Answer :

The current flowing through the network’s numerous branches is depicted in the diagram below:

Let I ₁ denote the current flowing through the outer circuit

Let I ₂ denote the current flowing through AB branch

Let I ₃ denote the current flowing through AD branch

Let I ₂ – I ₄ denote the current flowing through branch BC

Let I ₃ + I ₄ denote the current flowing through branch DC

Consider the closed-circuit ABDA, we know that potential is zero, therefore –

10 I ₂ + 5 I ₄ – 5 I ₃ = 0

Or, 2 I ₂ +  I ₄ –  I ₃ = 0

I ₃ = 2 I ₂ +  I ₄                                                                                  . . . . . . . . eq ( 1 )

Consider the closed circuit BCDB, since we know that potential is zero, we have-

5 ( I ₂ – I ₄ ) – 10 ( I ₃ + I ₄ ) – 5 I ₄ = 0

Or, 5 I ₂ – 5 I ₄ – 10 I ₃ – 10 I ₄ –  5 I ₄ = 5 I ₂ – 10 I ₃ –  20 I ₄ = 0

I ₂ = 2 I ₃ – 4 I ₄                                                                             . . . . . . .eq ( 2 )

Consider the closed-circuit ABCFEA, since we know that potential is zero –

– 10 + 10 ( I ₁ ) + 10 ( I ₂ ) + 5 ( I ₂ –  I ₄ ) = 0

Or, 10 = 15 I ₂ + 10 I ₁ – 5 I ₄

Therefore, we have –

3 I ₂ + 2 I ₂ – I ₄ = 2                                                                       . . . . . . . .  eq ( 3 )

Using equation ( 1 )  and ( 2 ), we get:

I ₃ = 2 ( 2 I ₃ + 4 I ₄ ) + I ₄ =  4 I ₃ + 8 I ₄ + I ₄

– 3 I ₃ = 9 I ₄

Or,   – 3 I ₄ = + I ₃                                                                          . . . . . . . . eq ( 4 )

Substituting equation  ( 4 ) in equation ( 1 ) , we get:

I ₃ = 2 I ₂ + I ₄

Or,  – 4 I ₄ = 2 I ₂

I ₂ = – 2 I ₄                                                                                 . . . . . . . . .eq ( 5 )

Using the above equation , we conclude that :

I ₁ = I ₃ + I ₂                                                                                   . . . . . . . .  .eq ( 6 )

Substituting equation ( 4 ) in equation ( 1 ) , we get –

3 I ₂ + 2 ( I ₃ + I ₂ ) – I ₄ = 2

Or, 5 I ₂ + 2 I ₃ – I ₄ = 2                                                                . . . . . . . . eq ( 7 )

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

5 ( – 2 I ₄ ) + 2 ( – 3 I ₄ ) – I ₄ = 2

Or, – 10 I ₄ –  6 I ₄ – I ₄ = 2

17 I ₄ = – 2

Therefore, we obtain –

\[{{I}_{4}}=\frac{-2}{17}A\]

Further, equation ( 4 ) reduces to

I ₃ = – 3 ( I ₄ )

Therefore, we obtain –

\[{{I}_{3}}=-3\left( \frac{-2}{17} \right)A=\frac{6}{17}A\]

Similarly, we have ——— I ₂ = – 2 ( I ₄ )

\[{{I}_{3}}=-2\left( \frac{-2}{17} \right)A=\frac{4}{17}A\]

Again, from the above equation we have –

\[{{I}_{2}}-{{I}_{4}}=\frac{4-(-2)}{17}=\frac{6}{17}A\]

Similary,

\[{{I}_{3}}+{{I}_{4}}=\frac{6-(-2)}{17}=\frac{4}{17}A\]

Also,

\[{{I}_{1}}={{I}_{3}}+{{I}_{2}}=\frac{6+4}{17}=\frac{10}{17}A\]

Therefore, current in each branch is given as :

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

The sum of all these currents give us the total current, which is 10/17 A.