Question

Question #7bd35

Answer 1

`A_1=19A`

`A_2=17A`

`A_3=14A`

`A_4=3A`

Explanation:

As shown in figure of the circuit the main current returning from cell is `19A` and emerging from cell is `A_1`. Hence `A_1=19A` by conservation of charge.

The emerging current `A_1=19A` is divided in two paths in the first branch as `A_2 and 2A` ,

So `A_2 +2A=19A=>A_2= (19-2)A=17A`

Similarly `A_2 =17A` is divided in two paths in the 2nd branch as

`A_3 and 3A`. So #A_3 +3A=17A=>A_3= (17-3)A=14A.

From the figure we see `3A` returns back passing through `R_3` as `A_4` So #A_4==3A