Question #7bd35

1 Answer
Dec 17, 2016

Answer:

#A_1=19A#

#A_2=17A#

#A_3=14A#

#A_4=3A#

Explanation:

drawn

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