Question E4B08:
Which of the following can be used to determine the Q of a series-tuned circuit?  

Choices:  
A: The ratio of inductive reactance to capacitive reactance  
B: The frequency shift  
C: The bandwidth of the circuit's frequency response  
D: The resonant frequency of the circuit    

Correct Answer is  C: The bandwidth of the circuit's frequency response  

To determine the Q of a series-tuned circuit, use its 3 dB bandwidth.

Why C is correct (bandwidth):
- For a resonant circuit, the loaded quality factor is given by:
  $$Q = \frac{f_0}{\text{BW}_{3\text{dB}}}$$
  where:
  - $f_0$ = resonant frequency
  - $\text{BW}_{3\text{dB}} = f_2 - f_1$ is the bandwidth between the half-power (−3 dB) frequencies.
- Practically: sweep frequency, find the two frequencies where the power is 3 dB down from peak (current down by $1/\sqrt{2}$), subtract to get bandwidth, then compute $Q$.

Why the other choices are incorrect:
- A: Ratio of inductive to capacitive reactance. At resonance, $X_L = X_C$, so $X_L/X_C = 1$ regardless of Q. Q depends on losses (resistance), not on the ratio of reactances. The relevant relation is $Q = X_L/R = X_C/R$ at resonance, not $X_L/X_C$.
- B: Frequency shift. “Frequency shift” is not a defined measure of Q. Q relates to fractional bandwidth or energy storage vs loss per cycle, not to an arbitrary shift.
- D: Resonant frequency alone. Knowing $f_0$ by itself does not give Q; you also need bandwidth (or resistance) to determine losses.

Useful formulas and rules of thumb:
- Resonant frequency:
  $$f_0 = \frac{1}{2\pi\sqrt{LC}}$$
- Series RLC quality factor:
  $$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 R C}, \quad \omega_0 = 2\pi f_0$$
- 3 dB bandwidth (series resonance):
  $$\text{BW}_{3\text{dB}} = \frac{R}{2\pi L} = \frac{f_0}{Q}$$
- At resonance (series RLC):
  $$X_L = X_C = Q\,R$$

Measuring bandwidth yields the loaded Q, which is typically what’s meant in practical, measurement-based questions.