For the resistance network shown in the figure, choose the correct...

Resistance of arm $P Q$ and $S T$ becomes ineffective as $P \& Q$ and $S \& T$ are at the same potential. The equivalent circuit is as shown in the figure.
The resistance of the upper arm $R_{1}=2 \Omega+2 \Omega+2 \Omega=6 \Omega$
The resistance of the lower arm $R_{2}=4 \Omega+4 \Omega+4 \Omega=12 \Omega$ Equivalent resistance of the circuit, $$\begin{array}{l} R_{\mathrm{eq}}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{(6 \Omega)(12 \Omega)}{6 \Omega+12 \Omega}=4 \Omega \\ \therefore I_{1}=\frac{12 \mathrm{~V}}{4 \Omega}=3 \mathrm{~A} \\ I_{2}=\left(\frac{12}{6+12}\right) \times 3=2 \mathrm{~A} \\ I_{3}=I_{1}-I_{2}=1 \mathrm{~A} \end{array}$$ Potential difference across A$and$P$,$$\begin{array}{l} V_{A}-V_{P}=I_{2} \times 2 \Omega=(2 A)(2 \Omega) \\ 12 V-V_{P}=4 V \text{ or } V_{P}=8 V \end{array}$$Potential difference across A$ and $Q$, $$\begin{array}{l} V_{A}-V_{Q}=I_{3} \times 2 \Omega=(1 A)(4 \Omega) \\ 12 V-V_{Q}=4 V \\ V_{Q}=12 \mathrm{~V}-4 \mathrm{~V}=8 \mathrm{~V} \end{array}$$ Potential difference across $P$ and $S$, $$\begin{array}{l} V_{P}-V_{S}=(2 A)(2 W)=4 V \\ 8 V-V_{S}=4 V \Rightarrow V_{S}=4 V \\ \therefore V_{S} < V_{Q} \end{array}$$