GCSE · AQA Combined Science · Paper 1 · P2 Electricity

Electricity, for the exam.

The whole of P2 — charge and current, resistance and potential difference, series and parallel circuits, I–V characteristics, mains and power. Built for both tiers.

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Both tiers in one booklet. Everything here is for Foundation and Higher. Anything that's Higher tier only sits in a purple HT box — Foundation students can skip those. Green boxes are required practicals. Do one topic at a time; each is about 10–15 minutes.

Topic 01 · 4.2.1 · Charge & current

Circuit symbols, charge & current

By the end of this topic you'll read a circuit diagram, say exactly what current is, and link current, charge and time with one equation you have to recall.

Part 1What a circuit is

An electric current is a flow of electrical charge. For current to flow you need two things: a complete (closed) circuit with no gaps, and a source of potential difference (such as a cell or battery) to push the charge round.

Engineers draw circuits using standard circuit symbols so anyone can read them. You need to recognise and draw the common ones.

Circuit symbols to know

Cell / battery
The source of potential difference. A battery is two or more cells.
Switch
Opens (breaks) or closes (completes) the circuit.
Resistor / variable resistor
Limits current; the variable one can be adjusted.
Lamp
Transfers energy as light when current flows.
Ammeter / voltmeter
Ammeter measures current (in series); voltmeter measures p.d. (in parallel).
Diode / LED · thermistor · LDR
Diode lets current flow one way; thermistor and LDR change resistance with temperature/light.
SIX SYMBOLS YOU MUST KNOW cell switch resistor lamp A ammeter V voltmeter
Standard AQA circuit symbols — learn to recognise and draw each one

⚠ Watch out — conventional current vs electron flow

Conventional current flows from + to − around the circuit (this is the direction you draw and use in the exam). The electrons themselves actually drift the other way (− to +), but unless a question asks specifically, always use conventional current: positive to negative.

Part 2Linking charge, current and time

Current is the rate of flow of charge — how much charge passes a point each second. Charge is measured in coulombs (C), current in amps (A), and 1 amp = 1 coulomb per second.

Equation

Q = I t recall
charge flow (C) = current (A) × time (s)

Worked example — charge through a lamp

A current of 0.5 A flows through a lamp for 120 s. Calculate the charge that passes through it.

EquationQ = I t
Sub in= 0.5 × 120
Answer= 60 C

⚠ Watch out — time must be in seconds

Time goes in as seconds, not minutes. A 2-minute flow is 120 s. If a question gives minutes or hours, convert first or your answer will be out by a factor of 60 (or 3600).

Quick check

A charge of 90 C flows through a wire in 30 s. What is the current?

  • A2700 A
  • B3 A
  • C0.33 A
  • D60 A
Show answer
B — 3 A. Rearrange Q = I t to I = Q ÷ t = 90 ÷ 30 = 3 A. Answer A is the slip of multiplying instead of dividing; C divides the wrong way round.
Topic 1 — quick quiz
Click to reveal · 4 questions
  1. What is an electric current?
    A flow of electrical charge. It needs a complete circuit and a source of potential difference to drive it.
  2. State the equation linking charge, current and time, with units.
    Q = I t. Charge in coulombs (C), current in amps (A), time in seconds (s).
  3. A current of 4 A flows for 5 minutes. Calculate the charge that flows.
    Time = 5 × 60 = 300 s. Q = I t = 4 × 300 = 1200 C.
  4. In which direction does conventional current flow?
    From the positive (+) terminal to the negative (−) terminal around the circuit.
Topic 02 · 4.2.1.3 · Resistance & p.d.

Resistance & potential difference

The most-used equation in the whole topic — V = I R — plus the required practical that shows resistance grows with length.

Part 1Potential difference, current and resistance

Potential difference (p.d.), measured in volts (V), is the energy transferred per unit charge as it moves between two points — it's the "push" driving the current. Resistance, measured in ohms (Ω), is how hard it is for current to flow. The bigger the resistance, the smaller the current for a given p.d.

