AQA GCSE Physics 8463: Electricity – Study Guide

This guide teaches section 4.2 Electricity of the AQA GCSE Physics (8463) specification, for teaching from September 2016 and exams from 2018 onwards. It covers every point from 4.2.1.1 to 4.2.5.2. Electricity is assessed on Paper 1 (topics 1 to 4), set at Foundation and Higher Tier, and Paper 2 may draw on its energy-transfer ideas. No content in section 4.2 is marked (HT only), so everything here applies to both tiers. Required practical activities 3 and 4 belong to this topic.

For quick recall, use the Electricity revision notes. To test yourself, use the Electricity practice questions. The course hub is AQA GCSE Physics, and the printable checklist lists every specification point. To find your weak spots quickly, try a free diagnostic.

What this unit covers

Spec point	What you must be able to do	Tier
4.2.1.1 Circuit symbols	Draw and interpret circuit diagrams	Both
4.2.1.2 Charge and current	Recall and apply Q = It; current is the same everywhere in a single loop	Both
4.2.1.3 Current, resistance, pd	Recall and apply V = IR; required practical 3	Both
4.2.1.4 Resistors	I–V graphs for resistor, lamp and diode; thermistors and LDRs; required practical 4	Both
4.2.2 Series and parallel	Rules for current, pd and resistance; R_total = R₁ + R₂; series circuit calculations	Both
4.2.3 Domestic uses and safety	ac and dc; 50 Hz, about 230 V; three-core cable and safety	Both
4.2.4 Energy transfers	Recall and apply P = VI, P = I²R, E = Pt, E = QV; the National Grid	Both
4.2.5 Static electricity	Charging by rubbing, forces between charges, sparking, electric fields	Both

4.2.1 Current, potential difference and resistance

Circuit symbols

Circuit diagrams use the standard symbols shown in the specification (cells, switches, resistors, lamps, diodes, sensors and meters). You must draw and interpret circuit diagrams with them.

Charge and current

For charge to flow round a closed circuit, the circuit must include a source of potential difference, such as a cell or battery. Electric current is the rate of flow of electrical charge.

charge flow = current × time          Q = I t
Q in coulombs (C), I in amperes (A), t in seconds (s)

A current has the same value at any point in a single closed loop.

Worked example 1. A current of 0.40 A flows for 2.5 minutes. Calculate the charge that flows.

t = 2.5 × 60 = 150 s
Q = I t = 0.40 × 150
Q = 60 C

Resistance and potential difference

The greater the resistance of a component, the smaller the current for a given potential difference (pd) across it. “Voltage” also gains credit.

potential difference = current × resistance        V = I R
V in volts (V), I in amperes (A), R in ohms (Ω)

Worked example 2. A resistor has a pd of 12 V across it and a current of 0.25 A through it. Calculate its resistance.

R = V ÷ I = 12 ÷ 0.25
R = 48 Ω

To measure the resistance of a component, put an ammeter in series with it and a voltmeter in parallel across it. Read I and V, then calculate R = V ÷ I.

Required practical 3 investigates the factors that affect resistance: the length of a wire at constant temperature, and combinations of resistors in series and parallel. For the wire, move a crocodile clip to measured lengths, read V and I, and calculate R. Switch off between readings so the wire does not heat up and change its resistance.

Resistors and I–V graphs

Component	I–V graph	What happens to resistance
Resistor at constant temperature (ohmic conductor)	Straight line through the origin	Constant; I is directly proportional to V
Filament lamp	Curve that gets less steep as V increases (in both directions)	Increases as the filament gets hotter
Diode	Current in one direction only, above a small pd	Very high resistance in the reverse direction

Required practical 4 measures the I–V characteristics of a resistor at constant temperature, a filament lamp and a diode. Vary the pd, reverse the connections for negative values, and plot I against V. A straight line through the origin shows a linear element; a curve shows a non-linear one.

Two components change resistance with conditions:

• Thermistor: resistance decreases as temperature increases. Used in thermostats.
• LDR (light-dependent resistor): resistance decreases as light intensity increases. Used to switch lights on when it gets dark.

4.2.2 Series and parallel circuits

	Series	Parallel
Current	Same through each component	Total current = sum of the branch currents
Potential difference	Supply pd is shared between components	Same pd across each branch
Resistance	R_total = R₁ + R₂	Total is less than the smallest individual resistor

Why adding resistors changes the total. In series, charge passes through every resistor in turn, so the total resistance increases. In parallel, each new resistor adds another path for the charge, so for the same pd more current flows in total and the total resistance decreases. You do not need to calculate the total resistance of two resistors in parallel.

Worked example 3 (series). A 6.0 V battery is connected to a 4.0 Ω resistor and an 8.0 Ω resistor in series. Find the current and the pd across each resistor.

R_total = 4.0 + 8.0 = 12 Ω
I = V ÷ R = 6.0 ÷ 12 = 0.50 A        (same through both)
V across 4.0 Ω = 0.50 × 4.0 = 2.0 V
V across 8.0 Ω = 0.50 × 8.0 = 4.0 V
Check: 2.0 + 4.0 = 6.0 V ✓

Worked example 4 (parallel). 6.0 Ω and 4.0 Ω resistors are in parallel across 12 V. Find the total current.

I₁ = 12 ÷ 6.0 = 2.0 A
I₂ = 12 ÷ 4.0 = 3.0 A
Total current = 2.0 + 3.0 = 5.0 A

Series circuits for measurement and testing

A sensor in series with a fixed resistor shares the supply pd. When the sensor’s resistance changes, its share of the pd changes, which a voltmeter can detect.

