Question

For the resistance network shown in the figure, choose the correct option(s).

• A. The current through PQ is zero.
• B. $I_1=3A.$
• C. The potential at S is less than that at Q.
• D. $I_2=2A.$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The current through PQ is zero.
B The potential at S is less than that at Q.
C $I_1=3A.$
D $I_2=2A.$

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
$I_1=\frac{12}{4}=3A$
$I_2=I_1\left(\frac{12}{6+12}\right)=2A$

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.