In the network shown in given figure the current `l_(1) ` through the 10 V battery .

Explanation

Applying junction rule to C and D
` l = l_(1) + l_(2)" "…..(i)`
10 V batttery is connected alone across CD so
`V_(D) - V_(C) = 10 V`
` therefore l_(2) = (V_(C) - V_(D))/(2 Omega) = (-10 V)/(2 Omega ) = - 5 A`
(Negative sign indicated current `l_(2)` is from D to C)
Applying ioop rule to CDEFGC.
`2l_(2) + 5l + 3l - 20 = 0`
` rArr 2(-5) + 8l - 20 = 0`
`rArr -10 + 8l - 20 = 0`
`rArr l =(30)/(8) =(15)/(4) A`
From equation (i) , we get
`l_(1) = l-l_(2) = (15)/(4) A - (-5A) = ((15)/(4) + 5) A = (35)/(4) A`