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)

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 $\begin{array}{l}{I}_1=\frac{12}{4}=3A\\ I_2=I_1\left(\frac{12}{6+12}\right)=2A\end{array}$ 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 $S$.