓ 1 | |
Exam tip: In a series circuit the same current flows through every component, so the lamp's current is also the resistor's current.
A train carriage has six identical 40 W lamps connected in parallel to a 230 V supply.
Calculate the current in one lamp. Give your answer to 2 significant figures.
Calculate the total current from the supply when all six lamps are on. Work from the data in the question, not from the rounded answer to (a). Give your answer to 3 significant figures.
The circuit may be protected by a 1 A, 3 A or 5 A fuse. State which fuse should be used and give a reason.
Explain why a fuse is connected in the live wire rather than the neutral wire.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 current = power ÷ potential difference | ✓ 1 | |
| A1 0.17 A | ✓ 1 | |
| Part (b) | ||
| M1 total power 240 W, current = 240 ÷ 230 | ✓ 1 | |
| A1 1.04 A | ✓ 1 | |
| Part (c) | ||
| 3 A | ✓ 1 | |
| it is greater than the normal current (1.04 A) so it does not melt in normal use, but is the lowest rating that protects the circuit | ✓ 1 | |
| Part (d) | ||
| if the fuse melts it disconnects the live wire from the supply | ✓ 1 | |
| so the appliance or wiring is not left connected to the mains p.d. | ✓ 1 | |
Common error: Choose a fuse rated just above the normal working current, not the largest one available.
An astronomer uses a 230 V heater to stop dew forming on a telescope mirror. The current in the heater is 0.50 A and it is on for 10 hours each night.
Calculate the power of the heater.
Calculate the energy transferred in one night, in kilowatt-hours.
Electricity costs 28p per kilowatt-hour. Calculate the cost of one night's use, to the nearest penny. Use the unrounded answer to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 power = potential difference × current | ✓ 1 | |
| A1 115 W | ✓ 1 | |
| Part (b) | ||
| M1 power in kW = 0.115 and time = 10 h | ✓ 1 | |
| A1 1.15 kWh , ECF from (a) | ✓ 1 | |
| Part (c) | ||
| M1 cost = 1.15 × 28 | ✓ 1 | |
| A1 32.2p, so 32p to the nearest penny , ECF | ✓ 1 | |
Exam tip: Energy in kWh is power in kW multiplied by time in hours; do not convert to joules unless asked.
A marine biologist measures water temperature with a thermistor probe. The thermistor is connected in series with a cell and an ammeter. The thermistor is an NTC type and the water is warmed. Which row describes the effect?
| Marking point | Mark | Notes |
|---|---|---|
| For an NTC thermistor, a rise in temperature releases more charge carriers, so its resistance falls. With the same cell p.d., I = V/R rises as R falls, so the ammeter reading increases. The other rows pair a change in resistance with a change in current that contradicts I = V/R, or give the wrong direction for resistance. | ✓ 1 | Answer A |
Answer: A
Exam tip: Link the two quantities with I = V ÷ R: if the p.d. stays the same, lower resistance means larger current.
A charger supplies a constant current of 2.0 A to a cordless drill battery for 30 minutes. How much charge flows through the battery?
| Marking point | Mark | Notes |
|---|---|---|
| Charge = current × time, and the time must be in seconds: 30 minutes = 1800 s, so Q = 2.0 × 1800 = 3600 C. The value 60 C uses the time in minutes. The value 1800 C is the time in seconds, which forgets to multiply by the current. The value 7200 C uses a time of one hour. | ✓ 1 | Answer D |
Answer: D
Common error: Convert minutes to seconds before using Q = I × t.
An electric blanket is rated at 50 W. It is switched on for 8.0 hours. How much energy does it transfer?
| Marking point | Mark | Notes |
|---|---|---|
| Energy = power × time, with the time in seconds: 8.0 h = 28 800 s, so E = 50 × 28 800 = 1.44 × 106 J. The value 400 J multiplies by hours instead of seconds. The value 24 000 J uses 8 × 60 s, which is minutes instead of seconds. The value 1440 J is a slip of three powers of ten, as if the answer was in kilojoules. | ✓ 1 | Answer B |
Answer: B
Exam tip: Check the size of your answer: 8 hours is tens of thousands of seconds, so the energy must be in the millions of joules.
A cordless drill battery in a logistics warehouse needs 9000 C of charge to be fully recharged. The charger supplies a constant current of 2.5 A at a p.d. of 20 V.
Calculate the time taken to recharge the battery.
Calculate the power of the charger.
Calculate the energy transferred by the charger during the recharging. Give your answer in joules.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| t = Q ÷ I | ✓ 1 | |
| 9000 ÷ 2.5 = 3600 s | ✓ 1 | |
| Part (b) | ||
| P = I × V | ✓ 1 | |
| 2.5 × 20 = 50 W | ✓ 1 | |
| Part (c) | ||
| E = P × t = 50 × 3600 | ✓ 1 | |
| 1.8 × 105 J (180 000 J) (accept ECF from (a) and (b)) | ✓ 1 | |
Exam tip: Use the values you have calculated, with units of seconds and watts, to get joules.
A thermistor in a marine temperature probe is connected to a 6.0 V cell. The thermistor has a resistance of 4800 Ω at 5 °C and 1200 Ω at 25 °C.
State how the resistance of the thermistor changes as its temperature rises.
Calculate the current in the thermistor at 5 °C.
Calculate the current in the thermistor at 25 °C. Give your answer in mA.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the resistance decreases | ✓ 1 | |
| Part (b) | ||
| I = V ÷ R = 6.0 ÷ 4800 | ✓ 1 | |
| 1.25 × 10−3 A (1.25 mA) | ✓ 1 | |
| Part (c) | ||
| I = 6.0 ÷ 1200 | ✓ 1 | |
| 5.0 mA | ✓ 1 | |
Exam tip: Keep ohms with volts to get amps; convert to milliamps by multiplying by 1000.
An electric blanket is rated at 90 W and works from the 230 V mains supply.
Calculate the current in the blanket. Give your answer to 2 significant figures.
The blanket is switched on for 6.0 hours. Calculate the energy transferred in kWh.
Electricity costs €0.30 per kWh. Calculate the cost of using the blanket for 6.0 hours. Give your answer to the nearest cent.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P ÷ V = 90 ÷ 230 | ✓ 1 | |
| 0.39 A | ✓ 1 | |
| Part (b) | ||
| 90 W = 0.090 kW | ✓ 1 | |
| 0.090 × 6.0 = 0.54 kWh | ✓ 1 | |
| Part (c) | ||
| cost = 0.54 × 0.30 (use the unrounded 0.54) | ✓ 1 | |
| €0.16 (accept 0.162) | ✓ 1 | |
Exam tip: Use power in kW and time in hours to find energy in kWh before working out the cost.
A metal fuel tanker is connected to the ground by a conducting cable before it unloads fuel. Why is this done?
| Marking point | Mark | Notes |
|---|---|---|
| Fuel flowing in the hose can charge the tanker by friction. A conducting cable lets the charge flow safely to earth, so the tanker stays at about the same potential as the ground and no spark can jump. A conductor does not insulate. Warming the fuel is not the purpose and would be dangerous. Positive charge is not supplied; it is the movement of electrons that removes the charge. | ✓ 1 | Answer C |
Answer: C
Exam tip: For static electricity dangers, state: charge builds up, a large p.d. results, a spark can occur, and earthing removes the charge.
A heater circuit in a space probe has a 12 V supply connected to a resistor of 8.0 Ω in series with a resistor of 4.0 Ω. What is the current in the circuit?
| Marking point | Mark | Notes |
|---|---|---|
| The resistances in series add: 8.0 + 4.0 = 12 Ω, so I = 12 ÷ 12 = 1.0 A. 1.5 A uses only the 8.0 Ω resistor. 3.0 A uses only the 4.0 Ω resistor. 4.5 A adds the two separate currents, which would be correct only if the resistors were in parallel. | ✓ 1 | Answer A |
Answer: A
Common error: In series, add the resistances first; the same current flows through every component.
A probe sent to Jupiter has a 28 V supply. A heater of resistance 14 Ω and a camera of resistance 56 Ω are connected in parallel across the supply.
Calculate the current in the heater.
Calculate the current in the camera.
State the total current from the supply.
Calculate the total power supplied.
Calculate the energy transferred in 1 hour. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = V ÷ R | ✓ 1 | |
| A1 2.0 A | ✓ 1 | |
| Part (b) | ||
| A1 0.50 A | ✓ 1 | |
| Part (c) | ||
| A1 2.5 A (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| M1 P = I V = 2.5 × 28 | ✓ 1 | |
| A1 70 W (ECF from (c)) | ✓ 1 | |
| Part (e) | ||
| M1 E = P t with t = 3600 s | ✓ 1 | |
| A1 2.5 × 105 J (252 000 J) (ECF from (d)) | ✓ 1 | |
Exam tip: In a parallel circuit every branch has the same p.d., so use I = V ÷ R for each branch and add the currents.
A coffee roasting machine has a heating element rated at 2760 W when connected to the 230 V mains supply.
Calculate the current in the element.
The plug can have a 3 A, 5 A or 13 A fuse. State which fuse should be used and give a reason.
Calculate the resistance of the element. Give your answer to 2 significant figures.
A batch takes 15 minutes to roast. Calculate the cost of the electricity at £0.30 per kWh. Give your answer to the nearest penny.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P ÷ V | ✓ 1 | |
| A1 12 A | ✓ 1 | |
| Part (b) | ||
| B1 13 A fuse | ✓ 1 | |
| B1 12 A is below 13 A so the fuse does not blow in normal use, but would exceed 3 A and 5 A | ✓ 1 | |
| Part (c) | ||
| M1 R = V ÷ I | ✓ 1 | |
| A1 19 Ω (ECF from (a)) | ✓ 1 | |
| Part (d) | ||
| M1 energy = 2.76 kW × 0.25 h | ✓ 1 | |
| A1 0.69 kWh | ✓ 1 | |
| A1 £0.21 | ✓ 1 | |
Exam tip: For kWh, use power in kW and time in hours.
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Identifies the correct highest or lowest setting, or makes isolated correct statements about series or parallel resistance, voltage or current, with little linking to power. |
| Level 2 (3–4) | Correctly explains one setting in terms of voltage, resistance and current leading to power, or identifies both settings correctly with partial reasoning for each. |
| Level 3 (5–6) | Logically linked explanation of both the series and the parallel settings, using the same mains voltage, the resistance and the current to conclude that setting 3 is highest and setting 2 lowest. |
A small satellite uses a 28 V battery to power two heaters, A and B, connected in parallel. The current from the battery is 4.0 A. The current in heater A is 1.5 A.
Calculate the current in heater B and explain your answer.
The satellite takes 90 minutes to complete one orbit. Heater B is on for the whole orbit.
Calculate the charge that flows through heater B during one orbit.
Calculate the energy transferred by heater B during one orbit.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4.0 − 1.5 = 2.5 A | ✓ 1 | |
| Charge is conserved / not used up, so the current into a junction equals the total current out | ✓ 1 | |
| Part (b) | ||
| t = 90 × 60 = 5400 s | ✓ 1 | |
| Q = 2.5 × 5400 = 13 500 C | ✓ 1 | |
| Part (c) | ||
| E = Q × V = 13 500 × 28 | ✓ 1 | |
| E = 3.8 × 105 J (378 000 J) | ✓ 1 | |
Common error: Each branch of a parallel circuit has the full battery voltage across it. Use 28 V for heater B, not half of it.
An electric drill used on a building site has a metal case. The drill is connected to the mains with a three-core cable, and the plug contains a 5 A fuse. The normal working current of the drill is 3.5 A.
A loose live wire inside the drill touches the metal case.
Explain how the earth wire and the fuse work together to protect the user.
Explain why the fuse rating is chosen to be only a little above the normal working current of 3.5 A.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The metal case is a conductor, so it becomes live (at 230 V) . | ✓ 1 | |
| The earth wire gives a low-resistance path from the case to earth, so a very large current flows . | ✓ 1 | |
| This current is much greater than the 5 A fuse rating, so the fuse wire heats up and melts . | ✓ 1 | |
| The live wire is disconnected, so the case is no longer live and the user does not get an electric shock . | ✓ 1 | |
| (Accept, in place of any one point: without the earth wire, current would flow through the user to earth.) | — | |
| Part (b) | ||
| A rating just above 3.5 A means the fuse does not melt during normal use . | ✓ 1 | |
| A fuse with a much higher rating might not melt for a smaller fault current, so a dangerous current could flow and cause overheating, a fire or a shock . | ✓ 1 | |
Exam tip: Give the full sequence in order: live touches case → large current to earth → current exceeds fuse rating → fuse melts → live supply cut off. Each link in the chain can earn a mark.
A satellite ground station has a large dish antenna. In winter, a heating element behind the dish melts snow and ice. The element has a resistance of 40 Ω and is connected to the 230 V mains supply.
Calculate the power of the heating element.
An engineer suggests using the same element on a 115 V supply.
Explain why its power would be only one quarter of the value in (a).
On a cold night, 2.4 MJ of energy must be transferred to clear the dish.
Calculate the time needed when the element is connected to the 230 V supply.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = V / R = 230 / 40 = 5.75 A | ✓ 1 | |
| P = I × V = 5.75 × 230 (or P = V2 / R = 2302 / 40) | ✓ 1 | |
| P = 1322.5 W ≈ 1300 W (accept 1300–1323 W) | ✓ 1 | |
| Part (b) | ||
| The resistance stays the same, so halving the voltage halves the current (I = V / R) . | ✓ 1 | |
| P = I × V, and both I and V are halved, so P becomes ½ × ½ = ¼ of the original value (or P = V2 / R, so halving V quarters P) . | ✓ 1 | |
| Part (c) | ||
| t = E / P (or E = I × V × t rearranged) | ✓ 1 | |
| t = 2.4 ×106 / 1322.5 (allow error carried forward from the power in (a)) | ✓ 1 | |
| t = 1815 s ≈ 1800 s (accept 1800–1850 s, e.g. 1846 s from using 1300 W) | ✓ 1 | |
Common error: Saying that halving the voltage halves the power is wrong. The current also halves, so the power falls to a quarter.
An electric bike battery has a voltage of 36 V. When fully charged, it can supply a current of 10 A for 1.0 hour.
How much energy can the fully charged battery transfer?
| Marking point | Mark | Notes |
|---|---|---|
| First find the charge: Q = It = 10 × 3600 = 36 000 C. Then E = QV = 36 000 × 36 = 1.3×106 J. The time must be in seconds, so 1.0 hour = 3600 s. 360 J leaves out the time completely. 2.2×104 J uses the time in minutes instead of seconds. 3.6×104 J is the charge in coulombs, not the energy in joules. | ✓ 1 | Answer A |
Answer: A
Common error: Always convert hours to seconds (× 3600) before using Q = I × t or E = P × t.