Equation

V = I R recall
potential difference (V) = current (A) × resistance (Ω)

Worked example — finding resistance

A p.d. of 6 V drives a current of 0.4 A through a resistor. Calculate its resistance.

EquationV = I R → R = V ÷ I
Sub in= 6 ÷ 0.4
Answer= 15 Ω

⚠ Watch out — rearrange before you panic

V = I R can find any of the three. To get current, I = V ÷ R; to get resistance, R = V ÷ I. The classic mistake is dividing the wrong way: resistance is the p.d. on top, divided by the current. Keep the units in: volts ÷ amps gives ohms.

Quick check

A 12 V supply drives 3 A through a heater. What is the heater's resistance?

  • A36 Ω
  • B0.25 Ω
  • C4 Ω
  • D15 Ω
Show answer
C — 4 Ω. R = V ÷ I = 12 ÷ 3 = 4 Ω. Answer A multiplies instead of dividing; B divides the wrong way round.

Part 2How resistance depends on length

If you make a wire longer, its resistance goes up. In fact resistance is directly proportional to length: double the length, double the resistance. The required practical lets you test this and get a straight-line graph through the origin.

Resistance of a wire vs its length

Aim: investigate how the resistance of a wire depends on its length.

  1. Set up a circuit with a battery, an ammeter in series with the test wire, and a voltmeter across the test wire.
  2. Tape the wire to a metre ruler and connect with a flying lead so you can choose the length.
  3. Set the length to 10 cm. Close the switch briefly, read the current (I) and p.d. (V), then open it again.
  4. Calculate resistance with R = V ÷ I for that length.
  5. Repeat for lengths up to 100 cm (e.g. every 10 cm). Plot resistance (y) against length (x).

Control / improve: only switch on for short bursts so the wire doesn't heat up (heating raises the resistance and spoils the pattern). A straight line through the origin shows resistance ∝ length.

RESISTANCE INCREASES WITH LENGTH length of wire (cm) resistance (Ω) straight line
A straight line through the origin means resistance is directly proportional to length
Quick check

A 20 cm length of a wire has a resistance of 4 Ω. What is the resistance of 60 cm of the same wire?

  • A4 Ω
  • B1.3 Ω
  • C12 Ω
  • D8 Ω
Show answer
C — 12 Ω. Resistance ∝ length. The wire is 3× longer (60 ÷ 20 = 3), so the resistance is 3 × 4 = 12 Ω.
Topic 2 — quick quiz
Click to reveal · 4 questions
  1. State the equation linking p.d., current and resistance.
    V = I R: potential difference (V) = current (A) × resistance (Ω).
  2. A 9 V battery drives 1.5 A through a bulb. Calculate the bulb's resistance.
    R = V ÷ I = 9 ÷ 1.5 = 6 Ω.
  3. In the wire required practical, why must the wire only be switched on briefly?
    A continuous current heats the wire, which raises its resistance and changes the readings — so switch on only long enough to take each reading.
  4. How does the resistance of a wire change as its length increases?
    It increases in direct proportion — double the length, double the resistance (a straight line through the origin).
Topic 03 · 4.2.2 · Series & parallel

Series & parallel circuits

The rules for current, p.d. and resistance in each type — and why adding a bulb in parallel doesn't dim the others.

Part 1Series circuits

In a series circuit there's just one loop — one path for the current. Components are connected end to end, so if one breaks, they all stop.

The current is the same everywhere in a series circuit. The total p.d. is shared between the components (it adds up to the supply p.d.). And the resistances add together: Rtotal = R1 + R2 + …, so adding a resistor always increases total resistance.

SERIES — ONE LOOP cell R₁ R₂ same current I everywhere · p.d. shared · R = R₁ + R₂
Series: one path — current the same, p.d. shared, resistances add

Worked example — total resistance in series

Two resistors, 4 Ω and 6 Ω, are connected in series to a 20 V supply. Find the total resistance and the current.

