In the circuit shown below, the current through 10Ω resistor is
A
Here, VA=VB=VC=0
VD=12V
VD−VE=12−V=V1
VE−VB=V
VE−VC=V
∴I1=12−V3,I2=V3,I3=V12
Apply Kirchhoff's junction rule, I1=I2+I3
i.e., 12−V3=V3+V12⇒163V
Hence, I3 = (163)12 = 49A