For the resistance network shown in the figure, choose the correct option(s). The current through PQ is zero. The potential at S is less than that at Q. I1=3A. I2=2A.

Node P and Q are equipotential and node S and T are equipotential due to formation of wheatstone bridge across them. Thus, no current passes through PQ and ST. I1=12/4=3A I2=I1(12/6+12)=2A now point P and Q are Equipotential and point S and T are equipotential. from resistor PS current flows from P to S hence voltage of point P is more than voltage of point S , also P is equipotential with Q that's why Q also has more potential than S.