I'm trying to calculate the approximate leakage current for some capacitors I have. They are 5F 2.7v supercapacitors. I charged them up to 2.69v using a precision bench supply through a 68 ohm resister, then I disconnected them and left them disconnected. I found that their voltage dropped from 2.69 to 2.40 volts in 3.5 days. That's a 10% drop in 3.5 days, extrapolating to a 63% drop (1TC), it would be 22 days (1.9E6 sec).
I'm trying to calculate leakage current by using the Time Constant formula to calculate the resistance of the RC circuit, assuming the leakage current is roughly equivalent to the current through a resistor shorting out the capacitor.
The capacitor lost 10% of its voltage in 3.5 days, I extrapolate the time constant to 22 days to lose 63% of its voltage (it may not be linear, but I'm only need an approximation). 22 days is 1.9 x 10^6 sec.
Given T=RC, then R=T/C
With T=1,900,000 sec, C=5F, that's 1.9E6/5 gives R of 380,000 ohms.
Given E=IR, then I=E/R, that's 2.7v / 380k gives a leakage current of 7 uA.
Is this reasonable? Is my method correct?
I looked at this existing question/answer but I don't see how to apply it here.