This time you add a 100 μF capacitor in series to the experiment 4. For this build the shown circuit.
Observe again the time it is required to load and to discharge the two capacitors. You notice that the LEDs light up half as long as only a 100 μF capacitor.
For two in series-connected capacitors, the total capacity Ctotal calculates with...
1 / Ctotal = 1 / C1 + 1 / C2
You can rewrite the formula to get...
Ctotal = ( C1 * C2 ) / ( C1 + C2 ) = ( 100 μF * 100 μF ) / ( 100 μF + 100 μF ) = 50 μF
The total capacity of a series circuit is therefore smaller than any of the individual capacities.