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Question 9

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

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

I1 = Current flowing through the outer circuit

I2 = Current flowing through branch AB

I3 = Current flowing through branch AD

I2 - I4 = Current flowing through branch BC

I3 + I4 = Current flowing through branch CD

I4 = Current flowing through branch BD

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

10I2 + 5I4 - 5I3 = 0

2I2 + I4 - I3 = 0

I3 = 2I2 + I4 … (1)

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

5(I2 - I4) - 10(I3 +I4) - 5I4 = 0

5I2 + 5I4 - 10I3 - 10I4 - 5I4 = 0

5I2 - 10I3 - 20I4 = 0

I2 = 2I3 + 4I4 … (2)

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

- 10 + 10 (I1) + 10(I2) + 5(I2 - I4) = 0

10 = 15I2 + 10I1 - 5I4

3I2 + 2I1 - I4 = 2 … (3)

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

I3 = 2(2I3 + 4I4) + I4

I3 = 4I3 + 8I4 + I4

- 3I3 = 9I4

- 3I4 = + I3 … (4)

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

I3 = 2I2 + I4

- 4I4 = 2I2

I2 = - 2I4 … (5)

It is evident from the given figure that,

I1 = I3 + I2 … (6)

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

3I2 +2(I3 + I2) - I4 = 2

5I2 + 2I3 - I4 = 2 … (7)

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

5( - 2 I4) + 2( - 3 I4) - I4 = 2

- 10I4 - 6I4 - I4 = 2

17I4 = - 2

Equation (4) reduces to

I3 = - 3(I4)

Therefore, current in branch

In branch BC =

In branch CD =