Consider the network as shown in Fig. Current is supplied to the network by two batteries as shown. Find the values of currents `I_(1), I_(2), I_(3)`. The direction of the currents are as indicated by the arrows.

Explanation

Applying Kirchhoff.s 1st law to junction C, we get
`I_(1)+I_(2)-I_(3)=0" ….(1)"`
Applying Kirchhoff.s Iind law to the closed meshes ACDA and BCDB, we get
`-5I_(1)-2I_(3)+12=0 rArr 5I_(1)+2I_(3)=12" ...(2)"`
`-3I_(1)-2I_(3)+6=0 rArr 3I_(1)+2I_(3)=6" ....(3)"`
On solving , `I_(1)=1.548A`.
`I_(2)=0.58A and I_(3)=I_(1)+I_(2)=2.3128A`.