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: (D)

$I_2=2A$

Answer 2

Due to symmetry on upper side and lower side, points P and Q are at same potentials. Similarly, points S and T are at same potentials.
Therefore, the simple circuit can be drawn as shown below
$I_2=\frac{12}{2+2+2}=2A$
$I_3=\frac{12}{4+4+4}=1A$
$\therefore I_1=I_2+I_3=3A$
$I_{PQ}=0$ because $V_P=V_Q$
Potential drop (from left to right) across each resistance is
$\frac{12}{3}=4V$
$\therefore V_{MS}=2\times4=8V$
$V_{NQ}=1\times4=4V$
or $V_S< V_Q$