iโ = 0.848 A
iโ = 2.142 A
iโ = 0.171 A
Explanation:
For 3 variables, you need 3 equations.
However, you can also label the currents in the other branches. ย This isn't required, but it can make the math easier to follow. ย In that case, you'll have 6 variables (one for each branch), so you'll need 6 equations.
Let's define these other currents:
iโ = the current in the top right corner (counterclockwise)
iโ
= the current in the top left corner (clockwise)
iโ = the current down the middle
Since there's nothing in the middle branch, we can ignore iโ. ย 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. ย Using the left node and right node:
iโ + iโ = iโ
iโ = iโ + iโ
According to Kirchoff's voltage law, the sum of the voltage gains and drops around a loop equals 0. ย Using the top left loop, the top right loop, and the bottom loop:
0 = -iโ + 12 โ 5iโ
0 = -iโ + 9 โ 8iโ
0 = -iโ + 12 โ 10iโ โ 9 + iโ
We now have a system of 5 equations and 5 variables.
Substituting for iโ
and iโ:
0 = -iโ + 12 โ 5(iโ โ iโ)
0 = -iโ + 9 โ 8(iโ + iโ)
0 = -iโ + 12 โ 10iโ โ 9 + iโ
We've now reduced this to a system of 3 equations and 3 variables. ย Simplifying and solving:
0 = 12 โ 6iโ + 5iโ
0 = 9 โ 9iโ โ 8iโ
0 = -iโ + 3 โ 10iโ + iโ
iโ = (12 + 5iโ) / 6
iโ = (9 โ 8iโ) / 9
0 = -(12 + 5iโ) / 6 + 3 โ 10iโ + (9 โ 8iโ) / 9
0 = -9 (12 + 5iโ) + 54 (3 โ 10iโ) + 6 (9 โ 8iโ)
0 = -108 โ 45iโ + 162 โ 540iโ + 54 โ 48iโ
0 = 108 โ 633iโ
iโ = 0.171 A
iโ = 2.142 A
iโ = 0.848 A