Total RR = R₁ + R₂ = 4 + 6 = 10 Ω
CurrentI = V ÷ R = 20 ÷ 10
AnswerR = 10 Ω, I = 2 A

Part 2Parallel circuits

In a parallel circuit there's more than one branch, so the current splits. Each component gets its own loop back to the cell, so one breaking doesn't stop the others — that's why home wiring and Christmas lights use parallel.

The p.d. is the same across each branch (equal to the supply). The total current is shared between the branches and the branch currents add up to the total. Adding a resistor in parallel actually lowers the total resistance, because you've given the current another path.

PARALLEL — TWO BRANCHES cell R₁ R₂ same p.d. across each branch · branch currents add to the total
Parallel: separate branches — p.d. the same, currents add, total resistance falls

⚠ Watch out — parallel resistance goes DOWN

The surprising one: adding a resistor in parallel reduces the total resistance below that of the smallest resistor, because you've opened up another route for the current. Don't add parallel resistances like you would in series — that only works for series.

Quick check

Three identical lamps are connected in parallel to a 6 V battery. What is the p.d. across each lamp?

  • A2 V
  • B18 V
  • C6 V
  • D0 V
Show answer
C — 6 V. In parallel the p.d. across each branch is the same as the supply, so every lamp gets the full 6 V. (In series the 6 V would be shared, giving 2 V each — answer A is that trap.)
Topic 3 — quick quiz
Click to reveal · 5 questions
  1. In a series circuit, what is true about the current?
    It is the same everywhere in the circuit (one path, so the same current flows through every component).
  2. How do you find the total resistance of resistors in series?
    Add them up: Rtotal = R1 + R2 + … The total is always bigger than any single resistor.
  3. What is the p.d. across each branch of a parallel circuit?
    The same for every branch, and equal to the supply p.d.
  4. Two 8 Ω resistors are connected in series to a 16 V supply. Find the current.
    R = 8 + 8 = 16 Ω. I = V ÷ R = 16 ÷ 16 = 1 A.
  5. Why does adding a resistor in parallel decrease the total resistance?
    It gives the current an extra path, so more current can flow overall for the same p.d. — which means a lower total resistance.
Topic 04 · 4.2.1.4 · I–V characteristics

I–V characteristics

Reading the shape of a current–p.d. graph for a resistor, lamp and diode — plus thermistors, LDRs, and the required practical that produces them.

Part 1Three classic graphs

An I–V graph (current against potential difference) shows how a component behaves. The shape tells you whether its resistance stays constant.

An ohmic conductor (a fixed resistor at constant temperature) gives a straight line through the origin — current is directly proportional to p.d., so the resistance is constant.

A filament lamp gives an S-shaped curve. As current increases the filament gets hotter, so its resistance increases and the line bends over.

A diode only lets current flow one way. It has a very high resistance in the reverse direction, so the graph is flat (no current) until the p.d. is forward and large enough.

THREE I–V CHARACTERISTICS resistor straight line filament lamp S-shaped curve diode one direction only
Each axis is current (y) against p.d. (x) — the shape reveals how resistance behaves

⚠ Watch out — the lamp's resistance rises

The filament lamp curve bends because the wire heats up, raising its resistance. A common slip is to read the flattening as the resistance falling — it's the opposite. A steeper-then-shallower curve means resistance is increasing with current.

Part 2Thermistors and LDRs

Two components change resistance with their surroundings, which makes them useful as sensors:

A thermistor's resistance falls as temperature rises. Used in thermostats and fire alarms — when it gets hot, more current flows.

An LDR (light-dependent resistor)'s resistance falls as light gets brighter. Used in automatic lights and camera exposure — in the dark its resistance is high.

Sensing components

Thermistor
Resistance decreases as temperature increases. Used in temperature sensors and thermostats.
LDR
Resistance decreases as light intensity increases. Used in light sensors and automatic street lights.

I–V characteristics of a resistor, a filament lamp and a diode

Aim: investigate how the current through a component varies with the p.d. across it.

