Potential difference and power
Cambridge A-Level Physics (9702) · Unit 9: Electricity · 8 flashcards
Potential difference and power is topic 9.2 in the Cambridge A-Level Physics (9702) syllabus , positioned in Unit 9 — Electricity , alongside Electric current and Resistance and resistivity. In one line: Potential difference (V) is the energy transferred per unit charge. It is the work done (W) in moving a unit charge (Q) between two points in a circuit. V = W / Q.
Marked as AS Level: examined at AS Level in Paper 1 (Multiple Choice), Paper 2 (AS Structured Questions) and Paper 3 (Advanced Practical Skills). The same content may also be assumed in Paper 4 (A Level Structured Questions).
The deck below contains 8 flashcards — 4 definitions and 4 calculations — covering the precise wording mark schemes reward. Use the 4 definition cards to lock down command-word answers (define, state), then move on to the concept and calculation cards to handle explain, describe, calculate and compare questions.
Potential difference across a component in terms of energy transfer
Potential difference (V) is the energy transferred per unit charge. It is the work done (W) in moving a unit charge (Q) between two points in a circuit. V = W / Q
What the Cambridge 9702 syllabus says
Official 2025-2027 spec · AS LevelThese are the exact learning outcomes Cambridge sets for this topic. The candidate is expected to be able to do each of these on the relevant paper.
- define the potential difference across a component as the energy transferred per unit charge
- recall and use V = W / Q
- recall and use P = VI, P = I 2R and P = V 2 / R
Cambridge syllabus keywords to use in your answers
These are the official Cambridge 9702 terms tagged to this section. Mark schemes credit responses that use the exact term — weave them into your answers verbatim rather than paraphrasing.
Tips to avoid common mistakes in Potential difference and power
- › Calculate the resistance of one strand using the resistivity formula, then divide by the number of strands to find the total resistance in parallel.
- › Always calculate resistance R as V/I for a specific point on a graph rather than using the gradient for non-ohmic components.
- › Use standard symbols provided in the syllabus glossary, such as ρ for resistivity and λ for wavelength.
- › Always record raw data to the full resolution of the measuring instrument (e.g., 0.1 cm for a meter rule).
- › Calculate thermal energy dissipation rate using P = Fv, where F is the sum of friction and air resistance.
Define potential difference across a component in terms of energy transfer.
Potential difference (V) is the energy transferred per unit charge. It is the work done (W) in moving a unit charge (Q) between two points in a circuit. V = W / Q
A 12V battery transfers 600J of energy to a component. Calculate the charge that flows through the component.
Using V = W / Q, rearrange to find Q = W / V. Therefore, Q = 600J / 12V = 50 Coulombs.
What is the relationship between power, potential difference, and current?
Power (P) is the product of potential difference (V) and current (I). P = VI. This equation shows the rate at which electrical energy is converted into other forms of energy.
A resistor has a potential difference of 6V across it and a current of 0.5A flowing through it. Calculate the power dissipated by the resistor.
Using P = VI, power P = 6V * 0.5A = 3 Watts. This represents the rate at which electrical energy is converted into heat in the resistor.
State the equation that relates power, current, and resistance.
Power (P) equals the square of the current (I) multiplied by the resistance (R): P = I²R. This equation is useful when the current and resistance are known.
A 2Ω resistor has a current of 3A flowing through it. Calculate the power dissipated.
Using P = I²R, the power dissipated is P = (3A)² * 2Ω = 18 Watts.
State the equation that relates power, potential difference and resistance.
Power (P) equals the square of the potential difference (V) divided by the resistance (R): P = V²/R. This equation is useful when voltage and resistance are known.
A 10Ω resistor is connected to a 5V power supply. Calculate the power dissipated in the resistor.
Using P = V²/R, the power dissipated is P = (5V)² / 10Ω = 2.5 Watts.
Review the material
Read full revision notes on Potential difference and power — definitions, equations, common mistakes, and exam tips.
Read NotesMore topics in Unit 9 — Electricity
Potential difference and power sits alongside these A-Level Physics decks in the same syllabus unit. Each uses the same spaced-repetition system, so progress in one informs the next.
Key terms covered in this Potential difference and power deck
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