How do I find currents i1, i2, i3? I can't figure how many equations needed here.. thanks

i₁ = 0.848 A

i₂ = 2.142 A

i₃ = 0.171 A

Explanation:

For 3 variables, you need 3 equations.

However, you can also label the currents in the other branches.  This isn't required, but it can make the math easier to follow.  In that case, you'll have 6 variables (one for each branch), so you'll need 6 equations.

Let's define these other currents:

i₄ = the current in the top right corner (counterclockwise)

i₅ = the current in the top left corner (clockwise)

i₆ = the current down the middle

Since there's nothing in the middle branch, we can ignore i₆.  So we need 5 equations instead of 6.

According to Kirchoff's current law, the sum of the currents entering a node equals the sum of the currents leaving the node.  Using the left node and right node:

i₁ + i₃ = i₄

i₂ = i₃ + i₅

According to Kirchoff's voltage law, the sum of the voltage gains and drops around a loop equals 0.  Using the top left loop, the top right loop, and the bottom loop:

0 = -i₂ + 12 − 5i₅

0 = -i₁ + 9 − 8i₄

0 = -i₂ + 12 − 10i₃ − 9 + i₁

We now have a system of 5 equations and 5 variables.

Substituting for i₅ and i₄:

0 = -i₂ + 12 − 5(i₂ − i₃)

0 = -i₁ + 9 − 8(i₁ + i₃)

0 = -i₂ + 12 − 10i₃ − 9 + i₁

We've now reduced this to a system of 3 equations and 3 variables.  Simplifying and solving:

0 = 12 − 6i₂ + 5i₃

0 = 9 − 9i₁ − 8i₃

0 = -i₂ + 3 − 10i₃ + i₁

i₂ = (12 + 5i₃) / 6

i₁ = (9 − 8i₃) / 9

0 = -(12 + 5i₃) / 6 + 3 − 10i₃ + (9 − 8i₃) / 9

0 = -9 (12 + 5i₃) + 54 (3 − 10i₃) + 6 (9 − 8i₃)

0 = -108 − 45i₃ + 162 − 540i₃ + 54 − 48i₃

0 = 108 − 633i₃

i₃ = 0.171 A

i₂ = 2.142 A

i₁ = 0.848 A

no. dot product   =0 when vecs at right angles