For the circuit shown in figure, the equivalent resistance between A and C is

At junction E of R:I1+I3+I4=I2−I3    or     I1−I2+2I3+I4=0Loop AFBA or loop ECDErI1=rI2+rI3          or   I1=I2+I3Loop BEFB or loop EDFEr(I2−I3)+rI4=rI3   or    I2-2I3+I4=0Loop AFDCMA  or   ABECMAV=rI1+r(I2−I3)+rI2=rI1+2rI2−rI3Solve to get I1=2V/5r,   I2=V/3rReq=VI=VI1+I2=V2V/5r+V/3r=15r11