Determine the current in each branch of the network shown in Fig. 3.30:

The current through the labelled diagram shows the current flowing the respective branches:

For the closed circuit ABDA, potential is zero i.e.,

10I₂ + 5I₄ − 5I₃ = 0

2I₂ + I₄ −I₃ = 0

I₃ = 2I₂ + I₄ ... (1)

For the closed circuit BCDB, potential is zero i.e.,

5(I₂ − I₄) − 10(I₃ + I₄) − 5I4 = 0

5I₂ + 5I₄ − 10I₃ − 10I₄ − 5I₄ = 0

5I₂ − 10I₃ − 20I₄ = 0

I₂ = 2I₃ + 4I₄ ... (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

10 = 15I₂ + 10I₁ − 5I₄

3I₂ + 2I₁ − I₄ = 2 ... (3)

From equations (1) and (2), we obtain

I₃ = 2(2I₃ + 4I₄) + I₄

I₃ = 4I₃ + 8I₄ + I₄

− 3I₃ = 9I₄

− 3I₄ = + I₃ ... (4)

Putting equation (4) in equation (1), we obtain

I₃ = 2I₂ + I₄

− 4I₄ = 2I₂

I₂ = − 2I₄ ... (5)

It is evident from the given figure that,

I₁ = I₃ + I₂ ... (6)

Putting equation (6) in equation (1), we obtain

3I₂ + 2(I₃ + I₂) − I₄ = 2

5I₂ + 2I₃ − I₄ = 2 ... (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2 I₄) + 2(− 3 I₄) − I₄ = 2

− 10I₄ − 6I₄ − I₄ = 2

17I₄ = − 2

I₄ = -2/17Ampere

Equation (4) reduces to

I₃ = − 3(I₄)

I₃ = -3 × -2/17

I₃ = 6/17Ampere

I₂ = -2(I4)

I₂ = -2 × -2/17

I₂ = 4/17Ampere

I₂ - I₄ = 6/17Ampere

I₃ + I₄ = 6/17Ampere

I₁ = I₂ + I₃ = 4/17 + 6/17 = 10/17 Ampere

Therefore, current in branch AB = 4/17Ampere

In branch BC = 6/17Ampere

In branch CD = -4/17Ampere

In branch AD = 6/17Ampere

In branch BD = -2/17Ampere

Total current = 4/17 + 6/17 + -4/17 + 6/17 + -2/17 = 10/17Ampere.