Find the currents going through the three resistors `R_(1),R_(2)` and `R_(3)` in the circuit of figure.

Explanation

Lets us take the potential of the point A to be zero. The potential at C will be (epsilon)_(1)and that at D will be `(epsilon)_(2)`.Let the potential at B be V . The currents through the three resistors are `i_(1),i_(2)and i_(1)+i_(2)`as shown in figure. Note that the current directed towards B equals the current directed away from B.
Applying Ohm's law to the three resistors `R_(1),R_(2)and R_(3),`
we get,
`(epsilon)_(1)-V=R_(1)i_(1) ...(i)`
`(epsilon)_(2)-V=R_(2)i_(2) ...(ii)`
`and V-0=R_(3)(i_(1)+I_(2)). ...(iii)`
Adding (i) and (iii),
`(epsilon)_(1)=R_(1)i_(1)+R_(3)(i_(1)+i_(2))`
`=(R_(1)+R_(3))i_(1)+R_(3)i_(2) ...(iv)`
`and adding (ii)and (iii)`
`(epsilon)_(2)=R_(2)i_(2)+R_(3)(i_(1)+I_(2))`
`=(R_(2)+R_(3))i_(2)+R_(3)i_(1). ...(v)`
Equations (iv) and (v) may be directly written from Kirchhoff's loop applied to the left half and the right half of the circuit.
Multiply (iv)by `(R_(2)+R_(3)), (v) by R_(3) and subtract to eliminate i_(2)`.This gives
`i_(1)=((epsilon_(1)(R_(2)+R_(3))-(epsilon_(2)R_(3))/(R_(1)+R_(3))(R_(2)+R_(3))-R_(3)^(2)).`
`((epsilon_1)(R_(2)+R_(3))-(epsilon_2)R_(3))/(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1)).`
Similarly, eliminating `i_(1)`from (iv) and (v) we obtain,
`i_(2)=((epsilon_2)(R_(1)+R_(3))-(epsilon_1)R_(3))/(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1))`
`And so`
`i_(1)+i_(2)=((epsilon_(1))R_(2)+(epsilon_(2)R_(1))/(R_(1)R_(2)+R_(2)R_(3)+R_(3)R_(1)).`