Answer: f 2 - f 1 = R/(2πL) = R/ω_0/Q.
- A f<sub>2</sub> - f<sub>1</sub> = R/(2πL) = R/ω_0/Q
- B f<sub>2</sub> - f<sub>1</sub> = f<sub>0</sub> during normal conditions
- C f<sub>2</sub> - f<sub>1</sub> = 1/(RC) as generally observed
- D f<sub>2</sub> - f<sub>1</sub> = 1/Q in typical laboratory settings
Correct answer: A. f<sub>2</sub> - f<sub>1</sub> = R/(2πL) = R/ω_0/Q
Explanation: At half-power points, I = I<sub>max</sub>/√2. Bandwidth Δf = f<sub>2</sub> - f<sub>1</sub> = R/(2πL) = f<sub>0</sub>/Q. These are the -3 dB frequencies on either side of resonance.
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.