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Parallel resonant circuit: f0, Q and bandwidth

Short answer

Resonance angular frequency ω0 = √(1/LC − R²/L²) = 9798 rad/s (f0 ≈ 1.56 kHz). Quality factor Q = ω0L/R = 4.9. Bandwidth BW = ω0/Q = 2000 rad/s ≈ 318 Hz.

In the parallel resonant circuit shown in the figure below, find resonance frequency, Q factor and band width. [Figure: a coil of 2 Ω resistance in series with 1 mH inductance, connected in parallel with a 10 μF capacitor]

Given: coil with R = 2 Ω and L = 1 mH in parallel with C = 10 μF.

Resonance frequency

For a coil in parallel with a capacitor:

ω0 = √(1/LC − R²/L²)
1/LC = 1/(1×10⁻³ × 10×10⁻⁶) = 1×10⁸
R²/L² = (2/10⁻³)² = 4×10⁶
ω0 = √(1×10⁸ − 4×10⁶) = √(9.6×10⁷) = 9798 rad/s
f0 = ω0/2π = 1559.4 Hz ≈ 1.56 kHz

(Approximation neglecting R: ω0 = 1/√LC = 10⁴ rad/s, f0 = 1591.5 Hz.)

Quality factor

Q0 = ω0L/R = 9798 × 10⁻³/2 = 4.90

(Using ω = 10⁴ gives Q = 5.)

Bandwidth

BW = ω0/Q0 = R/L = 2/10⁻³ = 2000 rad/s
BW (Hz) = f0/Q0 = 318.3 Hz

Dynamic impedance (extra)

Z0 = L/(CR) = 10⁻³/(10×10⁻⁶ × 2) = 50 Ω

Answer: f0 ≈ 1.56 kHz (ω0 = 9798 rad/s), Q ≈ 4.9, bandwidth ≈ 2000 rad/s ≈ 318 Hz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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