*Question E5A12:
What is the half-power bandwidth of a resonant circuit that has a resonant frequency of 3.7 MHz and a Q of 118?*  

*Correct Answer is  C: 31.4 kHz*  

### AI Response: (may contain errors)  

To determine the half-power bandwidth of a resonant circuit, we use the formula:

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

where:
- $BW$ is the bandwidth in hertz (Hz),
- $f_0$ is the resonant frequency in hertz (Hz),
- $Q$ is the quality factor (dimensionless).

Given:
- Resonant frequency, $f_0 = 3.7 \, \text{MHz} = 3,700,000 \, \text{Hz}$
- Quality factor, $Q = 118$

Plug these values into the formula:

$$
BW = \frac{3,700,000}{118} \approx 31,355 \, \text{Hz} \approx 31.4 \, \text{kHz}
$$

Thus, the correct answer is **31.4 kHz** (Choice C).

### Explanation of Choices:
- **Choice A: 436.6 kHz**  
  This value is much larger than the calculated bandwidth. It would imply a much lower $Q$ factor, which is not the case here.

- **Choice B: 218.3 kHz**  
  This is also larger than the calculated bandwidth. It suggests a $Q$ factor of about 17, which is incorrect.

- **Choice C: 31.4 kHz**  
  This matches our calculated result, making it the correct answer.

- **Choice D: 15.7 kHz**  
  This is smaller than the calculated bandwidth. It would imply a $Q$ factor of about 236, which is incorrect.

### Key Points:
- The half-power bandwidth is inversely proportional to the $Q$ factor.
- A higher $Q$ factor results in a narrower bandwidth, and vice versa.