The capacities of three capacitors are in the ratio of `1 : 2 : 3`. Their equivalent capacity in parallel is greater than their equivalent capacity in series by `60//11` `pF`. Calculate their individual capacitance.

Options

A. 4,6,7
B. 1,2,3
C. 2,3,4
D. 1,3,6

Answer

B

Explanation

Let, `C_(1) = C_(0), C_(2) = 2C_(0),C_(3) = 3C_(0)`
Their equivalent capacity, when they are connected in parallel
`C_(P) = C_(1) + C_(2) + C_(3) = 6C_(0)`
Their equivalent capacity, when they are connected in series
`((1)/(C_(S))) = (1)/(C_(0))[(1)/(1)+(1)/(2)+(1)/(3)] :. C_(S) = (6C_(0))/(11)`
`:. C_(P) - C_(S) = (60)/(11)C_(0)`
Given : `C_(p) - C_(S) = (60)/(11)`
From (i) and (ii), we get
`(60 C_(0))/(11) = (60)/(11)` or `C_(0) = 1`
Hence, `C_(1) = 1 mu F, C_(2) = 2 mu F, C_(3) = 3 mu F`