A timing gate at a sprint track uses a light-dependent resistor (LDR). The LDR is connected in series with a fixed resistor of 2.2 kΩ and a 6.0 V supply. When light falls on the LDR, its resistance is 0.80 kΩ.
Calculate the total resistance of the circuit when light falls on the LDR.
Calculate the current in the circuit.
Calculate the voltage across the LDR.
A runner passes through the gate and blocks the light reaching the LDR.
Explain what happens to the voltage across the LDR.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 2.2 + 0.80 = 3.0 kΩ (3000 Ω) | ✓ 1 | |
| Part (b) | ||
| I = V ÷ R = 6.0 ÷ 3000 | ✓ 1 | |
| = 2.0×10−3 A (2.0 mA) | ✓ 1 | |
| Part (c) | ||
| V = I × R = 2.0×10−3 × 800 | ✓ 1 | |
| = 1.6 V | ✓ 1 | |
| Part (d) | ||
| the resistance of the LDR increases in the dark | ✓ 1 | |
| the LDR's share of the total resistance increases / the current decreases, so the voltage across the fixed resistor decreases | ✓ 1 | |
| the supply voltage is shared, so the voltage across the LDR increases | ✓ 1 | |
Exam tip: For a series circuit, always state that the voltages across the components add up to the supply voltage. This links your reasoning to the conclusion.
A garden designer installs 20 identical outdoor lamps. Each lamp is rated at 12 V, 5.0 W. The lamps are powered by a 12 V supply.
Calculate the current in one lamp when it works normally.
Explain why the lamps should be connected in parallel and not in series, and why the main cable from the supply must be thicker than the wire to each lamp.
Refer to voltage, current and energy transferred per coulomb in your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P ÷ V = 5.0 ÷ 12 | ✓ 1 | |
| = 0.42 A | ✓ 1 | |
| Part (b) | ||
| in parallel, each lamp has the full 12 V across it | ✓ 1 | |
| so each coulomb transfers 12 J in each lamp and the lamp works at its rated power | ✓ 1 | |
| in series, the 12 V would be shared between 20 lamps, giving only about 0.6 V each, so each coulomb transfers only about 0.6 J per lamp and the lamps would be very dim | ✓ 1 | |
| in parallel, if one lamp fails the others stay lit / in series one failure breaks the whole circuit | ✓ 1 | |
| the supply current is the sum of the branch currents, 20 × 0.42 ≈ 8.3 A (accept 8.3–8.4 A), so the main cable carries far more current than any one lamp wire | ✓ 1 | |
| a thicker cable has lower resistance, so less energy is wasted by heating / the cable does not overheat / less voltage is lost in the cable | ✓ 1 | |
Exam tip: In extended answers, back up each idea with a number from the question, for example 0.6 V per lamp or about 8.3 A in total. This makes your reasoning precise.
A pottery studio has an electric kiln with a metal casing. The kiln is connected to the mains through a plug containing a fuse, and the metal casing is connected to the earth wire.
Explain how the earth wire and the fuse together protect a potter if a fault inside the kiln causes the live wire to touch the metal casing.
Explain how the earth wire and the fuse protect the potter. [6]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Award one mark for each point: | — | |
| • When the live wire touches the metal casing, the casing would become live, at a high voltage . | ✓ 1 | |
| • The earth wire connects the casing to the ground through a path of very low resistance . | ✓ 1 | |
| • A large current therefore flows from the live wire through the casing and the earth wire . | ✓ 1 | |
| • This current is larger than the rating of the fuse . | ✓ 1 | |
| • The fuse wire heats up, melts and breaks the circuit, disconnecting the live wire . | ✓ 1 | |
| • The casing is no longer live, so no current passes through the potter and the potter does not get an electric shock . | ✓ 1 | |
Exam tip: Give the steps in order: fault, casing live, low-resistance earth path, large current, fuse melts, live wire disconnected, no shock. Leaving out a link in this chain loses marks.
An electric shower in a gym changing room has a power of 9.5 kW and runs on the 230 V mains supply. It is protected by a circuit breaker on its own circuit.
Calculate the normal current in the shower.
Circuit breakers are available with ratings of 32 A, 40 A, 45 A and 50 A. Choose the most suitable rating and explain your choice.
The gym owner suggests saving money by using thinner cable for the shower circuit. Explain why this would be dangerous.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Uses P = I × V, rearranged to I = P / V . | ✓ 1 | |
| I = 9500 / 230 . | ✓ 1 | |
| I = 41 A . | ✓ 1 | |
| Part (b) | ||
| 45 A . | ✓ 1 | |
| It is just above the normal current of 41 A, so it will not trip in normal use, but it will trip quickly if a fault makes the current rise . | ✓ 1 | |
| Part (c) | ||
| Thinner cable has a higher resistance . The large current would cause more heating in the cable, so it could overheat, melt its insulation and start a fire . | ✓ 2 | |
Exam tip: For a fuse or circuit breaker, choose the rating just above the normal current. A rating below the normal current trips all the time, and one far above it gives poor protection.
A pottery studio uses an electric potter's wheel that has a metal casing. The wheel is connected to the mains supply through a three-pin plug containing a fuse. Inside the wheel, the insulation on the live wire wears through and the bare live wire touches the metal casing.
Explain how the earth wire and the fuse work together to protect a potter who touches the casing.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The metal casing is connected to the earth wire | ✓ 1 | |
| The earth wire provides a low-resistance path from the casing to the ground | ✓ 1 | |
| When the live wire touches the casing, a very large current flows from the live wire through the casing and down the earth wire | ✓ 1 | |
| This current is much larger than the fuse rating, so the fuse wire heats up and melts | ✓ 1 | |
| The fuse is in the live wire, so when it melts the circuit breaks and the appliance is disconnected from the live supply | ✓ 1 | |
| The casing is no longer at a high voltage, so current does not pass through the potter's body and the potter is not electrocuted | ✓ 1 | |
Exam tip: Write the protection as a chain of linked steps: earth connection, low-resistance path, large current, fuse melts, live supply cut off, casing safe. Each link is a separate marking point.
A student tests a 12 V filament lamp and records the following results.
| Potential difference (V) | 1.0 | 4.0 | 8.0 | 12.0 |
| Current (A) | 0.50 | 1.00 | 1.60 | 2.00 |
Calculate the resistance of the lamp when the pd is 12.0 V.
Calculate the resistance of the lamp when the pd is 1.0 V.
Evaluate the student's conclusion using the data.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| R = V/I = 12.0 ÷ 2.00 | ✓ 1 | |
| 6.0 Ω | ✓ 1 | |
| Part (b) | ||
| R = V/I = 1.0 ÷ 0.50 | ✓ 1 | |
| 2.0 Ω | ✓ 1 | |
| Part (c) | ||
| the resistance is not constant, it increases from 2.0 Ω to 6.0 Ω | ✓ 1 | |
| for Ohm's law the current must be directly proportional to the pd | ✓ 1 | |
| the pd triples from 4.0 V to 12.0 V but the current only doubles, so the conclusion is not supported | ✓ 1 | |
Common error: Current rising with pd is not enough for Ohm's law: the ratio V/I must stay constant.
In a wind farm access hut, a 12 V supply is connected to two parallel branches. Branch 1 contains a heater of resistance 10 Ω. The current from the supply is 2.0 A.
Calculate the current in branch 1.
Calculate the current in branch 2.
Calculate the resistance of the component in branch 2.
A third branch is added. The current from the supply rises to 3.0 A. Calculate the resistance of the component in the third branch.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = V/R = 12 ÷ 10 | ✓ 1 | |
| 1.2 A | ✓ 1 | |
| Part (b) | ||
| 2.0 − 1.2 = 0.80 A (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| R = V/I = 12 ÷ 0.80 (ECF from (b)) | ✓ 1 | |
| 15 Ω | ✓ 1 | |
| Part (d) | ||
| current in third branch = 3.0 − 2.0 = 1.0 A | ✓ 1 | |
| R = V/I = 12 ÷ 1.0 | ✓ 1 | |
| 12 Ω | ✓ 1 | |
Exam tip: Adding a parallel branch does not change the current in the existing branches; the extra current is the increase in the supply current.
A car-seat heater contains two heating elements connected in series to a 12 V supply. With both elements connected, the current in the heater is 2.0 A. One element has a resistance of 2.5 Ω.
Calculate the total resistance of the heater.
Calculate the resistance of the other element.
Calculate the potential difference across the 3.5 Ω element.
Explain why neither element heats if one of the elements breaks.
On a high setting, a switch connects the supply across the 2.5 Ω element only. Calculate the power of the heater on this setting.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| R = V ÷ I = 12 ÷ 2.0 | ✓ 1 | |
| 6.0 Ω | ✓ 1 | |
| Part (b) | ||
| 6.0 − 2.5 = 3.5 Ω (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| V = I R = 2.0 × 3.5 | ✓ 1 | |
| 7.0 V | ✓ 1 | |
| Part (d) | ||
| The elements are in series, so there is only one path for the current | ✓ 1 | |
| A break stops the current in the whole circuit | ✓ 1 | |
| Part (e) | ||
| Current = 12 ÷ 2.5 = 4.8 A | ✓ 1 | |
| P = 12 × 4.8 = 57.6 W | ✓ 1 | |
Common error: In series the p.d.s across the elements add up to the supply p.d. so a check is that your two p.d.s sum to the supply p.d..
A cyclist's hub dynamo supplies 6.0 V to a front lamp and a rear lamp connected in parallel. The front lamp has a power of 2.4 W and the rear lamp has a power of 0.60 W.
Calculate the current in each lamp.
Calculate the current in the dynamo.
Calculate the resistance of the rear lamp.
Calculate the combined resistance of the two lamps.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Front: I = P ÷ V = 2.4 ÷ 6.0 = 0.40 A | ✓ 1 | |
| Rear: 0.60 ÷ 6.0 = 0.10 A | ✓ 1 | |
| Part (b) | ||
| 0.40 + 0.10 = 0.50 A (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| R = V ÷ I = 6.0 ÷ 0.10 | ✓ 1 | |
| 60 Ω | ✓ 1 | |
| Part (d) | ||
| R = V ÷ I = 6.0 ÷ 0.50 (ECF from (b)) | ✓ 1 | |
| 12 Ω | ✓ 1 | |
Common error: In parallel, both lamps have the full p.d. across them and the combined resistance is smaller than either lamp's resistance.
A lighthouse lamp is marked 12 V, 24 W. It is operated at its rated values by connecting it in series with a resistor to a 20 V supply.
Calculate the current in the lamp.
Calculate the resistance of the resistor.
Calculate the power dissipated in the resistor.
Calculate the percentage of the power supplied by the 20 V supply that is transferred by the lamp.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = P / V = 24 / 12 | ✓ 1 | |
| 2.0 A | ✓ 1 | |
| Part (b) | ||
| p.d. across resistor = 20 − 12 = 8.0 V | ✓ 1 | |
| R = V / I = 8.0 / 2.0 | ✓ 1 | |
| 4.0 Ω | ✓ 1 | |
| Part (c) | ||
| P = V I = 8.0 × 2.0 | ✓ 1 | |
| 16 W | ✓ 1 | |
| Part (d) | ||
| power from supply = 20 × 2.0 = 40 W | ✓ 1 | |
| 24 / 40 × 100 = 60 % | ✓ 1 | |
Exam tip: In a series circuit, find the current first from the component whose rating you know, then use it for every other component.
An electric shower has two identical heating elements, each of resistance 11.5 Ω, connected in parallel to the 230 V mains supply on the high setting. On the low setting only one element is connected.
Calculate the current in one element.
State the total current drawn from the supply on the high setting.
Fuses of 13 A, 30 A, 45 A and 60 A are available. State which fuse should be fitted and explain why.
The shower is used on the high setting for 15 minutes. Electricity costs 28 p per kWh. Calculate the cost, in pence.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = V / R = 230 / 11.5 | ✓ 1 | |
| 20 A | ✓ 1 | |
| Part (b) | ||
| 40 A (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| 45 A | ✓ 1 | |
| It is the lowest rating above 40 A so it will not melt in normal use but protects the circuit | ✓ 1 | |
| Part (d) | ||
| power = 230 × 40 = 9200 W = 9.2 kW | ✓ 1 | |
| energy = 9.2 × 0.25 = 2.3 kWh | ✓ 1 | |
| cost = 2.3 × 28 = 64.4 p | ✓ 1 | |
Exam tip: Two parallel branches each carry their own current, and the supply provides the sum of the two.
A fashion workshop uses a steam press with a metal case. The press has a three-pin plug containing a fuse, and the case is connected to the earth pin. Explain how the fuse and the earth wire together protect the user if the live wire becomes loose inside the press and touches the metal case.
Explain how the fuse and the earth wire together protect the user. (6 marks)
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The case would become live (at mains p.d.) | ✓ 1 | |
| The earth wire provides a low-resistance path from the case to earth | ✓ 1 | |
| A large current flows through the live wire, case and earth wire | ✓ 1 | |
| The large current heats the fuse wire | ✓ 1 | |
| The fuse melts (blows) and disconnects the live supply | ✓ 1 | |
| The case is no longer live, so the user does not receive an electric shock | ✓ 1 | |
Exam tip: In a six-mark explanation, write the events in order: fault, large current, fuse blows, supply cut off, user safe.
An electrostatic spray gun coats metal panels with paint droplets. The gun delivers a steady current of 4.0 µA for 25 s. Each droplet carries a charge of 2.0 nC and has a mass of 4.0 × 10−8 kg. Only 80% of the droplets reach the panel.
Calculate the total charge sprayed in 25 s.
Calculate the number of droplets sprayed in 25 s.
Calculate the mass of paint that reaches the panel in 25 s. Give your answer in g.
A panel needs 6.4 g of paint. Calculate the time the gun must spray at this rate.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Q = I t = 4.0 × 10−6 × 25 | ✓ 1 | |
| Q = 1.0 × 10−4 C | ✓ 1 | |
| Part (b) | ||
| number = total charge ÷ charge per droplet = 1.0 × 10−4 ÷ 2.0 × 10−9 | ✓ 1 | |
| 5.0 × 104 droplets | ✓ 1 | |
| Part (c) | ||
| Droplets reaching panel = 0.80 × 5.0 × 104 = 4.0 × 104 | ✓ 1 | |
| Mass = 4.0 × 104 × 4.0 × 10−8 kg = 1.6 × 10−3 kg = 1.6 g | ✓ 1 | |
| Part (d) | ||
| Rate = 1.6 ÷ 25 = 0.064 g/s | ✓ 1 | |
| t = 6.4 ÷ 0.064 = 100 s | ✓ 1 | |
Exam tip: Break a long calculation into steps: charge, then number of droplets, then mass, then time. Write the unit at every step.
