NCERT Class XII Chapter
Current Electricity
Questions With Solutions
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Question : 9
Total: 24
Determine the current in each branch of the network shown in figure.
Solution:
Let us first distribute the current in different branches.
Now, equations for different loops using Kirchhoff’s II law,
Loop 1
ΣE = ΣIR
10 I 1 + 5 I g − 5 I 2
or2 I 1 + I g − I 2
Loop 2
ΣE = ΣIR
5 I g + 10 [ I 2 + I g ] − 5 [ I 1 − I g ]
10 I 2 + 20 I g − 5 I 1
or2 I 2 + 4 I g − I 1
Loop 3
5 I 2 + 10 ( I 2 + I g ) + 10 I
15 I 2 + 10 I g + 10 I
or3 I 2 + 2 I g + 2 I
Solving equations (i) and (ii)
2 I 1 + I g − I 2 [ − I i + 4 I g + 2 I 2 ]
or9 I g + 3 I 2 I 2 I g
In the loop ABCDA
10 I 1 + 5 [ I 1 − I g ] [ I 2 + I g ] 5 I 2
15 I 1 − 15 I 2 − 15 I g I 1 − I 2 − I g
Solving equations (ii) and (v)
2 I 2 + 4 I g − I 1 2 ( I 1 − I 2 − I g = 0 )
or2 I g + I 1 I 1 I g
Now using the result of (iv) and (vi) in equation (iii)
3 I 2 + 2 I g + 2 I 3 [ 3 I g ] + 2 I g 7 I g
Using Kirchhoff’s law, I =I 1 + I 2 5 I g
So, equation (vii)
2 [ − 5 I g ] − 7 I g − 17 I g
So, finallyI g
I 1
I 2
Current in branch AB =
Current in branch BD =−
Current in branch DC =
Now, equations for different loops using Kirchhoff’s II law,
Loop 1
ΣE = ΣIR
or
Loop 2
ΣE = ΣIR
or
Loop 3
or
Solving equations (i) and (ii)
or
In the loop ABCDA
Solving equations (ii) and (v)
or
Now using the result of (iv) and (vi) in equation (iii)
Using Kirchhoff’s law, I =
So, equation (vii)
So, finally
Current in branch AB =
Current in branch BD =
Current in branch DC =
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