A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Transformers and Power Transmission — mark scheme explained
The short answer
Transformers are essential devices in electrical engineering, used to step up or step down voltages in alternating current (AC) circuits. They play a crucial role in the efficient transmission of electrical power over long distances.
The question
A transformer has a primary coil with 100 turns and a secondary coil with 250 turns. If the input voltage is 120 V, what is the output voltage?
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For transformer questions, ensure you clearly show all steps in your calculations. Use the correct formulas and units. For conceptual questions, provide clear and concise explanations, using relevant physics principles and equations where appropriate.
What the command words demand
- Calculate
- Perform a numerical calculation using given data and appropriate formulas.
- Explain
- Provide a detailed account of the reasons or causes for a phenomenon, including relevant principles and equations.
- Describe
- Give a detailed account of the characteristics or features of a system or process without necessarily explaining why they occur.
- Compare
- Identify and explain similarities and differences between two or more concepts, systems, or processes.
- Determine
- Find out or establish something with certainty, often through calculation or measurement.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 5-7 minutes for each question on transformers and power transmission to ensure you have enough time to carefully read the question, perform necessary calculations, and write well-structured answers.
- Identify the given values: N p = 100, N s = 250, V p = 120 V.0 marks
- Use the transformer equation: V s / V p = N s / N p .1 mark
- Substitute the values into the equation: V s / 120 = 250 / 100.1 mark
- Solve for V s : V s = (250 / 100) × 120 = 300 V.2 marks
Final answer: The output voltage is 300 V.
Work through every step correctly and you earn all 4 marks.
Another worked example
A transformer has a primary coil with 500 turns and a secondary coil with 100 turns. If the input current is 2 A, what is the output current?
- Identify the given values: N p = 500, N s = 100, I p = 2 A.0 marks
- Use the power conservation equation: V p × I p = V s × I s .1 mark
- For currents, the ratio is inverted relative to turns: I p / I s = N s / N p .1 mark
- Rearrange for I s : I s = I p × N p / N s .1 mark
- Substitute the values: I s = 2 × (500 / 100).1 mark
- Compute: I s = 2 × 5.0 marks
- Solve for I s : I s = 2 × 5 = 10 A.2 marks
Final answer: The output current is 10 A.
Work through every step correctly and you earn all 6 marks.
Common mistakes
Confusing the transformer equation with Ohm's law
Why it happens: Students sometimes mix up the transformer equation (V s / V p = N s / N p ) with Ohm's law (V = IR). These are different equations and apply to different scenarios.
Fix: Review the transformer equation and understand that it relates voltages and turns, not resistance and current.
Forgetting to use the power conservation equation
Why it happens: Students often focus on the voltage relationship and forget that in an ideal transformer, power is conserved (P p = P s ). This can lead to incorrect calculations of current or power.
Fix: Always remember to use the power conservation equation when dealing with transformers: V p × I p = V s × I s .
Misinterpreting high-voltage transmission
Why it happens: Students sometimes think that high voltage is used to increase power, rather than reduce current and thus minimize power loss. This misunderstanding can lead to incorrect explanations of why high-voltage lines are used.
Fix: Understand that high voltage reduces the current in the transmission line, which in turn reduces power loss (P loss = I 2 R).
Confusing eddy currents with hysteresis losses
Why it happens: Both eddy currents and hysteresis losses are causes of inefficiency in transformers, but they have different mechanisms. Students sometimes mix them up.
Fix: Eddy currents are induced currents that cause heating in the core, while hysteresis losses are due to the magnetic properties of the core material. Review the definitions and mechanisms of both.
Incorrectly calculating transformer efficiency
Why it happens: Students sometimes use the wrong formula or forget to convert the result to a percentage when calculating transformer efficiency.
Fix: Use the correct formula for efficiency: Efficiency = (P s / P p ) × 100%. Ensure you multiply by 100 to get the percentage.
Forgetting to use laminations in the core
Why it happens: Students may not realize that using thin laminated sheets of iron is a crucial method for reducing eddy current losses. This can lead to incomplete or incorrect explanations of how transformers are made more efficient.
Fix: Always mention that the core is made from thin laminated sheets of iron to reduce eddy current losses when discussing transformer efficiency.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Transformer Calculation | Find the secondary voltage using the turns ratio and the primary voltage. | 4 |
| Transformer Current Calculation | Find the secondary current using the transformer turns ratio and given values. | 6 |
| Transformer Output Power | Find the output power of an ideal transformer given the input power. | 3 |
| Calculate Power Loss | Use P = I²R to find the power dissipated in a transmission line. | 4 |
| Total across these question types | 17 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.