Three capacitors `2, 3` and `6 mF` are joined in series with each other. What is the minimum effective capacitance

Options

A. `(1)/(2)muF`
B. `1 muF`
C. `2 muF`
D. `3 muF`

Answer

B

Explanation

`(1)/(C_(eq)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3))`
`= (1)/(2)+(1)/(3)+(1)/(6)`
`= (3+2+1)/(6) = (6)/(6) = 1 muF`