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GCSE · Physics

P2b Mains Electricity, Power and Energy

AQA 8464 · Calculator allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Using this pack: mains electricity, power and energy

Revision Library

This pack covers UK mains electricity (ac and dc, the three-pin plug, wiring colours and safety devices), electrical power (P = V I and P = I^2 R), energy transfers (E = P t and E = Q V), and the National Grid. Every equation you need is given in the question itself - you are never expected to recall a physics equation from memory in this pack. Always show your working: convert units before you substitute (especially minutes or hours into seconds, and kW into W), and give a correct unit with every final answer. Parts labelled 'Higher tier only' and flagged htOnly in the mark scheme can be skipped if you are studying the Foundation tier.

Electricity supplies can be either direct current (dc) or alternating current (ac).

(a)Define what is meant by direct current (dc).(1)

(b)Define what is meant by alternating current (ac).(1)

(c)State one reason why alternating current, rather than direct current, is used for the UK mains supply and for transmission through the National Grid.(1)

(Total for Question 1 is 3 marks)

For each quantity below, state its SI unit.

(a)State the SI unit of electrical power.(1)

(b)State the SI unit of energy.(1)

(c)State the SI unit of electric charge.(1)

(Total for Question 2 is 3 marks)

The UK mains electricity supply has a standard frequency and potential difference.

(a)State the frequency of the UK mains supply.(1)

(b)State the potential difference of the UK mains supply.(1)

(Total for Question 3 is 2 marks)

The diagram shows a UK three-pin plug with its cover removed. Three wires are connected to the pins: one with brown insulation, one with blue insulation and one with green-and-yellow insulation.

Figure (to be drawn): A UK three-pin plug, cover removed, showing three internal wires: one brown, one blue and one green-and-yellow, each connected to a separate pin.

(a)Name the wire with brown insulation and state its function.(2)

(b)Name the wire with blue insulation and state its function.(2)

(c)Name the wire with green-and-yellow insulation and state its function.(2)

(Total for Question 4 is 6 marks)

Some circuits include an extra safety device called a residual current device (RCD), in addition to a fuse.

(Total for Question 5 is 2 marks)

A vacuum cleaner is rated at 230 V and has a normal operating power of 966 W.

(a)Calculate the normal operating current of the vacuum cleaner. Use the equation P = V I.(3)

(b)Fuses are available rated at 3 A, 5 A and 13 A. State which fuse should be fitted in the plug, and explain your choice.(2)

(Total for Question 6 is 5 marks)

An appliance is connected to the mains supply through a switch that is wired into the live wire only. The appliance is switched off, but it is still plugged into the socket.

(a)Explain why the appliance can still be dangerous, even though it has been switched off.(2)

(b)State the normal potential difference between the neutral wire and earth.(1)

(Total for Question 7 is 3 marks)

A kettle's heating element has a resistance of 23 ohm. When the kettle is switched on, a current of 10 A flows through the element.

(a)Calculate the power output of the heating element. Use the equation P = I² R.(3)

(b)The kettle operates for 4 minutes to boil the water. Calculate the energy transferred to the water in this time. Give your answer in joules. Use E = P t, with t in seconds.(3)

(Total for Question 8 is 6 marks)

A 2 kW electric heater is used for 3 hours. Electricity costs 28p per kWh.

(a)Calculate the energy transferred by the heater, in kilowatt-hours (kWh). Use E = P t.(2)

(b)Calculate the cost of using the heater for this time.(2)

(Total for Question 9 is 4 marks)

A phone charger delivers a current of 1.5 A at a potential difference of 5 V for 2 hours while charging a phone's battery.

(a)Calculate the charge transferred to the battery. Give your answer in coulombs. Use Q = I t, with t in seconds.(3)

(b)Calculate the energy transferred to the battery. Use E = Q V.(2)

(Total for Question 10 is 5 marks)

Both fuses and circuit breakers are used to protect circuits from excessively large currents, but they work in different ways.

