AQA A-Level Physics Paper 3 (3BE), June 2025: Question 4

✅ Mark Scheme Model Answer

• Mark 1: A carrier wave is a continuous electromagnetic wave of fixed/constant frequency assigned to a radio station used to carry an information signal.
• Mark 2: A characteristic of the carrier wave (its amplitude or frequency) is varied / modulated to represent the information or audio signal being transmitted.

💡 Key Knowledge

• Why use a carrier wave? Audio frequencies (20 Hz – 20 kHz) have extremely long wavelengths and would require unrealistically massive antennae. They also cannot share transmission space without overlapping.
• Modulation: Superimposing an information signal onto a high-frequency carrier wave. In AM (amplitude modulation), the envelope reflects the audio wave; in FM (frequency modulation), frequency shifts encode the data.

🧠 Exam Technique

Notice the two-part prompt: "State what is meant..." followed by "Go on to explain how it is used..."

• Ensure you explicitly include the word "fixed frequency" or "constant frequency" for the first mark.
• State the specific word "modulated" or identify which parameter changes (amplitude or frequency) for the second mark.

❌ Common Errors

• Vague definitions: Saying simply "a wave that carries sound" without mentioning its electromagnetic nature or fixed frequency.
• Confusing carrier with signal: Thinking the carrier itself is the music/voice rather than the vehicle that is modulated.

✅ Mark Scheme Model Answer

Any two of the following points:

• The circuit resonates at a specific natural frequency determined by values of L and C (via f₀ = 1 / [2π√(LC)] ).
• It acts as a band-pass filter / produces a large output voltage only at the resonant frequency, rejecting frequencies further away.
• Varying the variable capacitor ( C ) alters the resonant frequency, allowing the user to tune into / select different stations.

💡 Key Knowledge

• The aerial receives hundreds of radio signals across many frequencies simultaneously.
• A parallel LC tank circuit provides maximum impedance at resonance:
  f₀ = 1 / (2π√(LC))
• Because voltage across parallel elements is V = I × Z , the maximum voltage developed across the circuit occurs for signals close to f₀ .

🧠 Exam Technique

Always connect the circuit components to physical actions:

• Point out the arrow through C : this indicates a variable capacitor, allowing adjustment of the tuning frequency.
• Use the term resonance—examiners actively look for this keyword when grading LC circuits.

❌ Common Errors

• Failing to mention the word resonate or resonance.
• Confusing the roles of components (e.g., claiming the inductor demodulates the signal).
• Not stating that different stations are selected by varying C.

📐 Step-by-Step Graph Analysis

1. Bandwidth definition: Measured across the frequency range where the response voltage is at or above V_max / √2 ≈ 0.707 × V_max (the -3 dB / half-power points).
2. Circuit A:
   • Peak voltage: V_max ≈ 90 mV
   • Half-power level: 0.707 × 90 mV ≈ 63.6 mV
   • Frequencies at 63.6 mV: approx 194 kHz and 206 kHz
   • Measured Bandwidth A: 206 − 194 = 12 kHz (± 2 kHz)
3. Circuit B:
   • Peak voltage: V_max ≈ 58 mV
   • Half-power level: 0.707 × 58 mV ≈ 41 mV
   • Frequencies at 41 mV: approx 188 kHz and 212 kHz
   • Measured Bandwidth B: 212 − 188 = 24 kHz (± 2 kHz)

✅ Mark Scheme Model Answer

• Mark 1 (Bandwidth Check): Both circuits have a bandwidth greater than or equal to the required 10 kHz (Bandwidth A ≈ 12 kHz, Bandwidth B ≈ 24 kHz), so both circuits can accommodate the signal without cutting sidebands.
• Mark 2 (Choice & Justification): Circuit A is chosen. It has a much sharper resonance peak (higher Q factor / narrower bandwidth), which means it provides greater selectivity and will better reject unwanted signals and interference from adjacent radio stations.

🧠 Exam Technique

• Must address both bullet points: The question asks for bandwidth AND station rejection. You cannot score full marks without discussing both.
• Show the numerical check: State the estimated bandwidth for at least one circuit (e.g., Bandwidth A ≈ 12 kHz > 10 kHz) to justify why Circuit A doesn't clip the transmission.
• "Q-factor" alone is NOT enough: The mark scheme explicitly notes: "Do not accept reference to Q factor alone." You must link it to the ability to avoid interference / reject neighbouring stations.

❌ Common Errors

• Choosing B incorrectly: Believing that having a wider bandwidth (24 kHz) is automatically "better", forgetting that excess bandwidth lets in noise and adjacent channel chatter.
• Reading bandwidth at the baseline: Measuring bandwidth at 0 mV or across the full base of the curve rather than at the standard 0.707 × V_max level.
• Forgetting to calculate: Simply stating "A is narrower" without proving it meets the minimum 10 kHz requirement.