Power — AQA GCSE Physics

SPEC 8463 • 4.2.4.1 • PAPER 1 ⭐

Electrical Power

Two recall equations. One squared. Three examiner traps. Calculation marks every series — and one error costs three at a time.

What is Electrical Power?

Definition

Electrical power: the rate at which a circuit device transfers energy from the electrical store to other stores (e.g. thermal, kinetic, light).

The bigger the rate of energy transfer, the bigger the power. A 60 W bulb transfers 60 J of energy every second; a 2000 W kettle transfers 2000 J every second — that's why the kettle heats water far faster.

Spec wording

"Students should be able to explain how the power transfer in any circuit device is related to the potential difference across it, the current through it, and the energy changes over time."

Units

Power is measured in watts (W). 1 W = 1 J/s — one joule of energy transferred per second.

Equation 1 — P = VI

The power transferred by a circuit device equals the potential difference across it multiplied by the current through it.

P = V × I

SymbolQuantityUnit

PPowerwatts (W)

VPotential differencevolts (V)

ICurrentamperes (A)

Why this works

V = energy transferred per coulomb of charge. I = charge passing per second. Multiply them: (energy/charge) × (charge/second) = energy per second = power. ✓

Equation 2 — P = I²R

The power dissipated as heat in a resistor equals current squared multiplied by resistance.

P = I² × R

SymbolQuantityUnit

PPowerwatts (W)

ICurrentamperes (A)

RResistanceohms (Ω)

The squared matters

The current is squared. Doubling the current quadruples the power. Forgetting the square is the single most common error in this topic — examiners flag it every series.

Why Both Equations?

The two equations describe the same physical quantity — they're algebraically linked through Ohm's law (V = IR).

The link

Start with P = VI. Substitute V = IR (since V = IR for any resistor):

P = (IR) × I = I² × R

Same power, expressed in terms of different given quantities.

Why you need both

Questions give you different data. If you're given V and I, use P = VI. If you're given I and R, use P = I²R. The right choice avoids extra steps and reduces errors.

AQA now provides an Equations Sheet in the exam (introduced 2023, ongoing) — but the spec still requires you to recall and apply both. Writing the equation from memory secures the recall mark even on calculation questions.

Decision Rule — Which Equation?

Given V and I?

Use P = V × I

Direct substitution. No intermediate steps.

Given I and R?

Use P = I² × R

Remember to square the current.

Given V and R only?

Two strategies:

Find I first using V = IR, then use either P equation.

OR use the derived form P = V²/R — but this is not on the AQA recall list (covered on Slide 15 as a stretch shortcut).

Worked Example — ESUA

A current of 2.0 A flows through a 5.0 Ω resistor. Calculate the power dissipated.

E
Equation: Given I and R → use P = I² × R
S
Substitute: P = (2.0)² × 5.0 = 4.0 × 5.0
U
Units: watts (W)
A
Answer: P = 20 W

The squared check

If you forget the square, you get P = 2.0 × 5.0 = 10 W — wrong by half. Same data, completely different answer. The squared is the difference between 3/3 and 0/3 marks.

Examiner Alert — Square the Current

Jun 2023 Paper 1H · Q01.3

Power calculation using P = I²R. Most students did fine, but the marker flagged a specific failure mode for those who didn't.

"Many students failed to square the current so scored zero for the substitution and subsequent rearrangement." — AQA Examiner Report, Jun 2023 Paper 1H

Cost of the error

Forgetting the square doesn't lose 1 mark — it cascades. The substitution is wrong, so the rearrangement is wrong, so the final answer is wrong. Three marks gone from one missed exponent.

Defence

Write out the substitution explicitly with the bracket: P = (I)² × R. Calculate (I)² FIRST, write down the squared value, THEN multiply by R. Forces you to square at the right step.

Why the Squared Matters — Visually

Power dissipated as heat scales with the square of the current. Small changes in I cause big changes in P.

Double the current (1 → 2) → 4× the power. Quadruple the current (1 → 4) → 16× the power.

Real-world consequence

This is exactly why the National Grid transmits at very high voltage and low current — keeping I small means keeping I²R energy losses tiny. Topic-linking to 2.09.

The algebra link — high-stakes exam reasoning

For a fixed resistance, if V doubles, then by V = IR I also doubles. Since P = V × I, both factors double — so P quadruples. This is the same conclusion as P = I²R. Jun 2023 Q06.4 examined this reasoning; only 2% of students scored 2 marks. Memorise the chain: "double V → double I → 4× P".

Examiner Alert — Milliwatts

Jun 2024 Paper 1H · Q05.2

A power calculation involving values in milliwatts (mW). Conversion to watts before substitution was the make-or-break step.

