Question

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

Solution

Consider the mesh ABDA,
Now, Applying Kirchhoff’s loop rule we get,
– 10I₁ – 5I_g + (I – I₁) 5 = 0

\Rightarrow

3I₁ – I + I_g = 0 ...(i)

Consider the mesh BDCB,
Again, applying Kirchhoff’s loop rule we get,
– 5I_g – 10(1 – I_{l +} I_g) + 5(I₁ – I_g) = 0

\Rightarrow

3I₁ – 21 – 4I_g = 0 ...(ii)

Applying Kirchhoff’s loop rule to the mesh ABCEA,
– 10I₁ – 5(I₁ – I_g) –10I + 10 = 0
or 3I₁ + 2I – I_g = 2 ...(iii)

Equations (i), (ii) and (iii) are simultaneous equations. On solving these equations, we will find the unknown values of current.

Adding (i) and (iii), we get
6I₁ + I = 2 ...(iv)

Multiplying (i) by 4 and adding in (ii), we get
15I₁ – 6I0 = 0 ...(v)

Solving equations (iv) and (v), we get
I₁ =

\frac{4}{17} A = 0.235 A

So, current in branch AB is 0.235 A.

Putting the value of I₁ in equation (v) and simplifying, we get
Total current,

I = \frac{10}{17} = 0.588 A

Putting the values of I and I₁ in equation (iii) and simplifying, we get

I_{g} = \frac{2}{17} A = - 0.118 A

The negative sign indicates that the direction of current is opposite to that shown in Fig. above.
So, current in branch BD is – 0.118 A.

Current in branch BC is (I₁-I_g) i.e.,

\frac{4}{17} - (- \frac{2}{17})

i.e.,

\frac{6}{17} or 0.353 A .

Current in branch AD is (I – I₁)
i.e.,

(\frac{10}{17} - \frac{4}{17}) A

i.e.,

\frac{6}{17} A

or 0.353 A

Current in branch DC is (I₁ – I₁ + I_g)
i.e.,

\frac{6}{17} + (- \frac{2}{17}) A

\frac{4}{17} A

or 0.235 A .

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

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