Question

Find the equivalent capacitance of FIVE capacitors each having 15.0 μF and connected: (a) all in series; (b) all in parallel.

Answer 1

(a) 3.0 μF

(b) 75.0 μF

Step-by-step explanation:

The equivalent capacitance when the capacitors are connected all in series is:

`(1)/(C_(T) ) =(1)/(C_(1) ) +(1)/(C_(2) ) +(1)/(C_(3) ) +(1)/(C_(4) ) +(1)/(C_(5) )`

Where
`C_(T)` is the equivalent capacitance,
`C_(1)`,
`C_(2)`,
`C_(3)`,
`C_(4)` and
`C_(5)` are the capacitance of each capacitors. In this case, they all have 15.0 μF. So we can replace as:

`(1)/(C_(T) ) =(1)/(15.0) +(1)/(15.0) +(1)/(15.0) +(1)/(15.0) +(1)/(15.0)`

`(1)/(C_(T) ) =(5)/(15.0)`

The solve for
`C_(T)`:

`C_(T)`=3.0 μF

The equivalent capacitance when the capacitors are connected all in parallel is:

`C_(T)=C_(1)+C_(2)+C_(3)+C_(4)+C_(5)`

Replacing values we get:

`C_(T)`=15.0 μF+15.0 μF+15.0 μF+15.0 μF+15.0 μF

`C_(T)`=75.0 μF