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

Explanation

In a parallel combination of capacitors, the capacitors are connected across a cell. Hence, the potential difference (V) across each each capacitor is the same. Let `C_(1), C_(2), C_(3)` be the capacitances of three capacitors connected in parallel and C be the equivalent capacitance of the combination. The charge Q supplied by the cell is distributed as `Q_(1), Q_(2) and Q_(3)` on the capacitors.

Since, capacitance `=("charge")/("potential difference" ')`
`C_(1)=(Q_(1))/(V), C_(2)=(Q_(2))/(V), C_(3)=(Q_(3))/(V) and C=(Q)/(V)`
`therefore Q_(1)=C_(1)V, Q_(2)=C_(2)V, Q_(3)=C_(3)V and Q=CV`
Now, `Q=Q_(1)+Q_(2)+Q_(3)`
`therefore CV=C_(1)V+C_(2)V+C_(3)V`
`therefore C=C_(1)+C_(2)+C_(3)`
[Note: In general if n capacitors of capacitances `C_(1), C_(2), ..., C_(n)` are connected in parallel, the equivalent capacitance of the equivalent capacitance of the combination is `C=C_(1)+C_(2)+..+C_(n)`]