Question

the current flowing in the branch BD is.​

Answer 1

Answer:

use Kirchoff's rule for this..

by using this

we get

l1 is the current from 15v

and l2 is the current from 30v

15 = 6l1 + 3(I1+l2)...

15 = 9l1 +3l2 ....(1)

30=3I2 + 3(I1+I2)

30=3l1 + 6l2 ....(2)

multiply ...(2) by 3 then subtract ...(1) from it

we get

15l2 = 75

l2 = 5A

put this in ...(1)

we get

l1= 0

and the current in BD is l1 + l2 = 5+0 = 5A