Answer:
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
