The 5.00V- battery in the figure is removed from the circuit and replaced by a 20.00V- battery, with its negative terminal next to point b . The rest of the circuit is as shown in the figure
http://i39.tinypic.com/kd4isl.jpg
http://i39.tinypic.com/kd4isl.jpg
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10 - i1*(2+3) = 20-i2*(1+4) = (i1+i2) * 10
i2 = i1+2
10-i1*5 = 20*i1-20
i1 = -0.4
i2 = +1.6
ie upper current 0.4A to the right, middle 1.6A to the left, bottom 1.2A to the right
i2 = i1+2
10-i1*5 = 20*i1-20
i1 = -0.4
i2 = +1.6
ie upper current 0.4A to the right, middle 1.6A to the left, bottom 1.2A to the right