(a)State one similarity between how a fuse and a circuit breaker protect a circuit.(1)

(b)State one difference between how a fuse and a circuit breaker work.(1)

(c)State one advantage a circuit breaker has over a fuse.(1)

(Total for Question 11 is 3 marks)

A student says: 'As long as an appliance has a fuse fitted, it is completely safe, so an earth wire is not really necessary.' Evaluate this statement.

(Total for Question 12 is 2 marks)

A toaster has a metal outer casing. A hairdryer has a plastic outer casing, no earth pin, and is described as 'double insulated'.

(a)Explain why the toaster needs an earth wire connected to its metal casing.(2)

(b)Explain why the hairdryer does not need an earth wire, even though a fault could still occur inside it.(2)

(Total for Question 13 is 4 marks)

Electricity generated at a power station is transmitted around the country by the National Grid at a very high potential difference, before being reduced for use in homes.

Figure (to be drawn): A simple schematic: power station, then a step-up transformer, then transmission cables carried on pylons, then a step-down transformer, then homes.

(a)State what is meant by the 'National Grid'.(1)

(b)Name the type of transformer used to increase the potential difference before transmission.(1)

(c)Name the type of transformer used to decrease the potential difference before it reaches homes.(1)

(d)State one reason why transmitting electrical power at a high potential difference is more efficient than at a low potential difference.(1)

(Total for Question 14 is 4 marks)

An LED lamp has an electrical power input of 50 W. It transfers 45 J of energy usefully as light every second it is used.

(a)Calculate the efficiency of the lamp. Give your answer as a percentage.(3)

(b)State what happens to the rest of the energy supplied to the lamp.(1)

(Total for Question 15 is 4 marks)

Required practical. A student investigates the specific heat capacity of water. She places 0.20 kg of water in an insulated cup and heats it using a low-voltage immersion heater connected to an ammeter and a voltmeter, as shown. The potential difference across the heater and the current through it both stay constant. She switches the heater on for 300 s (5 minutes) and uses a thermometer to record the water's temperature before and after heating. Her results: potential difference = 12.0 V; current = 5.0 A; time = 300 s; initial temperature = 20.0 degC; final temperature = 38.0 degC.

Figure (to be drawn): A low-voltage immersion heater dips into an insulated cup of water. An ammeter is connected in series with the heater and a voltmeter is connected across it. A thermometer and a stirrer sit in the water.

(a)In addition to the immersion heater, ammeter and voltmeter, state three other pieces of apparatus the student needs to carry out this investigation.(3)

(b)Calculate the energy transferred to the water by the immersion heater. Use E = V I t.(3)

(c)The mass of water heated was 0.20 kg and its temperature rose from 20.0 degC to 38.0 degC. Calculate the specific heat capacity of the water using the student's results. Use E = m c (delta)T and your answer to part (b).(3)

(d)The accepted value for the specific heat capacity of water is 4200 J/(kg degC). Suggest why the student's calculated value is higher than this accepted value.(1)

(e)Suggest one improvement to the method that would reduce this source of error.(1)

(f)Calculate the percentage difference between the student's calculated specific heat capacity and the accepted value of 4200 J/(kg degC). Give your answer to the nearest 1%.(2)

(g)Suggest why the student should stir the water gently while it is being heated.(1)

(Total for Question 16 is 14 marks)

A light bulb is rated at 230 V, 60 W.

(a)Show that the normal operating current of the light bulb is approximately 0.26 A. Use P = V I.(2)

(b)Calculate the resistance of the light bulb's filament when it is operating normally. Use V = I R.(3)

(Total for Question 17 is 5 marks)

A domestic ring circuit supplies several sockets around a house. Each plug contains a fuse of an appropriate rating; the consumer unit contains circuit breakers for each circuit; and the whole electrical installation is connected to an earth wire.

Evaluate how effectively fuses, circuit breakers and earthing work together to protect a user from electric shock and to protect the wiring from damage due to overheating, in the event of a fault.

(Total for Question 18 is 6 marks)

Higher tier only. A power station generates 100 MW (100,000,000 W) of electrical power for transmission through the National Grid. The transmission cables have a total resistance of 4 ohm. This power can be transmitted either at a current of 500 A, or, using transformers to increase the potential difference, at a current of only 50 A.

