Doing a quick refresher on Branch current method. I'm fine until I reach the final part of the lesson and he figures the current values from these final equations.
-1I1 + 1I2 - 1I3 = 0 KCL
4I1 + 2I2 + 0I3 = 28 KVL
0I1 - 2I2 - 1I3 = -7 KVL
I1 = 5A
I2 = 4A
I3 = -1A
Could someone show the algebraic equation that shows how we get 5 amps, 4 amps, and -1 amp from these equations above.
-1I1 + 1I2 - 1I3 = 0 KCL
4I1 + 2I2 + 0I3 = 28 KVL
0I1 - 2I2 - 1I3 = -7 KVL
I1 = 5A
I2 = 4A
I3 = -1A
Could someone show the algebraic equation that shows how we get 5 amps, 4 amps, and -1 amp from these equations above.