Question 15 MarksDetermine the current in each branch of the network shown in Fig.:AnswerApplying 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$$\quad$.....(2)Applying Kirchhoff"s second rule in branch ABCEA$-10 I _1-5\left( I _1- Ig \right)-10 I +10=0$$\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 $I=\frac{4}{17}+\frac{6}{17}=\frac{10}{17} A$View full question & answer→
