If `E_(1) = 3 V ,E_(2) = 2V,E_(3) = 1V`
`R= R_(1) = R_(2) = R_(3) = 1Omega` , find the potential difference between the points A and B and current through each branch.

Explanation

Labelling the circuit diagram , we get

Here , A and B are connected .
So , `I = I_(1) + l_(2) + l_(3) = 0`
`l_(3) = - l_(1) - l_(2)`
From PMNQP,
` - I_(1)R_(1) + E_(2) - E_(3) + l_(3)R_(3) =0`
`rArr " " l_(2) xx 1 + 2 - 1 + l_(3) = 0`
`rArr" " l_(3) - l_(2) + 1 =0`
`rArr " " - l_(1) - l_(2) - l_(2) + 1=0`
`rArr " " 2l_(1) + l_(2) - l = 0`
From Eqs. (i) and (ii) , we get
`l_(2) = 0` , putting this value in Eq. (i) we get
`l_(1) = 0 = 1`
`rArr " "l_(1) = 1A`
Also , `l_(1) + l_(2) + l_(3) = 0`
`rArr " " l_(3) = - l _(1) - l_(2) = = - 10 = - 1A`