Question

how will you arrange 3 capacitors each having the capacity of 2uf, to get a capacitor of capacity 3uf?​

Answer 1

We arrange one of the 2μF capacitors in parallel to the other two capacitors which will be arranged in series

Step-by-step explanation:

The number of capacitors in the network = 3

The capacitance of the each capacitor, C₁, C₂, and C₃ = 2 μF

The sum of capacitors in series = The inverse of the sum of the reciprocals of the capacitances of the capacitor

`C_(total \ series) = (1)/((1)/(C_a) +(1)/(C_b) )`

The sum of capacitances of capacitors arranged in parallel = The sum of the individual capacitances in parallel

`C_(total \ parallel)` = Cₐ + C
`_b`

By placing two of the capacitors in series, and the third in parallel to the first two, we get;

`C_(total) = C_(total \ series) + C_(total \ parallel)`

`C_(total \ series) =C_(1 \, and \, 2) = (1)/((1)/(2) +(1)/(2) ) = (1)/(1) = 1`

`C_(total \ parallel)` =
`C_(1 \, and \, 2)` + C₃

∴
`C_(total \ parallel)` = 1 μF + 2μF= 3 μF

Therefore, to get a capacitor of capacity of 3 μF from 3 capacitors of

2 μF, one of the capacitors is arranged in parallel across the other two capacitors arranged in series.