a)
Apply KVL in the upper loop,
12 - 21*I1 + 12 - 11*I2 = 0
21*I1 + 11*I2 = 24 ---(1)
Apply KVL in the lower loop,
12 - 11*I2 + 34*I3 - 6 = 0
11*I2 - 34*I3 = 6 --(2)
At the junction, I1 = I2 + I3 --(3)
on solving the equations 1,2 and 3
we get
I1 = 0.769 A
I2 = 0.714 A
I3 = 0.055 A
b) V(6)_terminal = 6 - I3*r
= 6 - 0.055*1
= 5.945 volts
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