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Series LC resonant frequency

At what frequency will this choke and this cell ring together?

The formula
f=12⁢π⁢L·C
f
Resonant frequency, Hz
L
Inductance, H
C
Capacitance, F
LaTeX 
f = \frac{1}{2\pi\sqrt{L \cdot C}}

Work it out

Fill from a part the library

mH

The total series inductance — in a VIC, the chokes in the charging path.

nF

The cell capacitance, water dielectric included.

Compare with a variation

Result

f Resonant frequency 805 Hz

Where inductive and capacitive reactance cancel and the loop rings.

ω Angular frequency 5.058 kHz

The same frequency in radians per second, which is the form the reactances use.

Z Characteristic impedance 1.977 Ω

The reactance of either element at resonance. Sets the current the drive must supply.

With your numbers
805 Hz=12⁢π⁢390.8 µH·100 µF
LaTeX 
805\,\mathrm{Hz} = \frac{1}{2\pi\sqrt{390.8\,\mathrm{µH} \cdot 100\,\mathrm{µF}}}

This result is a link — the address bar holds your numbers, so it can be pasted into a post and opened to the same answer.

What this looks like

Where the reactances cancel The choke's reactance rises with frequency; the cell's falls. Resonance is the one frequency where they are equal and opposite, and the loop looks purely resistive to the drive.
Where the reactances cancelTwo straight lines on logarithmic axes: inductive reactance rising with frequency, capacitive reactance falling, crossing at 805 Hz where both equal 1.977 Ω. A decade below resonance the cell dominates by a hundred to one; a decade above, the choke does.0.010.1110100100011e11e21e31e41e5805 Hz1.977 ΩFrequency (Hz)Reactance (Ω)Inductive, ωLCapacitive, 1 ÷ ωC
The formula behind the curve
f=12⁢π⁢L·C
f
Resonant frequency, Hz
L
Inductance, H
C
Capacitance, F
LaTeX 
f = \frac{1}{2\pi\sqrt{L \cdot C}}
Resonant frequency against inductance Inductance swept from 0.195 mH to 0.586 mH with everything else held at your numbers. The dashed lines cross where you are.
Resonant frequency against inductanceResonant frequency falls from 1139 Hz to 657 Hz as inductance rises from 0.195 mH to 0.586 mH. At your inductance of 0.391 mH it is 805 Hz.600800100012000.20.40.60.391 mH805 HzInductance (mH)Resonant frequency (Hz)
The formula behind the curve
f=12⁢π⁢L·C
f
Resonant frequency, Hz
L
Inductance, H
C
Capacitance, F
LaTeX 
f = \frac{1}{2\pi\sqrt{L \cdot C}}
What moves the answer Each input moved 10% either way, with the others held still, and the effect on resonant frequency.
What moves the answerResonant frequency is most sensitive to Inductance, which moves it by about 5.41% for a 10% change. The other input has exactly the same hold on it.Change in the answer when each input moves by 10%-10%-5%5%10%Inductance±5.41Capacitance±5.41
The formula behind the curve
f=12⁢π⁢L·C
f
Resonant frequency, Hz
L
Inductance, H
C
Capacitance, F
LaTeX 
f = \frac{1}{2\pi\sqrt{L \cdot C}}

Method

  1. Convert the inductance to henries and the capacitance to farads. Everything below is in SI; mixing millihenries with farads is the commonest way to get an answer that is wrong by three decades and looks plausible.
  2. Multiply them: L × C. The product has units of seconds squared, which is the first check that the inputs were what you meant.
  3. Take the square root. That is the time constant of the loop — very nearly the quarter-period of the ring.
  4. Multiply by 2π and take the reciprocal. That converts the time constant into a frequency in hertz.
  5. For the characteristic impedance, take √(L ÷ C). At resonance the inductive and capacitive reactances are equal and opposite, and this is the size of each.

Assumptions

  • The inductance and capacitance are constant with frequency. Neither is, in a real VIC: a choke has self-capacitance and a self-resonance of its own, and water’s permittivity falls with frequency (the Cole-Cole behaviour). This answer is the lossless ideal.
  • Series connection. A parallel LC has the same resonant frequency but behaves oppositely around it — high impedance rather than low.
  • Resistance is ignored. Real damping pulls the peak slightly below this frequency; the shift is negligible for Q above about 5 and is not for a cell full of tap water.
  • The cell is treated as a plain capacitor. It is not — it conducts, and its ESR is a function of water conductivity and frequency. Use this to find the neighbourhood, not the setpoint.