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

#### Solution

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

I_{1} = Current flowing through the outer circuit

I_{2} = Current flowing through branch AB

I_{3} = Current flowing through branch AD

I_{2} − I_{4} = Current flowing through branch BC

I_{3} + I_{4} = Current flowing through branch CD

I_{4} = Current flowing through branch BD

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

10 I_{2} + 5 I_{4} − 5 I_{3} = 0

2 I_{2} + I_{4} − I_{3} = 0

I_{3} = 2 I_{2} + I_{4} ….......(1)

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

5(I_{2} − I_{4}) − 10(I_{3} + I_{4}) − 5I_{4 }= 0

5 I_{2} + 5 I_{4} − 10 I_{3} − 10 I_{4 }− 5 I_{4 }= 0

5 I_{2} − 10 I_{3} − 20 I_{4 }= 0

I_{2} = 2 I_{3} + 4 I_{4 }…......(2)

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

−10 + 10 (I_{1}) + 10 (I_{2}) + 5(I_{2} − I_{4}) = 0

10 = 15 I_{2} + 10 I_{1 }− 5 I_{4}

3 I_{2} + 2 I_{1 }− I_{4} = 2 ….........(3)

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

I_{3} = 2(2 I_{3 }+ 4 I_{4}) + I_{4}

I_{3} = 4 I_{3 }+ 8 I_{4} + I_{4}

−3 I_{3} = 9 I_{4}

−3 I_{4 }= + I_{3 }…..........(4)

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

I_{3} = 2 I_{2 }+ I_{4}

− 4 I_{4} = 2 I_{2}

I_{2} = −2 I_{4}^{ }…..........(5)

It is evident from the given figure that,

I_{1} = I_{3 }+ I_{2}^{ }….........(6)

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

3 I_{2} + 2(I_{3 }+ I_{2}) − I_{4} = 2

5 I_{2} + 2 I_{3 }− I_{4} = 2 …........(7)

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

5(−2 I_{4}) + 2(−3 I_{4})_{ }− I_{4} = 2

− 10 I_{4} − 6 I_{4 }− I_{4} = 2

17 I_{4} = −2

I_{4} = `(-2)/17"A"`

Equation (4) reduces to

I_{3} = −3(I_{4})

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

= `6/17"A"`

I_{2} = −2(I_{4})

= `-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"`

I_{1} = I_{3} + I_{2}

= `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"`