Inactive Tutor answered 05/20/24
For each current loop, in, it will be the sum of all the loop resistances time in minus the sum of IR voltages of shared loop resistances. The sum will be equal to the given loop voltage.
[ 12 -5 0 0] [i1] = [22]
[ -5 12 -2 0] [i2] = [30]
[ 0 -2 10 -1] [i3] = [20]
[ 0 0 -1 8] [i4] = [22]
If A is the resistance matrix, C is the current vector and V is the voltage vector
AC=V
C=A-1V
Inactive Tutor
I was making assumptions without a circuit diagram, totally off the description given. That would be helpful to set up the equations for each loop.05/20/24
Ashyia W.
I tried this it was incorrect05/20/24