Find the current flowing through the resistance `R_(1)` of the circuit shown in Fig. If the resistances are equal to `R_(1) = 10 Omega. R_(2) = 20 Omega` and `R_(3) = 30 Omega`, and the potential of points 1,2 and 3 are equal to `varphi_(1) = 10 V, varphi_(2) = 6V`, and `varphi_(3) = 5 V`

Explanation

Let us distribute the currents in the paths as shown in the figure. Now, `varphi_(1) - varphi_(2) = i R_(1) + i_(1) R_(2)` ....(1)
and `varphi_(1) - varphi_(3) = i R_(1) + (i - i_(1)) R_(3)` ....(2)
Simpliflying Eqs. (1) and (2) we get
`i = (R_(3) (varphi_(1) - varphi_(2)) + R_(2) (varphi_(1) - varphi_(3)))/(R_(1) R_(2) + R_(2) R_(3) + R_(3) R_(1)) = 0.2 A`