Question
Determine the current in each branch of the network shown in Fig.:
✓
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 $I_1$ in equation (v) and simplifying,
we get Total current, $\text{I}=\frac{10}{17}=0.588\ \text{A}$
Putting the values of I and $I_1$ 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 $(I - I_1)$
$\text{i.e.,}\Big(\frac{10}{17}-\frac{4}{17}\Big)\text{A i.e.,}\ \frac{6}{17}\text{A or}\ 0.353\ \text{A}$
Current in branch DC is $ (I_1 - I_1 + I_g)$
$\text{i.e.,}\frac{6}{17}+\Big(-\frac{2}{17}\Big)\text{A or}\ \frac{4}{17}\text{A or}\ 0.235\ \text{A}.$
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 $I_1$ in equation (v) and simplifying,
we get Total current, $\text{I}=\frac{10}{17}=0.588\ \text{A}$
Putting the values of I and $I_1$ 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 $(I - I_1)$
$\text{i.e.,}\Big(\frac{10}{17}-\frac{4}{17}\Big)\text{A i.e.,}\ \frac{6}{17}\text{A or}\ 0.353\ \text{A}$
Current in branch DC is $ (I_1 - I_1 + I_g)$
$\text{i.e.,}\frac{6}{17}+\Big(-\frac{2}{17}\Big)\text{A or}\ \frac{4}{17}\text{A or}\ 0.235\ \text{A}.$
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