Find the branch currents, i_a - i_e =

i_d = I₂ - I₁

Next express the value of the dependent current source in terms of mesh currents.

4.3 i_d = I₃ - I₁

Combining the above two equations yields:

4.3 (I₂ - I₁) = I₃ - I₁

- 3.3 I₁ + 4.3 I₂ - I₃ = 0 (Eq. 1)

Removing the dependent current source leaves one supermesh (1,3) and one regular mesh (2).

KVL around mesh 2.

25 I₂ - 25 I₁ + 50 I₂ - 50 I₃ + 10 I₂ - 200 = 0

- 25 I₁ + 85 I₂ - 50 I₃ = 200 (Eq. 2)

KVL around supermesh 1-3.

100 I₃ + 50 I₃ - 50 I₂ + 25 I₁ - 25 I₂ + 10 I₁ = 0

35 I₁ - 75 I₂ + 150 I₃ = 0 (Eq. 3)

Solving the three equations yields:

I₁ = 5.7 A

I₂ = 4.6 A

I₃ = 970 mA

Therefore the branch currnets are:

i_a = I₁ = 5.7 A

i_b = I₂ = 4.6 A

i_c = I₃ = 970 mA

i_d = I₂ - I₁ = - 1.1 A

i_e = I₂ - I₃ = 3.63 A