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?  

Choices:  
A: 436.6 kHz  
B: 218.3 kHz  
C: 31.4 kHz  
D: 15.7 kHz    

Correct Answer is  C: 31.4 kHz  

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

Compute:
$$\text{BW} = \frac{3.7\,\text{MHz}}{118} \approx 31.36\,\text{kHz} \approx 31.4\,\text{kHz}$$

Why the correct answer (C) is correct:
- Using $BW = f_0/Q$, with $f_0 = 3.7\,\text{MHz}$ and $Q=118$, gives about $31.4\,\text{kHz}$.

Why the other choices are incorrect:
- A: 436.6 kHz — This comes from treating $Q$ as $0.118$ instead of $118$ (a decimal-place error). It’s too large.
- B: 218.3 kHz — Much too large; typical of mixing up $2\pi$ (using angular frequency) or other unit mistakes. Correct relation is $BW = f_0/Q$ without extra factors.
- D: 15.7 kHz — Exactly half the correct value; likely from mistakenly halving the bandwidth or confusing half-power with half-amplitude. The −3 dB points correspond to a voltage ratio of $1/\sqrt{2}$, not $1/2$.

Useful formulas and tips:
- $Q = \dfrac{f_0}{\text{BW}_{3\,\text{dB}}}$ so $\text{BW}_{3\,\text{dB}} = \dfrac{f_0}{Q}$.
- In angular frequency terms: $Q = \dfrac{\omega_0}{\Delta\omega}$ with $\omega_0 = 2\pi f_0$, but the $2\pi$ cancels, so the same $BW = f_0/Q$ applies in Hz.
- Rule of thumb: At a few MHz, $Q \approx 100$ gives bandwidth on the order of tens of kHz (e.g., $3.7\,\text{MHz}/100 \approx 37\,\text{kHz}$).