Calculate the capacitance of a supercapacitor
You can also solve the capacitance of a capacitor using the capacitor charge equation.

\$V = V_p{e^{-t/RC}}\$

Based on what you can see in the equation, you have to wire the resistor and the capacitor in parallel, and then put a voltmeter in the capacitor. Then, all you have to do is to have a supply voltage (Vp), the resistor value (R), find the voltage of the capacitor (V) at time (t), and then you'll get the Capacitance of the capacitor. Assuming that you have a very accurate resistor reading, an accurate voltage source that does not sag when a load is connected to it, and an accurate voltmeter, given time (t), you will get an accurate enough value of the capacitor.



Here's an example:



Let's say we have an unknown capacitor just like in the schematic. I grab my stopwatch, and start it the moment the switch is closed. I will then check the Voltmeter and take note of the time that passed to reach the capacitor voltage to, let's say 5 Volts. The capacitor reached this voltage in roughly 14 minutes and 30 seconds. Converting that to seconds and we will get 870 seconds. So we now have, V = 5V, Vp = 12V, t=870s, and R which is 20 Ohms. Plugging that into the equation will give me 49.69 F, rounding it up will give me 50 F.

Since you're dealing with a super capacitor here, charging times will be large enough to get a good reading as what I have demonstrated in the example above. However, if you need to measure a small capacitor, you can use a larger resistor that will charge the capacitor slower to prevent sub-second stopwatch readings, since we humans are limited to our ~20ms response time :)