(a)Calculate the power dissipated as heat in the cables when the transmission current is 500 A. Use P = I² R.(2)

(b)Calculate the power dissipated as heat in the cables when the transmission current is 50 A.(2)

(c)Use your answers to parts (a) and (b) to explain why the National Grid uses transformers to keep the transmission current as low as possible.(2)

(Total for Question 19 is 6 marks)

A kettle is rated at 2.3 kW (2300 W) and is 92% efficient. A student wants to heat 0.80 kg of water from 18 degC to 100 degC. The specific heat capacity of water is 4200 J/(kg degC).

(a)Calculate the useful energy required to heat the water from 18 degC to 100 degC. Use E = m c (delta)T.(3)

(b)The kettle is 92% efficient. Calculate the total electrical energy that must be supplied to the kettle to provide this useful energy.(2)

(Total for Question 20 is 5 marks)

Mark scheme · P2b Mains Electricity, Power and Energy

Question 1

(a) B1 current that flows in one direction only, oe
(a) Answer: Current that flows in one direction only.
(b) B1 current that repeatedly reverses direction (changes direction periodically), oe
(b) Answer: Current that repeatedly reverses direction.
(c) B1 ac can easily be stepped up or stepped down in potential difference using transformers, and transformers only work with a changing (alternating) current, oe
(c) Answer: Transformers (used to step the potential difference up or down for efficient transmission) only work with alternating current.

Question 2

(a) B1 watt (W)
(a) Answer: Watt (W).
(b) B1 joule (J)
(b) Answer: Joule (J).
(c) B1 coulomb (C)
(c) Answer: Coulomb (C).

Question 3

(a) B1 50 Hz
(a) Answer: 50 Hz.
(b) B1 230 V
(b) Answer: 230 V.

Question 4

(a) B1 live wire
(a) B1 carries the alternating current (potential difference) from the supply into the appliance; normally at about 230 V relative to earth, oe
(a) Answer: Live wire; it carries the current (at about 230 V) from the supply into the appliance.
(b) B1 neutral wire
(b) B1 completes the circuit, carrying current back to the supply; normally at, or very close to, 0 V (earth potential), oe
(b) Answer: Neutral wire; it completes the circuit and is normally at about 0 V.
(c) B1 earth wire
(c) B1 a safety wire that connects the metal casing of an appliance to earth, providing a low-resistance path for current if a fault occurs, so a large current flows and blows the fuse (or trips the circuit breaker), disconnecting the appliance before the casing can become dangerous, oe
(c) Answer: Earth wire; a safety wire connecting the metal casing to earth so a fault current blows the fuse (or trips the breaker) and disconnects the appliance.

Question 5

B1 an RCD detects/compares the current in the live and neutral wires and rapidly disconnects the circuit if it detects a difference (i.e. current leaking to earth, such as through a person), oe
B1 it operates (trips) much faster than a fuse and can detect much smaller fault currents, giving better protection against electric shock, oe
Answer: State what an RCD does, and state one advantage it has over a fuse. An RCD compares the current in the live and neutral wires and disconnects the circuit almost instantly if they differ (showing current is leaking to earth); it reacts far faster, and to much smaller currents, than a fuse, so it protects people from electric shock more effectively.

Question 6

(a) M1 rearranges to I = P / V
(a) M1 substitutes correctly: I = 966 / 230
(a) A1 I = 4.2 A (cao)
(a) Answer: 4.2 A
(b) B1 5 A
(b) B1 the fuse rating must be greater than the normal operating current (4.2 A) so it does not blow in normal use; 3 A is too low (would blow immediately) and 5 A is the smallest available rating above 4.2 A, giving protection without allowing an excessive current to flow for long before blowing, oe; ft from (a)
(b) Answer: 5 A fuse; it is the smallest available rating above the normal operating current of 4.2 A, so it will not blow in normal use but will still blow quickly if a fault causes an excessive current.

