Series resonance quality factor — voltage ratios: At resonance in a series RLC, is the quality factor given by Q = V_R / V_L, or is it correctly Q = V_L / V_R (which equals X_L / R)?

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:

• Series RLC circuit at resonance (X_L = X_C).
• Finite series resistance R representing all losses.
• Voltages and currents are RMS quantities.

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)