When a current of `0.5 A` is passed through two resistance in series, the potential difference between the ends of the series arrangement is `12.5V` .On connecting them in parallel and passing the current of `1.5 A`, the potential difference between their ends is 6 V, calculate the two resistances.

Answer

`20 Omega `, ` 5 Omega`

Explanation

`R_(s) = (R_(1) +R_(2)) = 12.5//0.5 = 25`……(i)
`R_(P) = (R_(1)R_(2))/((R_(1) + R_(2))) = (6)/(1.5) = 4`……(ii)
On solving (i) and (ii) , we get
`R_(1) = 20 Omega , R_(2) = 5 Omega`