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?  

Choices:  
A: 157.8 Hz  
B: 315.6 Hz  
C: 47.3 kHz  
D: 23.67 kHz    

Correct Answer is  C: 47.3 kHz  

To find the half-power (3 dB) bandwidth of a resonant circuit, use:
$$\text{BW}_{3\,\text{dB}} = \frac{f_0}{Q}$$
where:
- $f_0$ = resonant frequency (Hz)
- $Q$ = quality factor

Compute:
$$\text{BW} = \frac{7.1\times10^6}{150} \approx 47{,}333\ \text{Hz} \approx 47.3\ \text{kHz}$$
So the correct answer is 47.3 kHz (Choice C).

Why the other choices are incorrect:
- D: 23.67 kHz is $\frac{f_0}{2Q}$, which is the half-width from $f_0$ to one 3 dB point, not the full bandwidth.
- B: 315.6 Hz is roughly $\frac{f_0}{Q^2}$, an incorrect formula that results from dividing by $Q$ twice.
- A: 157.8 Hz is about $\frac{f_0}{2Q^2}$, also incorrect and half of the already-wrong value in B.

Useful rules of thumb:
- Full 3 dB bandwidth: $\text{BW} \approx \frac{f_0}{Q}$.
- For high $Q$, the half-power frequencies are approximately:
  $$f_{1,2} \approx f_0\left(1 \mp \frac{1}{2Q}\right)$$
  so the bandwidth is $f_2 - f_1 \approx \frac{f_0}{Q}$.