There are infinite number of a capacitance each of capacitance `1 muF`. They are connected in rows, such that number of capacitors in first, second, third rows are respectively `1,2,4,8 ………,` The rows of these capacitors are connected between `A` and `B` as shown in Fig. Determine teh equivalent capacitance of the network between points `A` and `B`

Explanation

Let `C_(1) , C_(2), C_(3)`, …….. be the capacitance of first, second, third row ….. Respectively. Then
`C_(1) = 1 muF`
`C_(2) = (1xx1)/(1+1) = (1)/(2) muF`
`(1)/(C_(3)) = (1)/(1) + (1)/(1) + (1)/(1) + (1)/(1) = 4`
`C_(3) = (1)/(4) muF`
Similarly `C_(4) = (1)/(8) muF` and so on,
So, equivalent capacitance capacitance `'C'` as they are connected in parallel is
`C = C_(1) + C_(2) + C_(3)` + .......
`= 1 + (1)/(2) + (1)/(4)` + ......
= sum of geometrical progression
`C = (a)/(1-r) = (1)/(1- 1//2) = 2 muF`