Question 7

(a) B1 the switch only breaks the connection between the switch and the appliance; the length of live wire between the socket and the switch remains at mains potential (about 230 V), oe
(a) B1 so touching this live wire/terminal (or a fault reaching it) can still give a person a dangerous electric shock, even with the appliance switched off, oe
(a) Answer: The wire between the socket and the switch is still live (at about 230 V) even when the appliance is switched off, so it can still give a dangerous shock if touched or if a fault occurs.
(b) B1 0 V (or very close to 0 V), oe
(b) Answer: 0 V (approximately).

Question 8

(a) M1 squares the current: I² = 10² = 100
(a) M1 multiplies by R: 100 x 23
(a) A1 P = 2300 W (2.3 kW) (cao)
(a) Answer: 2300 W (2.3 kW)
(b) M1 converts time to seconds: t = 4 x 60 = 240 s
(b) M1 substitutes: E = 2300 x 240 (ft from (a))
(b) A1 E = 552 000 J (5.52 x 10⁵ J) (awrt, ft)
(b) Answer: 552,000 J (5.52 x 10⁵ J)

Question 9

(a) M1 substitutes: E = 2 x 3
(a) A1 E = 6 kWh (cao)
(a) Answer: 6 kWh
(b) M1 substitutes: cost = 6 x 28 (ft from (a))
(b) A1 168p = GBP 1.68 (cao, ft; accept either unit)
(b) Answer: 168p (GBP 1.68)

Question 10

(a) M1 converts time to seconds: t = 2 x 3600 = 7200 s
(a) M1 substitutes: Q = 1.5 x 7200
(a) A1 Q = 10 800 C (1.08 x 10⁴ C) (cao)
(a) Answer: 10,800 C
(b) M1 substitutes: E = 10800 x 5 (ft from (a))
(b) A1 E = 54 000 J (5.4 x 10⁴ J) (cao, ft)
(b) Answer: 54,000 J

Question 11

(a) B1 both automatically break (disconnect) the circuit if the current becomes too large, protecting the wiring/appliance/user, oe
(a) Answer: Both automatically disconnect (break) the circuit if the current becomes too large.
(b) B1 a fuse contains a thin wire that heats up and melts (due to the heating effect of the current) when the current exceeds its rating, breaking the circuit; a circuit breaker is an electromagnetic switch that trips (opens) when the current exceeds its rating, oe; accept the converse
(b) Answer: A fuse contains a wire that melts (due to heating) when the current is too large; a circuit breaker uses an electromagnetic switch that trips instead.
(c) B1 a circuit breaker can be reset and reused, whereas a blown fuse must be replaced, oe; accept: a circuit breaker typically trips (acts) faster
(c) Answer: A circuit breaker can simply be reset and reused, whereas a fuse must be replaced once it has blown.

Question 12

B1 the statement is an oversimplification: a fuse alone only protects the wiring/circuit from a large current after a fault has occurred; without an earth wire, a fault could make the metal casing live without necessarily drawing enough current to blow the fuse, leaving the casing dangerous to touch, oe
B1 the earth wire works together with the fuse: it provides a low-resistance path that ensures a large enough fault current flows to blow the fuse quickly, and keeps the casing at (or very near) 0 V in the meantime, so the earth wire is still needed for full protection, oe
Answer: Not fully correct: a fuse alone cannot guarantee a fault current is big enough to blow it, or that the casing stays safe in the meantime. The earth wire works together with the fuse, giving a low-resistance path so a fault current blows the fuse quickly and keeps the casing near 0 V, so it is still needed.

Question 13

(a) B1 if a fault occurs (e.g. a live wire works loose and touches the casing), the metal casing could become live/reach a dangerous potential difference, oe
(a) B1 the earth wire provides a low-resistance path to earth, so a large fault current flows, blowing the fuse (or tripping the breaker) and disconnecting the appliance, keeping the casing safe to touch, oe
(a) Answer: The metal casing could become live if a fault occurs; the earth wire gives a low-resistance path to earth so a large fault current blows the fuse (or trips the breaker), disconnecting the appliance before the casing can shock someone.
(b) B1 the outer casing is plastic, which is an electrical insulator, so it cannot become live/cannot conduct a fault current to anyone touching it, oe
(b) B1 'double insulated' means the appliance has (at least) two independent layers of insulation between the user and any live parts, so an earth connection is not required for it to be safe, oe
(b) Answer: The plastic casing is an insulator, so it cannot become live even if a live wire touches it inside; double insulation means there are two independent layers of insulation protecting the user, so no earth wire is needed.

