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Solution :

a. `I_(1)` to the left thorugh the 10V battery `I_(2)` to the right through <br> 5V battery , and `I_(3)` to the right thorugh to `10Omega` resistor. <br> upper loop: anticlockwise <br> `10V - (2Omega +3 Omega)I_(1)` <br> `-(1Omega + 4Omega)I_(2) - 5V = 0` <br> or `5V - (5Omega)I_(1) - (5Omega)I_(2) =0` <br> or `I_(1)+I_(2) =1A` <br> Lower loop : antilockwise <br> `5V + (1Omega +4Omega)I_(2) (10Omega)I_(3) = 0` <br> or `5V +(5Omega)I_(2) - (10Omega)I_(3) = 0` <br> or `I_(2) - 2I_(3) = -1A` <br> Along with `I_(1) = I_(2) +I_(3),` we can solve for th three curents <br> and find `I_(1) = 0.8A, I_(2) = 0.2 A, I_(3)= 0.6A` <br> b. `V_(ab) = -(0.2A)(4Omega) - (0.8A) (3Omega) = 3.2V`