Matching.

Explanation

Applying Kirchhoff.s I law at B we get,
`I_1 - I_2 - I_3 = 0`
` :." "I_3 = I_1 - I_2" …(1)"`
Applying Kirchhoff.s II law to ABEFA
`100I_3 + 100I_2 = 15`
Using (1) we can write
`100 (I_1 - I_2) + 100 I_1 = 15`
` 100 I_2 - 100 I_2 + 100 I_1 = 15`
` :." "200 I_2 - 100I_2 = "15....(2)"`
Applying Kirchhoff.s II law to BCDEA we get,
`100I_2 - 100I_3 = -9`
Substitutin (1) in the abouve equation we get,
`100 I_2 - 100 (I_1 - I_2) = -9`
`100 I_2 - 100 I_1 + 100 I_2 = -9`
` 200I_2 -100I_2 " = -9 " ...(3)"`
Soving equations (2) and (3) we get,
`200I_1 - 100I_2 = -15`
`-200 I_1 + 400 I_2 = -18`
`:. 300 I_2 = -3`
`:. I_2 = (-3)/300 `
` = 0.01 A `
` :.I_2 = -0.01 A`
Substituting the values of `I_1 ` and ` I_2 ` in equation (3) we get,
`200I_2 - 100I_1 = -9`
`200 (-0.01) - 100I_1 = 9`
` -2 - 100 I_1 = -9`
` - 100 I_1 = - 7`
` :." I_1 = 0.07 A`
Substituting the values of `I_1` and `I_2` in equation (1) we get,
`I_3 = I_1 - I_2`
` = 0.07 + 0.01`
` = 0.08 A`
`:." "I_1 = 0.070 A I_2 = -.01 A " and " I_3 = 0.08A`