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Question
Determine the current in each branch of the network shown in figure.
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, the potential is zero i.e.,
10 I2 + 5 I4 − 5 I3 = 0
2 I2 + I4 − I3 = 0
I3 = 2 I2 + I4 ....(1)
For the closed-circuit BCDB, the potential is zero i.e.,
5(I2 − I4) − 10(I3 + I4) − 5I4 = 0
5 I2 + 5 I4 − 10 I3 − 10 I4 − 5 I4 = 0
5 I2 − 10 I3 − 20 I4 = 0
I2 = 2 I3 + 4 I4 .....(2)
For the closed circuit ABCFEA, potential is zero i.e.,
−10 + 10 (I1) + 10 (I2) + 5(I2 − I4) = 0
10 = 15 I2 + 10 I1 − 5 I4
3 I2 + 2 I1 − I4 = 2 ....(3)
From equations (1) and (2), we obtain
I3 = 2(2 I3 + 4 I4) + I4
I3 = 4 I3 + 8 I4 + I4
−3 I3 = 9 I4
−3 I4 = + I3 ....(4)
Putting equation (4) in equation (1), we obtain
I3 = 2 I2 + I4
− 4 I4 = 2 I2
I2 = −2 I4 ....(5)
It is evident from the given figure that,
I1 = I3 + I2 ....(6)
Putting equation (6) in equation (1), we obtain
3 I2 + 2(I3 + I2) − I4 = 2
5 I2 + 2 I3 − I4 = 2 .....(7)
Putting equations (4) and (5) in equation (7), we obtain
5(−2 I4) + 2(−3 I4) − I4 = 2
− 10 I4 − 6 I4 − I4 = 2
17 I4 = −2
I4 = `(-2)/17"A"`
Equation (4) reduces to
I3 = −3(I4)
= `-3((-2)/17)`
= `6/17"A"`
I2 = −2(I4)
= `-2((-2)/17)`
= `4/17"A"`
`"I"_2 - "I"_4 = 4/17 - ((-2)/17)`
= `6/17"A"`
`"I"_3 + "I"_4 = 6/17 + ((-2)/17)`
= `4/17 "A"`
I1 = I3 + I2
= `6/17 + 4/17`
= `10/17"A"`
Therefore, current in branch AB = `4/17"A"`
In branch BC = `6/17"A"`
In branch CD = `(-4)/17"A"`
In branch AD = `6/17"A"`
In branch BD = `((-2)/17)"A"`
Total current = `4/17 + 6/17 + (-4)/17 + 6/17 + (-2)/17`
= `10/17"A"`
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