Find the potential differnce across `R_(2)` in the circuit

Answer

`=(E_(1) R_(2) + E_(1) R_(3) - E_(2) R_(3))/(R_(1)R_(2) + R_(2) R_(3) + R_(1)R_(3))`

Explanation

Let the current through vation arms as per kirchhoff's first rule be

Using kirchhoff's voltage rule in the closed circuit `DABCD`
`(l_(1) + l_(2)) R_(3) + l_(1) R_(1) = E_(1)`
`l_(1) (R_(1) + R_(2)) + l_(2)R_(3) = E_(1)`....(i)
In the closed `ABFEA`
`(l_(1) + l_(2)) R_(3) + l_(2) R_(2) = E_(2)`
`l_(1) (R_(3) + l_(2)) + (R_(2)+ R_(3)) = E_(2)`....(ii)
Multipliying (i) by `(R_(2) + R_(3))` and (iii) by `R_(3)` and subrancting we get
`l_(1) = (E_(1) R_(2) + E_(1) R_(3) - E_(2) R_(3))/(R_(1)R_(2) + R_(2) R_(3) + R_(1)R_(3))`
potential difference across `R_(3) = (l_(1) + l_(3)) R_(3)`