Question

Determine the current in each branch of the network shown in figure.

Answer 1

Let us first distribute the current in different branches.

Now, equations for different loops using Kirchhoff’s second law,

Loop 1

∑E = ∑IR

10I₁ + 5I_g − 5I₂ = 0 

or 2I₁ + I_g − I₂ = 0     ...(i)

Loop 2

∑E = ∑IR

5I_g + 10[I₂ + I_g] − 5[I₁ − I_g] = 0

5I_g + 10I₂ + 10I_g − 5I₁ + 5I_g = 0

20I_g +10I₂ − 5I₁ = 0

or 4I_g + 2I₂ − I₁ = 0     ...(ii)

Loop 3

5I₂ + 10(I₂ + I_g) + 10I = 10

5I₂ + 10I₂ + 10I_g + 10I = 10

15I₂ + 10I_g + 10I = 10

or 3I₂ + 2I_g + 2I = 2     ...(iii)

Solving equations (i) and (ii)

2I₁ + I_g − I₂ = 0

[−I₁ + 4I_g + 2I₂ = 0]2

or 9I_g + 3I₂ = 0

or I₂ = −3I_g     ...(iv)

[−I₁ + 4I_g + 2I₂ = 0]2

or 9I_g + 3I₂ = 0 

or I₂ = −3I_g    ...(iv)

In the loop ABCDA

10I₁ + 5[I₁ − I_g] − 10[I₂ + I_g] − 5I₂ = 0

10I₁ + 5I₁ − 5I_g − 10I₂ − 10I_g − 5I₂ = 0

15I₁ −15I₂ −15I_g  = 0

or I₁ − I₂ − I_g = 0     ...(v)

Solving equations (ii) and (v)

2I₂ + 4I_g − I₁ = 0

or 2(I₁ − I₂ − Ig = 0)

or 2I_g + I₁ = 0

or I₁ = −2I_g    ...(vi)

Now using the result of (iv) and (vi) in equation (iii)

3I₂ + 2I_g + 2I = 2

−3[3I_g] + 2I_g + 2I = 2

−9I_g + 2I_g + 2I = 2

or 2I − 7I_g = 2    ...(vii)

Using Kirchhoff’s law, I = I₁ + I₂

I = −5I_g

So, equation (vii)

2[−5I_g] − 7I_g = 2

−10I_g − 7I_g = 2

or −17I_g = 2

So, finally I_g = `−2/17 A and I = (+10)/17 A`

Also `I_1 = 4/17 A`, `I_2 = 6/17 A`

Current in branch AB = `4/17 A`

Current in branch AD = `6/17 A`

Current in branch BD = `(-2)/17 A`

Current in branch BC = `6/17 A`

Current in branch DC = `4/17 A`