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Resonance of coil in parallel with capacitor

Short answer

For a practical coil (R in series with L) in parallel with C, resonance occurs when total susceptance is zero: ω0 = √(1/LC − R²/L²). At resonance, the circuit presents a maximum dynamic impedance Z0 = L/(CR). If R is small, f0 ≈ 1/(2π√LC).

Derive an expression with necessary diagrams for the resonance frequency of a circuit consisting of a coil in parallel with a capacitor excited by a sinusoidal AC voltage.

A practical coil has inductance L and a series resistance R. When it is connected in parallel with a capacitor C across a sinusoidal supply, the circuit is at resonance when the supply voltage and the line current are in phase (the total admittance is real).

         +--------+---------+
    I -->|        |         |
         |      (R+jωL)    (1/jωC)
  V      |      coil        C
         |        |         |
         +--------+---------+
 I_coil = V/(R + jωL)    I_C = jωCV

Derivation

The admittance of the coil branch is

Y_coil = 1/(R + jωL) = (R − jωL)/(R² + ω²L²)

Total admittance with the capacitor:

Y = R/(R² + ω²L²) + j[ ωC − ωL/(R² + ω²L²) ]

Resonance requires the imaginary part to be zero:

C = L/(R² + ω0²L²)
R² + ω0²L² = L/C
ω0² = 1/(LC) − R²/L²

ω0 = √(1/LC − R²/L²) rad/s
f0 = (1/2π)√(1/LC − R²/L²) Hz

Remarks

  • Resonance is possible only if R < √(L/C). Otherwise ω0 is imaginary and no resonance occurs.
  • If R << ω0L (high Q coil), f0 ≈ 1/(2π√LC).
  • At resonance the impedance is maximum: Z0 = (R² + ω0²L²)/R = L/(CR) (dynamic impedance), and the line current is minimum.

Answer: f0 = (1/2π)√(1/LC − R²/L²), dynamic impedance L/(CR).

✓ Answer checked by Claude Opus 5.5 (high effort)

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