Example (2):

Find I₁, I₂ and I₃ in the network shown in Fig. , using loop-current method. Numbers against resistances indicate their values in ohms.

Solution:

Different loops would be taken one after another.

Loop ABCDA:

- 10 I₁ – 20(I₁ – I₂) –10 = 0

or         3 I₁ –2I₂ = – 1              …(i)

Loop BEFCB:

            40 – 20 I₂+ 10 – 10 (I₂ – I₃) – 20(I₂ – I₁) = 0

or         2 I₁ –5I₂ + I₃ = – 5                   …(ii)

Loop EGHFE:

- 10 I₃+50 –10(I₃ – I₂) – 10 = 0

or         I₂– 2 I₃= – 4                             …(iii)

1-  Multiplying equation (ii) by 2 and adding it to equation (iii), we get

            4 I₁ – 9 I₂= – 14                       …(iv)

Solving for I₁ and I₂ from equation (i) and (iv), we get:

            I₁ = 1A and      I₂ = 2A

Solving these values in equation (iii), we have, I₃=3A