Considering the mesh ABDA, we have
100I1 + 15Ig – 60I2 = 0
or 20I1 + 3Ig – 12I2 = 0 ——(a)
Considering the mesh BCDB, we have
10 (I1 – Ig) – 5 (I2 + Ig) – 15Ig = 0
10I1 – 30Ig – 5I2 = 0
2I1 – 6Ig – I2 = 0 ——(b)
Considering the mesh ADCEA,
60I2 + 5 (I2 + Ig) = 10
65I2 + 5Ig = 10
13I2 + Ig = 2 ——(c)
Multiplying Eq. (b) by 10
20I1 – 60Ig – 10I2 = 0 ——(d)
From Eqs. (d) and (a) we have
63Ig – 2I2 = 0
I2 = 31.5Ig ——(e)
Substituting the value of I2 into Eq. [(c)], we get
13 (31.5Ig ) + Ig = 2
410.5 Ig = 2
Ig = 4.87 mA.