Long Answer Type Questions [5M]

Question

Determine the current in each branch of the network shown in Fig.:

Vidyadip · Accepted Answer

Consider the mesh ABDA,

Now, Applying Kirchhoff's loop rule we get,

$-10\text{I}_1-5\text{I}_\text{g}+(\text{I}-\text{I}_1)5=0$

$\Rightarrow\ \ 3\text{I}_1-\text{I}+\text{I}_\text{g}=0\ \ ...\text{(i)}$

Consider the mesh BDCB,

Again, applying Kirchhoff's loop rule we get,

$-5\text{I}_\text{g}-10(1-\text{I}_\text{l}+\text{I}_\text{g})+5(\text{I}_\text{I}-\text{I}_\text{g})=0$

$\Rightarrow\ \ 3\text{I}_1-2\text{I}-4\text{I}_\text{g}=0\ \ ...\text{(ii)}$

Applying Kirchhoff's loop rule to the mesh ABCEA,

$-10\text{I}_1-5(\text{I}_1-\text{I}_\text{g})-10\text{I}+10=0$

$\text{or}\ \ 3\text{I}_1+2\text{I}-\text{I}_\text{g}=2\ \ ...{(\text{iii})}$

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

$6\text{I}_\text{1}+\text{I}=2\ \ ...(\text{iv})$

Multiplying (i) by 4 and adding in (ii), we get

$15\text{I}_\text{1}-6\text{I}0=0\ \ ...\text{(v)}$

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

$\text{I}_1=\frac{4}{17}\text{A}=0.235\ \text{A}$

So, current in branch AB is 0.235 A.

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

Total current, $\text{I}=\frac{10}{17}=0.588\ \text{A}$

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

$\text{I}_\text{g}=\frac{2}{17}\text{A}=-0.118\ \text{A}$

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 $(\text{I}_1-\text{I}_\text{g})\text{i.e}.,\frac{4}{17}-\Big(-\frac{2}{17}\Big)$

$\text{i.e.,}\frac{6}{17}\ \text{or}\ 0.353\ \text{A}.$

Current in branch AD is (I - I₁)

$\text{i.e.,}\Big(\frac{10}{17}-\frac{4}{17}\Big)\text{A i.e.,}\ \frac{6}{17}\text{A or}\ 0.353\ \text{A}$

Current in branch DC is (I₁ - I₁ + I_g)

$\text{i.e.,}\frac{6}{17}+\Big(-\frac{2}{17}\Big)\text{A or}\ \frac{4}{17}\text{A or}\ 0.235\ \text{A}.$

Answer