Applying Kirchoff's law 10−6IB+14−4IA=0 . . .(loop 1) 24=6IB+4IA 12=3IB+2IA . . .(i) IA=IB+IC . . . (junction) 10−6IB+2IC=0 . . .(loop 2) 10=6IB−2IC 10=6IB−2(IA−IB).....(ii) On solving Eqs. (i) and (ii), we get IB=2A,IA=3A,IC=1A Now as per given circuit, I1=IB=2A,I2=−IA=−3A I3=−IC=−1A