A crop-spraying system gives each droplet of spray the same negative charge as it leaves the nozzle. The droplets are attracted to the leaves of a plant, which are earthed.
Explain why the droplets spread out as they leave the nozzle.
Explain why a negatively charged droplet is attracted to a leaf.
The crop-spraying aircraft is connected to earth by a metal cable before it is refuelled. Explain why.
The droplets carry away a total charge of 5.4 mC (5.4 × 10−3 C) in 90 s. Calculate the average current.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| all the droplets carry the same (negative) charge | ✓ 1 | |
| like charges repel, so the droplets move apart | ✓ 1 | |
| Part (b) | ||
| the droplet repels electrons in the leaf | ✓ 1 | |
| the surface of the leaf nearest the droplet becomes positively charged (as electrons are repelled away / to earth) | ✓ 1 | |
| opposite charges attract | ✓ 1 | |
| Part (c) | ||
| fuel flowing through the hose can cause a build-up of charge on the aircraft, which could produce a spark that ignites the fuel vapour | ✓ 1 | |
| the cable lets the charge flow safely to earth so no charge builds up | ✓ 1 | |
| Part (d) | ||
| I = Q ÷ t | ✓ 1 | |
| = 5.4 × 10−3 ÷ 90 = 6.0 × 10−5 A | ✓ 1 | |
Common error: In induction explanations, say that electrons are repelled or attracted. Do not say that positive charges move through a solid.
The heated seat in a tractor cab has two heating elements, X and Y, of resistance 4.0 Ω and 8.0 Ω. The seat is powered by the 12 V tractor battery. On the low setting the elements are connected in series.
Calculate the current from the battery on the low setting.
Show that the total power of the elements on the low setting is 12 W.
On the high setting the two elements are connected in parallel to the battery. Explain why more thermal energy is produced each second on this setting.
Calculate the total power of the elements on the high setting.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| total resistance = 4.0 + 8.0 = 12 Ω | ✓ 1 | |
| I = 12 ÷ 12 = 1.0 A | ✓ 1 | |
| Part (b) | ||
| P = I2R: X = 1.02 × 4.0 = 4.0 W and Y = 1.02 × 8.0 = 8.0 W | ✓ 1 | |
| total = 4.0 + 8.0 = 12 W (accept P = I V = 1.0 × 12 = 12 W with the substitution shown) | ✓ 1 | |
| Part (c) | ||
| each element now has the full 12 V across it, so the current in each element is larger | ✓ 1 | |
| the current from the battery is larger, so the power (P = I V) is greater | ✓ 1 | |
| Part (d) | ||
| IX = 12 ÷ 4.0 = 3.0 A and IY = 12 ÷ 8.0 = 1.5 A | ✓ 1 | |
| P = I V gives 36 W and 18 W | ✓ 1 | |
| total power = 54 W | ✓ 1 | |
Exam tip: Elements in parallel each have the full supply p.d. across them, so work out the current in each branch separately.
The lamp in a model lighthouse works normally with 3.0 V across it and a current of 0.60 A. It is connected in series with a 4.0 Ω resistor and a battery. The circuit is switched on for 5.0 minutes.
Calculate the potential difference of the battery needed for the lamp to work normally.
Calculate the energy transferred by the battery in 5.0 minutes.
Calculate the percentage of this energy that is transferred by the lamp. Give your answer to 3 significant figures. Use 972 J for the energy from the battery.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| pd across resistor = 0.60 × 4.0 = 2.4 V | ✓ 1 | |
| Use of V = IR (for the resistor) | ✓ 1 | |
| Battery pd = 3.0 + 2.4 = 5.4 V | ✓ 1 | |
| Part (b) | ||
| t = 5.0 × 60 = 300 s | ✓ 1 | |
| Use of E = VIt | ✓ 1 | |
| E = 5.4 × 0.60 × 300 = 972 J (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| Energy transferred by lamp = 3.0 × 0.60 × 300 = 540 J | ✓ 1 | |
| Efficiency = 540 ÷ 972 × 100 | ✓ 1 | |
| = 55.6 % | ✓ 1 | |
Exam tip: Energy transferred = V × I × t, with time in seconds; use each component's own pd to find its share of the energy.
A mains-powered lawnmower used by a bridge maintenance crew has a metal casing and a three-core cable. A fault makes the live wire touch the casing.
Explain how the earth wire and the fuse together protect a person holding the mower.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 the casing is connected to the earth wire | ✓ 1 | |
| B1 the earth wire gives a path of very low resistance to earth | ✓ 1 | |
| B1 so a very large current flows in the live wire | ✓ 1 | |
| B1 the current is larger than the fuse rating so the fuse melts | ✓ 1 | |
| B1 the circuit is broken and the casing is no longer live, so the person does not receive a shock | ✓ 1 | |
Exam tip: Write the explanation as a chain: fault, large current, fuse melts, casing safe.
A light-beam alarm uses a 6.0 V supply of negligible internal resistance, a 4.0 kΩ fixed resistor and a light-dependent resistor (LDR) connected in series. The alarm sounds when the potential difference across the LDR rises above 4.0 V. While the beam shines on the LDR its resistance is 2.0 kΩ. When the beam is blocked its resistance is 20 kΩ.
State how the resistance of an LDR depends on the intensity of light falling on it.
Calculate the current in the circuit while the beam is not blocked.
Calculate the potential difference across the LDR when the beam is blocked.
Explain why the alarm sounds when the beam is blocked.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 resistance decreases as light intensity increases (or increases as intensity decreases) | ✓ 1 | |
| Part (b) | ||
| M1 total resistance = 4.0 + 2.0 = 6.0 kΩ and I = V/R | ✓ 1 | |
| A1 1.0 mA (or 1.0 × 10−3 A) | ✓ 1 | |
| Part (c) | ||
| M1 total resistance = 4.0 + 20 = 24 kΩ | ✓ 1 | |
| M1 current = 6.0 ÷ 24 000 = 2.5 × 10−4 A and V = IR | ✓ 1 | |
| A1 5.0 V | ✓ 1 | |
| Part (d) | ||
| B1 the LDR resistance rises, so it takes a larger share of the supply potential difference | ✓ 1 | |
| B1 5.0 V (or the potential difference across it) is above 4.0 V, so the alarm is triggered; when unblocked the value is only 2.0 V | ✓ 1 | |
Exam tip: In a series circuit find the total resistance first, then the current, then the potential difference across the component you want.
An electric blanket contains two identical heating elements, each of resistance R. A switch connects them in parallel for the high setting, giving a power of 230 W, and in series for the low setting. The blanket is connected to the 230 V mains supply.
Calculate the current from the supply on the high setting.
Calculate the resistance R of one heating element.
Calculate the power of the blanket on the low setting.
Explain why the blanket transfers less power on the low setting than on the high setting.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P/V = 230 ÷ 230 | ✓ 1 | |
| A1 1.0 A | ✓ 1 | |
| Part (b) | ||
| M1 the identical elements share the current equally, so each carries 0.50 A, and R = V/I = 230 ÷ 0.50 (ECF from (a)) | ✓ 1 | |
| A1 460 Ω | ✓ 1 | |
| Part (c) | ||
| M1 total resistance in series = 2 × 460 = 920 Ω (ECF from (b)) | ✓ 1 | |
| M1 I = 230 ÷ 920 = 0.25 A, then P = IV | ✓ 1 | |
| A1 57.5 W | ✓ 1 | |
| Part (d) | ||
| B1 in series the total resistance is larger than in parallel | ✓ 1 | |
| B1 so the current from the same supply is smaller, and P = IV is smaller | ✓ 1 | |
Common error: Do not halve the resistance for the series arrangement: resistances add in series and combine to a smaller value in parallel.
A 12 V battery is connected to a 4.0 Ω resistor in series with two resistors, of 8.0 Ω and 24 Ω, connected in parallel with each other. These form the heating elements of a robot's grip pad. The battery supplies a total current of 1.2 A.
Calculate the pd across the 4.0 Ω resistor.
Calculate the pd across each of the two parallel resistors.
Calculate the current in the 24 Ω resistor.
Calculate the power of the 8.0 Ω resistor. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 V = I R, 1.2 × 4.0 | ✓ 1 | |
| A1 4.8 V | ✓ 1 | |
| Part (b) | ||
| M1 12 − 4.8 | ✓ 1 | |
| A1 7.2 V (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 I = V ÷ R, 7.2 ÷ 24 | ✓ 1 | |
| A1 0.30 A (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 current in 8.0 Ω resistor = 7.2 ÷ 8.0 | ✓ 1 | |
| M1 P = V I, 7.2 × 0.90 | ✓ 1 | |
| A1 6.5 W (ECF from (b)) | ✓ 1 | |
Common error: Parallel resistors share the same pd, but the current splits between them.
The rear window of a car has a demister made of three identical heating strips connected in parallel to a 12 V battery. Each strip has a resistance of 4.8 Ω.
State the potential difference across each strip.
Calculate the current in one strip.
Calculate the total current taken from the battery.
Calculate the energy transferred by the demister in 4.0 minutes. Use the unrounded values from earlier parts.
One strip breaks and stops conducting. Explain what happens to the current in each of the other strips and to the total current.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A1 12 V | ✓ 1 | |
| Part (b) | ||
| M1 I = V ÷ R = 12 ÷ 4.8 | ✓ 1 | |
| A1 2.5 A | ✓ 1 | |
| Part (c) | ||
| A1 3 × 2.5 = 7.5 A (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 P = I × V = 7.5 × 12 = 90 W | ✓ 1 | |
| M1 t = 4.0 × 60 = 240 s and E = P × t | ✓ 1 | |
| A1 21600 J | ✓ 1 | |
| Part (e) | ||
| the other strips still have 12 V across them, so each still carries 2.5 A | ✓ 1 | |
| the total current falls to 5.0 A because one branch no longer carries current | ✓ 1 | |
Common error: Convert the time to seconds when using E = P × t; 4.0 minutes is 240 s, not 4.0.
An electric fence energiser delivers pulses, each of energy E, at a rate of f pulses per second. It is then adjusted to deliver pulses of energy E/2 at a rate of 3f pulses per second. How does the mean power of the energiser change?
| Marking point | Mark | Notes |
|---|---|---|
| Mean power = energy per pulse × pulses per second. Originally this is E × f. After the change it is (E/2) × 3f = 1.5 Ef, so the power increases by a factor of 1.5. Considering only the pulse rate gives a factor of 3. Multiplying the rate by 3 and dividing by one half incorrectly gives a factor of 6. Considering only the lower pulse energy gives a halving. | ✓ 1 | Answer D |
Answer: D
Exam tip: Mean power for repeated pulses is energy per pulse multiplied by the number of pulses each second. Change both factors before you compare.
During a power cut, the radio equipment on a telecoms mast runs from a 48 V battery. The equipment draws a steady current of 2.5 A for 6.0 hours.
Calculate the charge that passes through the equipment in 6.0 hours.
Calculate the energy transferred by the battery in this time. Use the exact value from part (a). Give your answer to 3 significant figures.
Calculate the power of the equipment.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 t = 6.0 × 3600 = 21 600 s | ✓ 1 | |
| M1 Q = It = 2.5 × 21 600 | ✓ 1 | |
| A1 54 000 C | ✓ 1 | |
| Part (b) | ||
| M1 E = QV = 54 000 × 48 | ✓ 1 | |
| A1 2.59 × 106 J (2 592 000 J) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 P = VI = 48 × 2.5 | ✓ 1 | |
| A1 120 W | ✓ 1 | |
Exam tip: You can check an energy answer by comparing E = QV with E = Pt; both must agree.
An electric bicycle battery is charged in three stages. The charger holds the current constant within each stage, as shown in the table.
| Stage | Current / A | Duration |
|---|---|---|
| 1 | 4.0 | 2.0 h |
| 2 | 2.0 | 1.0 h |
| 3 | 0.50 | 0.50 h |
Calculate the charge that flows into the battery in stage 1.
Calculate the total charge that flows into the battery in the three stages.
The battery has a potential difference of 36 V. Use the exact value from part (b) to calculate the energy transferred to the battery. Give your answer to 3 significant figures.
Calculate the mean current over the whole charging time. Use the exact value from part (b). Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 Q = It = 4.0 × (2.0 × 3600) | ✓ 1 | |
| A1 28 800 C | ✓ 1 | |
| Part (b) | ||
| M1 stage 2: 2.0 × 3600 = 7200 C | ✓ 1 | |
| M1 stage 3: 0.50 × 1800 = 900 C and add all three stages | ✓ 1 | |
| A1 36 900 C | ✓ 1 | |
| Part (c) | ||
| M1 E = QV = 36 900 × 36 | ✓ 1 | |
| A1 1.33 × 106 J (1 328 400 J) (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 mean I = Q/t = 36 900 / (3.5 × 3600) | ✓ 1 | |
| A1 2.93 A (ECF from (b)) | ✓ 1 | |
Common error: The mean current is total charge divided by total time, not the average of the three current values.
An electric fence energiser produces one pulse every 1.2 s. In each pulse, 0.0020 C of charge is moved through a potential difference of 6000 V.
Calculate the energy transferred in one pulse.
Calculate the number of pulses in one hour.
Calculate the total energy transferred by the pulses in one hour.
Calculate the mean power of the energiser in one hour. Use the exact value from part (c).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 E = QV = 0.0020 × 6000 | ✓ 1 | |
| A1 12 J | ✓ 1 | |
| Part (b) | ||
| M1 3600 / 1.2 | ✓ 1 | |
| A1 3000 pulses | ✓ 1 | |
| Part (c) | ||
| M1 3000 × 12 | ✓ 1 | |
| A1 36 000 J (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| M1 P = E/t = 36 000 / 3600 | ✓ 1 | |
| A1 10 W (ECF from (c)) | ✓ 1 | |
Exam tip: Mean power over a long time is total energy divided by total time, even when the energy arrives in short pulses.
A clinic uses a portable air purifier rated 45 W at 230 V. It runs for 10 hours each day. Electricity costs 24 p per kilowatt-hour.
Calculate the current in the purifier. Give your answer to 3 significant figures.
Calculate the energy used in 30 days. Give your answer in kilowatt-hours.
Calculate the cost of running the purifier for 30 days. Give your answer in pounds (£) to 2 decimal places.
A second purifier is rated 60 W. One use costs 7.2 p. Calculate the time of that use, in hours.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P/V = 45 / 230 | ✓ 1 | |
| A1 0.196 A | ✓ 1 | |
| Part (b) | ||
| M1 0.045 kW × (10 × 30) h | ✓ 1 | |
| A1 13.5 kWh | ✓ 1 | |
| Part (c) | ||
| M1 13.5 × 24 p = 324 p | ✓ 1 | |
| A1 £3.24 (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 energy = 7.2 / 24 = 0.30 kWh | ✓ 1 | |
| M1 t = E/P = 0.30 / 0.060 | ✓ 1 | |
| A1 5.0 h | ✓ 1 | |
Exam tip: If the question gives the cost and the price per unit, work backwards to energy first, then to time.
