Figure shows the network of resistance . Name the circuit so formed . What is the curent flowing in the arm bd of this circuit ? State the two laws to fimd the current in different branches of circuit
Answers
Answer:
Current flowing through various branches of the circuit is represented in the given figure.
I
1
= Current flowing through the outer circuit
I
2
= Current flowing through branch AB
I
3
= Current flowing through branch AD
I
2
−I
4
= Current flowing through branch BC
I
3
+I
4
= Current flowing through branch CD
I
4
= Current flowing through branch BD
For the closed circuit ABDA, potential is zero i.e.,
10I
2
+5I
4
−5I
3
=0
2I
2
+I
4
−I
3
=0
I
3
=2I
2
+I
4
...(1)
For the closed circuit BCDB, potential is zero i.e.,
5(I
2
−I
4
)−10(I
3
+I
4
)−5/4=0
5I
2
+5I
4
−10I
3
−10I
4
−5I
4
=0
5I
2
−10I
3
−20I
4
=0
I
2
=2I
3
+4I
4
...(2)
For the closed circuit ABCFEA, potential is zero i.e.,
−10+10(I
1
)+10(I
2
)+5(I
2
−I
4
)=0
10=15I
2
+10I
1
−5I
4
3I
2
+2I
1
−I
4
=2 ...(3)
From equations (1) and (2), we obtain
I
3
=2(2I
3
+4I
4
)+I
4
I
3
=4I
3
+8I
4
+I
4
−3I
3
=9I
4
−3I
4
=+I
3
...(4)
Putting equation (4) in equation (1), we obtain
I
3
=2I
2
+I
4
−4I
4
=2I
2
I
2
=−2I
4
...(5)
It is evident from the given figure that,
I
1
=I
3
+I
2
...(6)
Putting equation (6) in equation (1), we obtain
3I
2
+2(I
3
+I
2
)−I
4
=2
5I
2
+2(I
3
+I
2
)−I
4
=2 ...(7)
Putting equations (4) and (5) in equation (7), we obtain
5(−2I
4
)+2(−3I
4
)−I
4
=2
−10I
4
−6I
4
−I
4
=2
17I
4
=−2
I
4
=
17
−2
A
Equation (4) reduces to
I
3
=−3(I
4
)
=−3(
17
−2
)=
17
6
A
I
2
=−2(I
4
)
=−2(
17
−2
)=
17
4
A
I
2
−I
4
=
17
4
−(
17
−2
)=
17
6
A
I
3
+I
4
=
17
6
+(
17
−2
)=
17
4
A
I
1
=I
3
+I
2
=
17
6
+
17
4
=
17
10
A
Therefore, Current in branch AB=
17
4
A
In branch BC =
17
6
A
In branch CD =
17
−4
A
In branch AD =
17
6
A
In branch BD =(
17
−2
)A
Total current ==
17
4
+
17
6
+
17
−4
+
17
6
+
17
−2
=
17
10
A