Question 14

(a) B1 the system of cables and transformers that transfers/distributes electrical power from power stations to consumers (homes, businesses, etc.) across the country, oe
(a) Answer: The system of cables and transformers that transfers electrical power from power stations to consumers across the country.
(b) B1 step-up transformer
(b) Answer: Step-up transformer.
(c) B1 step-down transformer
(c) Answer: Step-down transformer.
(d) B1 for the same power transmitted, a higher potential difference means a lower current is needed, and a lower current means less energy is wasted (dissipated as heat) in the transmission cables, oe
(d) Answer: A higher potential difference means a lower current is needed to transmit the same power, so less energy is wasted as heat in the cables.

Question 15

(a) M1 efficiency = useful output energy transferred / total input energy transferred = 45/50
(a) M1 multiplies by 100
(a) A1 90% (cao)
(a) Answer: 90%
(b) B1 it is transferred (wasted) as heat/thermal energy to the surroundings, oe
(b) Answer: It is transferred (wasted) as heat to the surroundings.

Question 16

(a) B1 a thermometer (to measure the water's temperature/temperature change)
(a) B1 a (mass/top-pan) balance (to measure the mass of the water)
(a) B1 a stopwatch/timer (to measure the heating time); accept an insulated container/lid or a stirrer as alternative correct answers, up to three in total
(a) Answer: A thermometer, a (mass/top-pan) balance, and a stopwatch/timer.
(b) M1 substitutes: E = 12.0 x 5.0 x 300
(b) M1 evaluates the product correctly
(b) A1 E = 18 000 J (18 kJ) (cao)
(b) Answer: 18,000 J (18 kJ)
(c) M1 calculates the temperature rise: (delta)T = 38.0 - 20.0 = 18.0 degC
(c) M1 rearranges and substitutes: c = E / (m (delta)T) = 18000 / (0.20 x 18.0), ft from (b)
(c) A1 c = 5000 J/(kg degC) (cao, ft)
(c) Answer: 5000 J/(kg degC)
(d) B1 some of the electrical energy supplied was transferred to the surroundings/the cup/the heater and thermometer rather than to the water (heat loss), so the water's actual temperature rise was smaller than it would have been if all the energy had gone into the water; using the full electrical energy in the calculation therefore overestimates the specific heat capacity, oe
(d) Answer: Some of the electrical energy was lost to the surroundings (not all of it heated the water), so the measured temperature rise was smaller than expected for that energy input, making the calculated specific heat capacity too high.
(e) B1 insulate the cup further (e.g. use a lid and/or lagging/polystyrene) to reduce energy transferred to the surroundings, oe; accept other valid improvements, e.g. using a more powerful heater for a shorter time to reduce the proportion of energy lost
(e) Answer: Insulate the cup better (e.g. add a lid and lagging) to reduce the energy transferred to the surroundings.
(f) M1 substitutes: percentage difference = (5000 - 4200)/4200 x 100, ft from (c)
(f) A1 19% (awrt, ft)
(f) Answer: 19% (awrt)
(g) B1 stirring ensures the water is heated evenly/reaches a uniform temperature throughout, so the thermometer gives an accurate reading of the whole sample's temperature, oe
(g) Answer: Stirring spreads the heat evenly through the water so the temperature is uniform, giving an accurate thermometer reading.

