# Determine the current in each branch of the network shown in fig 3.30:

Question:

Determine the current in each branch of the network shown in fig 3.30:

Solution:

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) − 5I= 0

5I2 + 5I4 − 10I3 − 10I− 5I= 0

5I2 − 10I3 − 20I= 0

I2 = 2I3 + 4I… (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

10 = 15I2 + 10I− 5I4

3I2 + 2I− I4 = 2 … (3)

From equations (1) and (2), we obtain

I3 = 2(2I+ 4I4) + I4

I3 = 4I+ 8I4 + I4

− 3I3 = 9I4

− 3I4 = + I… (4)

Putting equation (4) in equation (1), we obtain

I3 = 2II4

− 4I4 = 2I2

I2 = − 2I4 … (5)

It is evident from the given figure that,

I1 = II2 … (6)

Putting equation (6) in equation (1), we obtain

3I2 +2(II2) − I4 = 2

5I2 + 2I− I4 = 2 … (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2 I4) + 2(− 3 I4) − I4 = 2

− 10I4 − 6I− 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 $=\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17} \mathrm{~A}$