Find the current in every branch ?

Explanation

The above question could be solved by assuming potential x & y at the top junctions & zero potential at lower junctions At the junction 1 applying KCL,
`i_(1)+i_(2)+i_(3)=0`
`(x-4)/(2)+(x-2)/(2)+(x-y)/(2)=0`
`rArr3x-y=6` …. (1)

At the junction 2 applying KCL,
`i_(4)+i_(5)+i_(6)=0`
`(y-x)/(2)+(y-2)/(2)+(y)/(2)=0`
`rArr3y-x=2` ... (2)
Solving (1) & (2)
`9x-3y=18`
`rArr3y-x=2`
`rArr8x=20`
`x=(5)/(2),y=(3)/(2)`
Just put the values of x & y & then the evaluate the current in every branch anticlockwise direction , starting from point A , we have
`underset("In the opposite direction to current")("+IR")+underset("In the opposite direction to current")("Ir")- underset("From positive to negative terminal")("E")=`

Hence , `I=(E)/(R+r)`