ElectricityAQA GCSE Physics (8463)

Energy Transfers

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

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Master the topic

Worked examples7 Key formulae4 Definitions6 Common mistakes6

Step-by-step worked examples

01.Power of an iron
Foundation
Question
An iron draws 5.0 A from a 230 V mains supply. Calculate its power.
Solution
1. 1
  Apply P = VI.
  $P = 230 \times 5.0 = 1150\,\text{W}$
Answer
1150 W (1.15 kW).
02.Heat in a wire
Higher
Question
A wire of resistance 0.5 Ω carries 12 A. Calculate the power dissipated as heat.
Solution
1. 1
  Apply P = I²R.
  $P = 12^2 \times 0.5 = 144 \times 0.5 = 72\,\text{W}$
Answer
72 W.
03.Grid loss comparison
Higher
Question
A transmission cable has total resistance 5 Ω. Compare the power lost when transmitting 200 MW at 10 kV vs at 400 kV.
Solution
1. 1
  Current at 10 kV.
  $I_1 = 200{,}000{,}000 / 10{,}000 = 20{,}000\,\text{A}$
2. 2
  Loss at 10 kV.
  $P_1 = 20{,}000^2 \times 5 = 2 \times 10^{12}\,\text{W} \text{ — impossible}$
3. 3
  Current at 400 kV.
  $I_2 = 200{,}000{,}000 / 400{,}000 = 500\,\text{A}$
4. 4
  Loss at 400 kV.
  $P_2 = 500^2 \times 5 = 1{,}250{,}000\,\text{W} = 1.25\,\text{MW}$
Answer
At 10 kV the loss exceeds the supplied power (impossible). At 400 kV the loss is just 1.25 MW (~0.6 %). High-voltage transmission is essential.
Examiner note
Use the I² term — it makes the case dramatic.
04.Toaster energy
Foundation
Question
A 900 W toaster operates for 90 s. Calculate the energy it transfers.
Solution
1. 1
  Apply E = Pt.
  $E = 900 \times 90 = 81{,}000\,\text{J}$
Answer
81,000 J (81 kJ).
05.Energy via E = QV
Higher
Question
5.0 C of charge passes through a 12 V battery-powered drill. How much energy does the drill receive?
Solution
1. 1
  Apply E = QV.
  $E = 5.0 \times 12 = 60\,\text{J}$
Answer
60 J.
06.Describe transfers for an electric kettle
Foundation
Question
Describe the energy transfers in an electric kettle, naming the relevant stores.
Solution
1. 1
  Input.
  $\text{Electrical pathway from the mains.}$
2. 2
  Useful.
  $\text{Thermal store of the water.}$
3. 3
  Wasted.
  $\text{Thermal stores of the kettle body and surrounding air; small amount of sound radiation.}$
Answer
Electrical → useful thermal store of water; wasted thermal stores of kettle, air; small sound radiation.
07.Why high V transmission?
Higher
Question
100 MW must be transmitted along a 4 Ω cable. Calculate the power lost at (a) 10 kV, (b) 400 kV.
Solution
1. 1
  Current at 10 kV.
  $I = 10^8 / 10^4 = 10{,}000\,\text{A}$
2. 2
  Loss at 10 kV.
  $P = I^2 R = 10{,}000^2 \times 4 = 4 \times 10^8\,\text{W} = 400\,\text{MW (impossible)}$
3. 3
  Current at 400 kV.
  $I = 10^8 / (4 \times 10^5) = 250\,\text{A}$
4. 4
  Loss at 400 kV.
  $P = 250^2 \times 4 = 250{,}000\,\text{W} = 0.25\,\text{MW}$
Answer
10 kV: 400 MW lost (more than transmitted). 400 kV: 0.25 MW (~0.25 %). High voltage transmission is essential.
Examiner note
Emphasise I² in the explanation.

