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

This guide covers 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 carries a tier marking. Use it with the revision notes and the practice questions. The course hub, the printable checklist and the free diagnostics sit alongside it. For earlier energy ideas, see Energy.

What you need to be able to do

Section	What the specification asks for
3.6.3 a	Direct current (dc) from cells and batteries always flows the same way
3.6.3 b	Alternating current (ac) repeatedly changes direction. Find period and frequency from diagrams. Compare dc potential differences and ac peak potential differences from diagrams
3.6.3 c	Mains is an ac supply with a set frequency and voltage. Root mean square values are not required
3.6.3 d to f	Earthing, fuses and circuit breakers in the live wire, and double insulation
3.6.5 a	Give examples of electrical devices and identify their energy transfers
3.6.5 b	Power is the rate of energy transfer: P = E / t
3.6.5 c	P = I × V. Find the current from power and supply pd, then choose a fuse
3.6.5 d	E = Q × V
3.6.5 e, f	Energy in kilowatt-hours (kWh), E = P × t, cost from a price per kWh and from meter readings. No conversion between kWh and joules is required

Direct current and alternating current (3.6.3 a to c)

A cell or battery pushes charge round a circuit in one direction only. That is direct current (dc). On a graph of pd against time, a dc supply is a flat horizontal line at the value of its pd.

An alternating current (ac) repeatedly changes direction. The pd of an ac supply swings up to a maximum in one direction, falls through zero, reaches a maximum in the other direction, and returns. On an oscilloscope trace this shows as a wave that repeats.

Three readings matter on an ac trace.

• Peak potential difference: the greatest height of the wave above the zero line. Measure from the zero line to the top of a peak, not from trough to peak.
• Period, T: the time for one complete cycle, for example from one peak to the next peak.
• Frequency, f: the number of complete cycles each second, in hertz (Hz). The link is f = 1 / T, and so T = 1 / f.

To read a trace you need the scales. The vertical scale (volts per division) turns a height in squares into a pd. The horizontal scale (time per division) turns a width in squares into a time.

Worked example 1: reading a trace

An oscilloscope shows an ac supply. The vertical scale is 2 V per division and the time base is 2.5 ms per division. The peak is 3.0 divisions above the centre line. Two complete cycles fit across 8.0 divisions. A battery connected to the same screen gives a flat line 4.5 divisions above the centre line.

1. Peak pd of the ac supply = 3.0 × 2 = 6.0 V.
2. Width of one cycle = 8.0 / 2 = 4.0 divisions.
3. Period T = 4.0 × 2.5 ms = 10 ms = 0.010 s.
4. Frequency f = 1 / T = 1 / 0.010 = 100 Hz.
5. Battery pd = 4.5 × 2 = 9.0 V. The battery pd is 3.0 V greater than the ac peak pd (9.0 − 6.0).

Always change milliseconds to seconds before using f = 1 / T.

Mains electricity

Mains electricity is an ac supply with a set frequency and voltage. In the UK the mains is about 230 V at 50 Hz, so one cycle lasts 1 / 50 = 0.020 s. Other countries use other values, so a question will usually give the numbers you need. You do not need rms values: when a question gives a mains pd, use it directly in P = I × V.

Safety in the home (3.6.3 d to f)

A fault can make the metal case of an appliance live. Touching it could give someone a fatal shock. Three features reduce the danger.

Earthing. Appliances with metal cases are usually earthed. The earth wire joins the metal body to earth. If a fault makes the case live, the current is harmlessly conducted away through the earth wire instead of through the person.

Fuses. If a fault lets too great a current flow, a fuse in the live wire protects the circuit. The current heats the thin fuse wire until it melts. The break in the live wire disconnects the appliance.

Circuit breakers. A circuit breaker is also in the live wire. Too great a current makes it switch off, or trip. A circuit breaker works much faster than a fuse, and you reset it after the fault is fixed instead of replacing it.

How the features work together

Suppose a loose live wire touches the metal case of an earthed appliance.

1. The case is now live, but it is joined to earth by the earth wire.
2. A large current flows from the live wire, through the case and out through the earth wire.
3. That large current melts the fuse (or trips the circuit breaker) in the live wire.
4. The live wire is disconnected, so the case can no longer be live.

The fuse goes in the live wire so that, once it has melted, the appliance is cut off from the live supply. A fuse in the neutral wire would leave the appliance connected to the live supply.

Double insulation. Some appliances are double insulated. They have a casing and internal construction that keep the user away from live parts, so they have no earth wire connection. Many hair dryers and power tools with plastic bodies are examples. The metal parts a user can touch cannot become live.