A string of 20 identical decorative lamps is connected in series to a 230 V supply. Each lamp is rated 11.5 V, 0.20 A. One lamp fails and is replaced by a piece of wire, so only 19 working lamps remain in the string. Assume the resistance of each lamp stays constant.
Calculate the resistance of one lamp.
Calculate the current in the string with 19 lamps. Give your answer to 3 significant figures.
Explain why the remaining lamps are now more likely to fail.
Calculate the power of each remaining lamp. Use unrounded values from earlier parts. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 resistance = potential difference ÷ current | ✓ 1 | |
| A1 57.5 Ω | ✓ 1 | |
| Part (b) | ||
| M1 total resistance = 19 × 57.5 = 1092.5 Ω | ✓ 1 | |
| M1 current = 230 ÷ total resistance | ✓ 1 | |
| A1 0.211 A | ✓ 1 | |
| Part (c) | ||
| the p.d. across each lamp rises above 11.5 V (to about 12.1 V), so the current is larger than 0.20 A | ✓ 1 | |
| each lamp transfers more power and gets hotter, so the filament is more likely to melt | ✓ 1 | |
| Part (d) | ||
| M1 p.d. across each lamp = 230 ÷ 19 and power = V² ÷ R (or I²R) | ✓ 1 | |
| A1 2.55 W | ✓ 1 | |
Common error: Removing a component from a series string leaves the same supply p.d. shared among fewer components, so each gets more.
A logistics depot recharges 8 identical cordless drill batteries one after another from one charger. Each battery must receive 5000 C at a constant current of 2.0 A. The charger gives a p.d. of 20 V across each battery.
Calculate the time to recharge one battery. Give your answer in seconds.
Calculate the total energy transferred to the 8 batteries.
The charger takes 1.0 × 106 J from the mains to do this. Calculate the efficiency of the charger as a percentage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| t = Q ÷ I | ✓ 1 | |
| 5000 ÷ 2.0 = 2500 s | ✓ 1 | |
| Part (b) | ||
| energy for one battery = Q × V = 5000 × 20 = 1.0 × 105 J | ✓ 1 | |
| 8 × 1.0 × 105 = 8.0 × 105 J | ✓ 1 | |
| Part (c) | ||
| efficiency = useful energy ÷ total energy = 8.0 × 105 ÷ 1.0 × 106 (ECF from (b)) | ✓ 1 | |
| 80% | ✓ 1 | |
Common error: Use the energy transferred to the batteries as the useful energy, not the energy from the mains.
A student connects a 6.0 Ω resistor and a 12 Ω resistor in parallel. This pair is connected in series with a 4.0 Ω resistor and a 12 V battery of negligible internal resistance.
Calculate the combined resistance of the 6.0 Ω and 12 Ω resistors.
Calculate the total resistance of the circuit.
Calculate the current from the battery.
Calculate the p.d. across the parallel pair.
Calculate the current in the 12 Ω resistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 1 ÷ R = 1 ÷ 6.0 + 1 ÷ 12 | ✓ 1 | |
| A1 4.0 Ω | ✓ 1 | |
| Part (b) | ||
| A1 8.0 Ω (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 I = V ÷ R = 12 ÷ 8.0 | ✓ 1 | |
| A1 1.5 A (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 V = 12 − (1.5 × 4.0) or 1.5 × 4.0 | ✓ 1 | |
| A1 6.0 V (ECF from (c)) | ✓ 1 | |
| Part (e) | ||
| M1 I = V ÷ R = 6.0 ÷ 12 | ✓ 1 | |
| A1 0.50 A (ECF from (d)) | ✓ 1 | |
Common error: The p.d. across a parallel pair is the battery p.d. minus the p.d. across the series resistor, not the full 12 V.
In a science museum, a visitor stands on a plastic platform and places a hand on the metal dome of a Van de Graaff generator. The visitor's hair begins to stand on end.
The dome gives the visitor a negative charge.
State the name of the particles that move onto the visitor.
Explain why the visitor's hairs stand on end.
Suggest why the visitor must stand on an insulating platform.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Electrons | ✓ 1 | |
| Part (b) | ||
| Each hair gains the same (negative) charge | ✓ 1 | |
| Like charges repel, so the hairs push away from each other / from the head | ✓ 1 | |
| Part (c) | ||
| It stops the charge / electrons flowing through the visitor's feet to earth, so charge can build up on the visitor (accept either idea) | ✓ 1 | |
Exam tip: When explaining repulsion, say both that the objects have the same type of charge and that like charges repel.
A student investigates charging by friction. She rubs a polythene rod with a dry cloth.
State why a polythene rod becomes charged when rubbed but a metal rod held in the hand does not.
Describe how the student could use small pieces of paper to show that the rod is charged.
The polythene rod becomes negatively charged.
State the charge on the cloth and explain your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Polythene is an insulator so charge stays on it / in metal, charge flows away through the hand to earth | ✓ 1 | |
| Part (b) | ||
| Hold the rod close to (but not touching) the pieces of paper | ✓ 1 | |
| If the rod is charged, the pieces jump up / are attracted to the rod | ✓ 1 | |
| Part (c) | ||
| The cloth becomes positive | ✓ 1 | |
| Electrons were transferred from the cloth to the rod, so the cloth has lost electrons | ✓ 1 | |
Common error: Do not write that the cloth 'gains positive charge' or 'gains protons'. It becomes positive because it loses electrons.
An electric bicycle battery provides a steady current of 5.0 A at 36 V for 20 minutes.
Calculate the charge that flows. Use charge = current × time.
Calculate the energy transferred by the battery. Use the unrounded charge from (a) and energy = charge × p.d.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 20 min = 1200 s and Q = 5.0 × 1200 | ✓ 1 | |
| A1 6000 C | ✓ 1 | |
| Part (b) | ||
| M1 E = Q V = 6000 × 36 | ✓ 1 | |
| A1 216 000 J (accept 2.16 × 105 J) | ✓ 1 | |
Common error: Change minutes into seconds before using charge = current × time.
A copper wire in a mountain-hut lighting circuit carries a steady current of 3.0 A. The charge on one electron is 1.6 × 10−19 C.
State the name of the particles that move to make the current in the copper wire.
State the direction in which these particles move compared with the direction of the conventional current.
Calculate the charge that flows past a point in the wire in 40 s.
Calculate the number of electrons that pass the point in 40 s. Use the exact charge from part (c).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| (free / delocalised) electrons | ✓ 1 | |
| Part (b) | ||
| the opposite direction | ✓ 1 | |
| Part (c) | ||
| M1 Q = It = 3.0 × 40 | ✓ 1 | |
| A1 120 C | ✓ 1 | |
| Part (d) | ||
| M1 number = total charge ÷ charge on one electron | ✓ 1 | |
| M1 120 / (1.6 × 10−19) | ✓ 1 | |
| A1 7.5 × 1020 (ECF from (c)) | ✓ 1 | |
Common error: Count the electrons by dividing the total charge by the charge on one electron, not by multiplying.
A clear plastic document wallet is rubbed against a woollen jumper sleeve. The wallet becomes negatively charged.
Which statement explains why?
| Marking point | Mark | Notes |
|---|---|---|
| When two insulators are rubbed together, only electrons move from one to the other. Electrons are negatively charged. The wallet gains extra electrons from the wool, so it becomes negative and the wool is left positive. Protons are fixed in the nuclei and do not move, so the first option is wrong. If electrons left the wallet, it would become positive, so the third option is wrong. Charge is transferred, not created, so the fourth option is wrong. | ✓ 1 | Answer B |
Answer: B
Common error: Never say that positive charges or protons move when explaining charging by friction. Only electrons are transferred.
A child slides down a plastic playground slide. Afterwards the slide is negatively charged and the child is positively charged.
Which statement explains why the child is positively charged?
| Marking point | Mark | Notes |
|---|---|---|
| When two materials rub together, only electrons are transferred. The child lost electrons to the slide. The child is left with more protons than electrons, so the child is positive. The slide gained electrons, so it is negative. Protons are fixed in the nuclei and do not move. Charge is transferred, not created by friction. Electrons moving onto the child would make the child negative. | ✓ 1 | Answer B |
Answer: B
Exam tip: In any charging-by-friction answer, say that electrons are transferred and state the direction. Never say that protons or 'positive charges' move.
A workshop uses electrostatic powder coating to paint metal garden gates. A spray gun gives the paint powder particles a negative charge. The metal gate is connected to earth.
Explain why the powder particles spread out into a wide, even cloud after leaving the gun.
Explain, in terms of electrons, why the powder is attracted to the earthed gate.
Suggest one advantage of this method compared with ordinary spray painting.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| All the particles have the same (negative) charge | ✓ 1 | |
| Like charges repel, so the particles push each other apart | ✓ 1 | |
| Part (b) | ||
| The negative powder repels electrons in the gate, which flow away to earth through the earth connection | ✓ 1 | |
| The gate is left with a positive charge, and opposite charges attract, so the powder is pulled onto the gate | ✓ 1 | |
| Part (c) | ||
| Any one: less paint is wasted; gives an even coating; powder is also attracted to the back and edges of the bars | ✓ 1 | |
Exam tip: When explaining induced charge, say which way the electrons move and why. A clear answer states that electrons move and that protons stay fixed.
An electric fence energiser delivers pulses, each transferring energy E at a pd V. A different energiser uses pulses with twice the energy and twice the pd. How does the charge transferred in each pulse compare with the first energiser?
| Marking point | Mark | Notes |
|---|---|---|
| Charge = energy ÷ pd (Q = E/V). Doubling both E and V leaves Q unchanged, so A is correct. B ignores the doubled pd; C multiplies the two factors instead of dividing; D divides the wrong way round. | ✓ 1 | Answer A |
Answer: A
Common error: When two quantities change in a ratio equation, apply both changes: doubling the top and the bottom of E/V cancels.
An electric fence energiser transfers 0.90 J in each pulse and sends one pulse every 1.5 s. It draws electrical energy from its battery at a rate of 2.4 W.
Calculate the number of pulses sent in one hour.
Calculate the energy transferred by the pulses in one hour.
The battery supplies 8640 J to the energiser in one hour. Calculate the efficiency of the energiser as a percentage.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| time = 3600 s | ✓ 1 | |
| 3600 ÷ 1.5 = 2400 | ✓ 1 | |
| Part (b) | ||
| energy = 2400 × 0.90 (ECF from (a)) | ✓ 1 | |
| 2160 J | ✓ 1 | |
| Part (c) | ||
| efficiency = useful energy out ÷ total energy in | ✓ 1 | |
| (2160 ÷ 8640) × 100 = 25 % (ECF from (b)) | ✓ 1 | |
Exam tip: Efficiency compares energy (or power) out with energy (or power) in over the same time; check both use the same period.
An e-bike battery is charged in three stages.
Stage 1: the current is a steady 5.0 A for 40 minutes.
Stage 2: the current falls steadily from 5.0 A to 1.0 A over the next 30 minutes.
Stage 3: the current is a steady 1.0 A for the final 20 minutes.
Calculate the charge that flows in stage 1.
Calculate the charge that flows in stage 2.
Calculate the total charge that flows during the three stages.
The battery has a potential difference of 36 V. Calculate the energy transferred to the battery.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Q = I t = 5.0 × 2400 | ✓ 1 | |
| 12000 C | ✓ 1 | |
| Part (b) | ||
| Average current = (5.0 + 1.0) ÷ 2 = 3.0 A | ✓ 1 | |
| 3.0 × 1800 | ✓ 1 | |
| 5400 C | ✓ 1 | |
| Part (c) | ||
| Stage 3 charge = 1.0 × 1200 = 1200 C | ✓ 1 | |
| 12000 + 5400 + 1200 = 18600 C | ✓ 1 | |
| Part (d) | ||
| E = Q V = 18600 × 36 (ECF from (c)) | ✓ 1 | |
| 669600 J | ✓ 1 | |
Common error: For a current that changes steadily, use the average current. Time must be in seconds.
In a car factory, paint droplets are given a positive charge as they leave a spray gun. The car panel being painted is earthed.
Explain why the charged droplets spread out into a wide cloud after leaving the gun.
Explain why the earthed panel helps the paint to coat the panel evenly, including its edges.
Each droplet carries a charge of 2.4 × 10−12 C. The spray gun produces a current of 1.2 × 10−6 A in the droplets. Calculate the number of droplets sprayed each second.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| All the droplets carry the same (positive) charge | ✓ 1 | |
| Like charges repel each other, so the droplets move apart | ✓ 1 | |
| Part (b) | ||
| Electrons flow from earth to the panel, giving it a negative charge near the droplets | ✓ 1 | |
| Opposite charges attract, so the droplets are pulled towards the panel | ✓ 1 | |
| Droplets are drawn to all exposed surfaces, including edges and the back, rather than missing the panel | ✓ 1 | |
| Part (c) | ||
| Charge per second = 1.2 × 10−6 C, then 1.2 × 10−6 ÷ 2.4 × 10−12 | ✓ 1 | |
| 5.0 × 105 (= 500000) droplets per second | ✓ 1 | |
Exam tip: In a static electricity explanation, state who has what charge, then say whether the force is attraction or repulsion.
A rail signalling test rig has an LDR in series with a fixed resistor, connected to a 6.0 V supply of negligible internal resistance. A voltmeter is connected across the LDR and an ammeter is in the circuit. A lamp shining on the LDR is moved closer to it. What happens to the readings?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: B. More light means the LDR resistance falls, so the total resistance falls and the current rises. The fixed resistor now has a larger p.d. (V = IR), and since the supply p.d. is fixed the p.d. across the LDR falls. A claims more light gives a bigger p.d. across the LDR, forgetting the supply is shared. C has the resistance of the LDR rising, which is the effect of less light. D has a falling current, which does not follow from a falling total resistance. | ✓ 1 | Answer B |
Answer: B
Exam tip: In a series circuit with a fixed supply p.d., the component whose resistance falls takes a smaller share of the p.d.
A rail tanker is being filled with fuel through a plastic pipe. The flowing fuel rubs against the pipe and charge builds up on the tanker.
Explain how a plastic surface becomes negatively charged when it is rubbed.
A spark transfers 3.0 × 10−6 C of charge in 0.20 ms. Calculate the average current in the spark.
The p.d. between the tanker and earth just before the spark is 12 kV. Calculate the energy transferred in the spark.
