Three resistor 2 `Omega , 4 Omega and 6 Omega` are combined In parallel. What is the total resistance of the combination ?

Explanation

If equivalent resistance of parallel connection of given resistances is `R_(p)`, then according to law,
`(1)/(R_(p)) = (1)/(R_(1)) + (1)/(R_(2)) + (1)/(R_(3))`
= `(1)/(2) + (1)/(4) + (1)/(5)`
`= (10 + 5 + 4)/(20) `
`therefore (1)/(R_(p)) = (19)/(20)`
`therefore R_(p) = (20)/(19) Omega`