Key formulae

Power (V and I)
$P = V I$
- $P$
  — Power (W)
- $V$
  — p.d. (V)
- $I$
  — Current (A)
When to use
Whenever V and I are given. Recall — not on equation sheet.
Power (I and R)
$P = I^2 R$
- $P$
  — Power (W)
- $I$
  — Current (A)
- $R$
  — Resistance (Ω)
When to use
Use when I and R are known (e.g. power dissipated in a wire or resistor).
Energy (power and time)
$E = P t$
- $E$
  — Energy (J)
- $P$
  — Power (W)
- $t$
  — Time (s)
When to use
When P and t are given.
Energy (charge and p.d.)
$E = Q V$
- $E$
  — Energy (J)
- $Q$
  — Charge (C)
- $V$
  — p.d. (V)
When to use
When Q and V are given. Derived from 1 V = 1 J/C.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Electrical powerExaminer keyword: The rate at which electrical energy is transferred in a circuit (W).
Useful energyExaminer keyword: Energy transferred to the intended store (the appliance's purpose).
Wasted energyExaminer keyword: Energy dissipated to unintended stores, usually thermal store of the surroundings.
Step-up transformerExaminer keyword: A transformer that increases the AC voltage (and decreases the current). Has more turns on its secondary coil than primary.
Step-down transformerExaminer keyword: A transformer that decreases voltage (and increases current). Has fewer turns on secondary than primary.
National Grid: The UK-wide network of cables, pylons and substations that distributes electricity from power stations to consumers.

Common mistakes and misconceptions

Mistake
Forgetting to square I in P = I²R.
Why it happens
Quick substitution.
How to avoid it
Write P = I × I × R explicitly.
Mistake
Using P = VI when V is unknown but I and R are given.
Why it happens
Defaulting to the first formula.
How to avoid it
Choose the formula whose variables match what you have.
Mistake
Writing 'electrical energy store'.
Why it happens
Electrical is treated as a thing.
How to avoid it
Electrical is a transfer pathway, NOT a store. The relevant store is chemical (battery) or thermal/kinetic (in the appliance).
Mistake
Plugging time in minutes into E = Pt.
Why it happens
Forgetting to convert.
How to avoid it
Always convert to seconds first.
Mistake
Saying transformer increases the power.
Why it happens
Confusing V with P.
How to avoid it
Transformers conserve power (V_p I_p = V_s I_s). They TRADE voltage for current.
Mistake
Saying the National Grid is DC.
Why it happens
Confusion with HVDC links.
How to avoid it
GCSE answer is AC. Transformers need AC.

ElectricityAQA GCSE Physics (8463)

Energy Transfers

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

Previous Next

Master the topic

Worked examples7 Key formulae4 Definitions6 Common mistakes6

Step-by-step worked examples

01.Power of an iron
Foundation
Question
An iron draws 5.0 A from a 230 V mains supply. Calculate its power.
Solution
1. 1
  Apply P = VI.
  $P = 230 \times 5.0 = 1150\,\text{W}$
Answer
1150 W (1.15 kW).
02.Heat in a wire
Higher
Question
A wire of resistance 0.5 Ω carries 12 A. Calculate the power dissipated as heat.
Solution
1. 1
  Apply P = I²R.
  $P = 12^2 \times 0.5 = 144 \times 0.5 = 72\,\text{W}$
Answer
72 W.
03.Grid loss comparison
Higher
Question
A transmission cable has total resistance 5 Ω. Compare the power lost when transmitting 200 MW at 10 kV vs at 400 kV.
Solution
1. 1
  Current at 10 kV.
  $I_1 = 200{,}000{,}000 / 10{,}000 = 20{,}000\,\text{A}$
2. 2
  Loss at 10 kV.
  $P_1 = 20{,}000^2 \times 5 = 2 \times 10^{12}\,\text{W} \text{ — impossible}$
3. 3
  Current at 400 kV.
  $I_2 = 200{,}000{,}000 / 400{,}000 = 500\,\text{A}$
4. 4
  Loss at 400 kV.
  $P_2 = 500^2 \times 5 = 1{,}250{,}000\,\text{W} = 1.25\,\text{MW}$
Answer
At 10 kV the loss exceeds the supplied power (impossible). At 400 kV the loss is just 1.25 MW (~0.6 %). High-voltage transmission is essential.
Examiner note
Use the I² term — it makes the case dramatic.
04.Toaster energy
Foundation
Question
A 900 W toaster operates for 90 s. Calculate the energy it transfers.
Solution
1. 1
  Apply E = Pt.
  $E = 900 \times 90 = 81{,}000\,\text{J}$
Answer
81,000 J (81 kJ).
05.Energy via E = QV
Higher
Question
5.0 C of charge passes through a 12 V battery-powered drill. How much energy does the drill receive?
Solution
1. 1
  Apply E = QV.
  $E = 5.0 \times 12 = 60\,\text{J}$
Answer
60 J.
06.Describe transfers for an electric kettle
Foundation
Question
Describe the energy transfers in an electric kettle, naming the relevant stores.
Solution
1. 1
  Input.
  $\text{Electrical pathway from the mains.}$
2. 2
  Useful.
  $\text{Thermal store of the water.}$
3. 3
  Wasted.
  $\text{Thermal stores of the kettle body and surrounding air; small amount of sound radiation.}$
Answer
Electrical → useful thermal store of water; wasted thermal stores of kettle, air; small sound radiation.
07.Why high V transmission?
Higher
Question
100 MW must be transmitted along a 4 Ω cable. Calculate the power lost at (a) 10 kV, (b) 400 kV.
Solution
1. 1
  Current at 10 kV.
  $I = 10^8 / 10^4 = 10{,}000\,\text{A}$
2. 2
  Loss at 10 kV.
  $P = I^2 R = 10{,}000^2 \times 4 = 4 \times 10^8\,\text{W} = 400\,\text{MW (impossible)}$
3. 3
  Current at 400 kV.
  $I = 10^8 / (4 \times 10^5) = 250\,\text{A}$
4. 4
  Loss at 400 kV.
  $P = 250^2 \times 4 = 250{,}000\,\text{W} = 0.25\,\text{MW}$
Answer
10 kV: 400 MW lost (more than transmitted). 400 kV: 0.25 MW (~0.25 %). High voltage transmission is essential.
Examiner note
Emphasise I² in the explanation.