Suggest why a spark is dangerous in this situation and state one way of preventing it.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Electrons are transferred onto the plastic from the material rubbing it | ✓ 1 | |
| The plastic then has more negative charge than positive charge | ✓ 1 | |
| Part (b) | ||
| I = Q / t = 3.0 × 10−6 / 2.0 × 10−4 | ✓ 1 | |
| 0.015 A | ✓ 1 | |
| Part (c) | ||
| E = Q V = 3.0 × 10−6 × 12 000 | ✓ 1 | |
| 0.036 J | ✓ 1 | |
| Part (d) | ||
| The spark could ignite the fuel vapour | ✓ 1 | |
| Connect (earth) the tanker to the ground with a conducting wire so charge flows away safely | ✓ 1 | |
Common error: Convert 0.20 ms to 2.0 × 10⁻⁴ s and 12 kV to 12 000 V before substituting.
A diode in a robot's sensor board is tested in the forward direction. The results are shown in the table.
| V / V | 0.00 | 0.40 | 0.60 | 0.70 | 0.80 |
| I / mA | 0.0 | 0.0 | 5.0 | 20.0 | 80.0 |
Calculate the resistance of the diode when the p.d. across it is 0.70 V.
Explain how the resistance of the diode changes as the p.d. increases from 0.60 V to 0.80 V.
State the current in the diode when it is connected in the reverse direction.
The diode is connected in the forward direction in series with a resistor and a 5.0 V supply. The current must be 20 mA. Calculate the resistance of the resistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Use R = V ÷ I | ✓ 1 | |
| R = 0.70 ÷ 0.020 = 35 Ω | ✓ 1 | |
| Part (b) | ||
| Resistance decreases (from 120 Ω to 10 Ω) | ✓ 1 | |
| The current increases by a larger factor than the p.d. (so V ÷ I falls) | ✓ 1 | |
| Part (c) | ||
| Almost zero (or zero) | ✓ 1 | |
| Part (d) | ||
| p.d. across diode at 20 mA = 0.70 V (from table) | ✓ 1 | |
| p.d. across resistor = 5.0 − 0.70 = 4.3 V | ✓ 1 | |
| R = 4.3 ÷ 0.020 = 215 Ω | ✓ 1 | |
Common error: The p.d. across a series resistor is the supply p.d. minus the p.d. across the diode, not the whole supply p.d.
A light-beam alarm protects the door of a cold store. A 9.0 V battery is connected in series with a 3.0 kΩ fixed resistor and a light-dependent resistor (LDR). The alarm is connected across the LDR and sounds when the potential difference across the LDR is more than 5.0 V. A light beam shines on the LDR.
State how the resistance of an LDR changes when the light falling on it decreases.
When the beam falls on the LDR its resistance is 1.5 kΩ. Calculate the potential difference across the LDR.
A person blocks the beam and the resistance of the LDR rises to 6.0 kΩ. Calculate the new potential difference across the LDR and state whether the alarm sounds.
Suggest why the alarm also sounds if the lights in the cold store fail.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The resistance increases | ✓ 1 | |
| Part (b) | ||
| Total resistance = 3.0 + 1.5 = 4.5 kΩ | ✓ 1 | |
| I = 9.0 ÷ 4500 = 2.0 × 10−3 A | ✓ 1 | |
| V = I R = 2.0 × 10−3 × 1500 = 3.0 V | ✓ 1 | |
| Part (c) | ||
| Total resistance = 3.0 + 6.0 = 9.0 kΩ and I = 9.0 ÷ 9000 = 1.0 × 10−3 A | ✓ 1 | |
| V = 1.0 × 10−3 × 6000 = 6.0 V | ✓ 1 | |
| 6.0 V is more than 5.0 V so the alarm sounds | ✓ 1 | |
| Part (d) | ||
| Less light reaches the LDR so its resistance rises and the potential difference across it exceeds 5.0 V | ✓ 1 | |
Exam tip: In a potential-divider circuit the larger resistor takes the larger share of the supply voltage.
A student investigates how the current in a filament lamp depends on the potential difference across it. The results are shown.
Potential difference / V: 0, 2.0, 4.0, 6.0, 8.0, 10.0
Current / A: 0, 0.30, 0.48, 0.60, 0.68, 0.72
Describe a circuit the student could use to collect these results. Include the positions of the meters.
Calculate the resistance of the lamp when the potential difference is 6.0 V.
Describe how the resistance of the lamp changes as the potential difference increases and explain why this happens.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| lamp connected to a variable power supply (or cell with variable resistor) in a series circuit | ✓ 1 | |
| ammeter in series with the lamp | ✓ 1 | |
| voltmeter in parallel with the lamp | ✓ 1 | |
| change the potential difference in steps and record the current at each step | ✓ 1 | |
| Part (b) | ||
| R = V ÷ I | ✓ 1 | |
| 10 Ω | ✓ 1 | |
| Part (c) | ||
| resistance increases | ✓ 1 | |
| the filament gets hotter | ✓ 1 | |
| ions vibrate more so electrons collide with them more often | ✓ 1 | |
Exam tip: Name where each meter goes: ammeter in series, voltmeter in parallel with the component.
A charge of 2.5 C passes through an LED indicator in an aircraft cabin sign and 6.0 J of energy is transferred. What is the potential difference across the LED?
| Marking point | Mark | Notes |
|---|---|---|
| Potential difference = energy transferred ÷ charge = 6.0 ÷ 2.5 = 2.4 V. The value 0.42 V comes from dividing charge by energy. The value 3.5 V comes from subtracting the charge from the energy. The value 15 V comes from multiplying energy by charge. | ✓ 1 | Answer C |
Answer: C
Common error: Write E = Q V and rearrange it before substituting, so you do not invert the quotient.
An effects pedal for an electric guitar is powered by a 9.0 V battery. It draws a constant current of 40 mA.
Calculate the power of the pedal.
Calculate the charge that passes through the pedal in 5.0 minutes.
Use E = Q V to calculate the energy transferred by the battery in 5.0 minutes.
The battery can transfer a total of 3000 J before it is flat. Calculate how long the pedal can operate. Give your answer in hours to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P = I V | ✓ 1 | |
| = 0.040 × 9.0 = 0.36 W | ✓ 1 | |
| Part (b) | ||
| Q = I t = 0.040 × 300 | ✓ 1 | |
| = 12 C | ✓ 1 | |
| Part (c) | ||
| E = Q V = 12 × 9.0 | ✓ 1 | |
| = 108 J (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| time = energy ÷ power | ✓ 1 | |
| = 3000 ÷ 0.36 = 8333 s | ✓ 1 | |
| = 8333 ÷ 3600 = 2.31 hours | ✓ 1 | |
Common error: Keep the units consistent: use amperes and seconds in Q = I t, and divide by 3600 only at the end to convert seconds to hours.
The heater wire on a space probe has a resistance of 4.0 Ω and carries a current of 2.5 A.
Calculate the power of the heater.
The heater is on for 8.0 hours in each orbit. Calculate the energy transferred in joules.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 P = I2R | ✓ 1 | |
| A1 P = 2.52 × 4.0 = 25 W | ✓ 1 | |
| Part (b) | ||
| M1 E = P t = 25 × 8.0 × 3600 | ✓ 1 | |
| A1 E = 720 000 J (7.2 × 105 J) | ✓ 1 | |
Common error: Square only the current in P = I²R, not the resistance, and convert hours to seconds for joules.
A fuel tanker is unloading petrol through a long rubber hose. The tanker stands on rubber tyres on a dry concrete yard. Petrol vapour is present near the connection.
Explain how the tanker can become charged as the petrol flows.
Explain why the rubber tyres make the build-up of charge worse.
Explain why this charge is dangerous.
A conducting cable is used to connect the tanker to earth before unloading starts. Explain how this prevents the danger.
During unloading a current of 2.5 μA flows onto the tanker for 4.0 minutes. Calculate the charge that would collect on an unearthed tanker.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 friction between the flowing fuel and the hose or tank walls | ✓ 1 | |
| B1 electrons are transferred, so the tanker gains an overall charge | ✓ 1 | |
| Part (b) | ||
| B1 rubber is an insulator, so charge cannot flow away to earth | ✓ 1 | |
| B1 so the charge on the tanker, and its potential, keeps increasing | ✓ 1 | |
| Part (c) | ||
| B1 a spark can jump between the highly charged tanker and a nearby conductor or earthed object | ✓ 1 | |
| B1 the spark can ignite the petrol vapour | ✓ 1 | |
| Part (d) | ||
| B1 charge flows through the cable to earth as it builds up, so the tanker cannot reach a high potential | ✓ 1 | |
| Part (e) | ||
| M1 Q = It with time converted to seconds, 2.5 × 10−6 × 240 | ✓ 1 | |
| A1 6.0 × 10−4 C | ✓ 1 | |
Common error: An earthing cable does not remove the friction; it lets the charge leak away as fast as it is produced.
A hospital operating theatre lamp is connected to a 24 V supply. A steady current of 2.5 A flows in the lamp for 20 minutes.
Calculate the charge that flows through the lamp in 20 minutes.
Calculate the energy transferred by the lamp in this time.
Calculate the power of the lamp.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 Q = It with time in seconds, 2.5 × 1200 | ✓ 1 | |
| A1 3000 C | ✓ 1 | |
| Part (b) | ||
| M1 E = QV = 3000 × 24 (ECF from (a)) | ✓ 1 | |
| A1 72 000 J | ✓ 1 | |
| Part (c) | ||
| M1 P = IV = 2.5 × 24 (or energy ÷ time = 72 000 ÷ 1200) | ✓ 1 | |
| A1 60 W | ✓ 1 | |
Common error: Time must be in seconds in Q = It, E = QV and P = E/t.
A student investigates a length of thin metal wire from an aquarium heater. The potential difference across the wire and the current in it are measured.
Pd / V: 0.0, 1.0, 2.0, 3.0, 4.0, 5.0
Current / A: 0.00, 0.20, 0.36, 0.48, 0.57, 0.64
Calculate the resistance of the wire when the pd is 4.0 V. Give your answer to 2 significant figures.
Describe how the resistance of the wire changes as the pd increases. Support your answer with data.
Explain why the resistance changes in this way.
Describe how the ammeter and voltmeter are connected to measure the current in the wire and the pd across it.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V ÷ I, 4.0 ÷ 0.57 | ✓ 1 | |
| A1 7.0 Ω | ✓ 1 | |
| Part (b) | ||
| The resistance increases | ✓ 1 | |
| Data quoted, for example 5.0 Ω at 1.0 V and 7.8 Ω at 5.0 V | ✓ 1 | |
| Part (c) | ||
| The current heats the wire | ✓ 1 | |
| The metal ions vibrate more and electrons collide with them more often, so resistance increases | ✓ 1 | |
| Part (d) | ||
| Ammeter in series with the wire | ✓ 1 | |
| Voltmeter in parallel with (across) the wire | ✓ 1 | |
Exam tip: Resistance at a point is the pd divided by the current at that point, not the gradient of the line.
A filament lamp in a gaming computer case is tested. The table gives the potential difference V across the lamp and the current I in it.
V / V: 1.0, 2.0, 3.0, 4.0
I / A: 0.10, 0.18, 0.24, 0.28
Calculate the resistance of the lamp when the potential difference is 4.0 V. Give your answer to 3 significant figures.
Describe how the resistance of the lamp changes as the potential difference increases.
Explain why the resistance changes in this way.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V ÷ I = 4.0 ÷ 0.28 | ✓ 1 | |
| A1 14.3 Ω | ✓ 1 | |
| Part (b) | ||
| the resistance increases | ✓ 1 | |
| supported by values from the data, e.g. 10.0 Ω at 1.0 V and 14.3 Ω at 4.0 V | ✓ 1 | |
| Part (c) | ||
| as the current increases the filament gets hotter | ✓ 1 | |
| the metal ions vibrate more, so there are more collisions with the moving electrons and the current is opposed more | ✓ 1 | |
Common error: The graph of current against pd for a filament lamp is a curve because resistance changes; do not call it a straight line through the origin.
A data centre uses a copper busbar to carry a steady current of 250 A for 8.0 s to a rack of servers. The charge on one electron is 1.6 × 10−19 C.
Explain why copper can carry an electric current but a plastic cannot.
Calculate the charge that passes through the busbar in 8.0 s.
Calculate the number of electrons that pass through the busbar in 8.0 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Copper contains free (delocalised) electrons that can move through the metal | ✓ 1 | |
| Plastic has no free charge carriers, as its electrons are held in place, so no current can flow | ✓ 1 | |
| Part (b) | ||
| M1 charge = current × time | ✓ 1 | |
| A1 250 × 8.0 = 2000 C | ✓ 1 | |
| Part (c) | ||
| M1 number of electrons = total charge ÷ charge of one electron | ✓ 1 | |
| A1 2000 ÷ (1.6 × 10−19) = 1.25 × 1022 (ECF from (b)) | ✓ 1 | |
Exam tip: Check the power of ten on your calculator: dividing by a very small charge should give a very large number of electrons.
An electric shower has a power of 8.4 kW. Water flows through it at a rate of 0.10 kg every second. Assume that all of the electrical energy is transferred to the water. The specific heat capacity of water is 4200 J/kg °C.
Calculate the electrical energy transferred to the water each second.
Calculate the rise in temperature of the water.
The water enters the shower at 11 °C. Calculate the temperature of the water leaving the shower.
In practice the water leaves the shower cooler than this. Suggest one reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy = power × time with time = 1.0 s | ✓ 1 | |
| A1 8400 × 1.0 = 8400 J | ✓ 1 | |
| Part (b) | ||
| M1 temperature rise = energy ÷ (mass × specific heat capacity) | ✓ 1 | |
| A1 8400 ÷ (0.10 × 4200) = 20 °C (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| 11 + 20 = 31 °C (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| Some energy is transferred to the surroundings, casing or pipes | ✓ 1 | |
Common error: The mass in the equation is the mass of water heated in the same time as the energy used. Match 0.10 kg to 1.0 s.
An electrostatic sprayer used on a farm gives each droplet of pesticide spray the same positive charge as it leaves the nozzle. The spray is aimed at the leaves of a plant. The leaves are earthed and uncharged.
Explain why the droplets spread out in a cloud as they leave the nozzle.
Explain why the earthed leaves attract the droplets.
Suggest why this method can coat the underside of a leaf as well as the top.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the droplets have the same (positive) charge | ✓ 1 | |
| like charges repel, so the droplets push each other apart | ✓ 1 | |
| Part (b) | ||
| the positive droplets attract electrons (from earth) onto the leaf, so the leaf near the droplets becomes negative / charge is induced | ✓ 1 | |
| opposite charges attract (so the droplets are pulled to the leaf) | ✓ 1 | |
| Part (c) | ||
| the droplets are attracted to the leaf from all directions, so they curve round to the underside | ✓ 1 | |
Exam tip: For attraction to an uncharged object, say that charge is induced on it first.
A spark from a charged door handle transfers 0.60 μC of charge through a potential difference of 12 kV. The spark lasts 0.30 μs.
Calculate the energy transferred in the spark.
Calculate the average power of the spark. Use the energy given in the answer to (a).
State one situation in which a spark of this kind is a hazard.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 use of energy = charge × potential difference | ✓ 1 | |
| A1 7.2 × 10−3 J | ✓ 1 | |
| Part (b) | ||
| M1 use of power = energy ÷ time | ✓ 1 | |
| A1 2.4 × 104 W , ECF from (a) | ✓ 1 | |
| Part (c) | ||
| any one of: near flammable vapour or fuel (e.g. refuelling); near flammable gases or dust; near sensitive electronic components | ✓ 1 | |
Exam tip: A very large power can still transfer a tiny amount of energy if it lasts only a very short time.
An electric bicycle battery is charged for 50 minutes. For the first 20 minutes the current is a constant 6.0 A. Over the next 30 minutes the current falls steadily from 6.0 A to zero. The charger supplies the charge at an average potential difference of 40 V.
Calculate the charge that flows in the first 20 minutes.
Calculate the charge that flows in the next 30 minutes.
Calculate the total energy transferred by the charger.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 charge = 6.0 × 1200 | ✓ 1 | |
| A1 7200 C | ✓ 1 | |
| Part (b) | ||
| M1 average current = 3.0 A (or area of a triangle ½ × 6.0 × 1800) | ✓ 1 | |
| A1 5400 C | ✓ 1 | |
| Part (c) | ||
| M1 total charge = 7200 + 5400 = 12600 C | ✓ 1 | |
| M1 energy = charge × potential difference | ✓ 1 | |
| A1 5.04 × 105 J , ECF | ✓ 1 | |
Exam tip: When a current changes steadily, the charge is the average current multiplied by the time, which equals the area under the current–time graph.
A cordless drill has a rechargeable battery with a p.d. of 18 V. A charger supplies a constant current of 1.5 A for 80 minutes to charge the battery.
Calculate the charge delivered to the battery.
Calculate the energy transferred by this charge at 18 V. Give your answer to 2 significant figures.
Assume the fully charged battery can deliver the charge in (a). The drill draws a current of 6.0 A when in use. Calculate how many minutes the drill can run.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 convert 80 minutes to 4800 s | ✓ 1 | |
| M1 Q = I t = 1.5 × 4800 | ✓ 1 | |
| A1 7200 C | ✓ 1 | |
| Part (b) | ||
| M1 E = Q V | ✓ 1 | |
| A1 1.3 × 105 J (129 600 J) | ✓ 1 | |
| Part (c) | ||
| M1 t = Q ÷ I = 7200 ÷ 6.0 | ✓ 1 | |
| A1 1200 s = 20 minutes | ✓ 1 | |
Common error: Time must be in seconds when using Q = I t; 80 minutes is 4800 s.
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Isolated points, e.g. friction causes charge or sparks cause explosions, with little or no linking between how charge builds up and why the cable helps. |
| Level 2 (3–4) | Explains either the build-up of charge leading to a spark and ignition, or how the conductors remove charge, with some correct use of electron transfer. |
| Level 3 (5–6) | A logically linked explanation of both the hazard (electron transfer by friction, build-up on an insulator, large voltage, spark, ignition) and the precaution (conductors allow electrons to flow to/from earth so charge cannot build up), using correct terminology throughout. |
An aircraft is refuelled at an airport. Fuel is pumped at high speed from a tanker truck through a long hose into the aircraft's tanks.
Explain how static electricity can cause a hazard during refuelling, and describe how the risk is reduced.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Fuel rubs against the inside of the hose / friction between fuel and hose | ✓ 1 | |
| Electrons are transferred, so the fuel / aircraft / hose becomes charged | ✓ 1 | |
| The aircraft stands on rubber tyres (insulators), so charge builds up on it | ✓ 1 | |
| A large potential difference can build up, so a spark may jump | ✓ 1 | |
| The spark could ignite fuel vapour, causing a fire / explosion | ✓ 1 | |
| An earthing / bonding wire (or conducting hose) connects the aircraft and tanker to earth so charge flows away safely | ✓ 1 | |
Exam tip: For 6-mark explanations, build a chain: cause of charging → build-up of charge → spark → consequence → precaution. Each link is worth a mark.
In space, a spacecraft passes through a region of charged particles. Electrons from this region collect on the outside of the spacecraft. Some outer surfaces are metal panels and some are plastic insulating blankets. The spacecraft carries sensitive electronic equipment.
State the type of charge the spacecraft gains.
Explain why charge stays in separate patches on the insulating blankets but spreads out over the metal panels.
A sudden spark between a blanket and a panel transfers 3.0 μC of charge in 0.20 μs.
Calculate the average current in the spark.
Suggest why engineers coat the insulating blankets with a very thin conducting layer that is connected to the metal frame.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Negative | ✓ 1 | |
| Part (b) | ||
| In an insulator, electrons cannot move freely, so they stay where they land | ✓ 1 | |
| In a metal, electrons are free to move and repel each other, so they spread out over the surface | ✓ 1 | |
| Part (c) | ||
| I = Q ÷ t | ✓ 1 | |
| I = (3.0 × 10−6) ÷ (0.20 × 10−6) | ✓ 1 | |
| I = 15 A | ✓ 1 | |
| Part (d) | ||
| Charge can flow away / spread evenly, so a large charge (or potential difference) cannot build up | ✓ 1 | |
| So sparks do not happen, which prevents damage to the electronic equipment | ✓ 1 | |
Exam tip: Tiny charges can still give large currents if the time is very short. Convert μC and μs to C and s, and put each value in brackets on your calculator.
The energiser of an electric fence transfers 4.8 J of energy in each pulse. The pd across the fence wire is 6000 V and each pulse lasts 2.0 × 10−4 s.
Calculate the charge that passes in one pulse.
Calculate the average current during a pulse.
The farmer changes the energiser so that each pulse transfers twice as much energy at the same pd. Explain what happens to the charge in each pulse.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| E = QV, so Q = E/V | ✓ 1 | |
| Q = 4.8 ÷ 6000 | ✓ 1 | |
| 8.0 × 10−4 C | ✓ 1 | |
| Part (b) | ||
| I = Q/t | ✓ 1 | |
| (8.0 × 10−4 ÷ 2.0 × 10−4) = 4.0 A (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| the charge doubles | ✓ 1 | |
| because Q = E/V, and E doubles while V stays the same | ✓ 1 | |
Common error: Keep the time in seconds and the charge in coulombs; a pulse lasting 2.0 × 10⁻⁴ s is 0.00020 s, not 0.020 s.
A spray gun accelerates a paint droplet from rest towards a car panel through a potential difference of 8.0 kV. The droplet has a charge of 2.0 × 10−13 C and a mass of 5.0 × 10−11 kg.
Calculate the energy transferred to the droplet.
Assume all this energy becomes kinetic energy of the droplet. Calculate the speed of the droplet.
Suggest why the actual speed of the droplet is less than the speed in part (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| E = Q V = 2.0 × 10−13 × 8000 | ✓ 1 | |
| 1.6 × 10−9 J | ✓ 1 | |
| Part (b) | ||
| Ek = ½ mv2 | ✓ 1 | |
| v2 = 2 × 1.6 × 10−9 ÷ 5.0 × 10−11 = 64 (ECF from (a)) | ✓ 1 | |
| v = 8.0 m/s | ✓ 1 | |
| Part (c) | ||
| Some energy is transferred to the air (air resistance / collisions with air particles) so less becomes kinetic energy of the droplet | ✓ 1 | |
Common error: Convert kilovolts to volts before using E = QV (8.0 kV = 8000 V).
In a beehive security alarm, a light-dependent resistor (LDR) is connected in series with a fixed resistor across a battery. A siren switches on when the p.d. across the fixed resistor falls. A beam of light shining on the LDR is blocked. What happens to the p.d. across the LDR and to the p.d. across the fixed resistor?
| Marking point | Mark | Notes |
|---|---|---|
| Blocking the light raises the LDR resistance, so the current falls and the LDR takes a larger share of the battery p.d.; the two p.d.s always add to the battery p.d., so the fixed resistor p.d. falls. A reverses the effect of light on resistance. C breaks the rule that the p.d.s sum to the battery p.d. D confuses a fixed total with a fixed share. | ✓ 1 | Answer B |
Answer: B
Common error: In a series circuit the battery p.d. is shared in proportion to resistance, so when one resistance rises the other component's p.d. falls.
A beehive alarm uses a light beam shining on a light-dependent resistor (LDR). The LDR is connected in series with a 2.0 kΩ fixed resistor across a 12 V supply. When the beam shines on the LDR, its resistance is 1.0 kΩ. The siren sounds when the p.d. across the fixed resistor falls to 4.0 V or below.
Explain why the resistance of the LDR changes when an intruder blocks the beam.
Calculate the p.d. across the fixed resistor when the beam is shining on the LDR.
When the beam is blocked, the resistance of the LDR is 6.0 kΩ. Show that the siren sounds.
Calculate the resistance of the LDR when the p.d. across the fixed resistor is exactly 4.0 V.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 less light falls on the LDR so its resistance increases | ✓ 1 | |
| Part (b) | ||
| M1 I = 12 ÷ (1000 + 2000) = 4.0 mA (or p.d. share 2/3 of 12 V) | ✓ 1 | |
| A1 V = 0.0040 × 2000 = 8.0 V | ✓ 1 | |
| Part (c) | ||
| M1 I = 12 ÷ (6000 + 2000) = 1.5 mA | ✓ 1 | |
| A1 p.d. across fixed resistor = 0.0015 × 2000 = 3.0 V, which is less than 4.0 V | ✓ 1 | |
| Part (d) | ||
| M1 I = 4.0 ÷ 2000 = 2.0 mA and p.d. across LDR = 12 − 4.0 = 8.0 V | ✓ 1 | |
| M1 R = V ÷ I = 8.0 ÷ 0.0020 | ✓ 1 | |
| A1 4000 Ω (4.0 kΩ) | ✓ 1 | |
Exam tip: In a series circuit the current is the same everywhere. Find it from one component, then use it for the other.
A hospital heated blanket works from the 230 V mains supply. It contains two identical heating elements, each of resistance 52.9 Ω. On setting 1 the two elements are connected in series. On setting 2 only one element is connected.
Calculate the current in the blanket on setting 2.
Calculate the total resistance of the elements on setting 1.
Calculate the power of the blanket on setting 1.
Explain why the power on setting 1 is lower than on setting 2.
The blanket is switched to setting 2, which has a power of 1000 W. Calculate how many seconds are needed to transfer 1.8 MJ.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = V ÷ R = 230 ÷ 52.9 | ✓ 1 | |
| A1 4.3 A | ✓ 1 | |
| Part (b) | ||
| B1 52.9 + 52.9 = 105.8 Ω | ✓ 1 | |
| Part (c) | ||
| M1 P = V² ÷ R = 230² ÷ 105.8 (ECF from (b)) | ✓ 1 | |
| A1 500 W | ✓ 1 | |
| Part (d) | ||
| B1 the total resistance is greater on setting 1, so the current is smaller for the same p.d. | ✓ 1 | |
| B1 P = VI so the smaller current gives a smaller power | ✓ 1 | |
| Part (e) | ||
| M1 t = E ÷ P = 1 800 000 ÷ 1000 | ✓ 1 | |
| A1 1800 s | ✓ 1 | |
Exam tip: Write 1.8 MJ as 1 800 000 J before using E = Pt.
A telescope camera is cooled by a circuit in which a thermistor is in series with a 400 Ω fixed resistor and a 5.0 V cell. At the working temperature the ammeter in the circuit reads 8.0 mA.
Calculate the total resistance of the circuit.
Determine the resistance of the thermistor.
Calculate the p.d. across the thermistor.
The thermistor warms up and its resistance falls to 100 Ω. Calculate the new current in the circuit.
Show that the p.d. across the thermistor is now lower than before.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| R = V / I = 5.0 / 0.0080 | ✓ 1 | |
| 625 Ω | ✓ 1 | |
| Part (b) | ||
| 625 − 400 = 225 Ω (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| V = I R = 0.0080 × 225 | ✓ 1 | |
| 1.8 V | ✓ 1 | |
| Part (d) | ||
| total resistance = 500 Ω | ✓ 1 | |
| I = 5.0 / 500 = 0.010 A (10 mA) | ✓ 1 | |
| Part (e) | ||
| p.d. across fixed resistor = 0.010 × 400 = 4.0 V | ✓ 1 | |
| p.d. across thermistor = 5.0 − 4.0 = 1.0 V, which is less than 1.8 V | ✓ 1 | |
Exam tip: When one resistor in series gets smaller, the current rises and the other resistor takes a larger share of the supply p.d.
A fuel tanker unloads petrol into an underground storage tank at a service station through a long rubber hose. Petrol vapour is present near the connection.
Explain how the petrol flowing through the hose can become electrically charged.
Explain why this is dangerous.
Describe how the tanker can be made safe during unloading.
Charge builds up on the tanker at an average rate of 0.80 μA for 15 minutes. Calculate the total charge, in coulombs.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Friction between the petrol and the hose | ✓ 1 | |
| Electrons are transferred from one surface to the other, leaving the petrol and hose with opposite charges | ✓ 1 | |
| Part (b) | ||
| Charge builds up, creating a large p.d. between the tanker and earth (or nearby objects) | ✓ 1 | |
| A spark can jump and ignite the petrol vapour | ✓ 1 | |
| Part (c) | ||
| Connect the tanker to earth with a conducting cable (or earthing strap) | ✓ 1 | |
| Charge flows to earth so no p.d. builds up | ✓ 1 | |
| Part (d) | ||
| Time = 15 × 60 = 900 s | ✓ 1 | |
| Q = It = 0.80 × 10−6 × 900 = 7.2 × 10−4 C | ✓ 1 | |
Exam tip: Explain static danger in a chain: friction, charge builds up, large p.d., spark, ignition of vapour.
A fish-tank alarm uses a thermistor in series with a 2.0 kΩ fixed resistor across a 9.0 V supply. The thermistor resistance is 7.0 kΩ at 10 °C and 2.0 kΩ at 20 °C. The alarm sounds when the p.d. across the fixed resistor reaches 4.0 V.
Calculate the current in the circuit at 10 °C.
Calculate the p.d. across the fixed resistor at 10 °C.
Explain why the p.d. across the fixed resistor increases when the temperature rises.