Worked example 5. An LDR is in series with a 1.0 kΩ resistor across a 6.0 V supply. In bright light the LDR’s resistance is 500 Ω; in the dark it is 5.0 kΩ. Find the pd across the LDR in each case.

Bright: R_total = 1000 + 500 = 1500 Ω
        I = 6.0 ÷ 1500 = 0.0040 A
        V_LDR = 0.0040 × 500 = 2.0 V
Dark:   R_total = 1000 + 5000 = 6000 Ω
        I = 6.0 ÷ 6000 = 0.0010 A
        V_LDR = 0.0010 × 5000 = 5.0 V

The pd across the LDR rises in the dark, so it can switch on a lamp. Treating the pair as one equivalent resistance (1500 Ω or 6000 Ω) gives the current first.

4.2.3 Domestic uses and safety

Direct pd (dc) acts in one direction only, so the current always flows the same way, as from a cell or battery. Alternating pd (ac) keeps reversing direction, so the current keeps changing direction. Mains electricity is an ac supply. In the UK the domestic supply has a frequency of 50 Hz and is about 230 V.

Most appliances connect to the mains with three-core cable:

Wire	Colour	Job	Potential
Live	Brown	Carries the alternating pd from the supply	About 230 V relative to earth
Neutral	Blue	Completes the circuit	At, or close to, 0 V
Earth	Green and yellow stripes	Safety wire that stops the appliance becoming live	0 V; carries current only if there is a fault

Live wire with the switch open. It is still at about 230 V. Touch it and your body (at 0 V) has a large pd across it, so a current flows through you: an electric shock.

Live connected to earth. This low-resistance path lets a very large current flow, which can cause a fire, or a shock if the path is through a person.

4.2.4 Energy transfers

Power

The power of a device depends on the pd across it and the current through it.

power = potential difference × current        P = V I
power = current² × resistance                 P = I² R
P in watts (W), V in volts (V), I in amperes (A), R in ohms (Ω)

Energy transferred by electrical work

Work is done when charge flows in a circuit. The energy an appliance transfers depends on its power and how long it is switched on.

energy transferred = power × time                        E = P t
energy transferred = charge flow × potential difference  E = Q V
E in joules (J), P in watts (W), t in seconds (s), Q in coulombs (C), V in volts (V)

Appliances transfer energy from batteries or the ac mains to the kinetic energy of electric motors (a fan) or to heating devices (a kettle). A higher power rating means more energy transferred each second.

Worked example 6. A 2.3 kW kettle runs on 230 V for 3.0 minutes. Find the current and energy transferred.

I = P ÷ V = 2300 ÷ 230 = 10 A
t = 3.0 × 60 = 180 s
E = P t = 2300 × 180 = 414 000 J  (414 kJ)

Worked example 7. A current of 0.20 A passes through a 150 Ω resistor. Calculate the power.

P = I² R = 0.20² × 150 = 0.040 × 150 = 6.0 W

The National Grid

The National Grid is a system of cables and transformers linking power stations to consumers. Step-up transformers increase the pd from the power station to the transmission cables. Step-down transformers decrease the pd to a much lower value for domestic use. (How transformers work is in section 4.7.3, which is Higher tier only.)

Why it is efficient. For a given power, P = VI, so a higher pd means a smaller current. The heating effect in the cables depends on current (P = I²R), so a smaller current wastes much less energy.

Illustration: 1.0 MW sent through cables of total resistance 5.0 Ω
At 25 kV:  I = 1 000 000 ÷ 25 000 = 40 A;   loss = 40² × 5.0 = 8000 W
At 400 kV: I = 1 000 000 ÷ 400 000 = 2.5 A; loss = 2.5² × 5.0 ≈ 31 W

4.2.5 Static electricity

Static charge

When certain insulating materials are rubbed together, negatively charged electrons are rubbed off one and on to the other. The material that gains electrons becomes negative; the one that loses electrons is left with an equal positive charge.

Two objects with the same type of charge repel; objects with different types attract. These are non-contact forces. Evidence: a charged rod hung from a thread swings away from a rod with the same charge and towards one with the opposite charge, without touching.

Sparking. As charge builds up on an insulated object, the pd between it and a nearby earthed conductor increases and the electric field in the gap gets stronger. When it is strong enough, electrons jump the gap: a spark.

Electric fields

A charged object creates an electric field around itself, strongest close to the object. A second charged object in the field feels a force, stronger as the distance decreases. This explains the non-contact force.

For an isolated charged sphere, draw straight radial lines starting at the surface, evenly spaced. For a positive sphere the arrows point outwards; for a negative sphere they point inwards. The lines are closer together near the sphere, where the field is strongest.

Common errors

• Using minutes in Q = It or E = Pt. Convert to seconds.
• Writing “current is used up” in a series circuit. It is the same at every point in a single loop.
• Saying a filament lamp’s resistance falls as current rises. It increases, because the filament gets hotter.
• Mixing up thermistor and LDR: temperature controls a thermistor, light controls an LDR.
• Saying the earth wire carries current normally.
• Saying positive charges are rubbed off. Only electrons transfer.

Next, condense this with the revision notes and try the practice questions.

Official syllabus

AQA GCSE Physics (8463) specification, for teaching from September 2016 onwards, for exams in 2018 onwards (Version 1.1, 30 September 2019), published by AQA. This guide covers section 4.2 Electricity.