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)?

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:

  • 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:

Write Q = X_L / R at resonance.Using V = I*Z per element: V_L/V_R = (I X_L) / (I R) = X_L / R = Q.Therefore the correct voltage ratio is V_L / V_R, not V_R / V_L.


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:

Q = V_R / V_L is 1/Q, the inverse of the correct ratio.“Depends only on C” or “cannot be defined with voltages” contradict standard series-resonance relations.V_source/V_R equals 1 at resonance (since Z = R), so it is not Q.


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)

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