Two electrical conducting wires are connected in parallel. The resultant resistance of the combination is `0.5 Omega`. One of these has a resistance of `1 Omega`. The resistance of the other is

Options

A. `0.5 Omega`
B. `1 Omega`
C. `1.5 Omega`
D. `2 Omega`

Explanation

(b) As, `R_(P) = 0.6 Omega, R_(1) =1 Omega`
So, resistance, `(1)/( R_P) = (1)/( R_1) + (1)/( R_2)`
`rArr (1)/( R_P) = (1)/( R_1) + (1)/( R_2)`
`rArr (1)/( R_2) = (1)/( R_P) - (1)/( R_1) = (1)/(0.5) - (1)/(1) =2 -1 =1`
`therefore R_1 =1 Omega`