Question

14. In the following circuit,
ammeter reading will be

3V
K
10

15
O 0.2A
ОО.ЗА
O 0.5A
0.4 A​

Answer 1

Explanation:

From figure:-

In Loop ABDA:-

V

1

=R

1

i

2

+R

3

(i

1

+i

2

)

⟹3i

2

+6i

2

=15

⟹i

1

+3i

2

=5

⟹i

1

=5−3i

2

-------(a)

In Loop BCDB:-

V

2

=R

2

i

1

+R

3

(i

1

+i

2

)

⟹6i

1

+3i

2

=30

⟹2i

1

+i

2

=10

⟹2(5−3i

2

)+i

2

=10 from equation (a)

⟹10−5i

2

=10

⟹i

2

=0

from equation (a):-

i

1

=5−3i

2

=5A

Current in branch BD=i

1

+i

2

=5A

Hope it helps to u....

Answer 2

Answer:

sorry I not sure the answer