OxfordAQA International GCSE Physics: Mains electricity, safety, power and the cost of electricity (9203) – Revision Notes

These revision notes condense sections 3.6.3 (Using electricity in the home) and 3.6.5 (Transferring electrical energy) of the OxfordAQA International GCSE Physics (9203) specification, for exams from May/June 2018. The qualification is untiered and neither section has a tier marking. For full explanations, read the study guide. Test yourself with the practice questions. Also see the course hub, the printable checklist and the free diagnostics.

Key definitions

Term	Meaning
Direct current (dc)	Current that always passes in the same direction, from cells and batteries
Alternating current (ac)	Current that repeatedly changes direction
Peak potential difference	Greatest pd of an ac supply, measured from the zero line
Period, T	Time for one complete cycle of an ac supply
Frequency, f	Number of complete cycles per second, in hertz
Mains	An ac supply with a set frequency and voltage
Power	Rate of energy transfer, in watts (joules per second)
Kilowatt-hour	The energy transferred by a 1 kW appliance running for 1 hour

Equations

Quantity	Equation	Units
Frequency	f = 1 / T and T = 1 / f	Hz, s
Power from energy	P = E / t	W, J, s
Power from current	P = I × V	W, A, V
Energy from charge	E = Q × V	J, C, V
Energy from power	E = P × t	J, W, s, or kWh, kW, h
Cost	cost = kWh × price per kWh	pence or pounds

ac and dc from diagrams

• A dc trace is a flat line. Its height is the pd.
• An ac trace is a repeating wave. Peak pd is the height from the zero line to a peak.
• Read the vertical scale (V per division) and the time base (time per division).

Method for a trace

1. Count the divisions from the zero line to a peak, then multiply by V per division.
2. Count divisions for one full cycle, then multiply by time per division. This is T.
3. Change the time to seconds.
4. Calculate f = 1 / T.
5. To compare with a dc supply, subtract the two pd values and say which is larger.

Worked reminder. A trace has a peak of 2.5 divisions at 4 V per division, and 3 cycles across 12 divisions at 5 ms per division. Peak pd = 2.5 × 4 = 10 V. One cycle = 4 divisions = 20 ms = 0.020 s. f = 1 / 0.020 = 50 Hz.

Mains and safety

• Mains is ac with a set frequency and voltage. rms values are not required.
• If the metal case of an appliance becomes live through a fault, the earth wire carries the current harmlessly away.
• A fuse or circuit breaker in the live wire disconnects the circuit when the current is too great.
• A fuse wire overheats and melts. It must then be replaced.
• A circuit breaker trips. It works much faster than a fuse and can be reset.
• Appliances with metal cases are usually earthed. A fault sends a large current from the live wire to earth, which melts the fuse and disconnects the live wire.
• Double insulated appliances have no earth wire connection.

Sequence to quote for a fault in an earthed metal appliance

1. Live wire touches the case.
2. Large current flows through the earth wire.
3. Fuse melts (or circuit breaker trips) in the live wire.
4. Appliance is disconnected, so the case is no longer live.

Energy transfers

Device	Transfer
Kettle, toaster, heater	Electrical to thermal
Motor	Electrical to kinetic
Loudspeaker	Electrical to sound
Lamp	Electrical to light (some thermal wasted)
Battery charger	Electrical to chemical

Method: power, energy and charge

1. Decide which quantities you are given: energy and time, current and pd, or charge and pd.
2. Choose the matching equation: P = E / t, P = I × V or E = Q × V.
3. Rearrange before substituting, and write the substitution line.
4. Give the answer with its unit and a sensible number of significant figures.

Worked reminder. A hob ring takes 4.0 A from a 230 V supply. Power = 4.0 × 230 = 920 W. Over 300 s it transfers 920 × 300 = 276 000 J. The charge that passes in that time is 4.0 × 300 = 1200 C, and 1200 × 230 = 276 000 J gives the same energy.

Method: fuse choice

1. Write I = P / V.
2. Convert kilowatts to watts if needed.
3. Substitute the mains pd and calculate I.
4. Choose the smallest available fuse rating above I.

