If three resistors of resistance `2Omega, 4Omega` and `5 Omega` are connected in parallel then the total resistance of the combination will be

Options

A. `(20)/(19)Omega`
B. `(19)/(20)Omega`
C. `(10)/(20)Omega`
D. `(29)/(10)Omega`

Answer

A

Explanation

The resistance `2 Omega, 4 Omega` and `5 Omega` are connected in parallel combination. Therefore, resultant resistance is given by
`(1)/(R )=(1)/(2)+(1)/(4)+(1)/(5)=(19)/(20) rArr R = (20)/(19) Omega`