In the given circuit values are as follows
`epsi_1 = 2V, epsi_2 = 4V, R_1 = 1 Omega and R_2 = R_3 = 1 Omega`
Calculate the currents through `R_1 , R_2 and R_3`

Explanation

Potential difference across BE is equal to the e.m.f of `E_1`.
i) The current through `R_1 ` is zero because the resultant p.d across AG= 0 . The p.d. across BE due to `E_2 = -2 Omega` volts and `E_1 = +2` volts, hence , p.d. across AG= +2 - 2 =0
ii) p.d. across `R_3 =-2` volts.
`i_3 =(-2)/(R_3) = (-2)/1 = -2A`

iii) p.d. across BE = -2 volts.
`i_2 = (-2)/(R_2) = (-2)/(1) = - 2A`
Note :m You may solve this using Kirchoff.s Law as usual.