Worked reminder. A 690 W iron on 230 V draws 690 / 230 = 3.0 A. A 3 A fuse equals the working current and risks melting in normal use, so use the 5 A fuse from the choices 3 A, 5 A and 13 A.

Method: kWh and cost

1. Change power to kW (divide watts by 1000).
2. Change time to hours (minutes divided by 60).
3. Energy in kWh = kW × h.
4. Cost = kWh × price per kWh.
5. Change pence to pounds by dividing by 100 if asked for pounds.

Worked reminder. A 0.50 kW heater runs for 4.0 h at 31 p per kWh. Energy = 0.50 × 4.0 = 2.0 kWh. Cost = 2.0 × 31 = 62 p.

Meter readings. Units used = later reading minus earlier reading. A meter showing 10230 then 10265 means 35 kWh. At 30 p per kWh the cost is 35 × 30 = 1050 p = £10.50.

You will not need to convert between kWh and J. Stay in kWh, kW and h, or in J, W and s, for each calculation.

Must-know distinctions

• Period and frequency: period is a time per cycle, frequency is cycles per time. They are reciprocals.
• Peak pd and pd of a dc supply: compare like with like. The ac value is a peak, the dc value is steady.
• Fuse and circuit breaker: both go in the live wire. A fuse melts and is replaced. A circuit breaker trips and is reset.
• Earthing and double insulation: earthing gives fault current an escape route. Double insulation keeps live parts out of reach so no earth wire is needed.
• Watt and kilowatt-hour: watts measure power. Kilowatt-hours measure energy.
• E = Q × V and E = P × t: the first uses charge and pd. The second uses power and time. Both give energy.

Quick self-test

1. An ac supply has a period of 8.0 ms. Calculate its frequency.
2. A supply has a frequency of 40 Hz. Calculate its period in seconds.
3. A heater transfers 1500 W for 20 s. How much energy is transferred?
4. A 1200 W motor transfers 60 000 J. How long does that take?
5. A 12 V battery moves 8.0 C of charge round a circuit. Calculate the energy transferred.
6. What charge passes through a 230 V appliance that transfers 2300 J?
7. A 690 W appliance runs on 230 V. Calculate the current and state the fuse from 3 A, 5 A and 13 A.
8. A 250 W lamp runs for 3.0 h. Calculate the energy in kWh and the cost at 29 p per kWh.
9. State why a fuse is placed in the live wire.
10. State one difference between a fuse and a circuit breaker.

Answers

1. T = 8.0 ms = 0.0080 s. f = 1 / 0.0080 = 125 Hz.
2. T = 1 / 40 = 0.025 s.
3. E = P × t = 1500 × 20 = 30 000 J.
4. t = E / P = 60 000 / 1200 = 50 s.
5. E = Q × V = 8.0 × 12 = 96 J.
6. Q = E / V = 2300 / 230 = 10 C.
7. I = 690 / 230 = 3.0 A. Use the 5 A fuse.
8. P = 0.250 kW. E = 0.250 × 3.0 = 0.75 kWh. Cost = 0.75 × 29 = 21.75 p, so about 22 p.
9. So that when it melts, the appliance is disconnected from the live supply and the case cannot become live.
10. A circuit breaker works faster and can be reset. A fuse must be replaced after it melts.

Where marks are usually lost

• Reading peak pd from trough to peak instead of from the zero line.
• Leaving the period in milliseconds when using f = 1 / T.
• Writing “heat” where the question needs thermal energy, and “electricity” where it needs electrical energy.
• Putting the fuse in the neutral or earth wire, or giving a vague reason such as “it stops the shock”.
• Choosing a fuse that equals the working current or sits below it.
• Skipping the substitution line, so that one arithmetic slip hides the method.
• Dividing a power in watts by 1000 twice, or not at all, in a kWh calculation.
• Answering a cost question in pence when pounds were asked for, or the reverse.
• Subtracting meter readings the wrong way round.

Official syllabus

OxfordAQA International GCSE Physics (9203). For exams May/June 2018 onwards. Sections 3.6.3 and 3.6.5. Published by OxfordAQA.