In the given circuit values are as follows
`epsi=2V,epsi_(2)=4V,R_(1)=1Omega` and `R_(2)=R_(3)=1Omega`.
Calculate the current through `R_(1),R_(2)` and `R_(3)`

Explanation

Let `i_(1),i_(2)` are current across `R_(1)` and `R_(2)`
`(i_(1)+i_(2))` is current across `R_(2)`.
Their direction are taken as shown
From Kirchhoff's second law for `AGFBA` loop
`i_(1)R_(1)-(i_(1)+i_(2))R_(2)+E_(1)=0,i_(1)+i_(201)+i_(2)=2`
`2i_(1)+i_(2)=2to(1)`
From Kirchhoff's second law of `BCDEB` loop
`-i_(2)R_(3)-(i_(1)+i_(2))R_(2)+E_(2)=0,i_(2)+i_(1)+i_(2)=4`
`i_(2)+2i_(2)=4to(2)`
solving equation (1) and (2) we get `i_(1)=0A,i_(2)=2A`
Thus current across `R_(1)` is 0 while across `R_(3)` and `R_(2)` and 2A each.