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Figure 8 shows the first-stage filter circuit for a simple AM receiver - AQA - A-Level Physics - Question 3 - 2017 - Paper 8

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Figure 8 shows the first-stage filter circuit for a simple AM receiver. The circuit can be adjusted to resonate at 910 kHz so that it can receive a particular radio ... show full transcript

Worked Solution & Example Answer:Figure 8 shows the first-stage filter circuit for a simple AM receiver - AQA - A-Level Physics - Question 3 - 2017 - Paper 8

Step 1

Calculate the value of the capacitance when the circuit resonates at a frequency of 910 kHz.

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Answer

To find the capacitance, we use the resonance frequency formula:
f = rac{1}{2 \\pi \\sqrt{LC}}
Rearranging gives us:
C = rac{1}{(2 \\pi f)^2 L}
Substituting the given values:

  • Frequency, f=910 kHz=910×103 Hzf = 910 \text{ kHz} = 910 \times 10^3 \text{ Hz}
  • Inductance, L=1.1 mH=1.1×103 HL = 1.1 \text{ mH} = 1.1 \times 10^{-3} \text{ H}
    Thus,
    C=1(2pi(910×103))2(1.1×103)C = \frac{1}{(2 \\pi (910 \times 10^3))^2 (1.1 \times 10^{-3})}
    Calculating gives:
    C27.8 pFC \approx 27.8 \text{ pF}.

Step 2

Draw on Figure 9 an ideal response curve for the resonant circuit, labelling all relevant frequency values based upon a 10 kHz bandwidth.

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Answer

The ideal response curve is typically characterized by a sharp peak at the resonant frequency. For a resonance frequency of 910 kHz and a bandwidth of 10 kHz, the relevant frequency values are:

  • Lower Frequency: 905 kHz905 \text{ kHz}
  • Resonant Frequency: 910 kHz910 \text{ kHz}
  • Upper Frequency: 915 kHz915 \text{ kHz}
    The curve should show a peak gain of 1.0 at 910 kHz, gradually dropping to around 0.7 at 905 kHz and 915 kHz, creating a bell-shaped curve symmetric around the resonant frequency.
    These values should be clearly labelled on Figure 9.

Step 3

Discuss the changes the listener might notice when tuning to this station due to the practical Q-factor being smaller.

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Answer

A smaller Q-factor leads to several noticeable changes for the listener:

  1. Broader Bandwidth: The station will be broadcast over a wider frequency range, meaning the listener’s radio may pick up signals from adjacent frequencies, resulting in interference.
  2. Less Selectivity: The circuit becomes less effective in filtering out unwanted frequencies, which could cause a muddier sound with mixed signals from nearby stations.
  3. Increased Susceptibility to Crosstalk: With a lower Q-factor, the circuit is more prone to picking up signals from neighboring stations, leading to potential overlap and confusion in audio.
  4. Less Gain: As a result of decreased energy transfer, the listener may notice a reduction in the clarity and strength of the signal, making it more difficult to tune in on a clear transmission.

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