Key formulae

Power (V and I)
$P = V I$
- $P$
  — Power (W)
- $V$
  — p.d. (V)
- $I$
  — Current (A)
When to use
Whenever V and I are given. Recall — not on equation sheet.
Power (I and R)
$P = I^2 R$
- $P$
  — Power (W)
- $I$
  — Current (A)
- $R$
  — Resistance (Ω)
When to use
Use when I and R are known (e.g. power dissipated in a wire or resistor).
Energy (power and time)
$E = P t$
- $E$
  — Energy (J)
- $P$
  — Power (W)
- $t$
  — Time (s)
When to use
When P and t are given.
Energy (charge and p.d.)
$E = Q V$
- $E$
  — Energy (J)
- $Q$
  — Charge (C)
- $V$
  — p.d. (V)
When to use
When Q and V are given. Derived from 1 V = 1 J/C.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Electrical powerExaminer keyword: The rate at which electrical energy is transferred in a circuit (W).
Useful energyExaminer keyword: Energy transferred to the intended store (the appliance's purpose).
Wasted energyExaminer keyword: Energy dissipated to unintended stores, usually thermal store of the surroundings.
Step-up transformerExaminer keyword: A transformer that increases the AC voltage (and decreases the current). Has more turns on its secondary coil than primary.
Step-down transformerExaminer keyword: A transformer that decreases voltage (and increases current). Has fewer turns on secondary than primary.
National Grid: The UK-wide network of cables, pylons and substations that distributes electricity from power stations to consumers.

Common mistakes and misconceptions

Mistake
Forgetting to square I in P = I²R.
Why it happens
Quick substitution.
How to avoid it
Write P = I × I × R explicitly.
Mistake
Using P = VI when V is unknown but I and R are given.
Why it happens
Defaulting to the first formula.
How to avoid it
Choose the formula whose variables match what you have.
Mistake
Writing 'electrical energy store'.
Why it happens
Electrical is treated as a thing.
How to avoid it
Electrical is a transfer pathway, NOT a store. The relevant store is chemical (battery) or thermal/kinetic (in the appliance).
Mistake
Plugging time in minutes into E = Pt.
Why it happens
Forgetting to convert.
How to avoid it
Always convert to seconds first.
Mistake
Saying transformer increases the power.
Why it happens
Confusing V with P.
How to avoid it
Transformers conserve power (V_p I_p = V_s I_s). They TRADE voltage for current.
Mistake
Saying the National Grid is DC.
Why it happens
Confusion with HVDC links.
How to avoid it
GCSE answer is AC. Transformers need AC.

Energy Transfers

Master the topic

Step-by-step worked examples

01.Power of an iron

02.Heat in a wire

03.Grid loss comparison

04.Toaster energy

05.Energy via E = QV

06.Describe transfers for an electric kettle

07.Why high V transmission?

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions

Energy Transfers

Master the topic

Step-by-step worked examples

01.Power of an iron

02.Heat in a wire

03.Grid loss comparison

04.Toaster energy

05.Energy via E = QV

06.Describe transfers for an electric kettle

07.Why high V transmission?

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions