Previous Year Question

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 V_L = Iω₀L. Substituting the current gives V_L = (V/R)ω₀L.

Therefore, V_L/V = ω₀L/R = Q. Similarly, V_C = I/(ω₀C), so V_C/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: V_L and V_C 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 = V_L/V = V_C/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 V_L + V_C as though they were in phase. Also, the inductor voltage is not zero merely because net reactance is zero.