Derive an expression for the effective capacitance of three capacitors connected in series.

Explanation

In the series arrangement of capacitors, they are connected end to end and a cell is connected across their combination as shown in the figure.

Let `C_(1),C_(2),C_(3)` be the capacitances of the three capacitors connected in series and Q, the charge on each capacitor. Let `V_(1), V_(2), V_(3)` be the potential differences across the capacitors.
Now, charge = capacitance `xx` potential difference
`therefore Q=C_(1)V_(1) =C_(2)V_(2)=C_(3)V_(3)`
`therefore V_(1)=(Q)/(C_(1)), V_(2)=(Q)/(C_(2)) and V_(3)=(Q)/(C_(3))`
If V is the potential difference across the combination and C is the equivalent capacitance of the combination, we have,
`C=(Q)/(V)" " therefore V=(Q)/(C)`
But, `V=V_(1)+V_(2)+V_(3)`
`therefore (Q)/(C)=(Q)/(C_(1))+(Q)/(C_(2))+(Q)/(C_(3))`
`therefore (1)/(C)=(1)/(C_(1))+(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))`
[Note : In general, if n capacitances `C_(1),C_(2), C_(3), ..., C_(n)` are connected in series, the equivalent capacitance C of the combination is given by
`(1)/(C)=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))+...+(1)/(C_(n))]`