Mains electricity and electrical power | Pearson Edexcel GCSE Physics Notes

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The key idea

Electrical appliances transfer energy when charge moves through a potential difference.

Equation to know

power = current x potential difference

Mains Electricity And Power

Use the labels to explain the scientific relationship shown.

Revision notes

The bit that matters

Short notes first. Learn the idea, then use the worked example and questions to check it properly.

Mains supply and a.c.

) supply with a potential difference of about 230 V and a frequency of 50 Hz.) supplied by cells and batteries, which flows one way only. c. is produced by generators in power stations.

The three-core cable

A three-core mains cable has live (brown), neutral (blue) and earth (green and yellow stripes) wires. d.from the supply and is dangerous even when a device is switched off.The neutral wire completes the circuit at close to 0 V.The earth wire is a safety wire at 0 V that only carries current if there is a fault.

Electrical power

Electrical power = potential difference x current, written P = V x I.Power can also be found from power = current² x resistance, written P = I² x R.Power is the rate of energy transfer in watts. A higher power appliance transfers more energy each second.

Energy transferred and the National Grid

Energy transferred = power x time (E = P x t), and energy transferred = charge x potential difference (E = Q x V). d.d. to safe levels for use.

Key terms

Definitions to learn

Alternating current

Current that repeatedly reverses direction, as from the mains.

Direct current

Current that flows in one direction only, as from a cell.

Live wire

The wire carrying the alternating supply p.d.; always dangerous.

Earth wire

A safety wire at 0 V that carries current only during a fault.

National Grid

The network of cables and transformers that distributes electricity.

Worked example

A kettle operates at 230 V with a current of 9.0 A. Calculate its power.

Use P = IV.

Substitute the values.

Final answer

2070 W

Exam habit

For kWh questions, keep power in kW and time in hours — do not convert to SI units.For J/W/s questions, always convert first. State which form you are using before calculating.

Watch out

Do not use kilowatt-hours as a unit of power.

Examiner tips

How to score full marks

1Quote UK mains as 230 V and 50 Hz; do not confuse frequency with voltage.
2Choose P = V x I or P = I² x R depending on which quantities the question gives.
3Transformers step up p.d. to reduce current, which cuts heating losses in transmission cables.

Practice questions

Try these yourself

Start with the core skill, then open the answer only after you have attempted the full question.

1A 1.8 kW appliance operates for 45 minutes. Calculate the energy transferred in kWh.

Mark scheme

1.Convert 45 minutes to 0.75 hours.
2.Multiply power in kW by time in hours.

1.35 kWh

2Explain the purpose of the earth wire in a metal-cased appliance.

Mark scheme

1.Describe what happens if the live wire touches the case.

It provides a low-resistance path to ground, causing a large current so the fuse melts or circuit breaker trips.

3A device transfers 720 kJ in 5 minutes. Calculate its power.

Mark scheme

1.Convert to joules and seconds.
2.Use P = E / t.

2400 W

4State the potential difference and frequency of the UK mains supply.[2 marks]

Mark scheme

1.Recall the standard mains values.

About 230 V (1) and 50 Hz (1)

5An appliance has a p.d. of 230 V across it and draws a current of 5.0 A. Calculate its power.[2 marks]

Mark scheme

1.Use P = V x I.
2.Substitute V = 230 and I = 5.0.

P = V x I (1) = 230 × 5.0 = 1150 W (1)

6State the colour and purpose of the earth wire.[2 marks]

Mark scheme

1.Recall the colour code.
2.Recall the safety role.

Green and yellow striped (1); it is a safety wire that carries current away to earth only if a fault occurs, preventing the case from becoming live (1)

7A heater of resistance 20 ohms carries a current of 3.0 A. Calculate the power dissipated.[2 marks]

Mark scheme

1.Use P = I² x R.
2.Square the current first.
3.Multiply by resistance.

P = I² x R (1) = 3.0² × 20 = 9 × 20 = 180 W (1)

8Explain why the National Grid transmits electricity at very high potential difference, referring to current and energy losses.[4 marks]

Mark scheme

1.Link transmitting at high p.d. to power.
2.Link to current.
3.Link low current to reduced heating loss.
4.Mention transformers.

For a given power, raising the p.d. with a step-up transformer reduces the current (since P = V x I) (1); the energy wasted by heating in the cables depends on current squared (P = I² x R) (1); so a lower current greatly reduces the energy lost as heat in the transmission lines (1); step-down transformers then lower the p.d. to make it safe for homes (1)

9An electric shower is rated at 9200 W and runs at 230 V. Calculate the current it draws and the charge that flows in a 5-minute shower.[4 marks]

Mark scheme

1.Rearrange P = V x I to I = P / V.
2.Substitute P = 9200 and V = 230.
3.Use Q = I x t, converting 5 minutes to 300 s.

