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

Answer

Applying Kirchoff's second law in the branch ABCD
$\Rightarrow \quad-10 I _1-5 Ig +\left( I - I _1\right) 5=0$
$\Rightarrow \quad-10 I _1-5 Ig +5 I -5 I _1=0$
$\Rightarrow \quad-15 I _1-5 I g+5 I =0$
$3 I _1- I + I g=0$$\quad$.....(1)
Again, applying Kirchoff's second rule in the branch BDCB
$-5 Ig -10\left( I - I _1+ Ig \right)+5\left( I _1- Ig \right)=0$
$\Rightarrow \quad-5 Ig -10 I +10 I _1-10 Ig +5 I _1-5 Ig =0$
$\Rightarrow \quad 15 I _1-10 I -20 Ig =0$
or $\quad 3 I _1-2 I -4 Ig =0$$\quad$.....(2)
Image
Applying Kirchhoff"s second rule in branch ABCEA
$-10 I _1-5\left( I _1- Ig \right)-10 I +10=0$
$\Rightarrow \quad-10 I _1-5 I _1+5 Ig -10 I +10=0$
$\Rightarrow \quad-15 I_1-10 I+5 \lg =-10$
$\Rightarrow \quad 3 I _1+2 I - Ig =2$.....(3)
Adding equations (1) and (3).
$6 I_1+I=2$....(4)
Multiplying equation (1) by 4 and then on adding to equation (2)
$15 I_1-6 I=0$....(5)
On solving equations (4) and (5)
$I_1=\frac{4}{17} A$
Magnitude of current in $ABI ,=\frac{4}{17} A$
Putting value of $I_1$ in equation (5)
$15 \times \frac{4}{17}-6 I=0$
or$6 I=\frac{15 \times 4}{17}$
or$I=\frac{15 \times 4}{17 \times 6}=\frac{60}{102}=\frac{10}{7} A$
Putting the value of I and $I _1$ in equation (3)
$3 I _1+2 I - Ig =2$
$\Rightarrow \quad 3 \times \frac{4}{17}+\frac{2 \times 60}{102}- Ig =2$
$\Rightarrow \quad \frac{12}{17}+\frac{120}{102}- Ig =2$
$\Rightarrow \quad Ig =\frac{12}{17}+\frac{120}{102}-2$
$\Rightarrow \quad I _g=\frac{72+120-204}{102}$
$=\frac{-12}{102}=-\frac{2}{17} A$
Negative sign indicates that the direction of current is shown in opposite in the figure.
Therefore, direction of current in $BD = Ig =\frac{-2}{17} A$
Magnitude of current in $AB = I _1=\frac{4}{17} A$
Magnitude of current in $BC = I _1- Ig$
$=\frac{4}{17}-\left(\frac{-2}{17}\right)$
$=\frac{6}{17} A$
Magnitude of current in $AD = I - I _1$
$=\frac{10}{17}-\frac{4}{17}=\frac{6}{17} A$
Magnitude of current in $DC = I - I _1+ Ig$
$=\frac{16}{17}+\left(\frac{-2}{17}\right) A$ or $\frac{4}{17} A$
If DC is written as CD , then magnitude of current will be
$=-\frac{4}{17} A$
Total current $I=\frac{4}{17}+\frac{6}{17}=\frac{10}{17} A$

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