A technican has only two capacitors. By using them singly, in series kor in parallel, he is able to obtain th capacitane of 4,5,20 and 25 micro farad. What are the capacitance of the two capacitors ?

Explanation

In series grouping the resultant capacitance is minimum, i.e, `C_(s) = (C_(1) C_(2))/(C_(1) + C_(2)) = 4mu F`
In parallel grouping, the resultant capacitane is maximum, i.e, `C_(p) = C_(1) + C_(2) = 25 muF`
It implies that the remaining values `5 muF` and `20 muF` represent the individual capacitance values of two capcitors.