I = P / V = 9200 / 230 (1) = 40 A (1); Q = I x t = 40 × 300 (1) = 12000 C (1)

10Explain why the live wire is dangerous even when the appliance is switched off, while the neutral wire is not.[4 marks]

Mark scheme

1.The live wire is always at 230 V a.c. relative to earth.
2.The switch only breaks the live wire in a correctly wired circuit.
3.The neutral wire is at approximately 0 V.
4.Touching live could create a path to earth through the body.

The live wire is always at 230 V a.c. with respect to earth, even when the device is switched off (1); the switch interrupts the live wire so the appliance itself has no current, but the wire to the switch remains live (1); if a person touches the live wire, current can flow from it through their body to earth, causing an electric shock (1); the neutral wire is at approximately 0 V so there is no potential difference to drive a current through the body (1)

11A transmission cable has a resistance of 5 ohms and carries a current of 200 A. Calculate the power wasted as heat and the energy lost in one hour.[4 marks]

Mark scheme

1.Use P = I² x R.
2.E = P x t, converting 1 hour to 3600 s.

P = I² x R = 200² × 5 = 40000 × 5 (1) = 200 000 W = 200 kW (1); E = P x t = 200000 × 3600 (1) =

7.2 \times 10^{8}

J = 720 MJ (1)

12A 2.5 kW kettle is used for 4 minutes each morning for a week (7 days). Calculate the total energy transferred in kilowatt-hours (kWh) and the cost if electricity is 28p per kWh.[3 marks]

Mark scheme

1.Convert 4 minutes to hours: $\frac{4}{6}0$ .
2.Energy per day = power (kW) x time (hours).
3.Total energy = energy per day x 7.
4.Cost = energy in kWh x price per kWh.

Time per day = 4 / 60 = 0.0667 hours (1); energy per day = 2.5 × 0.0667 = 0.167 kWh; total energy = 0.167 × 7 = 1.17 kWh (accept 1.2 kWh) (1); cost = 1.17 × 28p = 32.7p (accept 32 to 34p) (1)

13Describe and explain, in detail, the function of each of the three wires in a mains plug, and explain what safety feature each one contributes to in a metal-cased appliance. Include in your answer what happens when a fault causes the live wire to touch the metal case.[4 marks]

Mark scheme

1.Brown live wire: carries the a.c. supply; always at 230 V.
2.Blue neutral wire: returns current to the supply at ~0 V.
3.Green-yellow earth wire: safety wire at 0 V connected to metal case.
4.Fault scenario: live touches case; current flows through earth wire; fuse or breaker trips.
5.Outcome: prevents electric shock.

The brown live wire carries the alternating potential difference (about 230 V) from the supply to the appliance, providing the energy for it to operate (1); the blue neutral wire completes the circuit at close to 0 V, allowing current to return to the supply — without it no current could flow (1); the green-and-yellow earth wire is a safety wire connected to the metal case of the appliance and to earth; under normal conditions it carries no current (1); if a fault causes the live wire to touch the metal case, the case rises to 230 V; current flows through the low-resistance earth wire to earth, causing a large current that blows the fuse or trips the circuit breaker, cutting off the supply before anyone receives a shock (1); this makes the appliance safe to touch after the fault has been detected (1) — award max 4

14A power station generates electricity at 25 000 V. A step-up transformer raises this to 400 000 V for transmission through cables with total resistance 20 ohms. The power station generates 800 MW. Calculate (a) the current in the transmission cables, (b) the power wasted in the cables, and (c) the efficiency of the transmission. Comment on why transmitting at high voltage is important.[4 marks]

Mark scheme

1.Current in cables I = P / V = $800 \times 10^{6}$ / 400 000.
2.Power wasted = I² x R.
3.Efficiency = (P output) / P input = (P input - P wasted) / P input.
4.Compare power wasted to total power to show why high V matters.

(a) I = P / V =

800 \times 10^{6}

/ 400 000 = 2000 A (1); (b) P wasted = I² x R = 2000² × 20 = 4 000 000 × 20 = 80 000 000 W = 80 MW (1); (c) efficiency = (800 - 80) / 800 = 720 / 800 = 0.90 = 90% (1); if the same 800 MW were transmitted at 25 000 V the current would be 32 000 A and the cable power loss would be 32000² × 20 = 20 480 MW — far exceeding the generated power — so transmitting at high voltage to keep current small is essential for efficient distribution (1)

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