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Academic session 2026-27, Chapter 3 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Physics - Chapter Current Electricity. This page offers a step-by-step solution to the specific question from Exercise 1, Question 9:
Determine the current in each branch of the network shown in figure

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

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

Equation (4) reduces to
I3 = - 3(I4)

Therefore, current in branch 
In branch BC = 
In branch CD = 
In branch AD 
In branch BD = 
Total current = 10/17 A
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 9: Determine the current in each branch of the network shown in figure....
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