Determine the current in each branch of the network shown in figure 3.30:

By Shivank Goel|Updated : August 17th, 2022

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

The current flowing through the various branches of the circuit is shown in the figure

Determine the current in each branch of the network shown in figure_1

I1 is the current flowing through the outer circuit

I2 is the current flowing through the branch AB

I3 is the current flowing through the branch AD

I2 - I4 is the current flowing through the branch BC

I3 + I4 is the current flowing through the branch CD

I4 is the current flowing through the branch BD

The potential is zero for the closed circuit ABDA

10I2 + 5I4 - 5I3 = 0

Dividing the equation by 5

2I2 + I4 - I3 = 0

I3 = 2I2 + I4 .... (1)

The potential is zero for the closed circuit BCDB

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

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

5I2 - 10I3 - 20I4 = 0

I2 = 2I3 + 4I4 .... (2)

The potential is zero for the closed circuit ABCFEA

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

10 = 15I2 + 10I1 - 5I4

3I2 + 2I1 - I4 = 2 .... (3)

Using the equation (1) and (2)

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

I3 = 4I3 + 8I4 + I4

-3I3 = 9I4

-3I4 = I3 .... (4)

Substituting equation (4) in (1)

I3 = 2I2 + I4

-4I4 = 2I2

I2 = -2I4 .... (5)

From the figure, it is evident that

I1 = I3 + I2 .... (6)

Substituting equation (6) in (1)

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

5I2 + 2I3 - I4 = 2 .... (7)

Substituting equations (4) and (5) in (7)

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

-10I4 -6I4 - I4 = 2

17I4 = -2

I4 = -2/17 A

So the equation (4) reduces to

I3 = -3I4 = -3 (-2/17) = 6/17 A

I2 = -2I4 = -2 (-2/17) = 4/17 A

I2 - I4 = 4/17 - (-2/17) = 6/17 A

I3 + I4 = 6/17 + (-2/17) = 4/17 A

I1 = I3 + I2 = 6/17 + 4/17 = 10/17 A

In branch BC = 6/17 A, CD = 14/17 A, AD = 6/17 A and BD = -2/17 A

So the total current at C = (I2 - I4) + (I3 + I4) = 6/17 + 4/17 = 10/17

Therefore, the current in each branch is BC = 6/17 A, CD = 14/17 A, AD = 6/17 A and BD = -2/17 A.

Summary:

Determine the current in each branch of the network shown in figure 3.30

The current in each branch of the network is BC = 6/17 A, CD = 14/17 A, AD = 6/17 A and BD = -2/17 A.

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