Calculate the p.d. across the fixed resistor at 20 °C and state whether the alarm sounds.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Total resistance = 7.0 + 2.0 = 9.0 kΩ | ✓ 1 | |
| I = 9.0 ÷ 9000 = 1.0 × 10−3 A (1.0 mA) | ✓ 1 | |
| Part (b) | ||
| Use V = IR | ✓ 1 | |
| V = 1.0 × 10−3 × 2000 = 2.0 V | ✓ 1 | |
| Part (c) | ||
| Thermistor resistance decreases as temperature rises | ✓ 1 | |
| Total resistance falls so the current increases, giving a larger p.d. across the fixed resistor | ✓ 1 | |
| Part (d) | ||
| Total resistance = 2.0 + 2.0 = 4.0 kΩ, current = 9.0 ÷ 4000 = 2.25 × 10−3 A | ✓ 1 | |
| p.d. across fixed resistor = 2.25 × 10−3 × 2000 = 4.5 V | ✓ 1 | |
| 4.5 V is above 4.0 V so the alarm sounds | ✓ 1 | |
Exam tip: For two resistors in series across a fixed supply, a larger share of the supply p.d. goes across the larger resistance.
A light-beam alarm uses a 6.0 V supply with a 2.0 kΩ fixed resistor in series with a light-dependent resistor (LDR). The alarm circuit monitors the potential difference across the fixed resistor. When the beam falls on the LDR its resistance is 1.0 kΩ. When a person blocks the beam the resistance rises to 10 kΩ. What is the potential difference across the fixed resistor when the beam is blocked?
| Marking point | Mark | Notes |
|---|---|---|
| With the beam blocked, total resistance = 2.0 + 10 = 12 kΩ, so I = 6.0 ÷ 12 000 = 0.50 mA and V across the fixed resistor = 0.50 mA × 2.0 kΩ = 1.0 V, so A is correct. 1.2 V comes from using 10 kΩ alone as the total resistance. 4.0 V is the potential difference across the fixed resistor when the beam is NOT blocked. 5.0 V is the potential difference across the LDR when blocked, not across the fixed resistor. | ✓ 1 | Answer A |
Answer: A
Exam tip: In a series circuit the supply voltage is shared in proportion to the resistances. Check that your two potential differences add up to the supply voltage.
A food scientist keeps a bowl of dough warm using a heated blanket. The blanket contains two identical heating elements, each of resistance 575 Ω, and is connected to the 230 V mains supply. A switch lets the elements be connected in parallel (high setting) or in series (low setting).
Calculate the current in one element when it is connected directly across the 230 V supply.
On the high setting the elements are in parallel across the supply. Calculate the total current and the total power.
On the low setting the elements are in series across the supply. Calculate the power of the blanket.
The blanket is used for 6.0 hours. Calculate how much less energy, in kWh, is transferred on the low setting than on the high setting. Give your answer to 3 significant figures and use the unrounded values.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| I = V ÷ R | ✓ 1 | |
| I = 230 ÷ 575 = 0.40 A | ✓ 1 | |
| Part (b) | ||
| Total current = 2 × 0.40 = 0.80 A (ECF from (a)) | ✓ 1 | |
| P = I V = 0.80 × 230 = 184 W | ✓ 1 | |
| Part (c) | ||
| Total resistance = 575 + 575 = 1150 Ω | ✓ 1 | |
| I = 230 ÷ 1150 = 0.20 A | ✓ 1 | |
| P = I V = 0.20 × 230 = 46 W | ✓ 1 | |
| Part (d) | ||
| High: 0.184 kW × 6.0 h = 1.104 kWh | ✓ 1 | |
| Low: 0.046 kW × 6.0 h = 0.276 kWh | ✓ 1 | |
| Difference = 0.828 kWh (ECF from (b) and (c)) | ✓ 1 | |
Exam tip: In parallel each branch has the full supply voltage and the currents add. In series the resistances add and the same current flows.
In a test on a filament lamp, a current of 0.20 A flows when the potential difference is 3.0 V, and a current of 0.40 A flows when the potential difference is 9.0 V. What is the value of (resistance at 9.0 V) ÷ (resistance at 3.0 V)?
| Marking point | Mark | Notes |
|---|---|---|
| R = V/I gives 3.0/0.20 = 15 Ω and 9.0/0.40 = 22.5 Ω, so the ratio is 22.5/15 = 1.5 (B). A is the ratio the wrong way round. C is the ratio of currents. D is the ratio of potential differences, which ignores that the current also changed. | ✓ 1 | Answer B |
Answer: B
Common error: Resistance is V ÷ I at each point, not the gradient of a line through the origin to a different point.
A cordless drill battery supplies a constant current of 12 A for 25 minutes before it is empty. The battery operates at 18 V.
Calculate the charge supplied by the battery.
Calculate the energy transferred by the battery. Use the value of charge from (a).
A charger supplies a constant current of 2.5 A. Calculate the time, in seconds, to restore the charge from (a).
While charging, the charger takes a constant current of 0.20 A from the 230 V mains for 7200 s. Calculate the efficiency of the charging process, using 324000 J as the energy stored. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| time = 1500 s | ✓ 1 | |
| Q = I × t = 18000 C | ✓ 1 | |
| Part (b) | ||
| E = Q × V | ✓ 1 | |
| 324000 J (accept 3.24 × 105 J) | ✓ 1 | |
| ECF from (a) | — | |
| Part (c) | ||
| t = Q ÷ I = 18000 ÷ 2.5 | ✓ 1 | |
| 7200 s | ✓ 1 | |
| ECF from (a) | — | |
| Part (d) | ||
| input energy = V × I × t = 230 × 0.20 × 7200 | ✓ 1 | |
| 331200 J | ✓ 1 | |
| efficiency = 324000 ÷ 331200 × 100 = 97.8 % | ✓ 1 | |
Exam tip: Efficiency = useful energy out ÷ total energy in; identify both energies before dividing.
A filament lamp is labelled 12 V, 36 W and is connected to a 12 V supply.
Calculate the current in the lamp when it is at its normal working temperature.
Calculate the resistance of the lamp at its normal working temperature.
Explain why filament lamps are most likely to fail at the moment they are switched on.
When the lamp is cold, its resistance is 0.50 Ω. Calculate the current at the moment of switching on.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P = I × V | ✓ 1 | |
| I = 36 ÷ 12 = 3.0 A | ✓ 1 | |
| Part (b) | ||
| R = V ÷ I | ✓ 1 | |
| 4.0 Ω | ✓ 1 | |
| ECF from (a) | — | |
| Part (c) | ||
| the cold filament has a low resistance so the initial current is much larger than the working current | ✓ 1 | |
| the extra heating can melt a weak part of the filament, breaking the circuit | ✓ 1 | |
| Part (d) | ||
| I = V ÷ R = 12 ÷ 0.50 | ✓ 1 | |
| 24 A | ✓ 1 | |
Common error: A filament lamp's resistance depends on temperature, so use the resistance for the stated condition.
A silicon diode in a data centre's power-protection circuit has a p.d. of 0.70 V across it when it carries a forward current of 10 mA.
Calculate the resistance of the diode at this current.
The diode is connected in series with a resistor and a 3.0 V cell, so that a forward current of 10 mA flows with 0.70 V across the diode. Calculate the resistance of the resistor.
The cell is now reversed. State what happens to the current and explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V ÷ I = 0.70 ÷ 0.010 | ✓ 1 | |
| A1 70 Ω | ✓ 1 | |
| Part (b) | ||
| M1 p.d. across resistor = 3.0 − 0.70 = 2.3 V | ✓ 1 | |
| M1 R = 2.3 ÷ 0.010 | ✓ 1 | |
| A1 230 Ω | ✓ 1 | |
| Part (c) | ||
| B1 the current is (almost) zero | ✓ 1 | |
| B1 a diode has a very high resistance in the reverse direction | ✓ 1 | |
Exam tip: In a series circuit, subtract the diode p.d. from the supply p.d. to find the p.d. across the resistor.
A heater used to dry out a storage tent at an archaeological dig has two identical heating elements. Each element has resistance 46 Ω. The heater is connected to the 230 V mains supply.
Calculate the power of one element when it is used alone.
The two elements are switched on in parallel. Calculate the current drawn from the supply.
The two elements are now connected in series. Calculate the power of the heater.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 P = V2 ÷ R = 2302 ÷ 46 | ✓ 1 | |
| A1 1150 W | ✓ 1 | |
| Part (b) | ||
| M1 current in each element = 230 ÷ 46 = 5.0 A | ✓ 1 | |
| A1 total current = 10 A | ✓ 1 | |
| Part (c) | ||
| M1 R = 46 + 46 = 92 Ω, P = 2302 ÷ 92 | ✓ 1 | |
| A1 575 W | ✓ 1 | |
Common error: Parallel elements each get the full supply p.d.; the series heater has twice the resistance and so a quarter of the parallel power.
A barn lamp switches on automatically when it gets dark. The control circuit has a light-dependent resistor (LDR) in series with a fixed resistor of resistance 2000 Ω, connected to a 12 V d.c. supply. A voltmeter is connected across the LDR. In the dark the LDR has a resistance of 6000 Ω.
Calculate the current in the circuit in the dark.
Calculate the reading on the voltmeter in the dark.
Explain why the voltmeter reading falls when light shines on the LDR.
In bright sunlight the voltmeter reads 2.0 V. Calculate the resistance of the LDR.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| total resistance = 2000 + 6000 = 8000 Ω | ✓ 1 | |
| I = 12 ÷ 8000 = 1.5 × 10−3 A | ✓ 1 | |
| Part (b) | ||
| V = I R = 1.5 × 10−3 × 6000 | ✓ 1 | |
| = 9.0 V (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| the resistance of the LDR decreases in the light | ✓ 1 | |
| the current in the circuit increases | ✓ 1 | |
| the p.d. across the fixed resistor increases, so the p.d. across the LDR (12 V minus this) decreases | ✓ 1 | |
| Part (d) | ||
| p.d. across the fixed resistor = 12 − 2.0 = 10 V | ✓ 1 | |
| I = 10 ÷ 2000 = 5.0 × 10−3 A | ✓ 1 | |
| R = 2.0 ÷ 0.0050 = 400 Ω | ✓ 1 | |
Exam tip: In a series circuit the p.d.s add up to the supply p.d. Find the current through the fixed resistor first, because it is the same current in the LDR.
A security alarm for a warehouse uses a light beam that shines on a light-dependent resistor (LDR). The LDR is in series with a fixed resistor of 6.0 kΩ across a 12 V supply. The alarm sounds when the potential difference across the LDR is greater than 8.0 V. The resistance of the LDR is 2.0 kΩ when the beam shines on it and 18 kΩ when the beam is blocked.
State how the resistance of the LDR changes when the beam is blocked.
Calculate the potential difference across the LDR when the beam is not blocked.
Calculate the potential difference across the LDR when the beam is blocked.
State whether the alarm sounds when the beam is blocked. Give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| It increases | ✓ 1 | |
| Part (b) | ||
| Total resistance = 6000 + 2000 = 8000 Ω | ✓ 1 | |
| I = 12 ÷ 8000 = 1.5 × 10−3 A | ✓ 1 | |
| V = 1.5 × 10−3 × 2000 = 3.0 V | ✓ 1 | |
| Part (c) | ||
| Total resistance = 6000 + 18 000 = 24 000 Ω | ✓ 1 | |
| I = 12 ÷ 24 000 = 5.0 × 10−4 A | ✓ 1 | |
| V = 5.0 × 10−4 × 18 000 = 9.0 V | ✓ 1 | |
| Part (d) | ||
| Yes | ✓ 1 | |
| 9.0 V is greater than 8.0 V (ECF from (c)) | ✓ 1 | |
Exam tip: In a series circuit the larger resistor takes the larger share of the supply pd; use the same current for both resistors.
A thermistor is used in a fire alarm in a barn. The thermistor is in series with a fixed resistor of 800 Ω and a 12 V cell of negligible internal resistance. The resistance of the thermistor at different temperatures is given below.
| Temperature / °C | 20 | 50 | 80 |
|---|---|---|---|
| Resistance / Ω | 3200 | 800 | 200 |
Calculate the current in the circuit at 50 °C. Give your answer in mA.
Calculate the pd across the thermistor at 50 °C.
Determine whether the alarm sounds when the temperature of the barn reaches 80 °C.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| total resistance = 800 + 800 = 1600 Ω | ✓ 1 | |
| I = V ÷ R = 12 ÷ 1600 | ✓ 1 | |
| = 7.5 mA | ✓ 1 | |
| Part (b) | ||
| V = I × R = 7.5×10−3 × 800 (ECF from (a)) | ✓ 1 | |
| = 6.0 V | ✓ 1 | |
| Part (c) | ||
| total resistance = 800 + 200 = 1000 Ω | ✓ 1 | |
| I = 12 ÷ 1000 = 12 mA | ✓ 1 | |
| pd across the fixed resistor = 0.012 × 800 = 9.6 V | ✓ 1 | |
| 9.6 V is greater than 9.0 V so the alarm sounds | ✓ 1 | |
Common error: The alarm is across the fixed resistor, not the thermistor, so find the pd across the correct component.
When an aircraft is refuelled, fuel flows through a rubber hose. Before the fuel starts to flow, an earth lead is connected between the aircraft and the ground.
Explain how the flow of fuel through the hose can cause charge to build up.
Explain why the earth lead is connected before refuelling.
During one discharge, a charge of 3.0 µC flows through the earth lead in 0.50 s. Calculate the average current.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| friction between the fuel and the hose transfers electrons | ✓ 1 | |
| so the fuel and hose gain opposite charges (charge builds up) | ✓ 1 | |
| Part (b) | ||
| charge builds up on the aircraft and its pd rises | ✓ 1 | |
| a spark could jump to a nearby object and ignite the fuel vapour | ✓ 1 | |
| the earth lead lets the charge flow safely to the ground | ✓ 1 | |
| Part (c) | ||
| I = Q ÷ t = 3.0×10−6 ÷ 0.50 | ✓ 1 | |
| = 6.0×10−6 A | ✓ 1 | |
Exam tip: Convert microcoulombs to coulombs (×10⁻⁶) before using I = Q ÷ t.
Each of two identical heating elements in a coffee roaster transfers 800 W when connected on its own to a 230 V supply. The two elements are now connected in series with each other across the same 230 V supply. What is the total power transferred?
| Marking point | Mark | Notes |
|---|---|---|
| Each element has resistance R = V2 / P = 2302 / 800. In series the total resistance is 2R, so the power is V2 / 2R = 2302 / (2 × 66.1) = 400 W, half of the single-element power. 1600 W is the total if the elements were in parallel. 800 W wrongly assumes the power does not change. 200 W comes from halving both the voltage and the current of one element and then also halving the power again. | ✓ 1 | Answer D |
Answer: D
Exam tip: For a fixed supply p.d., power is V² / R, so doubling the resistance halves the power.
