A current of 2 A gets divided into `I_1 and I_2` through resistances `20 Omega and 30 Omega`, respectively connected as shown in the figure. Calculate `I_1 and I_2`.

Explanation

Given, current, `I_(1) + I_(2) = 2A" "…(i)`
As two resistors are in parallel combination, so potential drop across them will be same.
i.e. `V_1 = V_2` or `I_1 xx 20 = I_2 xx 30`
`therefore I_(1) = (3)/(2) I_(2)`
By substituting `I_(1)` in Eq. (i), we get
`((3)/(2) + 1) I_(2) = 2`
`therefore I_(2) = 2 xx (2)/(5) = (4)/(5) = 0.8 A`
`therefore I_(1) = 2 - 0.8 =1.2 A`