For the resistor network shown in the potential drop between a and b s 12V.
a. Current through resistance of `6Omega` is
b. Current through resistance `2 Omega` is
c. Current through resistance of `8 Omega` is

Explanation

`V_(ab) = I_1(6+(4xx6)/(4+6)) = I_1(8.4) (i)`
`V_(ab) I_2(4 +(8xx8)/(8+8)) = I_2 (8) (ii)`
a. From Eq. (i) we get
`I_= (12)/(8.4) (10)/(7)A`
`I_(1)` is the current in the `6Omega` resistor.
b. `I_(2,4) = (4)/(6+4)xx(10)/(7) =(4)/(7)A`
c. From eq (ii) we get `I_(2) = 3//2 A,` so current in the `4Omega` resistacne
is 3//2A. It will be equally dicided inot `8Omega` resistance
which are in parallel. Therefore,
`I_(8) = (3)/(2xx2) = (3)/(4)A`