"The conversion of milliwatts to watts proved a problem for the vast majority of students." — AQA Examiner Report, Jun 2024 Paper 1H

Conversion rules

UnitConversion to Wmilliwatts (mW)÷ 1000 (e.g. 500 mW = 0.5 W)kilowatts (kW)× 1000 (e.g. 2 kW = 2000 W)megawatts (MW)× 10⁶ (e.g. 5 MW = 5,000,000 W)

When to convert

Convert at the substitution step, before any arithmetic. Never convert at the end — you'll forget, or your final answer will be wrong by a factor of 1000.

Practice Q1

A torch lamp draws a current of 0.50 A when connected to a 6.0 V battery. Calculate the power transferred by the lamp.3 marks

E

Given V and I → use P = V × I

S

P = 6.0 × 0.50

U

Power in watts (W)

A

P = 3.0 W

Practice Q2

A 3.0 A current flows through a heating element of resistance 40 Ω. Calculate the power dissipated by the element.3 marks

E

Given I and R → use P = I² × R

S

P = (3.0)² × 40 = 9.0 × 40

U

Power in watts (W)

A

P = 360 W

!

Common error: forgetting to square. 3.0 × 40 = 120 W is wrong by a factor of 3.

Practice Q3

A laptop charger is rated at 45 W and operates at a potential difference of 20 V. Calculate the current drawn by the charger.3 marks

E

Equation: P = V × I (rearrange for I: I = P ÷ V)

S

I = 45 ÷ 20

U

Current in amperes (A)

A

I = 2.25 A

!

Mark-scheme structure: 1 mark for the equation, 1 mark for the substitution, 1 mark for the rearranged answer with unit. Show all three steps.

Weak vs Strong — Two Classic Errors

Same questions. Wrong answers from skipping the squared, and from skipping the unit conversion.

Q: I = 3.0 A through R = 40 Ω. Find P.

Weak (0/3)

P = I × R

P = 3.0 × 40 = 120 W ❌

Skipped the squared. The substitution is wrong → no marks for the rearrangement either.

Strong (3/3)

P = I² × R

P = (3.0)² × 40 = 9.0 × 40

P = 360 W ✓

Q: P = 500 mW, V = 5.0 V. Find I.

Weak (1/3)

P = V × I, so I = P / V

I = 500 / 5.0 = 100 A ❌

Used mW as if they were W. Answer wrong by a factor of 1000.

Strong (3/3)

Convert: 500 mW = 0.5 W

I = P / V = 0.5 / 5.0

I = 0.10 A ✓

Mark Scheme Language

The exact phrasing AQA mark schemes credit on calculation questions.

StepWhat earns the mark

EquationWrite out "P = V × I" or "P = I² × R" — symbols or words both accepted. 1 mark.

SubstitutionNumerical substitution with correct values, with units converted (e.g. mW → W). 1 mark.

AnswerCorrect numerical answer with correct unit. 1 mark.

RearrangementFor "find I" or "find R" questions — show the algebra explicitly. Don't skip steps. 1 mark.

"Show your working"

Even if Equations Sheet is provided, write the equation symbolically before substituting. Mark schemes reward each step. Skipping straight to a number can cost the equation mark.

[STRETCH] — P = V² / R

A third form of the power equation, derived from the two recall ones.

[STRETCH] — NOT on AQA recall list

P = V² ÷ R

Useful when you're given V and R but not I. Derived by substituting I = V/R into P = VI: P = V × (V/R) = V²/R.

Why this is stretch only

The AQA spec lists only P = VI and P = I²R as the two recall equations for power. P = V²/R is consistent with the physics but NOT in the official recall list — you can use it, but examiners may prefer to see you find I first using V = IR, then apply P = VI.

Safe strategy

If you've memorised both recall equations, you can solve any power question by finding the missing quantity from V = IR first. P = V²/R is just a shortcut — useful, but not required by the spec.

Recap

P = V × I

Use when given V and I.

P = I² × R

Use when given I and R. Square the current.

Three traps to avoid

1

Forgetting the squared (Jun 2023 Q1.3) — three marks lost in one error.

2

Milliwatts substituted as watts (Jun 2024 Q5.2) — wrong by ×1000.

3

Skipping the equation line — losing the recall mark even when answer is correct.

Three calculation patterns to expect

Direct — given V and I, find P (or I and R, find P)

Rearrangement — given P and one of V/I, find the other

Reasoning — explain how P changes when V or I changes (the quadratic relationship)

ESUA every time

Equation → Substitute (with units converted) → Units → Answer. Write each step explicitly.

Checkpoint

You can now:

Define electrical power as the rate of energy transfer.

Recall and apply P = V × I and P = I² × R .

Choose the right equation based on the data given.

Remember to square the current in P = I²R.

Convert milliwatts to watts at the substitution step.

Use ESUA structure to secure all calculation marks.

Next lesson

2.09 — Energy transfers and the National Grid. Two more recall equations (E = Pt and E = QV) and the spec-favourite 6-mark question on why the Grid uses high voltage. This is where the squared-current rule from today pays off big-time.