Question

Determine the current in each branch of the network shown in figure:

Answer 1

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.,
10I2 + 5I4 − 5I3 = 0
2I2 + I4 −I3 = 0
I3 = 2I2 + I4 … (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
I4 =$\frac{-2}{17} A$

Equation (4) reduces to
I3 = − 3(I4)
I3 = $-3(\frac{-2}{17}) = \frac{6}{17} A$
I2 = -2(I4)
I2 = $-2(\frac{-2}{17}) =\frac{ 4}{17} A$
I2-I4 = $\frac{4}{17} - (\frac{-2}{17}) =\frac{ 6}{17} A$
I3+I4 = $\frac{6}{17}+(\frac{-2}{17}) =\frac{ 4}{17} A$
I1= I3 +I2
I1 = $\frac{6}{17} +\frac{ 4}{17} = \frac{10}{17} A$

Therefore,
current in branch AB =$\frac{4}{17} A$
current in branch BC = $\frac{6}{17} A$
current in branch CD = ($\frac{-4}{17}$) A
current in branch AD = $\frac{6}{17} A$
current in branch BD = ($\frac{-2}{17}$) A

Total Current =$\frac{4}{17} + \frac{6}{17} + (\frac{-4}{17}) + \frac{6}{17} + (\frac{-2}{17})$
$= \frac{10}{17} A$