*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 find the half-power bandwidth (HPBW) of a resonant circuit, we can use the formula:

$$
\text{HPBW} = \frac{f_0}{Q}
$$

where:
- $f_0$ = resonant frequency (in Hz)
- $Q$ = quality factor of the circuit

### Given:
- $f_0 = 3.7 \text{ MHz} = 3.7 \times 10^6 \text{ Hz}$
- $Q = 118$

### Calculation:
Substituting the values into the formula:

$$
\text{HPBW} = \frac{3.7 \times 10^6 \text{ Hz}}{118} \approx 31.36 \text{ kHz}
$$

Rounding this gives us approximately 31.4 kHz, which matches the correct answer.

### Explanation of the Correct Answer:
- The correct answer is **C: 31.4 kHz** because it accurately reflects the calculated half-power bandwidth using the resonant frequency and quality factor.

### Why the Other Choices are Incorrect:
- **A: 436.6 kHz**: This value is too high and does not correspond to the calculation based on the given $Q$.
- **B: 218.3 kHz**: This value is also too high and does not match the expected bandwidth for the given $Q$.
- **D: 15.7 kHz**: This value is too low and does not reflect the relationship between the resonant frequency and the quality factor.

### Summary:
- The half-power bandwidth is determined by the resonant frequency and the quality factor.
- The formula used is straightforward and essential for understanding the behavior of resonant circuits in radio applications.