A space probe has a sensor circuit powered by a 12 V supply of negligible internal resistance. An 8.0 Ω resistor is connected in series with a parallel combination of a 12 Ω resistor and a 6.0 Ω resistor.
Calculate the combined resistance of the 12 Ω and 6.0 Ω resistors in parallel.
Calculate the current supplied by the 12 V supply.
Calculate the p.d. across the 6.0 Ω resistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 1 / R = 1 / 12 + 1 / 6.0 | ✓ 1 | |
| M1 1 / R = 3 / 12 = 0.25 | ✓ 1 | |
| A1 R = 4.0 Ω | ✓ 1 | |
| Part (b) | ||
| M1 total resistance = 8.0 + 4.0 = 12 Ω (ECF from (a)) | ✓ 1 | |
| A1 I = 12 / 12 = 1.0 A | ✓ 1 | |
| Part (c) | ||
| M1 p.d. = 1.0 × 4.0 (ECF from (a) and (b)) | ✓ 1 | |
| A1 4.0 V | ✓ 1 | |
Exam tip: Both resistors in a parallel pair share the same p.d., so find that p.d. first, using the combined resistance.
Three identical plastic rods X, Y and Z are suspended in turn from a stand by insulating threads. X is rubbed with a dry cloth and becomes negatively charged.
Explain, in terms of electrons, how rubbing makes X negatively charged.
Rod Y is held near X and X swings away from Y. State the type of charge on Y.
Rod Z is held near X and X swings towards Z. A student concludes that Z must be positively charged. Explain why this conclusion may be wrong.
The charge on X is −2.4 nC. The charge of one electron is −1.6 × 10−19 C. Calculate the number of electrons that X has gained.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 electrons are transferred from the cloth to the rod | ✓ 1 | |
| B1 so the rod has more electrons than protons, giving an overall negative charge | ✓ 1 | |
| Part (b) | ||
| B1 negative | ✓ 1 | |
| Part (c) | ||
| B1 Z could be uncharged | ✓ 1 | |
| B1 the negative charge on X repels electrons in the nearer end of Z, leaving that end positive, so there is a net attraction | ✓ 1 | |
| Part (d) | ||
| M1 number = total charge ÷ charge of one electron = 2.4 × 10−9 ÷ 1.6 × 10−19 | ✓ 1 | |
| A1 1.5 × 1010 electrons | ✓ 1 | |
Exam tip: Repulsion proves both objects carry the same type of charge; attraction does not prove opposite charges, because an uncharged object is also attracted.
In a light-beam security alarm, an LDR is in series with a fixed resistor across a battery. A sounder connected across the LDR is triggered when the potential difference across the LDR rises above a set value. What happens when an intruder breaks the beam?
| Marking point | Mark | Notes |
|---|---|---|
| When light is blocked the resistance of an LDR increases. In a series circuit the battery's potential difference is shared in proportion to resistance, so the component whose resistance has increased takes a larger share, and the pd across the LDR rises above the trigger value. A decrease in resistance describes the LDR in bright light, and the pd across the LDR cannot fall when its resistance rises relative to the fixed resistor. | ✓ 1 | Answer D |
Answer: D
Exam tip: In a series potential divider, the larger resistance always gets the larger share of the supply pd.
A light-beam alarm in a robot assembly hall has an LDR in series with a fixed resistor of resistance 2.0 kΩ. They are connected across a 6.0 V battery. A sounder across the LDR is triggered when the pd across the LDR exceeds 4.0 V. When the beam is unbroken the LDR has a resistance of 1.0 kΩ.
State how the resistance of the LDR changes when the beam is broken.
Calculate the current in the circuit when the beam is unbroken. Give your answer in mA.
Explain why the pd across the LDR increases when the beam is broken.
With the beam broken, the LDR has a resistance of 10 kΩ. Calculate the pd across the LDR and show that the alarm sounds.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| It increases | ✓ 1 | |
| Part (b) | ||
| M1 total resistance 3000 Ω and I = V ÷ R, 6.0 ÷ 3000 | ✓ 1 | |
| A1 2.0 mA | ✓ 1 | |
| Part (c) | ||
| The total resistance of the circuit increases so the current decreases | ✓ 1 | |
| The pd across the fixed resistor decreases, so the pd across the LDR (supply minus fixed resistor pd) increases | ✓ 1 | |
| Part (d) | ||
| M1 total resistance 12 kΩ and current 6.0 ÷ 12 000 = 5.0 × 10−4 A | ✓ 1 | |
| M1 V = I R, 5.0 × 10−4 × 10 000 | ✓ 1 | |
| A1 5.0 V, which is greater than 4.0 V so the alarm sounds | ✓ 1 | |
Exam tip: Check the units of resistance: 2.0 kΩ is 2000 Ω before you use V = I R.
An electric heater has two heating elements, each rated at 1.15 kW when connected to the 230 V mains supply. The switch settings allow one element, both elements in parallel, or both elements in series.
Calculate the current in one element when it is connected alone to the mains supply.
Calculate the resistance of one element.
Calculate the total power of the heater when the two elements are connected in series to the mains supply.
Explain why the parallel setting gives the greatest power.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 I = P ÷ V = 1150 ÷ 230 | ✓ 1 | |
| A1 5.0 A | ✓ 1 | |
| Part (b) | ||
| M1 R = V ÷ I = 230 ÷ 5.0 | ✓ 1 | |
| A1 46 Ω (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| A1 total resistance = 46 + 46 = 92 Ω (ECF from (b)) | ✓ 1 | |
| M1 I = V ÷ R = 230 ÷ 92 = 2.5 A | ✓ 1 | |
| A1 P = I × V = 2.5 × 230 = 575 W | ✓ 1 | |
| Part (d) | ||
| each element has the full 230 V across it, so each carries 5.0 A | ✓ 1 | |
| the total current is greater than in any other setting, so P = I × V is greatest | ✓ 1 | |
Common error: Elements in series share the supply pd, so each gets less than the full 230 V and the total power is lower.
A pharmacy fridge uses a thermistor and a fixed resistor of 2.0 kΩ connected in series to a 6.0 V supply. At 4 °C the resistance of the thermistor is 4.0 kΩ.
Calculate the current in the circuit at 4 °C.
Calculate the potential difference across the thermistor at 4 °C. Use the unrounded value from (a).
The fridge warms up and the resistance of the thermistor falls. Explain what happens to the potential difference across the thermistor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A1 total resistance = 2000 + 4000 = 6000 Ω | ✓ 1 | |
| M1 I = V ÷ R = 6.0 ÷ 6000 | ✓ 1 | |
| A1 1.0 × 10−3 A | ✓ 1 | |
| Part (b) | ||
| M1 V = I × R = 1.0 × 10−3 × 4000 | ✓ 1 | |
| A1 4.0 V (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| the thermistor has a smaller share of the total resistance | ✓ 1 | |
| so the potential difference across it decreases | ✓ 1 | |
Exam tip: In a series circuit the pd across each component is shared in proportion to resistance.
The pulse from an electric fence energiser transfers 3.6 J of energy at a potential difference of 6000 V. One pulse is produced every 1.25 s.
Calculate the charge that passes through the fence wire in one pulse.
Each pulse lasts 0.30 ms. Calculate the mean current during a pulse.
Calculate the mean current averaged over the whole 1.25 s between pulses.
Explain why the current in (c) is much smaller than the current in (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy = charge × potential difference, so charge = energy ÷ p.d. | ✓ 1 | |
| A1 3.6 ÷ 6000 = 6.0 × 10−4 C | ✓ 1 | |
| Part (b) | ||
| M1 current = charge ÷ time with time = 3.0 × 10−4 s | ✓ 1 | |
| A1 6.0 × 10−4 ÷ 3.0 × 10−4 = 2.0 A (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 mean current = charge in one pulse ÷ time between pulses | ✓ 1 | |
| A1 6.0 × 10−4 ÷ 1.25 = 4.8 × 10−4 A (ECF from (a)) | ✓ 1 | |
| Part (d) | ||
| The current flows for only a very short part of each 1.25 s cycle and is zero for the rest of the time | ✓ 1 | |
Common error: Convert milliseconds to seconds before dividing: 0.30 ms is 3.0 × 10⁻⁴ s.
A negatively charged rod is held close to, but not touching, an uncharged metal sphere on an insulating stand. While the rod is still in place, the sphere is touched briefly with a finger and the finger is removed. The rod is then taken away. What is the charge on the sphere?
| Marking point | Mark | Notes |
|---|---|---|
| The rod repels electrons in the sphere to the far side. Touching the sphere lets these electrons flow away to earth through the finger. When the finger and then the rod are removed, the sphere has lost electrons and is left with an overall positive charge. Negative is wrong because the rod never touched the sphere. Zero ignores the electrons that left through the finger. The split charge only exists while the rod is present and the sphere is not earthed. | ✓ 1 | Answer B |
Answer: B
Common error: Charge is only left on the sphere if the earth connection is removed before the charged rod is taken away.
A heater contains two identical heating elements. Each element is rated 1000 W when connected to the 230 V mains supply. A switch allows the elements to be used on their own, connected in parallel, or connected in series with each other.
Calculate the resistance of one element. Give your answer to 3 significant figures.
Calculate the total resistance when the two elements are connected in series. Use the exact value from part (a).
Explain which of the three settings (one element, parallel or series) gives the greatest power, and state the value of that power.
Calculate the power of the heater when the elements are connected in series across the 230 V supply.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 R = V2/P = 2302 / 1000 (or I = P/V then R = V/I) | ✓ 1 | |
| A1 52.9 Ω | ✓ 1 | |
| Part (b) | ||
| M1 add the two resistances: 2 × 52.9 | ✓ 1 | |
| A1 106 Ω (105.8 Ω) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| parallel | ✓ 1 | |
| each element still has the full 230 V across it, so the powers add to give 2000 W | ✓ 1 | |
| Part (d) | ||
| M1 I = V/R = 230 / 105.8 = 2.17 A, or use P = V2/R | ✓ 1 | |
| M1 P = V2/R = 2302 / 105.8 | ✓ 1 | |
| A1 500 W | ✓ 1 | |
Exam tip: Power in a fixed-voltage appliance is V²/R, so doubling the resistance halves the power.
A model lighthouse has a lamp in series with a 10 Ω resistor connected to a 12 V supply. When the lamp is lit the current in the circuit is 0.40 A.
Calculate the potential difference across the lamp.
Calculate the resistance of the lamp.
A second identical lamp is now connected in parallel with the first lamp. Assume the resistance of each lamp stays at 20 Ω. Calculate the current in the 10 Ω resistor.
Explain why each lamp is dimmer in the circuit in (c) than it was in the circuit with one lamp.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 p.d. across resistor = 0.40 × 10 = 4.0 V | ✓ 1 | |
| M1 12 − 4.0 | ✓ 1 | |
| A1 8.0 V | ✓ 1 | |
| Part (b) | ||
| M1 resistance = potential difference ÷ current | ✓ 1 | |
| A1 20 Ω , ECF from (a) | ✓ 1 | |
| Part (c) | ||
| M1 resistance of the two lamps in parallel = 10 Ω | ✓ 1 | |
| M1 total resistance = 10 + 10 = 20 Ω | ✓ 1 | |
| A1 current = 12 ÷ 20 = 0.60 A | ✓ 1 | |
| Part (d) | ||
| the current in the resistor increases, so the p.d. across the resistor increases and the p.d. across the lamps falls | ✓ 1 | |
| each lamp has a smaller p.d. and a smaller current (0.30 A compared with 0.40 A), so its power is lower | ✓ 1 | |
Common error: The current in each parallel branch is only part of the total current; do not use the total current for one lamp.
A lift cable in an architect's tower contains a copper wire that carries a current of 0.80 A for 2.0 s. The charge on one electron is 1.6 × 10−19 C.
Calculate the charge that flows in this time.
Calculate the number of electrons that pass a point in the wire in this time.
Explain why a lamp in the circuit lights almost as soon as the switch is closed, even though electrons move slowly through the wire.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Q = I × t | ✓ 1 | |
| 0.80 × 2.0 = 1.6 C | ✓ 1 | |
| Part (b) | ||
| number = charge ÷ charge on one electron = 1.6 ÷ 1.6 × 10−19 | ✓ 1 | |
| 1.0 × 1019 | ✓ 1 | |
| Part (c) | ||
| free electrons are already present throughout the wire and the lamp filament | ✓ 1 | |
| all of them start to move at once when the circuit is completed | ✓ 1 | |
Exam tip: Divide the total charge by the charge on one electron to count the electrons.
A fuel tanker is unloading fuel through a plastic hose into an underground storage tank. The tanker is metal and stands on rubber tyres.
Explain how the tanker can become charged as fuel flows through the hose.
Explain why a charged tanker is a danger.
Describe how the danger can be removed.
Suggest why the metal tanker keeps its charge when it is not earthed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| friction between the fuel and the hose transfers electrons | ✓ 1 | |
| so the tanker gains a net (positive or negative) charge | ✓ 1 | |
| Part (b) | ||
| charge builds up so that the p.d. between the tanker and earth becomes large | ✓ 1 | |
| a spark can jump through the air and ignite the fuel vapour | ✓ 1 | |
| Part (c) | ||
| connect the tanker to earth with a metal (conducting) wire | ✓ 1 | |
| charge flows safely to earth so none builds up | ✓ 1 | |
| Part (d) | ||
| rubber tyres are insulators so charge cannot flow to the ground | ✓ 1 | |
Exam tip: Link each step: friction, charge builds up, large p.d., spark, ignition.
The lights on a bridge are connected to a 230 V supply by a long cable. The cable has a total resistance of 0.80 Ω and carries a current of 25 A.
Calculate the p.d. across the cable.
Calculate the power wasted in the cable.
State the p.d. across the lights.
Calculate the power transferred to the lights.
Calculate the percentage of the supplied power that is transferred to the lights. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 V = I R | ✓ 1 | |
| A1 20 V | ✓ 1 | |
| Part (b) | ||
| M1 P = I2R = 252 × 0.80 | ✓ 1 | |
| A1 500 W | ✓ 1 | |
| Part (c) | ||
| A1 210 V (ECF from (a)) | ✓ 1 | |
| Part (d) | ||
| M1 P = I V = 25 × 210 | ✓ 1 | |
| A1 5250 W (ECF from (c)) | ✓ 1 | |
| Part (e) | ||
| M1 efficiency = 5250 ÷ (230 × 25) × 100 | ✓ 1 | |
| A1 91 % | ✓ 1 | |
Exam tip: Total power from the supply is supply p.d. × current; the cable takes part and the lights take the rest.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy