Question 3.7: For the circuit in Fig. 3.24, find i1 to i4 using mesh analy...

Note that meshes 1 and 2 form a supermesh since they have an independent current source in common. Also, meshes 2 and 3 form another supermesh because they have a dependent current source in common. The two supermeshes intersect and form a larger supermesh as shown. Applying KVL to the larger supermesh,

2 i_{1}+4 i_{3}+8\left(i_{3}-i_{4}\right)+6 i_{2}=0

or

i_{1}+3 i_{2}+6 i_{3}-4 i_{4}=0                   (3.7.1)

For the independent current source, we apply KCL to node P:

i_{2}=i_{1}+5              (3.7.2)

For the dependent current source, we apply KCL to node Q:

i_{2}=i_{3}+3 i_{o}

But i_{o}=-i_{4}, hence,

i_{2}=i_{3}-3 i_{4}                   (3.7.3)

Applying KVL in mesh 4,

2 i_{4}+8\left(i_{4}-i_{3}\right)+10=0

or

5 i_{4}-4 i_{3}=-5                       (3.7.4)

From Eqs. (3.7.1) to (3.7.4),

i_{1}=-7.5  A , \quad i_{2}=-2.5  A , \quad i_{3}=3.93  A , \quad i_{4}=2.143  A