Energy transfers in appliances (3.6.5 a)

Electrical appliances are designed to transfer energy. For each device, say what electrical energy is transferred into.

Appliance	Useful transfer
Kettle	Electrical to thermal energy
Electric motor in a fan	Electrical to kinetic energy
Loudspeaker	Electrical to sound energy
Lamp	Electrical to light energy (and some wasted thermal energy)
Phone charger	Electrical to chemical energy stored in the battery

Write the energy forms in full. “Heat” alone is a weak answer: use thermal energy.

Power and energy (3.6.5 b)

The rate at which an appliance transfers energy is its power. The equation is

power (W) = energy transferred (J) / time (s), or P = E / t.

Rearranged, E = P × t and t = E / P. One watt is one joule per second.

Worked example 2: a kettle

A kettle has a power of 2400 W. How much energy does it transfer in 450 s?

E = P × t = 2400 × 450 = 1 080 000 J (1.08 MJ).

How long would it take to transfer 540 000 J? t = E / P = 540 000 / 2400 = 225 s.

Power, current and fuse choice (3.6.5 c)

The power transfer in a device depends on the current through it and the pd across it:

power (W) = current (A) × potential difference (V), or P = I × V.

To find the current, rearrange to I = P / V. A fuse should be rated just above the normal working current. A fuse rated below the working current melts in normal use. A fuse rated far above it will not protect the wiring. Questions give you the fuse ratings to choose from.

Worked example 3: choosing fuses

The mains pd is 230 V. The available fuses are 3 A, 5 A and 13 A.

Appliance	Power	Current I = P / V	Fuse
Table lamp	60 W	60 / 230 = 0.26 A	3 A
Toaster	800 W	800 / 230 = 3.5 A	5 A
Hair dryer	1400 W	1400 / 230 = 6.1 A	13 A

The toaster current of 3.5 A is above 3 A, so the 3 A fuse would melt. The next rating up is 5 A. The hair dryer needs 6.1 A, which is above 5 A, so 13 A is the correct choice.

Energy, charge and potential difference (3.6.5 d)

The potential difference across a component is the energy transferred per unit charge passing through it:

energy transferred (J) = charge (C) × potential difference (V), or E = Q × V.

So 12 V means 12 J for every coulomb.

Worked example 4: charge and energy

A 12 V lamp has 4.5 C of charge pass through it. E = Q × V = 4.5 × 12 = 54 J.

A different appliance is on the 230 V mains and carries 0.80 A for 300 s. Charge flowing is Q = I × t = 0.80 × 300 = 240 C. Energy transferred is E = 240 × 230 = 55 200 J. As a check, P = I × V = 0.80 × 230 = 184 W, and 184 × 300 = 55 200 J. Both routes agree.

Kilowatt-hours and the cost of electricity (3.6.5 e and f)

The joule is a small unit, so domestic energy use is measured in kilowatt-hours (kWh). One kWh is the energy transferred by a 1 kW appliance in 1 hour. Use the same equation with different units:

energy (kWh) = power (kW) × time (h), or E = P × t.

Convert watts to kilowatts by dividing by 1000, and minutes to hours by dividing by 60. You will not be asked to convert between kWh and joules, so do the whole calculation in kWh.

Cost = number of kWh × price per kWh.

Worked example 5: an oven

An oven has a power of 2.5 kW and is on for 1.5 h. Electricity costs 28.5 p per kWh.

1. Energy = 2.5 × 1.5 = 3.75 kWh.
2. Cost = 3.75 × 28.5 = 106.875 p, which is 107 p to the nearest penny (about £1.07).

Worked example 6: a meter bill

A meter reads 04817 kWh at the start of a quarter and 05103 kWh at the end. Electricity costs 28.5 p per kWh.

1. Units used = 05103 − 04817 = 286 kWh.
2. Cost = 286 × 28.5 = 8151 p.
3. Divide by 100 to change pence to pounds: £81.51.

Common errors

• Measuring ac peak pd from trough to peak. Peak pd is from the zero line.
• Using milliseconds in f = 1 / T. Convert to seconds first.
• Placing the fuse in the neutral or earth wire in an explanation. It is in the live wire.
• Choosing a fuse equal to or below the working current. Pick the next rating above it.
• Using 1.5 h as 1 h 50 min. Use decimal hours: 1 h 30 min is 1.5 h, and 20 min is one third of an hour.
• Mixing watts with hours, or kilowatts with seconds, in E = P × t.
• Forgetting to convert pence to pounds at the end.

Check yourself

Try the practice questions, then the revision notes. Take a free 10-minute diagnostic to find weak spots.

Official syllabus

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