If 3 capacitors of values 1, 2 and 3 `mu`F are available, the maximum and minimum values of capacitances one can obtain by different combinations of the 3 capacitors together are respectively

Options

A. `(6/5) mu` F
B. `(6 / 11) mu` F
C. `(11 / 6) mu` F
D. `(11/ 8) mu` F

Answer

(b)

Explanation

Maximum capacitance obtained in parallel order
`C_(max) = C_(1) + C_(2) + C_(3) = 1 + 2 + 3 = 6 mu`F
Mimimum capacitance will be ,
`1/C_(min) = 1/C_(1) + 1/C_(2) + 1/C_(3) = 1/1 + 1/2 + 1/3 = 11/6`
`:. C_(min) = 6/11 mu`F