Kirchhoff’s second rule for the closed loop ADCA gives,
10 – 4(I1– I2) + 2(I2 + I3 – I1) – I1 = 0
[3 61(a)]
that is, 7I1– 6I2 – 2I3 = 10
For the closed loop ABCA, we get
10 – 4I2– 2 (I2 + I3) – I1 = 0
that is, I1 + 6I2 + 2I3 =10
[3 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