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.