The currents in a circuit can be determined by solving a sys
Solution
apply kvl to loop current i1.
3i1+2(i1-i3)+4(i1-i2)=24
9i1-4i2-2i3=24 --------------------------------(1)
apply kvl to loop current i2.
4i2+16+4(i2-i1)+6(i2-i3)+3(i2-i4)=0
-4i1+17i2-6i3-3i4=-16 -------------------------------------(2)
apply kvl to loop current i3.
2(i3-i1)+6(i3-i4)+6(i3-i2)=0
-2i1-6i2+14i3-6i4=0 ------------------------------------(3)
apply kvl to loop current i4.
-18+2i4+3(i4-i2)+6(i4-i3)=0
-3i2-6i3+11i4=18 -----------------------------------(4).
from the above equations (1),(2),(3)&(4).
i1=4.034 amps.
i2=1.6545 amps.
i3=2.845 amps.
i4=3.6395 amps.
