Figure 8. Quiz: The three ideal capacitors are charged to the voltages shown. What is the final voltage across the bank after closing S1 and then S2? What is the final voltage across the bank if the order of switch closings is reversed?
If a capacitor, C, is charged to a voltage, V, the electric charge stored is given by Q = CV.
With the switches open, the individual capacitor charges are 1uF x 1V = 1uC, 1uF x 3V = 3uC, and 2uF x 6V = 12uC.
The total stored charge is therefore 1uC + 3uC + 12uC = 16uC.
After you close the switches, regardless of the order, the total charge is still 16uC. The equivalent capacitance of the three parallel capacitors is 1uF + 1uF + 2uF = 4uF. Therefore the final voltage across the bank is V = Q/C = 16uC/4uF = 4V.
It doesn't matter what order you close the two switches.
This assumes the capacitors are ideal and have no losses.
Note that the "C" in the unit "uC" stands for "coulomb". It's a little confusing working with equations where "C" by itself is capacitance.