Solution attempt:
I1 loop 9.00V-9.00(I1)=0 so I1=1A
I2 loop 12.00v-6.00(I2)=0 so I2=2A
For I3 loop 12.00V-10(I3)-8(I1)-1(I2)=0 by plugging in I1 and I2 I get I3 to be 5A.
[b]1. Calculate the three currents indicated in the circuit diagram below.
I have attached the diagram.
How would I use Kirchhoff's laws to find the currents I am getting thrown off by the third loop.