  1. Connect the component in series with an ammeter and a variable resistor, with a voltmeter across the component.
  2. Adjust the variable resistor to change the p.d. Record several pairs of current and p.d. readings.
  3. Reverse the connections to the component (or swap the cell) to get negative p.d. values too.
  4. Plot current (y) against p.d. (x) for each component.
  5. Repeat for the fixed resistor, the filament lamp and the diode.

Control / improve: only switch on briefly so components (especially the lamp) don't overheat between readings. Take repeats and average to reduce random error.

Quick check

A component gives a straight line through the origin on an I–V graph at constant temperature. What is it?

  • AA filament lamp
  • BA diode
  • CAn ohmic conductor (fixed resistor)
  • DA thermistor being heated
Show answer
C — an ohmic conductor. A straight line through the origin means current ∝ p.d., so the resistance is constant. The lamp curves (heating), the diode only conducts one way.
Topic 4 — quick quiz
Click to reveal · 5 questions
  1. Describe the I–V graph of an ohmic conductor.
    A straight line through the origin — current is directly proportional to p.d., so the resistance stays constant.
  2. Why does the filament lamp graph curve?
    As current increases the filament heats up, so its resistance increases and the line bends over (an S-shape).
  3. What does a diode do, and what does its I–V graph look like?
    It only lets current flow one way. The graph shows almost no current in reverse, then a sharp rise once the forward p.d. is large enough.
  4. What happens to a thermistor's resistance as it gets hotter?
    It decreases — so more current flows. (An LDR's resistance decreases as it gets brighter.)
  5. In the I–V required practical, what does the variable resistor do?
    It lets you change the p.d. across (and current through) the component, so you can take a range of readings.
Topic 05 · 4.2.4 · Mains electricity

Mains electricity

a.c. vs d.c., the UK mains numbers you must quote, and the job of each of the three wires in a plug — plus why the earth wire saves lives.

Part 1a.c., d.c. and the UK mains

Direct current (d.c.) flows in one direction only — it's what cells and batteries supply. Alternating current (a.c.) constantly changes direction, back and forth. Mains electricity is a.c., produced by generators.

You must remember the UK mains figures: a frequency of 50 Hz (it swaps direction 50 times a second) and a potential difference of about 230 V.

d.c. vs a.c. direct current (d.c.) steady — one direction cells & batteries alternating current (a.c.) swaps direction 50×/s UK mains: 230 V · 50 Hz
Mains is alternating current at 50 Hz and about 230 V

⚠ Watch out — learn the mains numbers

These two values are straight recall marks: 230 V and 50 Hz. Don't mix them up (50 V / 230 Hz is wrong). And remember a.c. = mains, d.c. = battery — not the reverse.

Part 2The three-pin plug

A mains cable has three wires, each with a job and a colour you must learn:

The live wire (brown) carries the alternating p.d. from the supply — it's at about 230 V and is the dangerous one. The neutral wire (blue) completes the circuit and stays at close to 0 V. The earth wire (green & yellow) is a safety wire — it carries no current normally and only does anything if there's a fault.

If a fault connects the live wire to a metal case, the earth wire carries the current safely away to the ground, and the large current blows the fuse, breaking the circuit so you don't get a shock.

The three wires

Live — brown
Carries the 230 V alternating p.d. from the supply. The dangerous wire.
Neutral — blue
Completes the circuit; near 0 V.
Earth — green & yellow
Safety wire to the case; carries current only in a fault, stopping the case becoming live.
THE THREE-PIN PLUG earth (green/yellow) live (brown) neutral (blue) fuse
Live (brown), neutral (blue) and earth (green & yellow); a fuse sits in the live wire
Quick check

Why is the earth wire connected to the metal case of an appliance?

