1.2V to 0.75V via 100Ω resister at ~40mA.
It's currently taking 4 seconds to charge and around 14 seconds to discharge
C=Ic dt/dV 10 * Tau=10* RC = 4s ( to 99% ) (est.)
Cch = 356 mF = 0.04A * 4s/0.45V , R = 10* Tau/C = 10 * 4s / 0.356 V = 110 with 100 in series. ESR = ? 10 ohms ? , ESR*C = Tau= 3.56 s to 63% target
Cdis = 14/4 * 0.356 = 14/4*.356 = 1.2 Farads
Double-electric layers have multiple ESR*C in parallel with short term and long term Tau's. With low current rates, they merge as one capacitance with an A-h or A-s value at constant current. C*dV=Ic*dt [A-s]
The sure sign of an undercharged battery is the rapid rise of ESR. A permanently damaged battery is similar but worse in that the ESR does not improve much due to the thickness of oxidation on the electrode-electrolyte interface. The oxidation rate increases when undercharged.
The best way to eliminate oxidation is to pulse high current without heat. Pavg <0.1W Yet heat damages the battery. Heat is the normal way to detect fully charged using a thermistor near the anode (?) to latch off a charger.
This would look like an inductive flyback pulse with the current battery polarity modulated by a vary low repetition rate to maintain a low avg. power dump
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This also creates a high enough voltage to bridge tiny slivers or dendrites in the electrolyte and possibly evaporate them. This can be detected by the change in flyback voltage. You can also use a very narrow current limited pulse from a slightly higher voltage such as 5V then monitor the change in average current using an RC filter to measure the average voltage rise.