Figure 4 shows the filter circuit that forms the first stage in an amplitude modulated (AM) radio receiver - AQA - A-Level Physics - Question 2 - 2020 - Paper 8

Looking at Table 2:

• Capacitor W (2-9 pF), suitable range and can adjust to the required 9.3 pF.
• Capacitor X (3-10 pF), suitable for slight adjustments but maximum is 10 pF.
• Capacitor Y (4.5-20 pF), covers larger ranges, suitable for tuning variations.
• Capacitor Z (10-50 pF), not suitable since it exceeds the required range.

Preference: I would prefer Capacitor Y as it provides the widest range and can be adjusted according to different frequency needs without exceeding the necessary values.

To find the bandwidth of the filter circuit from Figure 5, we look at the given output levels. The bandwidth can be identified as the range of frequencies between two points where the output voltage drops to half of its maximum value (cut-off frequencies).

From the graph:

• Lower cut-off: ~198 kHz
• Upper cut-off: ~220 kHz

Thus, the bandwidth (BW) can be calculated as: $BW = f_{upper} - f_{lower} = 220 kHz - 198 kHz = 22 kHz$.

The Q factor can be calculated using the formula:

$Q = \frac{f_0}{BW}$

where:

• $f_0$ is the resonant frequency (which can be estimated as the midpoint of the bandwidth).
• $BW$ is the bandwidth determined earlier.

Substituting in:

• Estimated $f_0 \approx \frac{220 kHz + 198 kHz}{2} = 209 kHz$.
• From above, $BW = 22 kHz$.

Calculating: $Q = \frac{209 kHz}{22 kHz} \approx 9.5$

Thus, the Q factor is approximately 9.5.

With a low Q factor, the filter has a wider bandwidth which can lead to:

• Overlapping stations: The listener may hear more interference from adjacent channels because the filter can pick up a wider range of frequencies. This can result in audio overlapping or a decrease in clarity, making it difficult to discern the desired station.