In the circuit shown in Fig. the emf of the sources is equal to `xi = 5.0 V` and the resistances are equal to `R_(1) = 4.0 Omega` and `R_(2) = 6.0 Omega`. The internal resistance of the source equals `R = 1.10 Omega`. Find the currents flowing through the resistances `R_(1)` and `R_(2)`.

Explanation

Let us make the current distribution, as shwon in the figure.
Current `i = (xi)/(R + (R_(1) R_(2))/(R_(1) + R_(2)))` (Using loop rule)
So, current through the resister `R_(1)`,
`i = (xi)/(R + (R_(1) R_(2))/(R_(1) + R_(2))) (R_(2))/(R_(1) + R_(2))`
`= (xi R_(2))/(R R_(1) + R R_(2) + R_(1) R_(2)) = 1.2 A`
and similarly, current through the resistance `R_(2)`
`i = (xi)/(R + (R_(1) R_(2))/(R_(1) + R_(2))) (R_(1))/(R_(1) + R_(2)) = (xi R_(1))/(R R_(1) + R_(1) R_(2) + R R_(2)) = 0.8 A`