Connect three capacitors of `3 muF, 3 muF` and `6 muF` such that their equivalent capacity becomes `5 muF`

Explanation

We know that, capacitors when connected in parallel give maximum capacity and in series given minimum capacity.
`therefore` In parallel, `C_P = 3 + 3 + 6 = 12 muF` (Maximum) and in series:
`1/C_(s) =1/3 + 1/3 + 1/6 =(2+2+1)/6 = 5/6`
or `C_(s) = 6/5 muF = 1.2 muF` (Minimum)
But required capacity is `5 muF`, let y it lies between maximum and minimum. Let it be C.
`therefore C = 5 muF =(3+2) muF`
or `C = 3muF +(1/(1/3 + 1/6)) muF`
`= 3muF + (1muF)/((2+1)/6)`
`=3 muF + 2muF`
Thus^.capacitors of 3 uF and 6 uF should be connected in series and another capacitor of 3 pF should be in parallel as shdwn (Fig.).