Question

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

Answer 1

*The current flowing through various branches of the network is shown in the upper figure*

Let ${\bf{I_{1}}}$ be the current flowing through the outer circuit.

Let ${\bf{I_{2}}}$ be the current flowing through AB branch.

Let ${\bf{I_{3}}}$ be the current flowing through AD branch.

Let ${\bf{I_{2} – I_{4}}}$ be the current flowing through branch BC.

Let $\bf{I_{3} + I_{4}}$ be the current flowing through branch BD.

Let us take closed-circuit ABDA into consideration, we know that potential is zero.

i.e , ${\bf{ : \implies 10 I_{2} + 5 I_{4} – 5 I_{3} = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀$\bf { : \implies 2 I_{2} + I_{4} – I_{3} = 0}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀${\bf{\red{ : \implies I_{3} = 2 I_{2} + l_{4}……….(1)}}}$

Let us take closed circuit BCDB into consideration, we know that potential is zero.

i.e , $\bf{ : \implies 5 ( I_{2} – I_{4} ) – 10 ( I_{3} + I_{4} ) – 5 I_{4} = 0}$

⠀⠀⠀⠀⠀⠀⠀⠀ $\bf{ : \implies 5 I_{2} + 5 I_{4} – 10 I_{3} – 10 I_{4} – 5 I_{4} = 0}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\bf{ : \implies 5 I_{2} – 10 I_{3} – 20 I_{4} = 0}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ $\bf{\red{ : \implies I_{2} = 2 I_{3} – 4 I_{4}……….(2)}}$

Let us take closed-circuit ABCFEA into consideration, we know that potential is zero.

i.e , ${\bf{ : \implies – 10 + 10 ( I_{1} ) + 10 ( I_{2} ) + 5 ( I_{2} – I_{4} ) = 0}}$

⠀⠀⠀⠀⠀⠀⠀⠀ ${\bf{ : \implies 10 = 15 I_{2} + 10 I_{1} – 5 I_{4}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ${\red{\bf{ : \implies 3 I_{2} + 2 I_{2} – I_{4} = 2}……….(3)}}$

From equation ( 1 ) and ( 2 ), we have :

⠀⠀⠀⠀⠀⠀⠀⠀ ${\bf{ : \implies I_{3} = 2 ( 2 I_{3} + 4 I_{4} ) + I_{4}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀${\bf{ : \implies I_{3} = 4 I_{3} + 8 I_{4} + I_{4}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies – 3 I_{3} = 9 I_{4}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\red{\bf{ : \implies – 3 I_{4} = + I_{3}}……….(4)}$

Putting equation ( 4 ) in equation ( 1 ) , we have :

⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies I_{3} = 2 I_{2} + I_{4}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies – 4 I_{4} = 2 I_{2}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{\red{ : \implies I_{2} = – 2 I_{4}……….(5)}}$

From the above equation , we infer that :

| ${\overline{\underline{\bf{\red{ : \implies I_{1} = I_{3} + I_{2} ……….(6)}}}}}$ |

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

⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies 3 I_{2} + 2 ( I_{3} + I_{2} ) – I_{4} = 2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\red{\bf{ : \implies 5 I_{2} + 2 I_{3} – I_{4} = 2}……….(7)}$

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

⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies 5 ( – 2 I_{4} ) + 2 ( – 3 I_{4} ) – I_{4} = 2 }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies – 10 I_{4} – 6 I_{4} – I_{4} = 2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies 17 I_{4} = – 2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\boxed{\bf{ : \implies l_{ 4 } = \frac{–2}{ 17} A}}$

Equation ( 4 ) reduces to

$\bf{ : \implies I_{3} = – 3 ( I_{4} )}$

⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies I_{ 3 } = – 3 ( \frac{ – 2 }{ 17 } ) }$

$\bf{ : \implies I_{ 2 } = – 2( l_{4}) }$

⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies I _{ 2 } – I _{ 4 } = \frac{ 4 }{ 17 } – \frac{ – 2 }{ 17 } = \frac{ 6 }{ 17 } A}$

