In the circuit shown in Fig, the sources have `emf's xi_(1) = 1.5 V, xi_(2) = 2.0 V, xi_(3) = 2.5 V` , and the resistances are equal to `R_(1) = 10 Omega, R_(2) = 20 Omega, R_(3) = 30 Omega`. The internal resistances of the sources are negligible. Find : ltbrtgt (a) the current flowing thorugh teh resistance `R_(1)` ,
(b) a potential differences `varphi_(A) - varphi_(B)` between the points `A` and `B` between the points `A` and `B`.

Explanation

At first indicate the currents in the branches using charge conservation (which also inclued the point rule).
In the loops 1 `BA` 61 and `B34AB` from teh loop rule, `-Delta varphi = 0`, we get respectively
`-xi_(2) + (i - i_(1)) R_(2) + xi_(1) - i, R_(i) = 0` ...(1)
`i R_(3) + xi_(3) - (i - i_(2)) R_(2) + xi_(2) = 0` ....(2)
On Solving Eqs. (1) and (2), we obtain
`i_(1) = ((xi_(1) - xi_(2)) R_(3) + R_(2) (xi_(1) + xi_(3)))/(R_(1) R_(2) + R_(2) R_(3) + R_(3) R_(1)) = 0.06 A`
Thus `varphi_(A) - varphi_(B) = xi_(2) - i_(2) R_(2) = 0.9 V`