Question 9

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

Answer

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, potential is zero i.e.,

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

2I_{2} + I_{4} - I_{3} = 0

I_{3} = 2I_{2} + I_{4} … (1)

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

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

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

5I_{2} - 10I_{3} - 20I_{4} = 0

I_{2} = 2I_{3} + 4I_{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 = 15I_{2} + 10I_{1} - 5I4

3I_{2} + 2I_{1} - I_{4} = 2 … (3)

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

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

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

- 3I_{3} = 9I_{4}

- 3I_{4} = + I_{3} … (4)

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

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

- 4I_{4} = 2I_{2}

I_{2} = - 2I_{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

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

5I_{2} + 2I_{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

- 10I_{4} - 6I_{4} - I_{4} = 2

17I_{4} = - 2

Equation (4) reduces to

I_{3} = - 3(I_{4})

Therefore, current in branch

In branch BC =

In branch CD =

In branch AD

In branch BD =

Total current = 10/17 A

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- NCERT Chapter

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