Three capacitors are connected in series. The equivalent capacitance of this combination is 3.18 µF. Two of the individual capacitances are 6.54 µF and 8.19 µF. What is the third capacitance (in µF)?

Answer 1

If 1/C1+1/C2+1/C3=1/Ceq and Ceq=3.18 mF, then replace the given variables as such:
1/6.54+1/8.19+1/C:unknown=1/3.18

Now it becomes an algebra problem:
Multiply both sides of the eqn by 3.18, then you'll have:

3.18/6.54 +3.18/8.18+3.18/Cunknown=1

Simplify:
.49+.39+3.18/C=1
1-(.49+.39)=3.18/C
.12=3.18/C

Solve for C:
C unknown= 3.18/.12= ~ 25 microF