Difficulty: Easy
Correct Answer: Q = V_L / V_R (correct for series resonance)
Explanation:
Introduction / Context:
Quality factor Q is a key figure of merit for resonant selectivity and voltage magnification. For series resonance, several equivalent expressions exist, including reactance-to-resistance ratios and voltage ratios at resonance.
Given Data / Assumptions:
Concept / Approach:
At series resonance, Q = X_L / R = V_L / V_R = V_C / V_R. The resistor drop V_R = IR and the inductor drop V_L = IX_L occur at the same current I, so their ratio equals X_L/R. In a high-Q circuit, V_L and V_C can greatly exceed the source voltage, even though the source sees only R at resonance.
Step-by-Step Solution:
Verification / Alternative check:
Measure V_R (across R) and V_L (across L) at the resonant peak; their ratio equals the independently measured Q from bandwidth (Q = f0/BW) for linear circuits.
Why Other Options Are Wrong:
Common Pitfalls:
Forgetting that large reactive voltages are internal and do not mean large source impedance; interchanging numerator and denominator when forming the ratio.
Final Answer:
Q = V_L / V_R (correct for series resonance)
Discussion & Comments