61(b)]
For the closed loop BCDEB, we get
5 – 2 (I2 + I3) – 2 (I2 + I3 – I1) = 0
that is, 2I1 – 4I2 – 4I3 = –5
[3 61(c)]
Equations (3 61 a, b, c) are three simultaneous equations in three
unknowns These can be solved by the usual method to give
I1 = 2