*Question E5A11:
What is the half-power bandwidth of a resonant circuit that has a resonant frequency of 7.1 MHz and a Q of 150?*  

*Correct Answer is  C: 47.3 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:
- $f_0 = 7.1 \text{ MHz} = 7.1 \times 10^6 \text{ Hz}$
- $Q = 150$

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

$$
\text{HPBW} = \frac{7.1 \times 10^6 \text{ Hz}}{150} \approx 47333.33 \text{ Hz} \approx 47.3 \text{ kHz}
$$

Thus, the half-power bandwidth of the resonant circuit is approximately 47.3 kHz, which corresponds to the correct answer.

### Explanation of Choices:
- **Choice A: 157.8 Hz**
  - This value is too small and does not reflect the relationship between frequency and Q factor.
  
- **Choice B: 315.6 Hz**
  - Similar to choice A, this value is also too small and does not align with the calculated HPBW.
  
- **Choice C: 47.3 kHz**
  - This is the correct answer, as shown in the calculation above.
  
- **Choice D: 23.67 kHz**
  - This value is incorrect and is less than the calculated HPBW, indicating a misunderstanding of the Q factor's effect on bandwidth.

### Summary:
- The half-power bandwidth is determined by the resonant frequency and the quality factor.
- The correct answer is derived from the formula, confirming that the bandwidth is inversely proportional to the Q factor.