Voltage Magnification at Series Resonance: Q = ω₀L/R
B — Q = ωL/R, with ω evaluated at resonance (ω = ω₀). The quality factor equals the voltage across either reactive component divided by the supply voltage at series resonance.
Step 1: Apply the resonance condition
For a series RLC circuit, the impedance is Z = R + j(ωL − 1/ωC). At resonance, inductive and capacitive reactances are equal: ω₀L = 1/(ω₀C). Their imaginary terms cancel, leaving Z = R. Here R is the total effective series resistance.
Step 2: Find current and inductor voltage
Because the impedance is purely resistive, the supply current is I = V/R. The inductor voltage magnitude is VL = Iω₀L. Substituting the current gives VL = (V/R)ω₀L.
Therefore, VL/V = ω₀L/R = Q. Similarly, VC = I/(ω₀C), so VC/V = 1/(ω₀CR) = Q. These are equivalent expressions at resonance.
Why component voltage can exceed supply voltage
If Q is greater than 1, each reactive-component voltage exceeds the supply voltage in magnitude. There is no violation of Kirchhoff’s voltage law: VL and VC are opposite in phase and cancel in the phasor sum. The supply balances the resistive voltage.
For example, if Q = 5 and the supply is 10 V RMS, each reactive component has 50 V RMS across it at resonance. This example illustrates the ratio; it is not additional data from the question.
Useful Formula
ω₀ = 1/√(LC); f₀ = 1/(2π√(LC)); Q = VL/V = VC/V = ω₀L/R = 1/(ω₀CR).
Why the Other Options Are Not the Answer
- A — ωCR: equals 1/Q for this series circuit at resonance, not Q.
- C — ωR/L: is not the required reactance-to-resistance ratio and is not dimensionless.
- D — ωC/R: is not 1/(ωCR) and is also dimensionally unsuitable for a voltage ratio.
Remember for the Exam
Q has no unit. For fixed L and C, increasing total series resistance lowers voltage magnification.
Common mistake: Do not add the magnitudes VL + VC as though they were in phase. Also, the inductor voltage is not zero merely because net reactance is zero.