  • ATo carry the working current during normal use
  • BTo stop the case becoming live in a fault, carrying current safely away
  • CTo increase the p.d. across the appliance
  • DTo turn the alternating current into direct current
Show answer
B. The earth wire carries no current in normal use. If the live wire touches the case, the earth wire carries the current safely to the ground and the large current blows the fuse, so the case can't give you a shock.
Topic 5 — quick quiz
Click to reveal · 4 questions
  1. State the difference between a.c. and d.c.
    a.c. (alternating current) keeps changing direction; d.c. (direct current) flows one way only. Mains is a.c.; cells give d.c.
  2. Give the frequency and potential difference of the UK mains supply.
    50 Hz and about 230 V.
  3. Name the three wires in a mains cable and their colours.
    Live (brown), neutral (blue) and earth (green & yellow).
  4. Why can the live wire be dangerous even when an appliance is switched off?
    It is still at the supply p.d. (about 230 V) relative to earth, so touching it could give a shock as your body provides a path to the ground.
Topic 06 · 4.2.4 · Energy & power

Energy & power in circuits

Four equations that connect power, current, p.d., charge and energy — two to recall, two to use — and how to pick the right one.

Part 1Power in a circuit

Power is the rate of energy transfer — energy transferred per second, in watts (W). The more powerful an appliance, the faster it transfers energy from the mains. Two equations give the power of a device in a circuit.

Power equations

P = V I recall
power (W) = potential difference (V) × current (A)
P = I² R recall
power (W) = current² (A²) × resistance (Ω) — handy when you know the current and resistance

Worked example — power of a kettle

A kettle runs from the 230 V mains and draws a current of 10 A. Calculate its power.

EquationP = V I
Sub in= 230 × 10
Answer= 2300 W (2.3 kW)

⚠ Watch out — square the current in P = I²R

In P = I² R it's the current that's squared, not the resistance. Square I first, then multiply by R. Because of the square, doubling the current gives four times the power dissipated — which is why thick, low-resistance cables matter for high currents.

Quick check

A current of 3 A flows through a 5 Ω heating element. What power is dissipated?

  • A15 W
  • B45 W
  • C75 W
  • D225 W
Show answer
B — 45 W. P = I² R = 3² × 5 = 9 × 5 = 45 W. Answer A is the mistake of not squaring the current (3 × 5 = 15).

Part 2Energy transferred

To find the total energy an appliance transfers, you can use either the power and time, or the charge and p.d. Both give energy in joules (J).

Energy equations

E = P t given
energy transferred (J) = power (W) × time (s)
E = Q V given
energy transferred (J) = charge flow (C) × potential difference (V)

Worked example — energy from charge and p.d.

A charge of 300 C flows through a 12 V motor. How much energy is transferred?

EquationE = Q V
Sub in= 300 × 12
Answer= 3600 J

Worked example — energy from power and time

A 2000 W kettle is switched on for 90 s. Calculate the energy transferred.

EquationE = P t
Sub in= 2000 × 90
Answer= 180 000 J (180 kJ)

⚠ Watch out — pick the equation that fits the data

Use E = P t when you're told the power; use E = Q V when you're told the charge. If you're given p.d. and current but not power, find power first with P = V I, then E = P t. Time always goes in as seconds.

Quick check

A 60 W lamp is left on for 5 minutes. How much energy does it transfer?

  • A300 J
  • B18 000 J
  • C12 J
  • D3600 J
Show answer
B — 18 000 J. First convert: 5 min = 300 s. E = P t = 60 × 300 = 18 000 J. Answer A forgets to convert minutes to seconds (60 × 5).
Topic 6 — quick quiz
Click to reveal · 5 questions
  1. Write the two power equations for a circuit.
    P = V I (power = p.d. × current) and P = I² R (power = current² × resistance). Both give power in watts.
  2. A 230 V hairdryer draws 5 A. Calculate its power.
    P = V I = 230 × 5 = 1150 W (1.15 kW).
  3. A 1500 W heater runs for 2 minutes. Calculate the energy transferred.
    Time = 120 s. E = P t = 1500 × 120 = 180 000 J (180 kJ).
  4. 240 C of charge flows through a 9 V battery. How much energy is transferred?
    E = Q V = 240 × 9 = 2160 J.
  5. A 2 A current flows through a 6 Ω resistor. Calculate the power dissipated.
    P = I² R = 2² × 6 = 4 × 6 = 24 W.
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