$\bf{ : \implies I _{ 3 } + I _{ 4 } = \frac{ 6 }{ 17 } – \frac{ – 2 }{ 17 } }$

⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies I _{ 3 } + I _{ 4 } = \frac{10}{17} A}$

Therefore, current in each branch is given as :

In branch $\bf{ AB = \frac{ 4 }{ 17 } A}$

In branch $\bf{ BC = \frac{ 6 }{ 17 } A}$

In branch $\bf{ CD = \frac{ – 4 }{ 17 } A}$

In branch $\bf{AD = \frac{ 6 }{ 17 } A}$

In branch $\bf{ BD = \frac{ – 2 }{ 17 } A}$

Total current $\bf{ \large = \frac{ 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 2 }{ 17}}$

$\bf{\red{ \Large ⇨ \: \frac{ 10 }{ 17 } A }}$

Answer 2

*The current flowing through various branches of the network is shown in the upper figure*

Let {\bf{I_{1}}}I

1

be the current flowing through the outer circuit.

Let {\bf{I_{2}}}I

2

be the current flowing through AB branch.

Let {\bf{I_{3}}}I

3

be the current flowing through AD branch.

Let {\bf{I_{2} – I_{4}}}I

2

–I

4

be the current flowing through branch BC.

Let \bf{I_{3} + I_{4}}I

3

+I

4

be the current flowing through branch BD.

Let us take closed-circuit ABDA into consideration, we know that potential is zero.

i.e , {\bf{ : \implies 10 I_{2} + 5 I_{4} – 5 I_{3} = 0}}:⟹10I

2

+5I

4

–5I

3

=0

⠀⠀⠀⠀⠀⠀⠀⠀\bf { : \implies 2 I_{2} + I_{4} – I_{3} = 0}:⟹2I

2

+I

4

–I

3

=0

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀{\bf{\red{ : \implies I_{3} = 2 I_{2} + l_{4}……….(1)}}}:⟹I

3

=2I

2

+l

4

……….(1)

Let us take closed circuit BCDB into consideration, we know that potential is zero.

i.e , \bf{ : \implies 5 ( I_{2} – I_{4} ) – 10 ( I_{3} + I_{4} ) – 5 I_{4} = 0}:⟹5(I

2

–I

4

)–10(I

3

+I

4

)–5I

4

=0

⠀⠀⠀⠀⠀⠀⠀⠀ \bf{ : \implies 5 I_{2} + 5 I_{4} – 10 I_{3} – 10 I_{4} – 5 I_{4} = 0}:⟹5I

2

+5I

4

–10I

3

–10I

4

–5I

4

=0

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ \bf{ : \implies 5 I_{2} – 10 I_{3} – 20 I_{4} = 0}:⟹5I

2

–10I

3

–20I

4

=0

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ \bf{\red{ : \implies I_{2} = 2 I_{3} – 4 I_{4}……….(2)}}:⟹I

2

=2I

3

–4I

4

……….(2)

Let us take closed-circuit ABCFEA into consideration, we know that potential is zero.

i.e , {\bf{ : \implies – 10 + 10 ( I_{1} ) + 10 ( I_{2} ) + 5 ( I_{2} – I_{4} ) = 0}}:⟹–10+10(I

1

)+10(I

2

)+5(I

2

–I

4

)=0

⠀⠀⠀⠀⠀⠀⠀⠀ {\bf{ : \implies 10 = 15 I_{2} + 10 I_{1} – 5 I_{4}}}:⟹10=15I

2

+10I

1

–5I

4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ {\red{\bf{ : \implies 3 I_{2} + 2 I_{2} – I_{4} = 2}……….(3)}}:⟹3I

2

+2I

2

–I

4

=2……….(3)

From equation ( 1 ) and ( 2 ), we have :

