Determine the current in each branch of the network shown in fig 3.30:
Current flowing through various branches of the circuit is represented in the given figure.
I1 = Current flowing through the outer circuit
I2 = Current flowing through branch AB
I3 = Current flowing through branch AD
I2 − I4 = Current flowing through branch BC
I3 + I4 = Current flowing through branch CD
I4 = Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
$10 l_{2}+5 l_{4}-5 l_{3}=0$
$2 I_{2}+I_{4}-I_{3}=0$
$I_{3}=2 I_{2}+I_{4} \ldots$ (1)
For the closed circuit BCDB, potential is zero i.e.,
5(I2 − I4) − 10(I3 + I4) − 5I4 = 0
5I2 + 5I4 − 10I3 − 10I4 − 5I4 = 0
5I2 − 10I3 − 20I4 = 0
I2 = 2I3 + 4I4 … (2)
For the closed circuit ABCFEA, potential is zero i.e.,
−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0
10 = 15I2 + 10I1 − 5I4
3I2 + 2I1 − I4 = 2 … (3)
From equations (1) and (2), we obtain
I3 = 2(2I3 + 4I4) + I4
I3 = 4I3 + 8I4 + I4
− 3I3 = 9I4
− 3I4 = + I3 … (4)
Putting equation (4) in equation (1), we obtain
I3 = 2I2 + I4
− 4I4 = 2I2
I2 = − 2I4 … (5)
It is evident from the given figure that,
I1 = I3 + I2 … (6)
Putting equation (6) in equation (1), we obtain
3I2 +2(I3 + I2) − I4 = 2
5I2 + 2I3 − I4 = 2 … (7)
Putting equations (4) and (5) in equation (7), we obtain
5(−2 I4) + 2(− 3 I4) − I4 = 2
− 10I4 − 6I4 − I4 = 2
17I4 = − 2
$I_{4}=\frac{-2}{17} \mathrm{~A}$
Equation (4) reduces to
$I_{3}=-3\left(I_{4}\right)$
$=-3\left(\frac{-2}{17}\right)=\frac{6}{17} \mathrm{~A}$
$I_{2}=-2\left(I_{4}\right)$
$=-2\left(\frac{-2}{17}\right)=\frac{4}{17} \mathrm{~A}$
$I_{2}-I_{4}=\frac{4}{17}-\left(\frac{-2}{17}\right)=\frac{6}{17} \mathrm{~A}$
$I_{3}+I_{4}=\frac{6}{17}+\left(\frac{-2}{17}\right)=\frac{4}{17} \mathrm{~A}$
$I_{1}=I_{3}+I_{2}$
$=\frac{6}{17}+\frac{4}{17}=\frac{10}{17} \mathrm{~A}$
Therefore, current in branch $\mathrm{AB}=\frac{4}{17} \mathrm{~A}$
In branch $B C=\frac{6}{17} A$
In branch $C D=\frac{-4}{17} A$
In branch $A D=\frac{6}{17} A$
In branch $B D=\left(\frac{-2}{17}\right) A$
Total current $=\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17} \mathrm{~A}$