Answer: Q × 100 V = 10 × 100 = 1000 V.
- A Q × 100 V = 10 × 100 = 1000 V
- B 100 V, equal to the source voltage with little quality-factor amplification
- C 0 V, as if the inductor carried no voltage drop at resonance whatsoever
- D 50 V, half of the applied source voltage with little resonance factor included
Correct answer: A. Q × 100 V = 10 × 100 = 1000 V
Explanation: f<sub>0</sub> = 1/(2π√(0.01×10⁻⁵)) = 1/(2π×10⁻³ · √10) ≈ 503 Hz. ω_0 ≈ 3162 rad/s. Q = ω_0 L/R = 3162 × 0.01/10 ≈ 3.16. V<sub>L</sub> = Q × V<sub>source</sub> = 3.16 × 100 ≈ 316 V.
In a pure inductor the current lags the voltage by 90° (energy is briefly stored in the magnetic field before current responds); in a pure capacitor the current leads the voltage by 90° (current must flow to build up charge before voltage rises) - the mnemonic "ELI the ICE man" keeps these straight.
Concept context
AC circuits, impedance, LCR series circuit, resonance, transformers, and power factor.