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Damping and ringdown, two ways

The same calculation run twice, side by side. Change anything on either side and the difference is shown output by output. This page is a link: both sets of numbers are in the address bar.

Side A open alone
mH

Everything inductive in the series loop.

nF

The cell.

Ω

Everything resistive in the loop — choke wire, cell ESR, wiring. This is what takes the energy out.

Side B open alone
mH

Everything inductive in the series loop.

nF

The cell.

Ω

Everything resistive in the loop — choke wire, cell ESR, wiring. This is what takes the energy out.

Swap sides

What changed

The two sides are the same. Change something on either.

Output A B B against A
ζ Damping ratio 0.05 0.05 same
τ Decay time constant 20 µs 20 µs same
d Ringing frequency 159 kHz 159 kHz same
m Cycles to 1 % energy 7.329 7.329 same

Side A, with its numbers

0.05=100 Ω21 nF1 mH20 µs=2·1 mH100 Ω159 kHz=12π11 mH·1 nF−120 µs27.329=ln10π·1100 Ω1 mH1 nF
LaTeX 
0.05 = \frac{100\,\mathrm{Ω}}{2} \sqrt{\frac{1\,\mathrm{nF}}{1\,\mathrm{mH}}} \qquad 20\,\mathrm{µs} = \frac{2 \cdot 1\,\mathrm{mH}}{100\,\mathrm{Ω}} \qquad 159\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{1\,\mathrm{mH} \cdot 1\,\mathrm{nF}} - \frac{1}{20\,\mathrm{µs}^{2}}} \qquad 7.329 = \frac{\ln 10}{\pi} \cdot \frac{1}{100\,\mathrm{Ω}} \sqrt{\frac{1\,\mathrm{mH}}{1\,\mathrm{nF}}}

Side B, with its numbers

0.05=100 Ω21 nF1 mH20 µs=2·1 mH100 Ω159 kHz=12π11 mH·1 nF−120 µs27.329=ln10π·1100 Ω1 mH1 nF
LaTeX 
0.05 = \frac{100\,\mathrm{Ω}}{2} \sqrt{\frac{1\,\mathrm{nF}}{1\,\mathrm{mH}}} \qquad 20\,\mathrm{µs} = \frac{2 \cdot 1\,\mathrm{mH}}{100\,\mathrm{Ω}} \qquad 159\,\mathrm{kHz} = \frac{1}{2\pi} \sqrt{\frac{1}{1\,\mathrm{mH} \cdot 1\,\mathrm{nF}} - \frac{1}{20\,\mathrm{µs}^{2}}} \qquad 7.329 = \frac{\ln 10}{\pi} \cdot \frac{1}{100\,\mathrm{Ω}} \sqrt{\frac{1\,\mathrm{mH}}{1\,\mathrm{nF}}}