Consider thr 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`
Applying Kirchhoff.s Iind law to the closed meshes ACDA and BCDB, we get
`5 I_1 + 2I_3 = 12" "…..(2)`
`3I_2 + 2I_3 + 6 " "….(3)`
Subtracting eq (3) from eq (2) we get
`5I_1 - 3I_2 = 6 " ".....(4)`
Multiplying eq. (1) by 2 adding with eq. (2) we get
`7I_2 + 2I_2 = 12 " "......(5)`
Multiplying eq (4) by 2 and eq. (5) by 3 and adding them we get
`31I_1 = 48 implies I_1 = 1.548 A.`

Putting value of `I_1 ` in eq. (5) we get `I_2 =0.58 A` and from eq. 1 we get `I_3 = I_1 + I_2 = 2.128 A`