⠀⠀⠀⠀⠀⠀⠀⠀ {\bf{ : \implies I_{3} = 2 ( 2 I_{3} + 4 I_{4} ) + I_{4}}}:⟹I

3

=2(2I

3

+4I

4

)+I

4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀{\bf{ : \implies I_{3} = 4 I_{3} + 8 I_{4} + I_{4}}}:⟹I

3

=4I

3

+8I

4

+I

4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies – 3 I_{3} = 9 I_{4}}:⟹–3I

3

=9I

4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\red{\bf{ : \implies – 3 I_{4} = + I_{3}}……….(4)}:⟹–3I

4

=+I

3

……….(4)

Putting equation ( 4 ) in equation ( 1 ) , we have :

⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies I_{3} = 2 I_{2} + I_{4}}:⟹I

3

=2I

2

+I

4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies – 4 I_{4} = 2 I_{2}}:⟹–4I

4

=2I

2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\bf{\red{ : \implies I_{2} = – 2 I_{4}……….(5)}}:⟹I

2

=–2I

4

……….(5)

From the above equation , we infer that :

| {\overline{\underline{\bf{\red{ : \implies I_{1} = I_{3} + I_{2} ……….(6)}}}}}

:⟹I

1

=I

3

+I

2

……….(6)

|

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

⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies 3 I_{2} + 2 ( I_{3} + I_{2} ) – I_{4} = 2}:⟹3I

2

+2(I

3

+I

2

)–I

4

=2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\red{\bf{ : \implies 5 I_{2} + 2 I_{3} – I_{4} = 2}……….(7)}:⟹5I

2

+2I

3

–I

4

=2……….(7)

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

⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies 5 ( – 2 I_{4} ) + 2 ( – 3 I_{4} ) – I_{4} = 2 }:⟹5(–2I

4

)+2(–3I

4

)–I

4

=2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies – 10 I_{4} – 6 I_{4} – I_{4} = 2}:⟹–10I

4

–6I

4

–I

4

=2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies 17 I_{4} = – 2}:⟹17I

4

=–2

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\boxed{\bf{ : \implies l_{ 4 } = \frac{–2}{ 17} A}}

:⟹l

4

=

17

–2

A

Equation ( 4 ) reduces to

\bf{ : \implies I_{3} = – 3 ( I_{4} )}:⟹I

3

=–3(I

4

)

⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies I_{ 3 } = – 3 ( \frac{ – 2 }{ 17 } ) }:⟹I

3

=–3(

17

–2

)

\bf{ : \implies I_{ 2 } = – 2( l_{4}) }:⟹I

2

=–2(l

4

)

⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies I _{ 2 } – I _{ 4 } = \frac{ 4 }{ 17 } – \frac{ – 2 }{ 17 } = \frac{ 6 }{ 17 } A}:⟹I

2

–I

4

=

17

4

–

17

–2

=

17

6

A

\bf{ : \implies I _{ 3 } + I _{ 4 } = \frac{ 6 }{ 17 } – \frac{ – 2 }{ 17 } }:⟹I

3

+I

4

=

17

6

–

17

–2

⠀⠀⠀⠀⠀⠀⠀⠀\bf{ : \implies I _{ 3 } + I _{ 4 } = \frac{10}{17} A}:⟹I

3

+I

4

=

17

10

A

Therefore, current in each branch is given as :

In branch \bf{ AB = \frac{ 4 }{ 17 } A}AB=

17

4

A

In branch \bf{ BC = \frac{ 6 }{ 17 } A}BC=

17

6

A

In branch \bf{ CD = \frac{ – 4 }{ 17 } A}CD=

17

–4

A

In branch \bf{AD = \frac{ 6 }{ 17 } A}AD=

17

6

A

In branch \bf{ BD = \frac{ – 2 }{ 17 } A}BD=

17

–2

A

Total current \bf{ \large = \frac{ 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 4 }{ 17 } + \frac{ 6 }{ 17 } + \frac{ – 2 }{ 17}}=

17

4

+

17

6

+

17

–4

+

17

6

+

17

–2

\bf{\red{ \Large ⇨ \: \frac{ 10 }{ 17 } A }}⇨

17

10

A