Using Kirchoff's rules, calculate the current through the `40Omega" and "20Omega` resistors in the following circit :

Explanation

Applying K.V.L. on loop ABCDA
`+80-20(I_(1)-I_(2))+40I_(2)=0`
`"or, "80-20I_(1)+60I_(2)=0`
`"or, "+20I_(1)-60I_(2)=80`
`"or, "I_(1)-3I_(2)=4" ...(i)"`
Applying K.V.L. on loop DCFED

`-40I_(2)-10I_(1)+40=0`
`10I_(1)+40I_(2)=40`
`I_(1)+4I_(2)=4" ...(2)"`
`I_(1)+4I_(2)=4" ...(2)"`
`I_(1)+4I_(2)=4" ...(2)"`
`-I_(1)-3I_(2)=4" ...(1)"`
`-+ -`
`7I_(2)=0`
`I_(2)=0`
Substituting `I_(2)=0` in equation no (1).
`I_(1)=4Amp`
Current in Arm AB `(20 Omega)`
`=I_(1)-I_(2)=4-0=4 Amp`
Current in DC `(40 Omega)`
`=I_(2)-0 Amp`