Question
What is the half-power bandwidth of a resonant circuit that has a resonant frequency of 3.7 MHz and a Q of 118?
Answer Options
- A) 436.6 kHz
- B) 218.3 kHz
- C) 31.4 kHz
- D) 15.7 kHz
Correct Answer: C
Explanation
The half-power bandwidth (\Delta F) of a resonant circuit defines the sharpness of its frequency response and is inversely proportional to the circuit’s Quality factor (Q). The formula used to determine the bandwidth is \Delta F = F_R / Q, where F_R is the resonant frequency. The bandwidth is the frequency difference between the -3 \text{ dB} points on the frequency response curve.
Using the given values, F_R = 3.7 \text{ MHz} and Q = 118, the calculation is: \Delta F = \frac{3,700,000 \text{ Hz}}{118} \approx 31,355 \text{ Hz}. This result means the half-power bandwidth is approximately 31.4 kHz. This level of selectivity is typical for a tuned circuit designed to operate in the 80-meter amateur band for general tuning or preselection.
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