Question

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

Answer 1

Current flowing through various branches of the circuit is represented in the given figure.

I₁ = Current flowing through the outer circuit

I₂ = Current flowing through branch AB

I₃ = Current flowing through branch AD

I₂ − I₄ = Current flowing through branch BC

I₃ + I₄ = Current flowing through branch CD

I₄ = Current flowing through branch BD

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

10 I₂ + 5 I₄ − 5 I₃ = 0

2 I₂ + I₄ − I₃ = 0

I₃ = 2 I₂ + I₄  ....(1)

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

5(I₂ − I₄) − 10(I₃ + I₄) − 5I₄ = 0

5 I₂ + 5 I₄ − 10 I₃ − 10 I₄ − 5 I₄ = 0

5 I₂ − 10 I₃ − 20 I₄ = 0

I₂ = 2 I₃ + 4 I₄  .....(2)

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

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

10 = 15 I₂ + 10 I₁ − 5 I₄

3 I₂ + 2 I₁ − I₄ = 2  ....(3)

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

I₃ = 2(2 I₃ + 4 I₄) + I₄

I₃ = 4 I₃ + 8 I₄ + I₄

−3 I₃ = 9 I₄

−3 I₄ = + I₃  ....(4)

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

I₃ = 2 I₂ + I₄

− 4 I₄ = 2 I₂

I₂ = −2 I₄  ....(5)

It is evident from the given figure that,

I₁ = I₃ + I₂  ....(6)

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

3 I₂ + 2(I₃ + I₂) − I₄ = 2

5 I₂ + 2 I₃ − I₄ = 2  .....(7)

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

5(−2 I₄) + 2(−3 I₄) − I₄ = 2

− 10 I₄ − 6 I₄ − I₄ = 2

17 I₄ = −2

I₄ = `(-2)/17"A"`

Equation (4) reduces to

I₃ = −3(I₄)

= `-3((-2)/17)`

= `6/17"A"`

I₂ = −2(I₄)

= `-2((-2)/17)`

= `4/17"A"`

`"I"_2 - "I"_4 = 4/17 - ((-2)/17)`

= `6/17"A"`

`"I"_3 + "I"_4 = 6/17 + ((-2)/17)`

= `4/17 "A"`

I₁ = I₃ + I₂

= `6/17 + 4/17`

= `10/17"A"`

Therefore, current in branch AB = `4/17"A"`

In branch BC = `6/17"A"`

In branch CD = `(-4)/17"A"`

In branch AD = `6/17"A"`

In branch BD = `((-2)/17)"A"`

Total current = `4/17 + 6/17 + (-4)/17 + 6/17 + (-2)/17`

= `10/17"A"`