KEAM 2017 Physics and Chemistry Paper

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