Question 17

(a) M1 rearranges and substitutes: I = P/V = 60/230
(a) A1 I = 0.26 A (awrt, showing at least 2 s.f. working)
(a) Answer: I = 0.26 A (awrt)
(b) M1 rearranges to R = V/I
(b) M1 substitutes using an unrounded current, ft from (a): R = 230/0.261
(b) A1 R = awrt 880 ohm (accept 860-900 ohm due to rounding in (a); ft)
(b) Answer: Approximately 880 ohm

Question 18

Level 3 (5-6): A clear, logically structured evaluation explaining how the earth wire, fuse and circuit breaker work together: the earth wire provides a low-resistance path so a fault produces a large current; this current is used by the fuse (melting) or circuit breaker (tripping) to disconnect the faulty circuit quickly, before the casing can remain dangerously live or the cables can overheat and start a fire. Includes a reasoned overall judgement on their combined effectiveness (e.g. noting limits, such as needing a correctly rated fuse, or that an RCD gives faster/more sensitive protection). Correct scientific terminology used throughout.
Level 2 (3-4): Relevant description of how one or two of the features (fuse, circuit breaker, earthing) reduce the risk of shock or fire, with some explanation of the underlying mechanism, but links between the features (or an overall judgement) are missing or unclear. Reasonably logical structure, mostly correct terminology.
Level 1 (1-2): Basic, isolated statement(s) about one safety feature, with little or no explanation of how it works or why it matters for safety. Limited or no use of correct scientific terminology.
Level 0 (0): No relevant content.
Indicative content:
- A fault (e.g. a loose live wire) can make an appliance's metal casing live/at a dangerous potential difference.
- The earth wire connects the casing to earth, providing a low-resistance path to earth.
- This causes a large current to flow when a fault occurs.
- The fuse contains a thin wire that heats up (heating effect of a current) and melts once the current exceeds its rating, breaking the live-wire connection.
- A circuit breaker instead uses an electromagnetic switch that trips (opens) once the current exceeds its rating, also breaking the live-wire connection.
- Both the fuse and circuit breaker disconnect the faulty circuit/appliance, isolating it from the mains supply.
- This prevents the casing remaining at a dangerous potential difference, reducing the risk of electric shock to the user.
- It also prevents a sustained excessive current overheating the cables, reducing the risk of fire.
- A circuit breaker can be reset and reused, and typically trips faster, than a fuse (which must be replaced).
- Overall judgement: used together, earthing with a fuse or circuit breaker gives fast, reliable protection at both appliance and circuit level, though protection depends on fitting a fuse of the correct rating, and an RCD would add faster, more sensitive protection against smaller fault/leakage currents than a fuse alone can detect.

Question 19

(a) M1 substitutes: P = 500² x 4
(a) A1 P = 1,000,000 W (1 MW / 1.0 x 10⁶ W) (cao)
(a) Answer: 1,000,000 W (1 MW)
(b) M1 substitutes: P = 50² x 4
(b) A1 P = 10,000 W (1.0 x 10⁴ W) (cao)
(b) Answer: 10,000 W
(c) B1 the power wasted as heat in the cables is proportional to the square of the current (P = I² R), so reducing the current (by using a higher potential difference for the same transmitted power) greatly reduces the energy wasted, oe; ft from (a)/(b)
(c) B1 e.g. reducing the current by a factor of 10 (500 A to 50 A) reduces the power loss by a factor of 100 (1,000,000 W to 10,000 W), so keeping the current low makes transmission far more efficient, oe; ft
(c) Answer: Because power loss = I^2R, a smaller current gives a loss reduced by the square of that factor; here a 10x smaller current (50 A vs 500 A) gives a 100x smaller power loss (10,000 W vs 1,000,000 W), so transmitting at low current (high potential difference) is far more efficient.

Question 20

(a) M1 calculates the temperature rise: (delta)T = 100 - 18 = 82 degC
(a) M1 substitutes: E = 0.80 x 4200 x 82
(a) A1 E = 275,520 J (awrt 275,000-276,000 J / 2.76 x 10⁵ J) (cao)
(a) Answer: 275,520 J (awrt 2.76 x 10⁵ J)
(b) M1 substitutes: total energy = useful energy / efficiency = 275520 / 0.92, ft from (a)
(b) A1 awrt 299,000-300,000 J (2.99 x 10⁵ J) (cao, ft)
(b) Answer: Approximately 299,478 J (awrt 2.99 x 10⁵ J)

Mains Electricity, Power and Energy - Worksheets, Questions and Revision